Study Guide and Solutions Manual to Accompany T.W. Graham Solomons / Craig B. Fryhle / Scott A. Snyder / Jon Antilla STUDY GUIDE AND SOLUTIONS MANUAL TO ACCOMPANY ORGANIC CHEMISTRY ELEVENTH EDITION This page is intentionally left blank STUDY GUIDE AND SOLUTIONS MANUAL TO ACCOMPANY ORGANIC CHEMISTRY ELEVENTH EDITION T. W. GRAHAM SOLOMONS University of South Florida CRAIG B. FRYHLE Pacific Lutheran University SCOTT A. SNYDER Columbia University ROBERT G. JOHNSON Xavier University JON ANTILLA University of South Florida Project Editor Jennifer Yee Senior Production Editor Elizabeth Swain Cover Image © Gerhard Schulz/Age Fotostock America, Inc. This book was set in 10/12 Times Roman by Aptara Delhi and printed and bound by Bind-Rite. The cover was printed by Bind-Rite. Copyright © 2014, 2011, 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, website http://www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review puposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. ISBN 978-1-118-14790-0 Binder-Ready version ISBN 978-1-118-63649-7 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ACKNOWLEDGMENTS We are grateful to those people who have made many helpful suggestions for various editions of this study guide. These individuals include: George R. Jurch, George R. Wenzinger, and J. E. Fernandez at the University of South Florida; Darell Berlin, Oklahoma State University; John Mangravite, West Chester State College; J. G. Traynham, Louisiana State University; Desmond M. S. Wheeler, University of Nebraska; Chris Callam, The Ohio State University; Sean Hickey, University of New Orleans; and Neal Tonks, College of Charleston. We are especially grateful to R.G. (Bob) Johnson for his friendship, dedication, and many contributions to this Study Guide and the main text over many years. T. W. Graham Solomons Craig B. Fryhle Scott A. Snyder Jon Antilla v CONTENTS To the Student INTRODUCTION “Solving the Puzzle” or “Structure Is Everything (Almost)” CHAPTER 1 THE BASICS: BONDING AND MOLECULAR STRUCTURE Solutions to Problems Quiz CHAPTER 2 FAMILIES OF CARBON COMPOUNDS: FUNCTIONAL GROUPS, INTERMOLECULAR FORCES, AND INFRARED (IR) SPECTROSCOPY Solutions to Problems Quiz CHAPTER 3 ACIDS AND BASES: AN INTRODUCTION TO ORGANIC REACTIONS AND THEIR MECHANISMS Solutions to Problems Quiz CHAPTER 4 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES Solutions to Problems Quiz CHAPTER 5 STEREOCHEMISTRY: CHIRAL MOLECULES Solutions to Problems Quiz vi xi xiii 1 1 15 18 18 30 34 34 44 46 46 62 65 65 82 CONTENTS CHAPTER 6 IONIC REACTIONS–NUCLEOPHILIC SUBSTITUTION AND ELIMINATION REACTIONS OF ALKYL HALIDES Solutions to Problems Quiz CHAPTER 7 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS. ELIMINATION REACTIONS OF ALKYL HALIDES Solutions to Problems Quiz CHAPTER 8 ALKENES AND ALKYNES II: ADDITION REACTIONS Solutions to Problems Quiz CHAPTER 9 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY: TOOLS FOR STRUCTURE DETERMINATION Solutions to Problems Quiz CHAPTER 10 RADICAL REACTIONS Solutions to Problems Quiz CHAPTER 11 ALCOHOLS AND ETHERS Solutions to Problems Quiz vii 85 85 103 106 106 128 130 130 153 157 157 180 182 182 200 203 203 225 viii CONTENTS CHAPTER 12 ALCOHOLS FROM CARBONYL COMPOUNDS: OXIDATIONREDUCTION AND ORGANOMETALLIC COMPOUNDS Solutions to Problems Quiz 227 227 257 ANSWERS TO FIRST REVIEW PROBLEM SET 259 (First Review Problem Set is available only in WileyPlus, www.wileyplus.com) CHAPTER 13 CONJUGATED UNSATURATED SYSTEMS Solutions to Problems Quiz 278 278 299 SUMMARY OF REACTIONS BY TYPE, CHAPTERS 1–13 301 METHODS FOR FUNCTIONAL GROUP PREPARATION, CHAPTERS 1–13 305 CHAPTER 14 AROMATIC COMPOUNDS Solutions to Problems Quiz CHAPTER 15 REACTIONS OF AROMATIC COMPOUNDS Solutions to Problems Quiz CHAPTER 16 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Solutions to Problems Quiz 308 308 323 325 325 355 357 357 386 CONTENTS CHAPTER 17 CARBOXYLIC ACIDS AND THEIR DERIVATIVES: NUCLEOPHILIC ADDITION-ELIMINATION AT THE ACYL CARBON Solutions to Problems Quiz CHAPTER 18 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS: ENOLS AND ENOLATES Solutions to Problems Quiz CHAPTER 19 CONDENSATION AND CONJUGATE ADDITION REACTIONS OF CARBONYL COMPOUNDS: MORE CHEMISTRY OF ENOLATES Solutions to Problems Quiz CHAPTER 20 AMINES Solutions to Problems Quiz CHAPTER 21 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION ix 389 389 418 421 421 446 448 448 482 488 488 526 Solutions to Problems Quiz 530 530 547 ANSWERS TO SECOND REVIEW PROBLEM SET 550 (Second Review Problem Set is available only in WileyPlus, www.wileyplus.com) CHAPTER 22 CARBOHYDRATES Solutions to Problems Quiz 566 567 592 x CONTENTS CHAPTER 23 LIPIDS Solutions to Problems Quiz CHAPTER 24 AMINO ACIDS AND PROTEINS Solutions to Problems Quiz CHAPTER 25 NUCLEIC ACIDS AND PROTEIN SYNTHESIS Solutions to Problems 596 596 607 610 610 625 626 626 Special Topics A–F and H are available only in WileyPlus, www.wileyplus.com. Solutions to problems in the Special Topics are found on the following pages: Special Topic A 13 C NMR Spectroscopy 634 Special Topic B Chain-Growth Polymers 636 Special Topic C Step-Growth Polymers 637 Special Topic D Thiols, Sulfur Ylides, and Disulfides 643 Special Topic E Thiol Esters and Lipid Biosynthesis 645 Special Topic F Alkaloids 646 Special Topic G Carbon Carbon Bond-Forming and Other Reactions of Transition Metal Organometallic Compounds 651 Electrocyclic and Cycloaddition Reactions 654 Special Topic H Problems Additional Problems Solutions to Problems of Appendix A 659 661 662 663 APPENDIX B ANSWERS TO QUIZZES 667 APPENDIX C MOLECULAR MODEL SET EXERCISES 682 APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS To the Student Contrary to what you may have heard, organic chemisty does not have to be a difficult course. It will be a rigorous course, and it will offer a challenge. But you will learn more in it than in almost any course you will take—and what you learn will have a special relevance to life and the world around you. However, because organic chemistry can be approached in a logical and systematic way, you will find that with the right study habits, mastering organic chemistry can be a deeply satisfying experience. Here, then, are some suggestions about how to study: 1. Keep up with your work from day to day-never let yourself get behind. Organic chemistry is a course in which one idea almost always builds on another that has gone before. It is essential, therefore, that you keep up with, or better yet, be a little ahead of your instructor. Ideally, you should try to stay one day ahead of your instructor’s lectures in your own class preparations. The lecture, then, will be much more helpful because you will already have some understanding of the assigned material. Your time in class will clarify and expand ideas that are already familiar ones. 2. Study material in small units, and be sure that you understand each new section before you go on to the next. Again, because of the cumulative nature of organic chemistry, your studying will be much more effective if you take each new idea as it comes and try to understand it completely before you move on to the next concept. 3. Work all of the in-chapter and assigned problems. One way to check your progress is to work each of the in-chapter problems when you come to it. These problems have been written just for this purpose and are designed to help you decide whether or not you understand the material that has just been explained. You should also carefully study the Solved Problems. If you understand a Solved Problem and can work the related in-chapter problem, then you should go on; if you cannot, then you should go back and study the preceding material again. Work all of the problems assigned by your instructor from the end of the chapter, as well. Do all of your problems in a notebook and bring this book with you when you go to see your instructor for extra help. 4. Write when you study. Write the reactions, mechanisms, structures, and so on, over and over again. Organic chemistry is best assimilated through the fingertips by writing, and not through the eyes by simply looking, or by highlighting material in the text, or by referring to flash cards. There is a good reason for this. Organic structures, mechanisms, and reactions are complex. If you simply examine them, you may think you understand them thoroughly, but that will be a misperception. The reaction mechanism may make sense to you in a certain way, but you need a deeper understanding than this. You need to know the material so thoroughly that you can explain it to someone else. This level of understanding comes to most of us (those of us without photographic memories) through writing. Only by writing the reaction mechanisms do we pay sufficient attention to their details, such as which atoms are connected to which atoms, which bonds break in a reaction and which bonds form, and the threedimensional aspects of the structures. When we write reactions and mechanisms, connections are made in our brains that provide the long-term memory needed for success in organic chemistry. We virtually guarantee that your grade in the course will be directly proportional to the number of pages of paper that you fill with your own writing in studying during the term. 5. Learn by teaching and explaining. Study with your student peers and practice explaining concepts and mechanisms to each other. Use the Learning Group Problems and other exercises your instructor may assign as vehicles for teaching and learning interactively with your peers. xi xii TO THE STUDENT 6. Use the answers to the problems in the Study Guide in the proper way. Refer to the answers only in two circumstances: (1) When you have finished a problem, use the Study Guide to check your answer. (2) When, after making a real effort to solve the problem, you find that you are completely stuck, then look at the answer for a clue and go back to work out the problem on your own. The value of a problem is in solving it. If you simply read the problem and look up the answer, you will deprive yourself of an important way to learn. 7. Use molecular models when you study. Because of the three-dimensional nature of most organic molecules, molecular models can be an invaluable aid to your understanding of them. When you need to see the three-dimensional aspect of a particular topic, use the Molecular VisionsTM model set that may have been packaged with your textbook, or buy a set of models separately. An appendix to the Study Guide that accompanies this text provides a set of highly useful molecular model exercises. 8. Make use of the rich online teaching resources in WileyPLUS (www.wileyplus.com) and do any online exercises that may be assigned by your instructor. INTRODUCTION “Solving the Puzzle” or “Structure Is Everything (Almost)” As you begin your study of organic chemistry it may seem like a puzzling subject. In fact, in many ways organic chemistry is like a puzzle—a jigsaw puzzle. But it is a jigsaw puzzle with useful pieces, and a puzzle with fewer pieces than perhaps you first thought. In order to put a jigsaw puzzle together you must consider the shape of the pieces and how one piece fits together with another. In other words, solving a jigsaw puzzle is about structure. In organic chemistry, molecules are the pieces of the puzzle. Much of organic chemistry, indeed life itself, depends upon the fit of one molecular puzzle piece with another. For example, when an antibody of our immune system acts upon a foreign substance, it is the puzzle-piece-like fit of the antibody with the invading molecule that allows “capture” of the foreign substance. When we smell the sweet scent of a rose, some of the neural impulses are initiated by the fit of a molecule called geraniol in an olfactory receptor site in our nose. When an adhesive binds two surfaces together, it does so by billions of interactions between the molecules of the two materials. Chemistry is truly a captivating subject. As you make the transition from your study of general to organic chemistry, it is important that you solidify those concepts that will help you understand the structure of organic molecules. A number of concepts are discussed below using several examples. We also suggest that you consider the examples and the explanations given, and refer to information from your general chemistry studies when you need more elaborate information. There are also occasional references below to sections in your text, Solomons, Fryhle, and Snyder Organic Chemistry, because some of what follows foreshadows what you will learn in the course. SOME FUNDAMENTAL PRINCIPLES WE NEED TO CONSIDER What do we need to know to understand the structure of organic molecules? First, we need to know where electrons are located around a given atom. To understand this we need to recall from general chemistry the ideas of electron configuration and valence shell electron orbitals, especially in the case of atoms such as carbon, hydrogen, oxygen, and nitrogen. We also need to use Lewis valence shell electron structures. These concepts are useful because the shape of a molecule is defined by its constituent atoms, and the placement of the atoms follows from the location of the electrons that bond the atoms. Once we have a Lewis structure for a molecule, we can consider orbital hybridization and valence shell electron pair repulsion (VSEPR) theory in order to generate a three-dimensional image of the molecule. Secondly, in order to understand why specific organic molecular puzzle pieces fit together we need to consider the attractive and repulsive forces between them. To understand this we need to know how electronic charge is distributed in a molecule. We must use tools such as formal charge and electronegativity. That is, we need to know which parts of a molecule xiii xiv INTRODUCTION are relatively positive and which are relatively negative—in other words, their polarity. Associations between molecules strongly depend on both shape and the complementarity of their electrostatic charges (polarity). When it comes to organic chemistry it will be much easier for you to understand why organic molecules have certain properties and react the way they do if you have an appreciation for the structure of the molecules involved. Structure is, in fact, almost everything, in that whenever we want to know why or how something works we look ever more deeply into its structure. This is true whether we are considering a toaster, jet engine, or an organic reaction. If you can visualize the shape of the puzzle pieces in organic chemistry (molecules), you will see more easily how they fit together (react). SOME EXAMPLES In order to review some of the concepts that will help us understand the structure of organic molecules, let’s consider three very important molecules—water, methane, and methanol (methyl alcohol). These three are small and relatively simple molecules that have certain similarities among them, yet distinct differences that can be understood on the basis of their structures. Water is a liquid with a moderately high boiling point that does not dissolve organic compounds well. Methanol is also a liquid, with a lower boiling point than water, but one that dissolves many organic compounds easily. Methane is a gas, having a boiling point well below room temperature. Water and methanol will dissolve in each other, that is, they are miscible. We shall study the structures of water, methanol, and methane because the principles we learn with these compounds can be extended to much larger molecules. Water HOH Let’s consider the structure of water, beginning with the central oxygen atom. Recall that the atomic number (the number of protons) for oxygen is eight. Therefore, an oxygen atom also has eight electrons. (An ion may have more or less electrons than the atomic number for the element, depending on the charge of the ion.) Only the valence (outermost) shell electrons are involved in bonding. Oxygen has six valence electrons—that is, six electrons in the second principal shell. (Recall that the number of valence electrons is apparent from the group number of the element in the periodic table, and the row number for the element is the principal shell number for its valence electrons.) Now, let’s consider the electron configuration for oxygen. The sequence of atomic orbitals for the first three shells of any atom is shown below. Oxygen uses only the first two shells in its lowest energy state. 1s, 2s, 2px , 2p y , 2pz , 3s, 3px , 3p y , 3pz The p orbitals of any given principal shell (second, third, etc.) are of equal energy. Recall also that each orbital can hold a maximum of two electrons and that each equal energy orbital must accept one electron before a second can reside there (Hund’s rule). So, for oxygen we place two electrons in the 1s orbital, two in the 2s orbital, and one in each of the 2p orbitals, for a subtotal of seven electrons. The final eighth electron is paired with another in one of the 2p orbitals. The ground state configuration for the eight electrons of oxygen is, therefore 1s2 2s2 2px 2 2p y 1 2pz 1 INTRODUCTION xv where the superscript numbers indicate how many electrons are in each orbital. In terms of relative energy of these orbitals, the following diagram can be drawn. Note that the three 2p orbitals are depicted at the same relative energy level. 2px 2py 2pz 2s 1s Energy Now, let’s consider the shape of these orbitals. The shape of an s orbital is that of a sphere with the nucleus at the center. The shape of each p orbital is approximately that of a dumbbell or lobe-shaped object, with the nucleus directly between the two lobes. There is one pair of lobes for each of the three p orbitals (px, p y, pz ) and they are aligned along the x, y, and z coordinate axes, with the nucleus at the origin. Note that this implies that the three p orbitals are at 90◦ angles to each other. y x z an s orbital px, py, pz orbitals Now, when oxygen is bonded to two hydrogens, bonding is accomplished by the sharing of an electron from each of the hydrogens with an unpaired electron from the oxygen. This type of bond, involving the sharing of electrons between atoms, is called a covalent bond. The formation of covalent bonds between the oxygen atom and the two hydrogen atoms is advantageous because each atom achieves a full valence shell by the sharing of these electrons. For the oxygen in a water molecule, this amounts to satisfying the octet rule. A Lewis structure for the water molecule (which shows only the valence shell electrons) is depicted in the following structure. There are two nonbonding pairs of electrons around the oxygen as well as two bonding pairs. H x O x O H H H xvi INTRODUCTION In the left-hand structure the six valence electrons contributed by the oxygen are shown as dots, while those from the hydrogens are shown as x’s. This is done strictly for bookkeeping purposes. All electrons are, of course, identical. The right-hand structure uses the convention that a bonding pair of electrons can be shown by a single line between the bonded atoms. This structural model for water is only a first approximation, however. While it is a proper Lewis structure for water, it is not an entirely correct three-dimensional structure. It might appear that the angle between the hydrogen atoms (or between any two pairs of electrons in a water molecule) would be 90◦ , but this is not what the true angles are in a water molecule. The angle between the two hydrogens is in fact about 105◦ , and the nonbonding electron pairs are in a different plane than the hydrogen atoms. The reason for this arrangement is that groups of bonding and nonbonding electrons tend to repel each other due to the negative charge of the electrons. Thus, the ideal angles between bonding and nonbonding groups of electrons are those angles that allow maximum separation in three-dimensional space. This principle and the theory built around it are called the valence shell electron pair repulsion (VSEPR) theory. VSEPR theory predicts that the ideal separation between four groups of electrons around an atom is 109.5◦ , the so-called tetrahedral angle. At an angle of 109.5◦ all four electron groups are separated equally from each other, being oriented toward the corners of a regular tetrahedron. The exact tetrahedral angle of 109.5◦ is found in structures where the four groups of electrons and bonded groups are identical. In water, there are two different types of electron groups—pairs bonding the hydrogens with the oxygen and nonbonding pairs. Nonbonding electron pairs repel each other with greater force than bonding pairs, so the separation between them is greater. Consequently, the angle between the pairs bonding the hydrogens to the oxygen in a water molecule is compressed slightly from 109.5◦ , being actually about 105◦ . As we shall see shortly, the angle between the four groups of bonding electrons in methane (CH4 ) is the ideal tetrahedral angle of 109.5◦ . This is because the four groups of electrons and bound atoms are identical in a methane molecule. O H 105° H Orbital hybridization is the reason that 109.5◦ is the ideal tetrahedral angle. As noted earlier, an s orbital is spherical, and each p orbital is shaped like two symmetrical lobes aligned along the x, y, and z coordinate axes. Orbital hybridization involves taking a weighted average of the valence electron orbitals of the atom, resulting in the same number of new hybridized orbitals. With four groups of valence electrons, as in the structure of water, one s orbital and three p orbitals from the second principal shell in oxygen are hybridized (the 2s and 2px , 2p y , and 2pz orbitals). The result is four new hybrid orbitals of equal energy designated as sp3 orbitals (instead of the original three p orbitals and one s orbital). Each of the four sp3 orbitals has roughly 25% s character and 75% p character. The geometric result is that the major lobes of the four sp3 orbitals are oriented toward the corners of a tetrahedron with an angle of 109.5◦ between them. INTRODUCTION xvii sp3 hybrid orbitals (109.5° angle between lobes) In the case of the oxygen in a water molecule, where two of the four sp3 orbitals are occupied by nonbonding pairs, the angle of separation between them is larger than 109.5◦ due to additional electrostatic repulsion of the nonbonding pairs. Consequently, the angle between the bonding electrons is slightly smaller, about 105◦ . More detail about orbital hybridization than provided above is given in Sections 1.9– 1.15 of Organic Chemistry. With that greater detail it will be apparent from consideration of orbital hybridization that for three groups of valence electrons the ideal separation is 120◦ (trigonal planar), and for two groups of valence electrons the ideal separation is 180◦ (linear). VSEPR theory allows us to come to essentially the same conclusion as by the mathematical hybridization of orbitals, and it will serve us for the moment in predicting the three-dimensional shape of molecules. Methane CH4 Now let’s consider the structure of methane (CH4 ). In methane there is a central carbon atom bearing four bonded hydrogens. Carbon has six electrons in total, with four of them being valence electrons. (Carbon is in Group IVA in the periodic table.) In methane each valence electron is shared with an electron from a hydrogen atom to form four covalent bonds. This information allows us to draw a Lewis structure for methane (see below). With four groups of valence electrons the VSEPR theory allows us to predict that the three-dimensional shape of a methane molecule should be tetrahedral, with an angle of 109.5◦ between each of the bonded hydrogens. This is indeed the case. Orbital hybridization arguments can also be used to show that there are four equivalent sp3 hybrid orbitals around the carbon atom, separated by an angle of 109.5◦ . H .x . H x. C x H . x H H H C H H H H H C H All H-C-H angles are 109.5° The structure at the far right above uses the dash-wedge notation to indicate three dimensions. A solid wedge indicates that a bond projects out of the paper toward the reader. A dashed bond indicates that it projects behind the paper away from the viewer. Ordinary lines represent bonds in the plane of the paper. The dash-wedge notation is an important and widely used tool for depicting the three-dimensional structure of molecules. xviii INTRODUCTION Methanol CH3 OH Now let’s consider a molecule that incorporates structural aspects of both water and methane. Methanol (CH3 OH), or methyl alcohol, is such a molecule. In methanol, a central carbon atom has three hydrogens and an O–H group bonded to it. Three of the four valence electrons of the carbon atom are shared with a valence electron from the hydrogen atoms, forming three C H bonds. The fourth valence electron of the carbon is shared with a valence electron from the oxygen atom, forming a C–O bond. The carbon atom now has an octet of valence electrons through the formation of four covalent bonds. The angles between these four covalent bonds is very near the ideal tetrahedral angle of 109.5◦ , allowing maximum separation between them. (The valence orbitals of the carbon are sp3 hybridized.) At the oxygen atom, the situation is very similar to that in water. The oxygen uses its two unpaired valence electrons to form covalent bonds. One valence electron is used in the bond with the carbon atom, and the other is paired with an electron from the hydrogen to form the O–H bond. The remaining valence electrons of the oxygen are present as two nonbonding pairs, just as in water. The angles separating the four groups of electrons around the oxygen are thus near the ideal angle of 109.5◦ , but reduced slightly in the C–O–H angle due to repulsion by the two nonbonding pairs on the oxygen. (The valence orbitals of the oxygen are also sp3 hybridized since there are four groups of valence electrons.) A Lewis structure for methanol is shown below, along with a three-dimensional perspective drawing. H H C H O H H H H C O H THE ”CHARACTER” OF THE PUZZLE PIECES With a mental image of the three-dimensional structures of water, methane, and methanol, we can ask how the structure of each, as a “puzzle piece,” influences the interaction of each molecule with identical and different molecules. In order to answer this question we have to move one step beyond the three-dimensional shape of these molecules. We need to consider not only the location of the electron groups (bonding and nonbonding) but also the distribution of electronic charge in the molecules. First, we note that nonbonding electrons represent a locus of negative charge, more so than electrons involved in bonding. Thus, water would be expected to have some partial negative charge localized in the region of the nonbonding electron pairs of the oxygen. The same would be true for a methanol molecule. The lower case Greek δ (delta) means “partial.” H δ− δ − H O H H H H C δ− O δ− INTRODUCTION xix Secondly, the phenomenon of electronegativity influences the distribution of electrons, and hence the charge in a molecule, especially with respect to electrons in covalent bonds. Electronegativity is the propensity of an element to draw electrons toward it in a covalent bond. The trend among elements is that of increasing electronegativity toward the upper right corner of the periodic table. (Fluorine is the most electronegative element.) By observing the relative locations of carbon, oxygen, and hydrogen in the periodic table, we can see that oxygen is the most electronegative of these three elements. Carbon is more electronegative than hydrogen, although only slightly. Oxygen is significantly more electronegative than hydrogen. Thus, there is substantial separation of charge in a water molecule, due not only to the nonbonding electron pairs on the oxygen but also to the greater electronegativity of the oxygen with respect to the hydrogens. The oxygen tends to draw electron density toward itself in the bonds with the hydrogens, leaving the hydrogens partially positive. The resulting separation of charge is called polarity. The oxygen–hydrogen bonds are called polar covalent bonds due to this separation of charge. If one considers the net effect of the two nonbonding electron pairs in a water molecule as being a region of negative charge, and the hydrogens as being a region of relative positive charge, it is clear that a water molecule has substantial separation of charge, or polarity. δ+ δ− δ− O H H δ+ An analysis of polarity for a methanol molecule would proceed similarly to that for water. Methanol, however, is less polar than water because only one O–H bond is present. Nevertheless, the region of the molecule around the two nonbonding electron pairs of the oxygen is relatively negative, and the region near the hydrogen is relatively positive. The electronegativity difference between the oxygen and the carbon is not as large as that between oxygen and hydrogen, however, so there is less polarity associated with the C–O bond. Since there is even less difference in electronegativity between hydrogen and carbon in the three C–H bonds, these bonds contribute essentially no polarity to the molecule. The net effect for methanol is to make it a polar molecule, but less so than water due to the nonpolar character of the CH3 region of the molecule. H H H H C δ+ O δ− δ− Now let’s consider methane. Methane is a nonpolar molecule. This is evident first because there are no nonbonding electron pairs, and secondly because there is relatively little electronegativity difference between the hydrogens and the central carbon. Furthermore, what little electronegativity difference there is between the hydrogens and the central carbon atom is negated by the symmetrical distribution of the C–H bonds in the tetrahedral shape of methane. The slight polarity of each C–H bond is canceled by the symmetrical xx INTRODUCTION orientation of the four C–H bonds. If considered as vectors, the vector sum of the four slightly polar covalent bonds oriented at 109.5◦ to each other would be zero. H H C H Net dipole is zero. H The same analysis would hold true for a molecule with identical bonded groups, but groups having electronegativity significantly different from carbon, so long as there were symmetrical distribution of the bonded groups. Tetrachloromethane (carbon tetrachloride) is such a molecule. It has no net polarity. Cl Cl Cl C Net dipole is zero. Cl INTERACTIONS OF THE PUZZLE PIECES Now that you have an appreciation for the polarity and shape of these molecules it is possible to see how molecules might interact with each other. The presence of polarity in a molecule bestows upon it attractive or repulsive forces in relation to other molecules. The negative part of one molecule is attracted to the positive region of another. Conversely, if there is little polarity in a molecule, the attractive forces it can exert are very small [though not completely nonexistent, due to van der Waals forces (Section 2.13B in Organic Chemistry)]. Such effects are called intermolecular forces (forces between molecules), and strongly depend on the polarity of a molecule or certain bonds within it (especially O H, N H, and other bonds between hydrogen and more electronegative atoms with nonbonding pairs). Intermolecular forces have profound effects on physical properties such as boiling point, solubility, and reactivity. An important manifestation of these properties is that the ability to isolate a pure compound after a reaction often depends on differences in boiling point, solubility, and sometimes reactivity among the compounds present. Boiling Point An intuitive understanding of boiling points will serve you well when working in the laboratory. The polarity of water molecules leads to relatively strong intermolecular attraction between water molecules. One result is the moderately high boiling point of water (100 ◦ C, as compared to 65 ◦ C for methanol and −162 ◦ C for methane, which we will discuss shortly). Water has the highest boiling point of these three example molecules because it will strongly associate with itself by attraction of the partially positive hydrogens of one molecule (from the electronegativity difference between the O and H) to the negatively charged region in another water molecule (where the nonbonding pairs are located). INTRODUCTION O xxi O δ+ H H H δ− δ+ O δ− + δ H H δ− O H H hydrogen bonds H The specific attraction between a partially positive hydrogen atom attached to a heteroatom (an atom with both nonbonding and bonding valence electrons, e.g., oxygen or nitrogen) and the nonbonding electrons of another heteroatom is called hydrogen bonding. It is a form of dipole-dipole attraction due to the polar nature of the hydrogen–heteroatom bond. A given water molecule can associate by hydrogen bonding with several other water molecules, as shown above. Each water molecule has two hydrogens that can associate with the non-bonding pairs of other water molecules, and two nonbonding pairs that can associate with the hydrogens of other water molecules. Thus, several hydrogen bonds are possible for each water molecule. It takes a significant amount of energy (provided by heat, for example) to give the molecules enough kinetic energy (motion) for them to overcome the polarityinduced attractive forces between them and escape into the vapor phase (evaporation or boiling). Methanol, on the other hand, has a lower boiling point (65 ◦ C) than water, in large part due to the decreased hydrogen bonding ability of methanol in comparison with water. Each methanol molecule has only one hydrogen atom that can participate in a hydrogen bond with the nonbonding electron pairs of another methanol molecule (as compared with two for each water molecule). The result is reduced intermolecular attraction between methanol molecules and a lower boiling point since less energy is required to overcome the lesser intermolecular attractive forces. H H H C O δ− H δ H H H C H H H O C + O δ− H H + δ The CH3 group of methanol does not participate in dipole–dipole attractions between molecules because there is not sufficient polarity in any of its bonds to lead to significant partial positive or negative charges. This is due to the small electronegativity difference between the carbon and hydrogen in each of the C–H bonds. Now, on to methane. Methane has no hydrogens that are eligible for hydrogen bonding, since none is attached to a heteroatom such as oxygen. Due to the small difference in electronegativity between carbon and hydrogen there are no bonds with any significant polarity. Furthermore, what slight polarity there is in each C–H bond is canceled due to the tetrahedral symmetry of the molecule. [The minute attraction that is present between xxii INTRODUCTION methane molecules is due to van der Waals forces, but these are negligible in comparison to dipole–dipole interactions that exist when significant differences in electronegativity are present in molecules such as water and methanol.] Thus, because there is only a very weak attractive force between methane molecules, the boiling point of methane is very low (−162 ◦ C) and it is a gas at ambient temperature and pressure. H H H C H Solubility An appreciation for trends in solubility is very useful in gaining a general understanding of many practical aspects of chemistry. The ability of molecules to dissolve other molecules or solutes is strongly affected by polarity. The polarity of water is frequently exploited during the isolation of an organic reaction product because water will not dissolve most organic compounds but will dissolve salts, many inorganic materials, and other polar byproducts that may be present in a reaction mixture. As to our example molecules, water and methanol are miscible with each other because each is polar and can interact with the other by dipole–dipole hydrogen bonding interactions. Since methane is a gas under ordinary conditions, for the purposes of this discussion let’s consider a close relative of methane–hexane. Hexane (C6 H14 ) is a liquid having only carbon—carbon and carbon—hydrogen bonds. It belongs to the same chemical family as methane. Hexane is not soluble in water due to the essential absence of polarity in its bonds. Hexane is slightly soluble in methanol due to the compatibility of the nonpolar CH3 region of methanol with hexane. The old saying “like dissolves like” definitely holds true. This can be extended to solutes, as well. Very polar substances, such as ionic compounds, are usually freely soluble in water. The high polarity of salts generally prevents most of them from being soluble in methanol, however. And, of course, there is absolutely no solubility of ionic substances in hexane. On the other hand, very nonpolar substances, such as oils, would be soluble in hexane. Thus, the structure of each of these molecules we’ve used for examples (water, methanol, and methane) has a profound effect on their respective physical properties. The presence of nonbonding electron pairs and polar covalent bonds in water and methanol versus the complete absence of these features in the structure of methane imparts markedly different physical properties to these three compounds. Water, a small molecule with strong intermolecular forces, is a moderately high boiling liquid. Methane, a small molecule with only very weak intermolecular forces, is a gas. Methanol, a molecule combining structural aspects of both water and methane, is a relatively low boiling liquid, having sufficient intermolecular forces to keep the molecules associated as a liquid, but not so strong that mild heat can’t disrupt their association. Reactivity While the practical importance of the physical properties of organic compounds may only be starting to become apparent, one strong influence of polarity is on the reactivity of molecules. It is often possible to understand the basis for a given reaction in organic INTRODUCTION xxiii chemistry by considering the relative polarity of molecules and the propensity, or lack thereof, for them to interact with each other. Let us consider one example of reactivity that can be understood at the initial level by considering structure and polarity. When chloromethane (CH3 Cl) is exposed to hydroxide ions (HO− ) in water a reaction occurs that produces methanol. This reaction is shown below. CH3 Cl + HO− (as NaOH dissolved in water) → HOCH3 + Cl− This reaction is called a substitution reaction, and it is of a general type that you will spend considerable time studying in organic chemistry. The reason this reaction occurs readily can be understood by considering the principles of structure and polarity that we have been discussing. The hydroxide ion has a negative charge associated with it, and thus should be attracted to a species that has positive charge. Now recall our discussion of electronegativity and polar covalent bonds, and apply these ideas to the structure of chloromethane. The chlorine atom is significantly more electronegative than carbon (note its position in the periodic table). Thus, the covalent bond between the carbon and the chlorine is polarized such that there is partial negative charge on the chlorine and partial positive charge on the carbon. This provides the positive site that attracts the hydroxide anion! H H − HO + + δ C H H − Cl δ HO + C H − Cl H The intimate details of this reaction will be studied in Chapter 6 of your text. Suffice it to say for the moment that the hydroxide ion attacks the carbon atom using one of its nonbonding electron pairs to form a bond with the carbon. At the same time, the chlorine atom is pushed away from the carbon and takes with it the pair of electrons that used to bond it to the carbon. The result is substitution of OH for Cl at the carbon atom and the synthesis of methanol. By calculating formal charges (Section 1.5 in the text) one can show that the oxygen of the hydroxide anion goes from having a formal negative charge in hydroxide to zero formal charge in the methanol molecule. Similarly, the chlorine atom goes from having zero formal charge in chloromethane to a formal negative charge as a chloride ion after the reaction. The fact that the reaction takes place at all rests largely upon the complementary polarity of the interacting species. This is a pervasive theme in organic chemistry. Acid-base reactions are also very important in organic chemistry. Many organic reactions involve at least one step in the overall process that is fundamentally an acid-base reaction. Both Brønsted-Lowry acid-base reactions (those involving proton donors and acceptors) and Lewis acid-base reactions (those involving electron pair acceptors and donors, respectively) are important. In fact, the reaction above can be classified as a Lewis acid-base reaction in that the hydroxide ion acts as a Lewis base to attack the partially positive carbon as a Lewis acid. It is strongly recommended that you review concepts you have learned previously regarding acid-base reactions. Chapter 3 in Organic Chemistry will help in this regard, but it is advisable that you begin some early review about acids and bases based on your previous studies. Acid–base chemistry is widely applicable to understanding organic reactions. xxiv INTRODUCTION JOINING THE PIECES Finally, while what we have said above has largely been in reference to three specific compounds, water, methanol, and methane, the principles involved find exceptionally broad application in understanding the structure, and hence reactivity, of organic molecules in general. You will find it constantly useful in your study of organic chemistry to consider the electronic structure of the molecules with which you are presented, the shape caused by the distribution of electrons in a molecule, the ensuing polarity, and the resulting potential for that molecule’s reactivity. What we have said about these very small molecules of water, methanol, and methane can be extended to consideration of molecules with 10 to 100 times as many atoms. You would simply apply these principles to small sections of the larger molecule one part at a time. The following structure of Streptogramin A provides an example. A region with trigonal planar bonding δ− O OH A few of the partially positive NH CH3 CH3 O H3C O CH O O H3C A region with tetrahedral bonding N δ+ N δ+ O and partially negative regions are shown, as well as regions of tetrahedral and trigonal planar geometry. See if you can identify more of each type. δ− Streptogramin A A natural antibacterial compound that blocks protein synthesis at the 70S ribosomes of Gram-positive bacteria. We have not said much about how overall shape influences the ability of one molecule to interact with another, in the sense that a key fits in a lock or a hand fits in a glove. This type of consideration is also extremely important, and will follow with relative ease if you have worked hard to understand the general principles of structure outlined here and expanded upon in the early chapters of Organic Chemistry. An example would be the following. Streptogramin A, shown above, interacts in a hand-in-glove fashion with the 70S ribosome in bacteria to inhibit binding of transfer RNA at the ribosome. The result of this interaction is the prevention of protein synthesis in the bacterium, which thus accounts for the antibacterial effect of Streptogramin A. Other examples of hand-in-glove interactions include the olfactory response to geraniol mentioned earlier, and the action of enzymes to speed up the rate of reactions in biochemical systems. FINISHING THE PUZZLE In conclusion, if you pay attention to learning aspects of structure during this initial period of “getting your feet wet” in organic chemistry, much of the three-dimensional aspects INTRODUCTION xxv of molecules will become second nature to you. You will immediately recognize when a molecule is tetrahedral, trigonal planar, or linear in one region or another. You will see the potential for interaction between a given section of a molecule and that of another molecule based on their shape and polarity, and you will understand why many reactions take place. Ultimately, you will find that there is much less to memorize in organic chemistry than you first thought. You will learn how to put the pieces of the organic puzzle together, and see that structure is indeed almost everything, just applied in different situations! This page is intentionally left blank 1 THE BASICS: BONDING AND MOLECULAR STRUCTURE SOLUTIONS TO PROBLEMS Another Approach to Writing Lewis Structures When we write Lewis structures using this method, we assemble the molecule or ion from the constituent atoms showing only the valence electrons (i.e., the electrons of the outermost shell). By having the atoms share electrons, we try to give each atom the electronic structure of a noble gas. For example, we give hydrogen atoms two electrons because this gives them the structure of helium. We give carbon, nitrogen, oxygen, and fluorine atoms eight electrons because this gives them the electronic structure of neon. The number of valence electrons of an atom can be obtained from the periodic table because it is equal to the group number of the atom. Carbon, for example, is in Group IVA and has four valence electrons; fluorine, in Group VIIA, has seven; hydrogen, in Group 1A, has one. As an illustration, let us write the Lewis structure for CH3 F. In the example below, we will at first show a hydrogen’s electron as x, carbon’s electrons as o’s, and fluorine’s electrons as dots. Example A 3 H , C , and F are assembled as H H H C F or H C F H H If the structure is an ion, we add or subtract electrons to give it the proper charge. As an example, consider the chlorate ion, ClO3 − . Example B Cl , and O and an extra electron × are assembled as − Ο Ο Cl Ο − or Ο Ο Cl Ο 1 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.1 14 N, 7 protons and 7 neutrons; 15 N, 7 protons and 8 neutrons 1.2 (a) one (b) seven 1.3 (a) O (b) N 1.4 (a) ionic (c) four (c) Cl (e) eight (f ) five (c) covalent (d) covalent Cl F 1.5 (a) H (d) three (d) S (b) covalent C (b) H F H C Cl Cl H 1.6 H O C H H 1.7 H H O C C H •• •• H O •• (d) H •• O •• N (g) H O • • F • •• •• •• • 1.8 (a) H •• O •• • • •• •• (b) •• F •• (e) H • F • •• •• • • F •• C O P O (f ) H B H •• •• H 1.9 − O S O − O – H H (c) H O H H (h) H •• O •• •• P O •• O •• H H •• • • O − 2 O C •• O •• H THE BASICS: BONDING AND MOLECULAR STRUCTURE O H 1.10 (a) H C − O (c) − C N C (e) H O (f ) H C O H O (b) H − N (d) H C O H H C+ C 1.11 (a) H O (d) H H H (b) H O H + (e) H H C H 1.12 H C H C N H H (f ) H C H C C H H + + H (h) H C H H H H or (CH3)2CHCH(CH3)CH(CH3)2 CH3 CH3 CH3 1.14 (a) CH CH3 CH3 = CH2 = CH2 CH3 (b) CH CH3 CH2 OH N H CH3 1.13 CH3CHCHCHCH3 C H H O C H H O (g) H − O H H H H O (c) H H H − − − C C ΟΗ + N N − 3 4 THE BASICS: BONDING AND MOLECULAR STRUCTURE CH3 H C (c) C = CH3 CH2 CH3 CH2 (d) CH 3 CH2 = CH3 CH2 CH2 CH2 (e) CH3 CH3 CH = ΟΗ OH CH2 CH3 CH2 CH3 C (f ) CH2 = O (g) C CH2 CH3 CH3 CH2 CH2 (h) Cl CH3 CH CH CH3 Ο = Cl = CH3 CH2 1.15 (a) and (d) are constitutional isomers with the molecular formula C5 H12 . (b) and (e) are constitutional isomers with the molecular formula C5 H12 O. (c) and (f) are constitutional isomers with the molecular formula C6 H12 . H 1.16 (a) H H Cl H H O H C C C C C C H H H H H H H (c) H H O H H H (b) H H H C H C C C H H H H H H C C C H C C H H H H C H H H C C C C C C C H H H H H THE BASICS: BONDING AND MOLECULAR STRUCTURE 5 Cl 1.17 (a) C H (Note that the Cl atom and the three H atoms may be written at any of the four positions.) H H Cl Cl (b) or C Cl and so on C H H H H Cl Cl (c) H (d) and others C Br C H H Cl C H H H O O 1.18 (a) H C H − C H O and others O − (b) and (c). Since the two resonance structures are equivalent, each should make an equal contribution to the overall hybrid. The C—O bonds should therefore be of equal length (they should be of bond order 1.5), and each oxygen atom should bear a 0.5 negative charge. O 1.19 (a) H C Ο H H O (b) H − C C C + − H O H C H − C H H H H (c) H + C N H H H H H H − H N H H (d) C + C C C N H C N − 6 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.20 (a) CH3CH CH + CH OH CH3CH + CH3CH δ+ CH CH3CH (b) CH2 CH CH CH + δ+ δ+ CH OH + CH2 CH2 CH2 δ+ CH2 δ+ CH CH CH + CH CH OH CH CH OH CH CH CH CH CH CH δ+ CH2 + (c) + + δ+ δ+ (d) CH2 CH δ+ − Br δ− CH2 CH Br + δ+ Br CH + CH2+ CH2 CH2 CH2 CH2 (e) + + δ+ δ+ CH2 δ+ δ+ O O (f ) C C − H2C − CH3 H2C CH3 O δ− δ− H2C (g) CH3 S C CH3 CH2+ CH3 CH3 δ+ δ+ S CH2 + S CH2 CH2 + CH2 THE BASICS: BONDING AND MOLECULAR STRUCTURE (h) CH3 O O + N CH3 N CH3 − O − + O − O − 7 2+ N O (minor) O δ− + N CH3 − Oδ CH3 CH3 1.21 (a) H N + H C N CH3 + C + CH2 CH 3 N(CH3)2 because all atoms + N C C O+ H H O CH3 C CH3 NH2 O N− C •• C (c) NH2 O O O (b) CH3 •• H H have a complete octet (rule 3), and there are more covalent bonds (rule 1). NH2 C because it has no charge separation (rule 2). H N because it has no charge separation (rule 2). 1.22 (a) Cis-trans isomers are not possible. CH3 (b) CH3 C and C H H CH3 C C CH3 H H (c) Cis-trans isomers are not possible. Cl (d) CH3CH2 C and C H H H CH3CH2 C H C Cl 1.23 sp 3 1.24 sp 3 1.25 sp 2 1.27 (a) B H H (b) − H 1.26 sp F Be H F There are four bonding pairs. The geometry is tetrahedral. There are two bonding pairs about the central atom. The geometry is linear. 8 THE BASICS: BONDING AND MOLECULAR STRUCTURE (c) + H There are four bonding pairs. The geometry is tetrahedral. N H H H (d) There are two bonding pairs and two nonbonding pairs. The geometry is tetrahedral and the shape is angular. S H H H (e) There are three bonding pairs. The geometry is trigonal planar. B H H F (f) There are four bonding pairs around the central atom. The geometry is tetrahedral. C F F F F (g) There are four bonding pairs around the central atom. The geometry is tetrahedral. Si F F F − (h) Cl C Cl Cl F 120° F C 1.28 (a) There are three bonding pairs and one nonbonding pair around the central atom. The geometry is tetrahedral and the shape is trigonal pyramidal. F (b) CH3 C 120° trigonal planar at each carbon atom F 180° C C CH3 linear 180° (c) H C N linear Problems Electron Configuration 1.29 (a) Na+ has the electronic configuration, 1s 2 2s 2 2 p 6 , of Ne. (b) Cl− has the electronic configuration, 1s 2 2s 2 2 p 6 3s 2 3 p 6 , of Ar. (c) F+ and (h) Br+ do not have the electronic configuration of a noble gas. (d) H− has the electronic configuration, 1s 2 , of He. THE BASICS: BONDING AND MOLECULAR STRUCTURE 9 (e) Ca2+ has the electronic configuration, 1s 2 2s 2 2 p 6 3s 2 3 p 6 , of Ar. (f) S2− has the electronic configuration, 1s 2 2s 2 2 p 6 3s 2 3 p 6 , of Ar. (g) O2− has the electronic configuration, 1s 2 2s 2 2 p 6 , of Ne. Lewis Structures Cl O O Cl 1.30 (a) Cl (b) S P Cl (c) Cl P Cl Cl Cl Cl O (d) H + O N O Cl O 1.31 (a) CH3 O S O O − − (c) S O − S (b) CH3 O CH3 + − O O O O (d) CH3 S O − O 1.32 N + − N N + H − O (a) H O N (b) O + (c) O N F Br S N Cl (d) O O (e) Br B − O S + N H Structural Formulas and Isomerism 1.33 (a) (CH3)2CHCH2OH (c) H2C HC CH2 CH O (b) (CH3)2CHCCH(CH3)2 (d) (CH3)2CHCH2CH2OH Cl − 10 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.34 (a) C4H10O (c) C4H6 (b) C7H14O (d) C5H12O 1.35 (a) Different compounds, not isomeric (i) Different compounds, not isomeric (b) Same compound (j) Same compound (c) Same compound (k) Constitutional isomers (d) Same compound (l) Different compounds, not isomeric (e) Same compound (m) Same compound (f) Constitutional isomers (n) Same compound (g) Different compounds, not isomeric (o) Same compound (h) Same compound (p) Constitutional isomers ΟΗ Ο 1.36 (a) (d) Ο (e) (b) or Ο ΟΗ (c) (f ) 1.37 H 1.38 H C H O + N H O H − C H O + N O O H O C N H H H H C H O N O (Other structures are possible.) − THE BASICS: BONDING AND MOLECULAR STRUCTURE 11 Resonance Structures Ο Ο − 1.39 + H2N H2N Ν 1.40 − N Ν + − O 1.41 (a) Ο+ − Ο Ο (b) Ο − + + + + (c) NH2 NH2 NH2 + − − + (d) Ο Ο (e) − Ο − + − Ο − + + Ο (f ) − Ο − + Ο Ο (g) − + Ο Ο Ο (h) Ο N Ο + Ν Ο − Ο − Ο+ Ο − + Ν Ν O (i) + H 1.42 (a) While the structures differ in the position of their electrons, they also differ in the positions of their nuclei and thus they are not resonance structures. (In cyanic acid the hydrogen nucleus is bonded to oxygen; in isocyanic acid it is bonded to nitrogen.) 12 THE BASICS: BONDING AND MOLECULAR STRUCTURE (b) The anion obtained from either acid is a − resonance hybrid of the following structures: O H 1.43 H C H (a) A +1 charge. (F 4 − 6 /2 − 2 = +1) (b) A +1 charge. (It is called a methyl cation.) (c) Trigonal planar, that is, H C + H H (d) sp 2 H 1.44 C H H (a) A −1 charge. (F = 4 − 6/2 − 2 = −1) (b) A −1 charge. (It is called a methyl anion.) (c) Trigonal pyramidal, that is H C− H H (d) sp 3 H 1.45 H C H (a) No formal charge. (F = 4 − 6/2 − 1 = 0) (b) No charge. (c) sp 2 , that is, H C H H C N O C N − THE BASICS: BONDING AND MOLECULAR STRUCTURE 13 1.46 (a) H2 CO or CH2 O H C O H sp2 (b) H2 C CHCH CH2 H H H C C H C C H sp2 (c) H2 C sp2 C H C CH2 sp H H C C C C H H sp2 1.47 (a) and (b) + − O O O O O + O − (c) Because the two resonance structures are equivalent, they should make equal contributions to the hybrid and, therefore, the bonds should be the same length. (d) Yes. We consider the central atom to have two groups or units of bonding electrons and one unshared pair. 1.48 + N N N 2− − + N N N − 2− + N N N B A C Structures A and C are equivalent and, therefore, make equal contributions to the hybrid. The bonds of the hybrid, therefore, have the same length. 1.49 (a) ΟΗ OH OH OH O O O 14 THE BASICS: BONDING AND MOLECULAR STRUCTURE (b) (CH3)2NH CH3CH2NH2 (c) (CH3)3N CH3CH2NHCH3 CH3CHCH3 CH3CH2CH2NH2 NH2 (d) 1.50 (a) constitutional isomers (b) the same (c) resonance forms (d) constitutional isomers (e) resonance forms (f) the same Challenge Problems 1.51 (a) + O N O (b) Linear (c) Carbon dioxide Br 1.52 Set A: Br Br Br Br Br Br Set B: H2N H2N OH N OH OH N N O H NH2 O NH2 OH Br O N O H H H O Set C: H O c O [and unstable enol forms of a, b, and c] Set D: N + NH3 H + H − Set E: − OH H H C b a OH O O − − (i.e., CH3CH2CH2 and CH3CHCH3) THE BASICS: BONDING AND MOLECULAR STRUCTURE 15 1.53 (a) No, a carbon atom in its ground state would have 2 electrons in the 1s orbital, 2 electrons in the 2s orbital, and only 2 unpaired electrons in the degenerate 2 px ,2 p y , and 2 pz orbitals. So the two unpaired electrons can pair with only 2 hydrogen atoms with their one unpaired electron, respectively to form the compound CH2 , which would be divalent and have 180 degree bond angles. (b) In this case 4 unpaired electrons can combine with 4 hydrogen atoms to give CH4 , the correct bonding for methane, a tetravalent compound. However, the tetrahedral geometry known to exist for methane would not result from bonding at the 2s and three 2 p orbitals in the excited state. Hybridized sp 3 orbitals are required for tetrahedral geometry. 1.54 (a) Dimethyl ether Dimethylacetylene O CH3 CH3 CH3 C cis-1,2-Dichloro-1,2-difluoroethene Cl CH3 C Cl C C F (b) F Cl Cl F F O H (c) H O C C H H H H H H C H C C C H Cl Cl C H F H C F Cl Cl or F C C 1.55 The large lobes centered above and below the boron atom represent the 2 p orbital that was not involved in hybridization to form the three 2sp2 hybrid orbitals needed for the three boron-fluorine covalent bonds. This orbital is not a pure 2 p atomic orbital, since it is not an isolated atomic p orbital but rather part of a molecular orbital. Some of the other lobes in this molecular orbital can be seen near each fluorine atom. − CH2 CH O and CH2 CH O − . 1.56 The two resonance forms for this anion are The MEP indicates that the resonance contributor where the negative charge on the anion is on the oxygen is more important, which is what we would predict based on the fact that oxygen is more electronegative than carbon. δ− δ− Resonance hybrid, CH2 CH O QUIZ 1.1 Which of the following is a valid Lewis dot formula for the nitrite ion (NO− 2 )? (a) − O N (b) O O N O − (c) O N (e) None of the above 1.2 What is the hybridization state of the boron atom in BF3 ? (a) s (b) p (c) sp (d) sp 2 (e) sp 3 − O (d) Two of these F 16 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.3 BF3 reacts with NH3 to produce a compound, F of B is (a) s (b) p (d) sp 2 (c) sp F H B N F H H . The hybridization state (e) sp 3 1.4 The formal charge on N in the compound given in Problem 1.3 is (a) −2 (b) −1 (c) 0 (d) +1 (e) +2 1.5 The correct bond-line formula of the compound whose condensed formula is CH3 CHClCH2 CH(CH3 )CH(CH3 )2 is Cl Cl Cl (a) (b) Cl (c) Cl (d) (e) 1.6 Write another resonance structure for the acetate ion. Ο − O Acetate ion 1.7 In the boxes below write condensed structural formulas for constitutional isomers of CH3 (CH2 )3 CH3 . 1.8 Write a three-dimensional formula for a constitutional isomer of compound A given below. Complete the partial structure shown. H C Cl H C H3C H H A C H H H C H3C Constitutional isomer of A THE BASICS: BONDING AND MOLECULAR STRUCTURE 17 1.9 Consider the molecule (CH3 )3 B and give the following: (a) Hybridization state of boron (b) Hybridization state of carbon atoms (c) Formal charge on boron (d) Orientation of groups around boron (e) Dipole moment of (CH3)3B 1.10 Give the formal charge on oxygen in each compound. (a) CH3 O CH3 (c) CH3 O − O (b) 1.11 Write another resonance structure in which all of the atoms have a formal charge of zero. O − + H N H H 1.12 Indicate the direction of the net dipole moment of the following molecule. Cl H3C C H3C F 1.13 Write bond-line formulas for all compounds with the formula C3 H6 O. 2 FAMILIES OF CARBON COMPOUNDS: FUNCTIONAL GROUPS, INTERMOLECULAR FORCES, AND INFRARED (IR) SPECTROSCOPY SOLUTIONS TO PROBLEMS 2.1 The four carbon atoms occupy different positions in the two representations (cf. rule 2, Sec. 1.8A). 2.2 (a) H (b) I F or Br or δ+ δ− (c) Br Br µ=0D F H δ+ δ− I Br (d) F F µ=0D 2.3 VSEPR theory predicts a trigonal planar structure for BF3 . F B F F µ=0D The vector sum of the bond moments of a trigonal planar structure would be zero, resulting in a prediction of µ = 0 D for BF3 . This correlates with the experimental observation and confirms the prediction of VSEPR theory. 2.4 The shape of CCl2 moments is zero. Cl bond Cl Cl C Cl CCl2 (below) is such that the vector sum of all of the C C Cl 2.5 The fact that SO2 has a dipole moment indicates that the molecule is angular, not linear. S O O not O S O µ = 1.63 D µ=0D An angular shape is what we would expect from VSEPR theory, too. 18 FAMILIES OF CARBON COMPOUNDS 19 2.6 VSEPR theory predicts the following. net dipole δ+ H δ− O H3Cδ+ 2.7 In CFCl3 the large C F bond moment opposes the C Cl moments, leading to a net dipole moment in the direction of the fluorine. Because hydrogen is much less electronegative than fluorine, no such opposing effect occurs in CHCl3 ; therefore, it has a net dipole moment that is larger and in the direction of the chlorine atoms. F (b) C H H F C Cl C Cl Cl Larger net dipole moment C F H Br C Br Br F C µ=0D C F H net dipole moment C Br cis Br C Cl F Br C Br C trans µ=0D H Cis-trans isomers C Br net dipole moment H H C Br µ=0D C F (d) H H Br net dipole moment C C (b) F C Cl F H net dipole moment H 2.9 (a) Cl C Cl Cl Smaller net dipole moment net dipole moment H (c) H F C 2.8 (a) F Cl C cis Cl net dipole moment Cl Cis-trans isomers Br C Br C trans µ=0D Cl 20 FAMILIES OF CARBON COMPOUNDS and Br 2.10 (a) Br Br (b) 2.11 (a) (c) Br (b) F (c) Propyl bromide Cl (d) Isopropyl fluoride (e) Phenyl iodide and OH 2.12 (a) OH OH (c) (b) OH OH OH 2.13 (a) (b) 2.14 (a) O (b) (c) O (d) Methyl propyl ether (e) Diisopropyl ether (f) Methyl phenyl ether O ether 2.15 OCH3 OH alkene phenol 2.16 (a) Isopropylpropylamine (b) Tripropylamine (c) Methylphenylamine (d) Dimethylphenylamine (e) (f) CH3 NH2 N CH3 (g) N CH3 CH3 or (CH3)3N FAMILIES OF CARBON COMPOUNDS 2.17 (a) (e) only (b) (a, c) (c) (b, d, f, g) CH3 2.18 (a) CH3 CH3 + N H Cl N CH3 CH3 (b) sp 21 + H + − Cl CH3 3 O O + − 2.19 O 2.20 (a) O H O O H H O O H O (b) O 2.21 O O H O 2.22 O O O O CH3 O O O CH3 C O CH2CH3 + O CH3 C NH2 − C NH2 + CH3 + − C O O 2.24 CH3 H O O O 2.23 CH3 O O H CH2CH3 others H 22 FAMILIES OF CARBON COMPOUNDS 2.25 (a) OH would boil higher because its molecules can form hydrogen bonds to each other through the O H group. CH3 N would boil higher because its molecules can form hydrogen bonds to (b) H each other through the N H group. OH because by having two O H groups, it can form more hydrogen (c) HO bonds. 2.26 Cyclopropane would have the higher melting point because its cyclic structure gives it a rigid compact shape that would permit stronger crystal lattice forces. 2.27 d < a < b < c (c) has the highest boiling point due to hydrogen bonding involving its O (b) is a polar molecule due to its C non-polar (a) and (d). H group. O group, hence higher boiling than the essentially (a) has a higher boiling point than (d) because its unbranched structure permits more van der Waals attractions. 2.28 If we consider the range for carbon-oxygen double bond stretching in an aldehyde or ketone to be typical of an unsubstituted carbonyl group, we find that carbonyl groups with an oxygen or other strongly electronegative atom bonded to the carbonyl group, as in carboxylic acids and esters, absorb at somewhat higher frequencies. On the other hand, if a nitrogen atom is bonded to the carbonyl group, as in an amide, then the carbonyl stretching frequency is lower than that of a comparable aldehyde or ketone. The reason for this trend is that strongly electronegative atoms increase the double bond character of the carbonyl, while the unshared electron pair of an amide nitrogen atom contributes to the carbonyl resonance hybrid to give it less double bond character. Functional Groups and Structural Formulas 2.29 (a) Ketone (e) Alcohol (b) Alkyne (f) Alkene (c) Alcohol (d) Aldehyde 2.30 (a) Three carbon-carbon double bonds (alkene) and a 2◦ alcohol (b) Phenyl, carboxylic acid, amide, ester, and a 1◦ amine (c) Phenyl and a 1◦ amine (d) Carbon-carbon double bond and a 2◦ alcohol (e) Phenyl, ester, and a 3◦ amine (f) Carbon-carbon double bond and an aldehyde (g) Carbon-carbon double bond and 2 ester groups Br Br 2.31 1° Alkyl bromide Br 2° Alkyl bromide 1° Alkyl bromide Br 3° Alkyl bromide FAMILIES OF CARBON COMPOUNDS OH 2.32 OH OH 1° Alcohol 2° Alcohol O Ether O Ether 1° Alcohol (b) 2◦ (c) 3◦ (d) 3◦ (e) 2◦ 2.34 (a) 2◦ (b) 1◦ (c) 3◦ (d) 2◦ (e) 2◦ O Me O (f) 3◦ O OH OH (b) 3° Alcohol O Ether 2.33 (a) 1◦ 2.35 (a) Me OH OH OH OH OH (c) (d) O O Me (e) O H O (f) Br Br Br Br (g) Br Br Br (h) O (i) O H (j) Br O O H O H O 23 24 FAMILIES OF CARBON COMPOUNDS NH2 (k) (l) NH2 Me N H (m) Me3 N O (n) H O Me N H NH2 2.36 Crixivan has the following functional groups: 3°Amine 2°Alcohol Phenyl H OH H N HN HO H NH N H O N Aromatic amine C6H5 O C(CH3)3 Amide 2.37 The following formula is for Taxol, a natural compound with anticancer activity. Taxol has the following functional groups. Ester Phenyl Ketone O Alkene Ester O O Alcohol (2º) OH O O N H O OH HO O Alcohol (3º) Alcohol (2º) Ketone Phenyl Amide H O O Ether O Ester Phenyl Physical Properties 2.38 (a) The O H group of Vitamin A is the hydrophilic portion of the molecule, but the remainder of the molecule is not only hydrophobic but much larger. Attractive dispersion forces between the hydrophobic region of one Vitamin A molecule and another outweighs the effect of hydrogen bonding to water through a single hydroxyl group. Hence, Vitamin A is not expected to be water soluble. (b) For Vitamin B3 , there are multiple hydrophilic sites. The carbonyl oxygen and the O H of the acid function as well as the ring nitrogen can all hydrogen bond to water. Since the hydrophobic portion (the ring) of the molecule is modest in size, the molecule is expected to be water soluble. FAMILIES OF CARBON COMPOUNDS 25 2.39 The attractive forces between hydrogen fluoride molecules are the very strong dipole-dipole attractions that we call hydrogen bonds. (The partial positive charge of a hydrogen fluoride molecule is relatively exposed because it resides on the hydrogen nucleus. By contrast, the positive charge of an ethyl fluoride molecule is buried in the ethyl group and is shielded by the surrounding electrons. Thus the positive end of one hydrogen fluoride molecule can approach the negative end of another hydrogen fluoride molecule much more closely, with the result that the attractive force between them is much stronger.) 2.40 The cis isomer is polar while the trans isomer is nonpolar (µ = 0 D). The intermolecular attractive forces are therefore greater in the case of the cis isomer, and thus its boiling point should be the higher of the two. 2.41 Because of its ionic character—it is a salt—the compound is water-soluble. The organic cation and the bromide ion are well-solvated by water molecules in a fashion similar to sodium bromide. The compound also is soluble in solvents of low polarity such as diethyl ether (though less so than in water). The hydrophobic alkyl groups can now be regarded as lipophilic—groups that seek a nonpolar environment. Attractive forces between the alkyl groups of different cations can be replaced, in part, by attractive dispersion forces between these alkyl groups and ether molecules. 2.42 (a) and (b) are polar and hence are able to dissolve ionic compounds. (c) and (d) are non-polar and will not dissolve ionic compounds. H C 2.43 (a) H F H (b) C (e) H H F (f ) F F H3C No dipole moment B Cl H3C C O (i) H Cl F (c) H O (h) Cl C H H3C Cl H H (g) F F Be No dipole F moment C ( j) O H F F (d) F No dipole F moment C F 2.44 (a) Dimethyl ether: There are four electron pairs around the central oxygen: two bonding pairs and two nonbonding pairs. We would expect sp 3 hybridization of the oxygen with a bond angle of approximately 109.5◦ between the methyl groups. H3C H3C O μ> 0 D 26 FAMILIES OF CARBON COMPOUNDS (b) Trimethylamine: There are four electron pairs around the central nitrogen: three bonding pairs and one nonbonding pair. We would expect sp 3 hybridization of the nitrogen with a bond angle of approximately 109.5◦ between the methyl groups. H3C H3C N µ>0D CH3 (c) Trimethylboron: There are only three bonding electron pairs around the central boron. We would expect sp 2 hybridization of the boron with a bond angle of 120◦ between the methyl groups. CH3 µ=0D B CH3 H3C (d) Dimethylberyllium: There are only two bonding electron pairs around the central beryllium atom. We would expect sp hybridization of the beryllium atom with a bond angle of 180◦ between the methyl groups. H3C Be CH3 µ=0D 2.45 Without one (or more) polar bonds, a molecule cannot possess a dipole moment and, therefore, it cannot be polar. If the bonds are directed so that the bond moments cancel, however, the molecule will not be polar even though it has polar bonds. OH 2.46 (a) O (b) because its molecules can form hydrogen bonds to each other through its H group. HO OH because with two hydrogen bonds with each other. H groups, its molecules can form more OH because its molecules can form hydrogen bonds to each other. (c) OH [same reason as (c)]. (d) NH (e) because its molecules can form hydrogen bonds to each other through the N (f) O F F H group. because its molecules will have a larger dipole moment. (The trans compound will have µ = 0 D.) FAMILIES OF CARBON COMPOUNDS 27 O (g) OH [same reason as (c)]. (h) Nonane, because of its larger molecular weight and larger size, will have larger van der Waals attractions. O because its carbonyl group is far more polar than the double bond of (i) . IR Spectroscopy 2.47 (a) The alcohol would have a broad absorption from the O H group in the 3200 to 3500 cm−1 region of its IR spectrum. The ether would have no such absorption. (c) The ketone would have a strong absorption from its carbonyl group near 1700 cm−1 in its IR spectrum. The alcohol would have a broad absorption due to its hydroxyl group in the 3200 to 3500 cm−1 region of its IR spectrum. (d) Same rationale as for (a). (e) The secondary amine would have an absorption near 3300 to 3500 cm−1 arising from N H stretching. The tertiary amine would have no such absorption in this region since there is no N H group present. (g) Both compounds would exhibit absorptions near 1710 to 1780 cm−1 due to carbonyl stretching vibrations. The carboxylic acid would also have a broad absorption somewhere between 2500 and 3500 cm−1 due to its hydroxyl group. The ester would not have a hydroxyl absorption. (i) The ketone would have a strong absorption from its carbonyl group near 1700 cm−1 in its IR spectrum. The alkene would have no such absorption but would have an absorption between 1620 and 1680 cm−1 due to C C stretching. Hydrogen bond 2.48 O CH3CH2 H O CH2CH3 C C O H O Hydrogen bond 2.49 There are two peaks as a result of the asymmetric and symmetric stretches of the carbonyl groups. O C O O C C O asymmetric O C O symmetric 28 FAMILIES OF CARBON COMPOUNDS Multiconcept Problems 2.50 Any four of the following: O CH3CCH3 Ketone CH2 H2C O CH3CH2CH Aldehyde H2C O Ether O H3C HC CH2 Ether H2C CH2 CHCH2OH CH2 Alkene, alcohol CH O CHOH CH3 H2C Alcohol Alkene, ether The ketone carbonyl absorption is in the 1680−1750 cm−1 range; that for the aldehyde is found in the 1690–1740 cm−1 region. The C O absorption for the ethers is observed at about 1125 cm−1 . The C C absorption occurs at approximately 1650 cm−1 . Absorption for the (hydrogen-bonded) O H group takes the form of a broad band in the 3200–3550 cm−1 region. O 2.51 (a) CH3CH2CNH2 O O CH3CNCH3 HCNCH2CH3 H A O H B HCNCH3 CH3 C D (b) D, because it does not have a hydrogen that is covalently bonded to nitrogen and, therefore, its molecules cannot form hydrogen bonds to each other. The other molecules all have a hydrogen covalently bonded to nitrogen and, therefore, hydrogen-bond formation is possible. With the first molecule, for example, hydrogen bonds could form in the following way: H O H N CH3CH2C CCH2CH3 N H O H (c) All four compounds have carbonyl group absorption at about 1650 cm−1 , but the IR spectrum for each has a unique feature. A shows two N H bands (due to symmetrical and asymmetrical stretching) in the 3100–3400 cm−1 region. B has a single stretching absorption band in that same region since it has only a single N H bond. C has two absorption bands, due to the H C bond of the aldehyde group, at about 2820 cm−1 and 2920 cm−1 , as well as one for the N H bond. D does not absorb in the 3100–3500 cm−1 region, as the other compounds do, since it does not possess a N H bond. FAMILIES OF CARBON COMPOUNDS 29 2.52 The molecular formula requires unsaturation and/or one or more rings. The IR data exclude O C C C and the functional groups: OH, . Oxygen (O) must be present in an ether linkage and there can be either a triple bond or two rings present to account for the low hydrogen-to-carbon ratio. These are the possible structures: HC CCH2OCH3 O COCH2CH3 O CH3 HC O CH3C COCH3 O O CH3 O 2.53 C C O C (a cyclic ester) (CH2)n Challenge Problems O O O 2.54 H A B′ B The 1780 cm−1 band is in the general range for C O stretching so structure B′ is considered one of the possible answers, but only B would have its C O stretch at this high frequency (B′ would be at about 1730 cm−1 ). 2.55 (a) HO OH H H cis H HO H OH trans (b) The cis isomer will have the 3572 cm−1 band because only in it are the two hydroxyl groups close enough to permit intramolecular hydrogen-bonding. (Intermolecular hydrogen-bonding is minimal at high dilution.) 2.56 OH CH3 CH3 C 30 FAMILIES OF CARBON COMPOUNDS 2.57 The helical structure results from hydrogen bonds formed between amide groups— specifically between the carbonyl group of one amide and the N H group of another. H O C N C C H C H O C R H R N O H R N C O H C H R N H O C H O H C C C R N H C N H C R H R QUIZ 2.1 Which of the following pairs of compounds is not a pair of constitutional isomers? O (a) O and H (b) and O O (c) and OH (d) CH3CH2C CH (e) CH3CHCH(CH3)2 HO H and CH3CH C CH2 and (CH3)2CHCH(CH3)2 CH3 2.2 Which of the answers to Problem 2.1 contains an ether group? FAMILIES OF CARBON COMPOUNDS 2.3 Which of the following pairs of structures represents a pair of isomers? (a) and (b) and CH3 (c) CH3CH2CHCH2CH3 and CH3CH2CHCH3 CH2CH3 (d) More than one of these pairs are isomers. 2.4 Give a bond-line formula for each of the following: (a) A tertiary alcohol with the formula C5 H12 O (b) An N ,N -disubstituted amide with the formula C4 H9 NO (c) The alkene isomer of C2 H2 Cl2 that has no dipole moment (d) An ester with the formula C2 H4 O2 (e) The isomer of C2 H2 Cl2 that cannot show cis-trans isomerism 31 32 FAMILIES OF CARBON COMPOUNDS (f) The isomer of C3 H8 O that would have the lowest boiling point (g) The isomer of C4 H11 N that would have the lowest boiling point 2.5 Write the bond-line formula for a constitutional isomer of the compound shown below that does not contain a double bond. CH3 CH2 CH CH2 2.6 Circle the compound in each pair that would have the higher boiling point. O OH (a) or H (b) H or CH3 or N N O O (c) O (d) CH 3 O OH OH or CH3 O O CH3 O O (e) CH3 CH3 N H or N CH3 CH3 FAMILIES OF CARBON COMPOUNDS 2.7 Give an acceptable name for each of the following: O (a) C6H5 CH3 (b) N C6H5 (c) NH2 33 3 ACIDS AND BASES: AN INTRODUCTION TO ORGANIC REACTIONS AND THEIR MECHANISMS SOLUTIONS TO PROBLEMS O 3.1 O O H H + O − O O O O − + H H O O (a) + O O O H + S (b) O O H + F + B Cl + Al CH3 F O B− F H F Cl + CH3 Cl Cl Cl CH3 F + + B F CH3 F 34 Al − Cl Cl F O S O Cl (c) CH3 O O F (b) CH3 O − + H F 3.2 (a) CH3 H O 3.3 (a) Lewis base (d) Lewis base (b) Lewis acid (e) Lewis acid (c) Lewis base (f) Lewis base O B− F CH3 F O H ACIDS AND BASES H 3.4 CH3 + N B CH3 F F CH3 Lewis base 3.5 (a) K a = H F + 35 F − N B CH3 F F Lewis acid [H3 O+ ][HCO2 − ] = 1.77 × 10−4 [HCO2 H] Let x = [H3 O+ ] = [HCO2 − ] at equilibrium then, 0.1 − x = [HCO2 H] at equilibrium but, since the K a is very small, x will be very small and 0.1 − x ≃ 0.1 Therefore, (x)(x) = 1.77 × 10−4 0.1 x 2 = 1.77 × 10−5 x = 0.0042 = [H3 O+ ] = [HCO2 − ] (b) % Ionized = = [H3 O+ ] × 100 0.1 [HCO2 − ] × 100 0.1 or .0042 × 100 = 4.2% 0.1 3.6 (a) pK a = − log 10−7 = −(−7) = 7 (b) pK a = − log 5.0 = −0.7 (c) Since the acid with a K a = 5 has a larger K a , it is the stronger acid. H 3.7 (a) (b) O − (c) − (d) N HO O − O 3.8 The pK a of the methylaminium ion is equal to 10.6 (Section 3.6C). Since the pK a of the anilinium ion is equal to 4.6, the anilinium ion is a stronger acid than the methylaminium ion, and aniline (C6 H5 NH2 ) is a weaker base than methylamine (CH3 NH2 ). 3.9 + NH2− − + NH3 36 ACIDS AND BASES O O 3.10 R Na+ + C O H − C O O O R OH Na+ C O + C HO − OH 3.11 (a) Negative. Because the atoms are constrained to one molecule in the product, they have to become more ordered. (b) Approximately zero. (c) Positive. Because the atoms are in two separate product molecules, they become more disordered. 3.12 (a) If K eq = 1 then, log K eq = 0 = G ◦ = 0 −G ◦ 2.303RT (b) If K eq = 10 then, −G ◦ 2.303RT G ◦ = −(2.303)(0.008314 kJ mol−1 K−1 )(298 K) = −5.71 kJ mol−1 log K eq = 1 = (c) G ◦ = H ◦ − T S ◦ G ◦ = H ◦ = −5.71 kJ mol−1 if S ◦ = 0 3.13 Structures A and B make equal contributions to the overall hybrid. This means that the carbon-oxygen bonds should be the same length and that the oxygens should bear equal negative charges. O O CH3 CH3 C O A O B O CH3 C − − δ− C − Oδ hybrid 3.14 (a) CHCl2 CO2 H would be the stronger acid because the electron-withdrawing inductive effect of two chlorine atoms would make its hydroxyl proton more positive. The electronwithdrawing effect of the two chlorine atoms would also stabilize the dichloroacetate ion more effectively by dispersing its negative charge more extensively. (b) CCl3 CO2 H would be the stronger acid for reasons similar to those given in (a), except here there are three versus two electron-withdrawing chlorine atoms involved. (c) CH2 FCO2 H would be the stronger acid because the electron-withdrawing effect of a fluorine atom is greater than that of a bromine atom (fluorine is more electronegative). (d) CH2 FCO2 H is the stronger acid because the fluorine atom is nearer the carboxyl group and is, therefore, better able to exert its electron-withdrawing inductive effect. (Remember: ACIDS AND BASES 37 Inductive effects weaken steadily as the distance between the substituent and the acidic group increases.) 3.15 All compounds containing oxygen and most compounds containing nitrogen will have an unshared electron pair on their oxygen or nitrogen atom. These compounds can, therefore, act as bases and accept a proton from concentrated sulfuric acid. When they accept a proton, these compounds become either oxonium ions or ammonium ions, and having become ionic, they are soluble in the polar medium of sulfuric acid. The only nitrogen compounds that do not have an electron pair on their nitrogen atom are quaternary ammonium compounds, and these, already being ionic, also dissolve in the polar medium of concentrated sulfuric acid. 3.16 (a) CH3O H Stronger acid pKa = 16 + + (b) CH3CH2O H Stronger acid pKa = 16 (c) H N H − H Stronger base (from NaH) − NH2 Stronger base (from NaNH2) − + H Stronger acid pK a = 38 methanol CH2CH3 ethanol CH3O − Weaker base CH3CH2O Weaker base NH2− hexane Stronger base (from CH3CH2Li) − + H2 Weaker acid pKa = 35 + NH3 Weaker acid pKa = 38 + CH3CH3 Weaker acid pKa = 50 Weaker base H N H + H Stronger acid pK a = 9.2 (from NH4Cl) (e) H O H Stronger acid pK a = 15.7 − NH2 NH3 liq. NH3 Stronger base (from NaNH2) + − OC(CH3)3 Stronger base [from (CH3)3CONa] Weaker base H2O •• − + NH3 Weaker acid pKa = 38 •• (d) H + + HOC(CH3)3 Weaker base Weaker acid pKa = 18 H O •• (f) No appreciable acid-base reaction would occur because HO− is not a strong enough base to remove a proton from (CH3 )3 COH. 38 ACIDS AND BASES 3.17 (a) HC CH (b) HC CNa + D2O (c) CH3CH2Li + D2O + (d) CH3CH2OH NaH + (e) CH3CH2ONa (f ) CH3CH2CH2Li + H2 CD + NaOD CH3CH2D + LiOD HC hexane NaH + CNa HC hexane hexane CH3CH2ONa + H2 CH3CH2OT + NaOT T2O + D2O + CH3CH2CH2D hexane LiOD Problems Brønsted-Lowry Acids and Bases 3.18 (a) − NH2 (b) H O (d) H (the amide ion) − (the hydroxide ion) (c) H − (the hydride ion) 3.19 − NH2 > H − > H C (e) CH3O − C − (the ethynide ion) (the methoxide ion) (f ) H2O (water) C C − 3.20 (a) H2 SO4 > CH3O − ≈ H O − > H2O (d) NH3 + (b) H3 O (e) CH3 CH3 (c) CH3 NH3 + (f) CH3 CO2 H 3.21 H2 SO4 > H3 O+ > CH3 CO2 H > CH3 NH3 + > NH3 > CH3 CH3 Lewis Acids and Bases Cl 3.22 (a) CH3CH2 Cl + AlCl3 Lewis base CH3CH2 Cl + Al − Cl Lewis acid Cl F (b) CH3 OH + Lewis base BF3 CH3 Lewis acid C+ B− F F CH3 + H2O CH3 Lewis acid + H CH3 (c) CH3 O Lewis base CH3 C CH3 + OH2 ACIDS AND BASES 39 Curved-Arrow Notation 3.23 (a) CH3 + OH H CH3 I O + − H + I H + Cl H + F H H (b) CH3 NH2 + H CH3 Cl N + − H H H (c) C 3.24 (a) + C H H H H + F C H H − H − BF3 + O C O + BF3 − BF3 (b) (c) O+ + BF3 O + O H + H H − H Cl Cl O O H + CH3CH2CH2CH2 O (d) O Li O (a) CH3CH2 O C H + − O H − +H O H + H O H O O S O C CH3CH2 O (b) C6H5 Li+ + CH3CH2CH2CH3 O O 3.25 − H + − O O S C6H5 H O − O (c) No appreciable acid-base reaction takes place because CH3CH2ONa is too weak a base to remove a proton from ethyne. (d) H C C H + − CH2CH3 (from LiCH2CH3) hexane H C C − + CH3CH3 40 ACIDS AND BASES (e) CH3 O CH2 H + − CH2CH3 hexane CH3 CH2 O − + CH3CH3 (from LiCH2CH3) Acid-Base Strength and Equilibrium 3.26 Because the proton attached to the highly electronegative oxygen atom of CH3 OH is much more acidic than the protons attached to the much less electronegative carbon atom. 3.27 CH3CH2 O + H − C C H liq. NH3 CH3CH2 − O +H C C H 3.28 (a) pK a = − log 1.77 × 10−4 = 4 − 0.248 = 3.75 (b) K a = 10−13 3.29 (a) HB is the stronger acid because it has the smaller pK a . (b) Yes. Since A− is the stronger base and HB is the stronger acid, the following acid-base reaction will take place. A − + H B Stronger Stronger base acid pKa =10 3.30 (a) C6H5 then C C C6H5 C C (b) CH3 CH H + NaNH2 − O B− A H + Weaker Weaker acid base pKa =20 C6H5 C C Na+ + T2O C6H5 C C H + NaH CH3 CH ether − Na+ + NH3 T + NaOT O− Na+ + H2 CH3 CH3 then CH3 CH O− Na+ + D2O CH3 (c) CH3CH2CH2OH + NaH then CH3CH2CH2O− Na+ + D2O CH3 CH O D + NaOD CH3 + CH3CH2CH2O− Na + H2 CH3CH2CH2OD + NaOD ACIDS AND BASES 41 3.31 (a) CH3 CH2 OH > CH3 CH2 NH2 > CH3 CH2 CH3 Oxygen is more electronegative than nitrogen, which is more electronegative than carbon. The O-H bond is most polarized, the N-H bond is next, and the C-H bond is least polarized. (b) CH3 CH2 O− < CH3 CH2 NH− < CH3 CH2 CH2 − The weaker the acid, the stronger the conjugate base. CH > CH3CH 3.32 (a) CH3C CH2 > CH3CH2CH3 (b) CH3CHClCO2H > CH3CH2CO2H > CH3CH2CH2OH (c) CH3CH2OH2 + > CH3CH2OH > CH3OCH3 − + 3.33 (a) CH3NH3 < CH3NH2 < CH3NH (b) CH3O − < CH3NH − < CH3CH2 − CH − < CH3CH2CH2 − C − < CH3CH (c) CH3C General Problems 3.34 The acidic hydrogens must be attached to oxygen atoms. In H3 PO3 , one hydrogen is bonded to a phosphorus atom: O H O O P O H H O H P O O H H O 3.35 (a) H O + C O − O H H H O O O (b) H + C O O (c) H + C − O H H CH3 − H − H − C O H O CH3 O C O O CH3 H H + C O H − O O CH3 ACIDS AND BASES (d) H O − + CH3 H I CH3 + I O CH3 (e) H − O + CH2 H C − CH3 CH2 Cl + Cl C − CH3 + H CH3 O H C CH2 N H Stronger base O H •• O •• H H N + CH2 H Weaker acid Stronger acid •• O •• C − •• •• H •• •• O •• 3.36 (a) Assume that the acidic and basic groups of glycine in its two forms have acidities and basicities similar to those of acetic acid and methylamine. Then consider the equilibrium between the two forms: •• 42 Weaker base We see that the ionic form contains the groups that are the weaker acid and weaker base. The equilibrium, therefore, will favor this form. (b) The high melting point shows that the ionic structure better represents glycine. 3.37 (a) The second carboxyl group of malonic acid acts as an electron-withdrawing group and stabilizes the conjugate base formed (i.e., HO2 CCH2 CO2 − ) when malonic acid loses a proton. [Any factor that stabilizes the conjugate base of an acid always increases the strength of the acid (Section 3.11C).] An important factor here may be an entropy effect as explained in Section 3.10. (b) When − O2 CCH2 CO2 H loses a proton, it forms a dianion, − O2 CCH2 CO2 − . This dianion is destabilized by having two negative charges in close proximity. 3.38 HB is the stronger acid. 3.39 ∆G ° log Keq pKa pKa pKa ∆ H °− T∆S ° 6.3 kJ mol−1 − (298 K)(0.0084 kJ mol−1K−1) 3.8 kJ mol−1 log Ka = −pKa = − ∆G° 2.303RT ∆G° = 2.303RT 3.8 kJ mol −1 = (2.303)(0.008314 kJ mol−1K−1)(298 K) = = = = = 0.66 3.40 The dianion is a hybrid of the following resonance structures: O O O O − − − O O O O − − − O O O O − O O O O − ACIDS AND BASES 43 If we mentally fashion a hybrid of these structures, we see that each carbon-carbon bond is a single bond in three structures and a double bond in one. Each carbon-oxygen bond is a double bond in two structures and a single bond in two structures. Therefore, we would expect all of the carbon-carbon bonds to be equivalent and of the same length, and exactly the same can be said for the carbon-oxygen bonds. Challenge Problems 3.41 (a) A is CH3 CH2 S− B is CH3 OH − C is CH3 CH2 SCH2 CH2 O D is CH3 CH2 SCH2 CH2 OH E is OH− (b) CH3CH2 H + CH3 S O − − CH3CH2 S + CH2 CH2 O CH3CH2 S CH2CH2 O − + H O CH3CH2 CH3CH2 S CH3CH2 S − + CH3 H O CH2CH2 O − H S CH2CH2 O H + H O − CH3(CH2)8O− Li + + CH3(CH2)8 D 3.42 (a) CH3(CH2)8OD + CH3(CH2)8Li Hexane could be used as solvent. Liquid ammonia and ethanol could not because they would compete with CH3 (CH2 )8 OD and generate mostly non-deuterio-labelled CH3 (CH2 )7 CH3 . (b) NH2− + CH3C CH NH3 + CH3C C − Hexane or liquid ammonia could be used; ethanol is too acidic and would lead to CH3 CH2 O− (ethoxide ion) instead of the desired alkynide ion. (c) HCl + + NH3 + Cl − NH2 Hexane or ethanol could be used; liquid ammonia is too strong a base and would lead to NH4 + instead of the desired anilinium ion. O 3.43 (a,b) O CH3 C H N CH3 C H − + CH3 N CH3 The uncharged structure on the left is the more important resonance form. 44 ACIDS AND BASES (c) Since DMF does not bind with (solvate) anions, their electron density remains high and their size small, both of which make nucleophiles more reactive. 3.44 (a) O − O + O (b) O − − O (c) O NH2− + C CH3 H3C C H3C H3C + NH3 O O C CH2− CH2− + D2O + OD − C CH2D H3C 3.45 The most acidic hydrogen atoms in formamide are bonded to the nitrogen atom. They are acidic due to the electron-withdrawing effect of the carbonyl group and the fact that the resulting conjugate base can be stabilized by resonance delocalization of the negative charge into the carbonyl group. The electrostatic potential map shows deep blue color near the hydrogen atoms bonded to the nitrogen atom, consistent with their relative acidity. QUIZ 3.1 Which of the following is the strongest acid? (a) CH3 CH2 CO2 H (b) CH3 CH3 (c) CH3 CH2 OH (d) CH2 CH2 3.2 Which of the following is the strongest base? (a) CH3 ONa (b) NaNH2 (c) CH3 CH2 Li (d) NaOH (e) CH3 CO2 Na 3.3 Dissolving NaNH2 in water will give: (a) A solution containing solvated Na+ and NH2 − ions. (b) A solution containing solvated Na+ ions, HO− ions, and NH3 . (c) NH3 and metallic Na. (d) Solvated Na+ ions and hydrogen gas. (e) None of the above. 3.4 Which base is strong enough to convert (CH3 )3 COH into (CH3 )3 CONa in a reaction that goes to completion? (a) NaNH2 (b) CH3 CH2 Na (e) More than one of the above. (c) NaOH (d) CH3 CO2 Na ACIDS AND BASES 45 3.5 Which would be the strongest acid? (a) CH3 CH2 CH2 CO2 H (b) CH3 CH2 CHFCO2 H (d) CH2 FCH2 CH2 CO2 H (e) CH3 CH2 CH2 CH2 OH (c) CH3 CHFCH2 CO2 H 3.6 Which would be the weakest base? (a) CH3 CO2 Na (b) CF3 CO2 Na (c) CHF2 CO2 Na (d) CH2 FCO2 Na 3.7 What acid-base reaction (if any) would occur when NaF is dissolved in H2 SO4 ? 3.8 The pK a of CH3 NH3 + equals 10.6; the pK a of (CH3 )2 NH2 + equals 10.7. Which is the stronger base, CH3 NH2 or (CH3 )2 NH? 3.9 Supply the missing reagents. (a) CH3CH2C hexane CH + CH3CH2C C − Li + + CH3CH3 (b) CH3CH2C CD + LiOD 3.10 Supply the missing intermediates and reagents. (a) CH3Br + 2 Li + (b) CH3 CH3CHCH2OT (c) + LiOT T2O LiBr 4 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES SOLUTIONS TO PROBLEMS 4.1 or CH3(CH2)5CH3 Heptane or (CH3)2CHCH2CH2CH2CH3 2-Methylhexane or CH3CH2CH(CH3)CH2CH2CH3 3-Methylhexane or (CH3)3CCH2CH2CH3 2,2-Dimethylpentane or (CH3CH2)2C(CH3)2 3,3-Dimethylpentane or (CH3)2CHCH(CH3)CH2CH3 2,3-Dimethylpentane or (CH3)2CHCH2CH(CH3)2 2,4-Dimethylpentane or (CH3CH2)3CH 3-Ethylpentane or (CH3)3CCH(CH3)2 2,2,3-Trimethylbutane 46 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.2 (d); it represents 3-methylpentane 4.3 CH3 CH2 CH2 CH2 CH2 CH3 hexane 4.4 (a,b) 2-Methylheptane 2,2-Dimethylhexane 3-Methylheptane 4-Methylheptane 2,3-Dimethylhexane 2,4-Dimethylhexane 2,5-Dimethylhexane 3,3-Dimethylhexane 3-Ethylhexane 2,2,3-Trimethylpentane 2,2,4-Trimethylpentane 2,3,3-Trimethylpentane 2,3,4-Trimethylpentane 3-Ethyl-2-methylpentane 3-Ethyl-3-methylpentane 2,2,3,3-Tetramethylbutane 3,4-Dimethylhexane 47 48 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.5 (a) Pentyl 1-Methylbutyl 1-Ethylpropyl 2-Methylbutyl 3-Methylbutyl 1,2-Dimethylpropyl 1,1-Dimethylpropyl (b) See the answer to Review Problem 4.1 for the formulas and names of C7 H16 isomers. 4.6 (a) Cl Cl 1-Chloro-2-methylpropane 1-Chlorobutane Cl Cl 2-Chloro-2-methylpropane 2-Chlorobutane Br (b) 1-Bromopentane Br 1-Bromo-3-methylbutane Br Br 2-Bromopentane 1-Bromo-2-methylbutane Br Br 3-Bromopentane 2-Bromo-3-methylbutane Br Br 1-Bromo-2,2-dimethylpropane 2-Bromo-2-methylbutane NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.7 (a) OH OH 2-Methyl-1-propanol 1-Butanol OH OH 2-Methyl-2-propanol 2-Butanol (b) OH 1-Pentanol OH 3-Methyl-1-butanol OH OH 2-Pentanol 2-Methyl-1-butanol OH 3-Pentanol OH 3-Methyl-2-butanol OH OH 2,2-Dimethyl-1-propanol 2-Methyl-2-butanol 4.8 (a) 1,1-Dimethylethylcyclopentane or tert-butylcyclopentane (b) 1-Methyl-2-(2-methylpropyl)cyclohexane or 1-isobutyl-2-methylcyclohexane (c) Butylcyclohexane (d) 1-Chloro-2,4-dimethylcyclohexane (e) 2-Chlorocyclopentanol (f) 3-(1,1-Dimethylethyl)cyclohexanol or 3-tert-butylcyclohexanol 4.9 (a) 2-Chlorobicyclo[1.1.0]butane (e) 2-Methylbicyclo[2.2.2]octane (b) Bicyclo[3.2.1]octane (c) Bicyclo[2.1.1]hexane (f) Bicyclo[3.1.0]hexane or (d) 9-Chlorobicyclo[3.3.1]nonane bicyclo[2.1.1]hexane 49 50 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.10 (a) trans-3-Heptene (d) 3,5-Dimethylcyclohexene (b) 2,5-Dimethyl-2-octene (e) 4-Methyl-4-penten-2-ol (c) 4-Ethyl-2-methyl-l-hexene (f) 2-Chloro-3-methylcyclohex-3-en-1-ol 4.11 (a) (c) (b) Cl (d) Br (e) Br (f) Br Cl (g) (h) (i) Cl Cl (j) Cl 4.12 1-Hexyne 3-Methyl-1-pentyne 2-Hexyne 4-Methyl-1-pentyne 3-Hexyne 4-Methyl-2-pentyne 3,3-Dimethyl1-butyne NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.13 H3C H3C CH3 H Potential energy HH CH3 H3C H3C 3 3 H H3C H H CH3 H H H H3C CH3 H3C CH3 H3C CH3 H3C 120° 180° Rotation H H H H 60° H H H CH H CH H 0° H3C CH3 H C 3 H H3C H H 240° 300° 360° −7600 J G ◦ = = 1.32 −2.303RT (−2.303)(8.314 J K−1 )(298 K) K eq = 21.38 Let e = amount of equatorial form 4.14 log K eq = and a = amount of axial form e then, K eq = = 21.38 a e = 21.38a 21.38a × 100 = 95.5% %e= a + 21.38a Br Br 4.15 (a) H H Cl Cl (cis) Cl H (b) Br H Cl (trans) Br (cis) (trans) (c) No 4.16 (a-d) More stable because larger group is equatorial and so preferred at equilibrium Less stable because larger group is axial 51 52 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES F 4.17 Br Cl Br F Cl all axial (less stable) all equatorial (more stable) 4.18 (a,b) Less stable because the large tert-butyl group is axial (more potential energy) 4.19 More stable because the large tert-butyl group is equatorial (less potential energy) H2 H2 Pd, Pt or Ni pressure Pd, Pt or Ni pressure H2 Pd, Pt or Ni pressure H2 Pd, Pt or Ni pressure 4.20 (a) C6 H14 = formula of alkane C6 H12 = formula of 2-hexene H2 = difference = 1 pair of hydrogen atoms Index of hydrogen deficiency = 1 (b) C6 H14 = formula of alkane C6 H12 = formula of methylcyclopentane H2 = difference = 1 pair of hydrogen atoms Index of hydrogen deficiency = 1 (c) No, all isomers of C6 H12 , for example, have the same index of hydrogen deficiency. (d) No NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 53 (e) C6 H14 = formula of alkane C6 H10 = formula of 2-hexyne H4 = difference = 2 pairs of hydrogen atoms Index of hydrogen deficiency = 2 (f) C10 H22 (alkane) C10 H16 (compound) H6 = difference = 3 pairs of hydrogen atoms Index of hydrogen deficiency = 3 The structural possibilities are thus 3 double bonds 1 double bond and 1 triple bond 2 double bonds and 1 ring 1 double bond and 2 rings 3 rings 1 triple bond and 1 ring 4.21 (a) C15 H32 = formula of alkane C15 H24 = formula of zingiberene H8 = difference = 4 pairs of hydrogen atoms Index of hydrogen deficiency = 4 (b) Since 1 mol of zingiberene absorbs 3 mol of hydrogen, one molecule of zingiberene must contain three double bonds. (We are told that molecules of zingiberene do not contain any triple bonds.) (c) If a molecule of zingiberene has three double bonds and an index of hydrogen deficiency equal to 4, it must have one ring. (The structural formula for zingiberene can be found in Review Problem 23.2.) 4.22 CH3O molecular formula C10 H16 O2 O C10 H22 = formula for alkane C10 H16 = formula for unsaturated ester (ignoring oxygens) H6 = difference = 3 pairs of hydrogen atoms IHD = 3 CH3O molecular formula C10 H18 O2 Ο C10 H22 = formula for alkane C10 H18 = formula for hydrogenation product (ignoring oxygens) H4 = difference = 2 pairs of hydrogen atoms IHD = 2 54 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES Nomenclature and Isomerism 4.23 Cl (a) Cl (c) Br (b) (d) Cl (e) (f ) H H (g) CH3 OH (k) (m) H CH3 CH3 H (h) CH3 (i) Cl OH ( j) (l) or (n) (o) 4.24 (a) 5-Ethyl-7-isopropyl-2,3-dimethyldecane (b) 2,2,5-Trimethylhexane (c) 4-Bromo-6-chloro-3-methyloctane (d) 2-Methyl-1-butanol (e) 2-Bromobicyclo[3.3.1]nonane OH NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 55 (f) trans-1-Bromo-3-fluorocyclopentane (g) 5,6-Dimethyl-2-heptane (h) 7-Chlorobicyclo[2.2.1]heptane 4.25 The two secondary carbon atoms in sec-butyl alcohol are equivalent; however, there are three five-carbon alcohols (pentyl alcohols) that contain a secondary carbon atom. H3C 4.26 (a) CH3 C CH3 (b) CH3 C H3C CH3 2,2,3,3-Tetramethylbutane Cyclohexane CH3 (c) (d) CH3 1,1-Dimethylcyclobutane Bicyclo[2.2.2]octane CH3 4.27 CH3 C CH2CHCH3 CH3 or 2,2,4-trimethylpentane or 2,3,3-trimethylpentane or 2,2,3-trimethylpentane CH3 CH3 CH3CH2 C H3C CHCH3 CH3 CH3 CH3 C H3C CHCH2CH3 CH3 4.28 Cyclopentane Methylcyclobutane trans-1,2-Dimethylcyclopropane cis-1,2-Dimethylcyclopropane 1,1-Dimethylcyclopropane Ethylcyclopropane 56 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES H Cl 4.29 (a) (b) (c) CH3 (d) 4.30 S − A + 1 = N For cubane, S = 12 and A = 8. Thus 12 − 8 + 1 = N; N = 5 rings in cubane. 4.31 (a) (b) 4.32 A homologous series is one in which each member of the series differs from the one preceding it by a constant amount, usually a CH2 group. A homologous series of alkyl halides would be the following: CH3 X CH3 CH2 X CH3 (CH2 )2 X CH3 (CH2 )3 X CH3 (CH2 )4 X etc. Hydrogenation 4.33 H2 Pd, Pt, or Ni pressure H2 Pd, Pt, or Ni pressure H2 Pd, Pt, or Ni pressure H2 Pd, Pt, or Ni pressure NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 57 4.34 (a) Each of the desired alkenes must have the same carbon skeleton as 2-methylbutane, C C—C—C—C; they are therefore Pd, Pt, or Ni + H2 C2H5OH pressure (b) 4.35 2,3-Dimethylbutane From two alkenes H2 Pd, Pt, or Ni pressure Conformations and Stability 4.36 < < Least stable Most stable CH3 CH3 4.37 H H CH3 C C CH3 CH3 This conformation is less stable because 1,3-diaxial interactions with the large tert-butyl group cause considerable repulsion. H H CH3 CH3 CH3 This conformation is more stable because 1,3-diaxial interactions with the smaller methyl group are less repulsive. NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.38 HH HCH3 (a) Potential Energy H H3C H3C H3C H H CH3 H3C H3C CH3 H3C CH3 CH3 H3C H 120° 180° Rotation H3C CH3 H C 3 CH3 CH3 H3C CH3 0° CH3 CH3 H3C 240° 300° H CH3 CH3 H3C CH3 H3C 120° 180° Rotation CH3 CH3 360° H3C CH3 H3C CH3 H C 3 CH3 CH3 60° CH3 H H3C CH3 CH3 H3C H3C H H H 60° H3C H3C CH3 H3C H3C H3C H3C CH3 H3C H CH3 CH3 0° (b) CH3 CH3 CH3 CH3 H H3C Potential Energy 58 CH3 CH3 CH3 CH3 H3C CH3 H3C CH3 240° 300° CH3 CH3 CH3 CH3 360° 4.39 (a) Pentane would boil higher because its chain is unbranched. Chain-branching lowers the boiling point. (b) Heptane would boil higher because it has the larger molecular weight and would, because of its larger surface area, have larger van der Waals attractions. (c) 2-Chloropropane because it is more polar and because it has a larger molecular weight. NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 59 (d) 1-Propanol would boil higher because its molecules would be associated with each other through hydrogen-bond formation. (e) Propanone (CH3 COCH3 ) would boil higher because its molecules are more polar. H 4.40 C4H6 1-Butyne Bicyclo[1.1.0]butane The IR stretch at ∼ 2250 cm−1 for the alkyne C pounds. C bond distinguishes these two com- 4.41 trans-1,2-Dimethylcyclopropane would be more stable because there is less crowding between its methyl groups. H3C CH3 H3C H H H CH3 H Less stable More stable 4.42 For 1,2-disubstituted cyclobutanes, the trans isomer is e,e; the cis isomer is a,e, and so the less stable of the two. For 1,3-disubstituted cyclobutanes, the cis isomer is e,e and more stable than the a,e trans isomer. a e e Trans a e e Cis e e Cis Trans C(CH3)3 CH 3 4.43 (a) (CH3)3C CH3 More stable conformation because both alkyl groups are equatorial C(CH3)3 (b) (CH3)3C CH3 CH3 More stable because larger group is equatorial 60 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES (c) C(CH3)3 (CH3)3C CH3 CH3 More stable conformation because both alkyl groups are equatorial (d) CH3 C(CH3)3 CH3 (CH3)3C More stable because larger group is equatorial 4.44 Certainly it is expected that the alkyl groups would prefer the equatorial disposition in a case such as this, and indeed this is true in the case of all-trans-1,2,3,4,5,6-hexaethylcyclohexane, which does have all ethyl groups equatorial. Apparently, the torsional and steric effects resulting from equatorial isopropyl groups destabilize the all-equatorial conformation to a greater degree than axial isopropyl groups destabilize the all-axial conformation. The fully axial structure assigned on the basis of X-ray studies is also supported by calculations. 4.45 If the cyclobutane ring were planar, the C—Br bond moments would exactly cancel in the trans isomer. The fact that trans-1,3-dibromocyclobutane has a dipole moment shows the ring is not planar. Br Br Br Br Planar form µ=0D Bent form µ≠0D Synthesis 4.46 (a) H2 ; Pd, Pt, or Ni catalyst, pressure (b) H2 ; Pd, Pt, or Ni catalyst, pressure (c) Predicted IR absorption frequencies for reactants in parts (a) and (b) are the following: (1) trans-5-Methyl-2-hexene has an absorption at approximately 964 cm−1 , characteristic of C—H bending in a trans-substituted alkene. (2) The alkene double bond of the reactant is predicted to have a stretching absorption between 1580 and 1680 cm−1 . NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 61 Challenge Problems 4.47 If trans-1,3-di-tert-butylcyclohexane were to adopt a chair conformation, one tert-butyl group would have to be axial. It is, therefore, more energetically favorable for the molecule to adopt a twist boat conformation. 4.48 (a) More rules are needed (see Chapter 7) to indicate relative stereochemistry for these 1-bromo-2-chloro-1-fluoroethenes. (b) Bicyclo[4.4.0]decane (or decalin) (c) Bicyclo[4.4.0]dec-2-ene (or 1 -octalin) (d) Bicyclo[4.4.0]dec-1-ene (or 1(8a) -octalin) NOTE: The common name decalin comes from decahydronaphthalene, the derivative of 8 7 naphthalene 6 8a 4a 5 1 2 3 that has had all of its five double bonds converted 4 to single bonds by addition of 10 atoms of hydrogen. Octalin similarly comes from octahydronaphthalene and contains one surviving double bond. When using these common names derived from naphthalene, their skeletons are usually numbered like that of naphthalene. When, as in case (d), a double bond does not lie between the indicated carbon and the next higher numbered carbon, its location is specified as shown. Also, the symbol  is one that has been used with common names to indicate the presence of a double bond at the position specified by the accompanying superscript number(s). 4.49 The cyclohexane ring in trans-1-tert-butyl-3-methylcyclohexane is in a chair conformation. The ring in trans-1,3-di-tert-butylcyclohexane is in a twist-boat conformation. In trans-1tert-butyl-3-methylcyclohexane the tert-butyl group can be equatorial if the methyl group is axial. The energy cost of having the methyl group axial is apparently less than that for adopting the twist-boat ring conformation. In trans-1,3-di-tert-butylcyclohexane the potential energy cost of having one tert-butyl group equatorial and the other axial is apparently greater than having the ring adopt a twist-boat conformation so that both can be approximately equatorial. 4.50 All of the nitrogen-containing five-membered rings of Vitamin B12 contain at least one sp2 -hybridized atom (in some cases there are more). These atoms, because of the trigonal planar geometry associated with them, impose some degree of planarity on the nitrogencontaining five-membered rings in B12 . Furthermore, 13 atoms in sequence around the cobalt are sp2 -hybridized, a fact that adds substantial resonance stabilization to this part of the ring system. The four five-membered nitrogen-containing rings that surround the cobalt in roughly a plane and whose nitrogens lend their unshared pairs to the cobalt comprise what is called a corrin ring. The corrin ring may look familiar to you, and for good reason, because it is related to the porphyrin ring of heme. (Additional question: What geometry is associated with the cobalt atom and its six bound ligands? Answer: octahedral, or square bipyramidal.) 4.51 The form of a benzene ring occurs 16 times in buckminsterfullerene. The other eight facets in the 24 faces of a “buckyball” are five-membered rings. Every carbon of buckminsterfullerene is approximately sp2 -hybridized. 62 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES QUIZ 4.1 Consider the properties of the following compounds: NAME Ethane Fluoromethane Methanol FORMULA CH3 CH3 CH3 F CH3 OH BOILING POINT (◦ C) MOLECULAR WEIGHT −88.2 30 −78.6 34 +64.7 32 Select the answer that explains why methanol boils so much higher than ethane or fluoromethane, even though they all have nearly equal molecular weights. (a) Ion-ion forces between molecules. (b) Weak dipole-dipole forces between molecules. (c) Hydrogen bonding between molecules. (d) van der Waals forces between molecules. (e) Covalent bonding between molecules. 4.2 Select the correct name of the compound whose structure is (a) 2,5-Diethyl-6-methyloctane (b) 4,7-Diethyl-3-methyloctane (c) 4-Ethyl-3,7-dimethylnonane (d) 6-Ethyl-3,7-dimethylnonane (e) More than one of the above CH3 4.3 Select the correct name of the compound whose structure is CH3CHCH2Cl . (a) Butyl chloride (b) Isobutyl chloride (c) sec-Butyl chloride (d) tert-Butyl chloride (e) More than one of the above 4.4 The structure shown in Problem 4.2 has: (a) 1◦ , 2◦ , and 3◦ carbon atoms (b) 1◦ and 2◦ carbon atoms only (c) 1◦ and 3◦ carbon atoms only (d) 2◦ and 3◦ carbon atoms only (e) 1◦ , 3◦ , and 4◦ carbon atoms only NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.5 How many isomers are possible for C3 H7 Br? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 4.6 Which isomer of 1,3-dimethylcyclohexane is more stable? (a) cis (b) trans (c) Both are equally stable (d) Impossible to tell 4.7 Which is the lowest energy conformation of trans-1,4-dimethylcyclohexane? H (a) CH3 CH3 CH3 (b) H H H CH3 H (c) H CH3 CH3 (d) CH 3 CH3 H H (e) More than one of the above 4.8 Supply the missing structures (a) ring flip Cl Cl (b) 2-Bromobicyclo[2.2.1]heptane (c) Newman projection for a gauche form of 1,2-dibromoethane 63 64 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 4.9 Supply the missing reagents in the box: 4.10 The most stable conformation of trans-1-isopropyl-3-methylcyclohexane: 5 STEREOCHEMISTRY: CHIRAL MOLECULES SOLUTIONS TO PROBLEMS 5.1 (a) Achiral (c) Chiral (e) Chiral (g) Chiral (b) Achiral (d) Chiral (f) Chiral (h) Achiral (b) No (c) No 5.2 (a) Yes 5.3 (a) They are the same molecule. (b) They are enantiomers. 5.4 (a), (b), (e), (g), and (i ) do not have chirality centers. CH3 CH3 (c) H C Cl Cl C CH3 CH2OH HOCH2 CH2 CH3 C (1) CH3 (2) CH3 Br Br C CH2 CH2 CH2 CH2 CH3 H C CH2 CH3 (f) H (2) CH3 (1) CH3 (d) H H CH2 CH2 CH3 C (1) CH3 H (2) 65 66 STEREOCHEMISTRY: CHIRAL MOLECULES (h) H CH3 CH3 CH2 CH2 CH3 C H3C CH2 CH2 CH2 CH2 CH3 CH3 (1) CH3 (j) H H C CH3 CH2Cl C ClCH2 CH2 CH3 (2) C H CH2 (1) CH3 (2) O 5.5 (a) (b) * N O * N O Limonene O H Thalidomide O O N N H H O (R) O H (S ) O O N N H H O (S ) O (R) H STEREOCHEMISTRY: CHIRAL MOLECULES OH 5.6 (a) OH O * (c) HO * O O * OH HO H * (b) HO OH OH (d) O OH * * * * * HO 5.7 The following items possess a plane of symmetry, and are, therefore, achiral. (a) A screwdriver (b) A baseball bat (ignoring any writing on it) (h) A hammer 5.8 In each instance below, the plane defined by the page is a plane of symmetry. (a) (b) H C H3C H3C F (g) (e) CH3 H H3C C C H3C H C CH3CH2 C CH3 CH3CH2 CH3 5.9 H Br F F C C Cl (S ) 5.10 (c) (1) is (S ) (2) is (R) (d) (1) is (S) (2) is (R) Cl (R) H Br H CH2CH3 (i) CH3 H H C CH2CH3 C CH3 67 68 STEREOCHEMISTRY: CHIRAL MOLECULES (f) (1) is (S) (2) is (R) (h) (1) is (S) (2) is (R) (j) (1) is (S) (2) is (R) SH > Cl > 5.11 (a) > H CH2Cl > (b) CH2Br (c) OH (d) C(CH3)3 > (e) OCH3 > (f ) OPO3H2 > 5.12 (a) (S ) OH > CHO > CH CH2OH > CH3 > CH2 > N(CH3)2 > > OH > (b) (R ) CH3 H CH(CH3)2 > CH3 > CHO > (c) (S ) H H H (d) (R ) 5.13 (a) Enantiomers (b) Two molecules of the same compound (c) Enantiomers 5.14 O O H H (R)-(−)-Carvone (S )-(+)-Carvone 5.15 (a) Enant. excess = observed specific rotation × 100 specific rotation of pure enantiomer +1.151 × 100 +5.756 = 20.00% = (b) Since the (R) enantiomer (see Section 5.8C) is +, the (R) enantiomer is present in excess. STEREOCHEMISTRY: CHIRAL MOLECULES O O 5.16 (a) HO CH2 CH3 C CH3 NH2 HO CH3 C CH2 C OH C H H CH3 (S )-Ibuprofen 5.17 HO H (R) configuration 5.18 O N C OH C NH2 (S )-Penicillamine O CH3 C SH H (S )-Methyldopa (c) CH3 (b) OH C O (R) (S) 5.19 (a) Diastereomers. (b) Diastereomers in each instance. (c) No, diastereomers have different melting points. (d) No, diastereomers have different boiling points. (e) No, diastereomers have different vapor pressures. 69 70 STEREOCHEMISTRY: CHIRAL MOLECULES STUDY AID An Approach to the Classification of Isomers We can classify isomers by asking and answering a series of questions: Do the compounds have the same molecular formula? Yes No Are they different? They are not isomers. Yes No They are isomers. They are identical. Do they differ in connectivity? Yes No They are constitutional isomers. They are stereoisomers. Are they mirror images of each other? Yes No They are enantiomers. They are diastereomers. 5.20 (a) It would be optically active. (b) It would be optically active. (c) No, because it is a meso compound. (d) No, because it would be a racemic form. 5.21 (a) Represents A (b) Represents C (c) Represents B STEREOCHEMISTRY: CHIRAL MOLECULES H 5.22 (a) Cl H H Cl (1) H H Cl Cl H H Cl (2) (3) Enantiomers (b) Meso compound HO H OH H HO (1) H H OH (3) Meso compound Enantiomers F F Cl (c) Cl H H F F (2) H HO H (1) Meso compound H Cl (2) F (3) (4) Enantiomers H F HO H H OH H Cl Enantiomers H H (3) Enantiomers H OH Cl Cl Cl H (1) F H Cl Cl F (d) H OH HO (2) H Cl H F F H (e) H Br (1) Br H (2) Enantiomers H Br Br (3) Enantiomers H (4) Cl H 71 72 STEREOCHEMISTRY: CHIRAL MOLECULES OH H (f ) HO2C H H HO HO2C CO2H CO2H HO OH (1) H (2) Enantiomers HO HO2C H H CO2H OH (3) Meso compound 5.23 B is (2S,3S)-2,3-Dibromobutane C is (2R,3S)-2,3-Dibromobutane 5.24 (a) (1) is (2S,3S)-2,3-Dichlorobutane (2) is (2R,3R)-2,3-Dichlorobutane (3) is (2R,3S)-2,3-Dichlorobutane (b) (1) is (2S,4S)-2,4-Pentanediol (2) is (2R,4R)-2,4-Pentanediol (3) is (2R,4S)-2,4-Pentanediol (c) (1) is (2R,3R)-1,4-Dichloro-2,3-difluorobutane (2) is (2S,3S)-1,4-Dichloro-2,3-difluorobutane (3) is (2R,3S)-1,4-Dichloro-2,3-difluorobutane (d) (1) is (2S,4S)-4-Chloro-2-pentanol (2) is (2R,4R)-4-Chloro-2-pentanol (3) is (2S,4R)-4-Chloro-2-pentanol (4) is (2R,4S)-4-Chloro-2-pentanol (e) (1) is (2S,3S)-2-Bromo-3-fluorobutane (2) is (2R,3R)-2-Bromo-3-fluorobutane (3) is (2S,3R)-2-Bromo-3-fluorobutane (4) is (2R,3S)-2-Bromo-3-fluorobutane (f) (1) is (2R,3R)-Butanedioic acid (2) is (2S,3S)-Butanedioic acid (3) is (2R,3S)-Butanedioic acid STEREOCHEMISTRY: CHIRAL MOLECULES NO2 5.25 HO C H C H NHCOCHCl2 CH2OH Chloramphenicol 5.26 (a) A C1, R; C2, R B C1, S; C2, S (b,c) CH3 H Br H Br optically inactive, a meso compound CH3 C 5.27 (a) No (b) Yes (c) No (d) No (e) Diastereomers (f) Diastereomers 5.28 Me Me Me H H Me H H H Me Me H Meso compound Enantiomers 73 74 STEREOCHEMISTRY: CHIRAL MOLECULES H 5.29 (a) Cl Cl Br = H H H Br H Cl Cl Br H H H (1R,2R) Br Cl Br Cl = (1S,2S) = (1R,2S) Br H H H Br H Cl Br Cl Br Cl Br Cl H H H Br Cl H = H Enantiomers (both cis) (1S,2R) Br Cl (b) Enantiomers (both trans) H Cl H Cl = H Br Cl Br Cl Br Br Cl Cl H = Br H H Cl H H (1R,3S) Br Cl Cl = H Br Enantiomers (both cis) H H Br (1S,3R) (1S,3S) H Br Br H Cl Enantiomers (both trans) Cl = Br H H (1R,3R) Cl Br STEREOCHEMISTRY: CHIRAL MOLECULES H Br (c) H Cl Br H = Br H Cl Cl Cl Achiral (trans) Br H Br 75 H Cl = Br H H Cl Achiral (cis) 5.30 See Problem 5.29. The molecules in (c) are achiral, so they have no (R, S ) designation. HO 5.31 H HO HgO O HO OH (R)-(+)-3-Bromo2-hydroxypropanoic acid CHO R S OH HO CH2OH (R)-Glyceraldehyde CH2OH (S )-Glyceraldehyde CO2H CO2H R H S OH HO R HO H H S H CO2H (+)-Tartaric acid H OH CO2H (−)-Tartaric acid (S )-(−)-Isoserine (see the following reaction also) HO O Br CHO (b) H HBr OH (S )-(−)-Isoserine 5.32 (a) H H2O HO HNO2 O HNO2 OH (S)-(+)-Glyceric acid H H2N O HO H (S)-(−)-Glyceraldehyde HO H Zn H3O+ H O OH (S )-(+)-Lactic acid 76 STEREOCHEMISTRY: CHIRAL MOLECULES CO2H CO2H R (c) H S OH HO CH3 H CH3 (S)-Lactic acid (R)-Lactic acid Problems Chirality and Stereoisomerism 5.33 (a), (b), (f), and (g) only 5.34 (a) Seven. (b) (R)- and (S)-3-Methylhexane and (R)- and (S)-2,3-dimethylpentane. S Cl F Cl Br S F H2N Cl 5.35 R R S SH Br S S Cl SH S S H H H 5.36 H H H H N O H H O H 5.37 O (a) Me H Me Me H (R) configuration Cl (b) Two, indicated by asterisks in (a) Cl (c) Four (d) Because a trans arrangement of the one carbon bridge is structurally impossible. Such a molecule would have too much strain. STEREOCHEMISTRY: CHIRAL MOLECULES 77 5.38 (a) A is (2R,3S)-2,3-dichlorobutane; B is (2S,3S)-2,3-dichlorobutane; C is (2R,3R)-2,3dichlorobutane. (b) A CH3 5.39 (a) or (b) CH3 CH2CH3 and CH3 CH3 or CH2 or CH3 and CH3 CH3 H3C CH3 etc. (other answers are possible) and (c) CH3 CH3 CH3 CH3 (other answers are possible) CH3 (d) CH3 and C C H CH CH3CH2 CH2 CH2 CH CH2CH2CH3 (e) H3C C H C H CH2CH3 C and H H (other answers are possible) 5.40 (a) Same: (S) (b) Enantiomers: left (S); right (R) (c) Diastereomers: left (1S, 2S); right (1R, 2S) (d) Same: (1S, 2S) (e) Diastereomers: left (1S, 2S); right (1S, 2R) (f) Constitutional isomers: both achiral (g) Diastereomers: left, cis (4S); right, trans (4R) (h) Enantiomers: left (1S, 3S); right (1R, 3R) (i) Same: no chirality centers H H3C C CH2CH2CH3 CH2 CH3 CH 3 78 STEREOCHEMISTRY: CHIRAL MOLECULES (j) Different conformers of the same molecule (interconvertable by a ring flip): (1R, 2S) (k) Diastereomers: left (1R, 2S); right (1R, 2R) (l) Same: (1R, 2S) (m) Diastereomers: no chirality center in either (n) Constitutional isomers: no chirality center in either (o) Diastereomers: no chirality center in either (p) Same: no chirality center (q) Same: no chirality center 5.41 All of these molecules are expected to be planar. Their stereochemistry is identical to that of the corresponding chloroethenes. (a) can exist as cis and trans isomers. Only one compound exists in the case of (b) and (c). 5.42 (a) diastereomers (c) enantiomers (b) enantiomers (d) same compound H H 5.43 CH2 C CH2 CH2CH3 C C CH3 CH2CH3 C H H CH3 D (racemic) CH3CH2 CH2CH3 C CH3 H E (achiral) 5.44 CH3 CH3 C C H2 C H H (or enantiomer) F 5.45 CH3 (or enantiomer) H or CH2CH3 (or enantiomer) CH3CH2CH2CH2CH3 Pd, Pt, or Ni pressure (achiral) G CH3 H2 Pd, Pt, or Ni pressure (achiral) I CH2CH3 H2 Pd, Pt, or Ni pressure (achiral) STEREOCHEMISTRY: CHIRAL MOLECULES 5.46 79 (S) Ο + H3N O N H − CH3 Ο O (S) Ο Aspartame CH3 CH3 CH3 5.47 (a) CH3 (1) (2) CH3 (3) CH3 (4) CH3 CH3 (b) (3) and (4) are chiral and are enantiomers of each other. (c) Three fractions: a fraction containing (1), a fraction containing (2), and a fraction containing (3) and (4) [because, being enantiomers, (3) and (4) would have the same vapor pressure]. (d) None R S Cl S 5.48 S Br Cl R CH3 H3C Br Enantiomer (mirror image) all centers changed Diastereomer (one center changed) H H 5.49 (a) Et Et Et Et H H (b) No, they are not superposable. (c) No, and they are, therefore, enantiomers of each other. Et Et (d) H Et H H Et H S 80 STEREOCHEMISTRY: CHIRAL MOLECULES (e) No, they are not superposable. (f) Yes, and they are, therefore, just different conformations of the same molecule. H H 5.50 (a) Et Et Et Et H H (b) Yes, and, therefore, trans-1,4-diethylcyclohexane is achiral. (c) No, they are different orientations of the same molecule. (d) Yes, cis-1,4-diethylcyclohexane is a stereoisomer (a diastereomer) of trans-1, 4-diethylcyclohexane. Et Et H H cis-1,4-Diethylcyclohexane (e) No, it, too, is superposable on its mirror image. (Notice, too, that the plane of the page constitutes a plane of symmetry for both cis-1,4-diethylcyclohexane and for trans-1, 4-diethylcyclohexane as we have drawn them.) 5.51 trans-1,3-Diethylcyclohexane can exist in the following enantiomeric forms. Et Et H H H Et Et trans-1,3-Diethylcyclohexane enantiomers H cis-1,3-Diethylcyclohexane consists of achiral molecules because they have a plane of symmetry. [The plane of the page (below) is a plane of symmetry.] Et Et HH cis-1,3-Diethylcyclohexane (meso) STEREOCHEMISTRY: CHIRAL MOLECULES 81 Challenge Problems 5.52 (a) Since it is optically active and not resolvable, it must be the meso form: CO2H H H C C (b) OH H OH HO C C OH H CO2H CO2H (R, R) (meso) (c) No CO2H CO2H HO H C C H OH CO2H (S, S) (d) A racemic mixture 5.53 (a) [α]D = −30 = −300 (0.10 g/mL)(1.0 dm) +165 = +3300 (0.05 g/mL)(1.0 dm) The two rotation values can be explained by recognizing that this is a powerfully optically active substance and that the first reading, assumed to be −30, was really +330. Making this change the [α]D becomes +3300 in both cases. (b) [α]D = (c) No, the apparent 0 rotation could actually be + or −360 (or an integral multiple of these values). 5.54 Yes, it could be a meso form or an enantiomer whose chirality centers, by rare coincidence, happen to cancel each other’s activities. 5.55 A compound C3 H6 O2 has an index of hydrogen deficiency of 1. Thus, it could possess a carbon-carbon double bond, a carbon-oxygen double bond, or a ring. The IR spectral data rule out a carbonyl group but indicate the presence of an —OH group. No stable structure having molecular formula C3 H6 O2 with a C C bond can exist in stereoisomeric forms but 1,2-cyclopropanediol can exist in three stereoisomeric forms. Only ethylene oxide (oxirane) derivatives are possible for Y. CH2OH O HOCH2 O 82 STEREOCHEMISTRY: CHIRAL MOLECULES QUIZ 5.1 Describe the relationship between the two structures shown. CH3 H C H Br CH3 Cl Br C Cl (a) Enantiomers (b) Diastereomers (c) Constitutional isomers (d) Conformations (e) Two molecules of the same compound 5.2 Which of the following molecule(s) possess(es) a plane of symmetry? F (a) H H F Cl C (b) F Cl C Cl (c) Br Br Br H (e) None of these (d) More than one of these 5.3 Give the (R, S) designation of the structure shown: O CH3 C HO H C Cl (a) (R) (b) (S) (c) Neither, because this molecule has no chirality center. (d) Impossible to tell 5.4 Select the words that best describe the following structure: CH3 H H Cl C C Cl CH3 (a) Chiral (b) Meso form (e) More than one of these (c) Achiral (d) Has a plane of symmetry STEREOCHEMISTRY: CHIRAL MOLECULES 83 5.5 Select the words that best describe what happens to the optical rotation of the alkene shown when it is hydrogenated to the alkane according to the following equation: H H CH3CH2 C (R) CH CH3 H2 CH3CH2 Ni pressure C CH3 CH2CH3 CH2 (a) Increases (b) Changes to zero (c) Changes sign (d) Stays the same (e) Impossible to predict 5.6 There are two compounds with the formula C7 H16 that are capable of existing as enantiomers. Write three-dimensional formulas for the (S) isomer of each. 5.7 Compound A is optically active and is the (S ) isomer. H2 Ni pressure CH3CHCH2CH3 CH2CH3 A 5.8 Compound B is a hydrocarbon with the minimum number of carbon atoms necessary for it to possess a chirality center and, as well, alternative stereochemistries about a double bond. B 84 STEREOCHEMISTRY: CHIRAL MOLECULES 5.9 Which is untrue about the following structure? Cl Cl (a) It is the most stable of the possible conformations. (b) µ = 0 D (c) It is identical to its mirror image. (d) It is optically active. (e) (R,S ) designations cannot be applied. CH2OH 5.10 H C OH HO C H H C OH is a Fischer projection of one of stereoisomers. CH3 (a) 2 (b) 3 (c) 4 (d) 7 (e) 8 6 IONIC REACTIONS — NUCLEOPHILIC SUBSTITUTION AND ELIMINATION REACTIONS OF ALKYL HALIDES SOLUTIONS TO PROBLEMS 6.1 (a) cis-1-Bromo-2-methylcyclohexane (b) cis-1-Bromo-3-methylcyclohexane (c) 2,3,4-Trimethylheptane 6.2 (a) 3◦ 6.3 (a) CH3 + CH3CH2 I Substrate (b) I (c) 2◦ (b) alkenyl (vinylic) O (e) 1◦ (d) aryl − CH3 O CH2CH3 + Nucleophile − + CH3CH2 Nucleophile Nucleophile Cl − Leaving group Br CH3CH2 I + Br Substrate (c) 2 CH3OH + (CH3)3C I − Leaving group (CH3)3C O CH3 + Substrate Cl − + + CH3OH2 Leaving group N Br (d) + Substrate (e) C N Nucleophile Br Substrate − + 2 NH3 Nucleophile C + Br − Leaving group NH2 + Br − + + NH4 Leaving group 85 IONIC REACTIONS Transition state CH2CH2CH3 δ− δ− CH2 I Cl 6.4 ∆G ‡ Free energy of activation Cl − I + Free energy 86 Reactants Freeenergy change ∆G° I + Cl− Products Reaction coordinate I − I 6.5 (CH3)3C (CH3)3C Br + Br − 6.6 (a) We know that when a secondary alkyl halide reacts with hydroxide ion by substitution, the reaction occurs with inversion of configuration because the reaction is SN 2. If we know that the configuration of (−)-2-butanol (from Section 5.8C) is that shown here, then we can conclude that (+)-2-chlorobutane has the opposite configuration. H OH − HO SN2 Cl H (S )-(+)-2-Chlorobutane [α] 25° = +36.00 D (R)-(− )-2-Butanol [α] 25° = − 13.52 D (b) Again the reaction is SN 2. Because we now know the configuration of (+)-2-chlorobutane to be (S) [cf., part (a)], we can conclude that the configuration of (−)-2-iodobutane is (R). Cl H (S )-(+)-2-Chlorobutane I− SN2 H I (R)-(−)-2-Iodobutane (+)-2-Iodobutane has the (S ) configuration. IONIC REACTIONS 6.7 (b) < (c) < 87 (a) in order of increasing stability 6.8 (a, b) CH3 (CH3)3C (b) H2O I (CH3)3C SN1 + OH2 CH3 + A− OH CH3 (CH3)3C + OH (CH3)3C CH3 By path (a) 6.9 (a) − HA By path (b) OCH3 CH3 (CH3)3C and OCH3 (CH3)3C CH3 6.10 (c) is most likely to react by an SN 1 mechanism because it is a tertiary alkyl halide, whereas (a) is primary and (b) is secondary. 6.11 (a) Being primary halides, the reactions are most likely to be SN 2, with the nucleophile in each instance being a molecule of the solvent (i.e., a molecule of ethanol). (b) Steric hindrance is provided by the substituent or substituents on the carbon β to the carbon bearing the leaving group. With each addition of a methyl group at the β carbon (below), the number of pathways open to the attacking nucleophile becomes fewer. Nu H H Nu H3C C Br H H H H Nu H3C C Br H H H H Nu H3C C Br H H3C 6.12 CN− > CH3 O− > CH3 CO− 2 > CH3 CO2 H > CH3 OH Order of decreasing nucleophilicity in methanol H H H3C H C Br H3C 88 IONIC REACTIONS 6.13 CH3 S− > CH3 O− > CH3 CO− 2 > CH3 SH > CH3 OH Order of decreasing nucleophilicity in methanol 6.14 Protic solvents are those that have an H bonded to an oxygen or nitrogen (or to another O strongly electronegative atom). Therefore, the protic solvents are formic acid, HCOH ; O formamide, HCNH2 ; ammonia, NH3 ; and ethylene glycol, HOCH2 CH2 OH. Aprotic solvents lack an H bonded to a strongly electronegative element. Aprotic solO vents in this list are acetone, CH3CCH3 ; acetonitrile, CH3C trimethylamine, N(CH3 )3 . N ; sulfur dioxide, SO2 ; and 6.15 The reaction is an SN 2 reaction. In the polar aprotic solvent (DMF), the nucleophile (CN− ) will be relatively unencumbered by solvent molecules, and, therefore, it will be more reactive than in ethanol. As a result, the reaction will occur faster in N ,N -dimethylformamide. 6.16 (a) CH3 O− (b) H2 S (c) (CH3 )3 P 6.17 (a) Increasing the percentage of water in the mixture increases the polarity of the solvent. (Water is more polar than methanol.) Increasing the polarity of the solvent increases the rate of the solvolysis because separated charges develop in the transition state. The more polar the solvent, the more the transition state is stabilized (Section 6.13D). (b) In an SN 2 reaction of this type, the charge becomes dispersed in the transition state: I− + CH3CH2 δ− I Cl CH3 C + + δ− Cl ICH2CH3 + Cl − H H Reactants Charge is concentrated Transition state Charge is dispersed Increasing the polarity of the solvent increases the stabilization of the reactant I− more than the stabilization of the transition state, and thereby increases the free energy of activation, thus decreasing the rate of reaction. 6.18 CH3OSO2CF3 > CH3I > CH3Br (Most reactive) > CH3Cl > CH3F > 14CH OH 3 (Least reactive) IONIC REACTIONS CH3 6.19 (a) CH3CH2O − Na + + C CH3CH2 O Br (b) CH3CO + Na + Br (c) HS + Na + C CH3CH2 (d) CH3S Na + inversion SN2 Br + C CH3CH2 CH3CO Br + Na+ Br − C inversion SN2 H (S ) CH2CH3 CH3 HS + Na+ Br − C (R) H H (S ) + Na+ Br − CH2CH3 CH3 (R) H inversion SN2 CH3 − (R) H H (S ) CH3 − C CH3CH2O O C CH3CH2 CH3 H (S ) CH3 − inversion SN2 89 CH2CH3 CH3 CH3S C (R) H + Na+ Br − CH2CH3 Relative Rates of Nucleophilic Substitution 6.20 (a) 1-Bromopropane would react more rapidly because, being a primary halide, it is less hindered. (b) 1-Iodobutane, because iodide ion is a better leaving group than chloride ion. (c) 1-Chlorobutane, because the carbon bearing the leaving group is less hindered than in 1-chloro-2-methylpropane. (d) 1-Chloro-3-methylbutane, because the carbon bearing the leaving group is less hindered than in 1-chloro-2-methylbutane. (e) 1-Chlorohexane because it is a primary halide. Phenyl halides are unreactive in SN 2 reactions. 6.21 (a) Reaction (1) because ethoxide ion is a stronger nucleophile than ethanol. (b) Reaction (2) because the ethyl sulfide ion is a stronger nucleophile than the ethoxide ion in a protic solvent. (Because sulfur is larger than oxygen, the ethyl sulfide ion is less solvated and it is more polarizable.) (c) Reaction (2) because triphenylphosphine [(C6 H5 )3 P] is a stronger nucleophile than triphenylamine. (Phosphorus atoms are larger than nitrogen atoms.) (d) Reaction (2) because in an SN 2 reaction the rate depends on the concentration of the substrate and the nucleophile. In reaction (2) the concentration of the nucleophile is twice that of the reaction (1). 90 IONIC REACTIONS 6.22 (a) Reaction (2) because bromide ion is a better leaving group than chloride ion. (b) Reaction (1) because water is a more polar solvent than methanol, and SN 1 reactions take place faster in more polar solvents. (c) Reaction (2) because the concentration of the substrate is twice that of reaction (1). The major reaction would be E2. (However, the problem asks us to consider that small portion of the overall reaction that proceeds by an SN 1 pathway.) (d) Considering only SN 1 reactions, as the problem specifies, both reactions would take place at the same rate because SN 1 reactions are independent of the concentration of the nucleophile. The predominant process in this pair of reactions would be E2, however. (e) Reaction (1) because the substrate is a tertiary halide. Phenyl halides are unreactive in SN 1 reactions. Synthesis 6.23 (a) Br + NaOH OH + NaBr (b) Br + NaI I + NaBr (c) Br + O + NaBr (d) Br + ONa S CH3SNa CH3 + NaBr O (e) Br + O ONa + NaBr O (f ) Br + (g) Br + (h) Br + NaCN (i) Br + NaSH N3 NaN3 + NaBr + N(CH3)3 Br − N(CH3)3 N SH + NaBr + NaBr 6.24 Possible methods are given here. (a) CH3Cl I− CH3OH SN2 CH3I (b) Cl I− CH3OH SN2 I IONIC REACTIONS (c) CH3Cl HO − CH3OH/H2O SN2 91 CH3OH − (d) Cl (e) CH3Cl (f ) Cl HO CH3OH/H2O SN2 HS − CH3OH SN2 HS − CH3OH SN2 −CN (g) CH3I OH CH3SH SH CH3CN DMF −CN (h) Br CN DMF (i) CH3OH NaH (−H2) ( j) NaH (−H2) OH CH3ONa CH3I CH3OCH3 CH3OH CH3I OMe ONa Cl ONa (k) OH 6.25 (a) The reaction will not take place because the leaving group would have to be a methyl anion, a very powerful base, and a very poor leaving group. (b) The reaction will not take place because the leaving group would have to be a hydride ion, a very powerful base, and a very poor leaving group. (c) The reaction will not take place because the leaving group would have to be a carbanion, a very powerful base, and a very poor leaving group. HO − + − ΟΗ ΟΗ 92 IONIC REACTIONS (d) The reaction will not take place by an SN 2 mechanism because the substrate is a tertiary halide, and is, therefore, not susceptible to SN 2 attack because of the steric hindrance. (A very small amount of SN 1 reaction may take place, but the main reaction will be E2 to produce an alkene.) (e) The reaction will not take place because the leaving group would have to be a CH3 O− ion, a strong base, and a very poor leaving group. NH3 + CH3 OCH3 CH3NH 3+ + CH3O − CH3NH2 + CH3OH (f) The reaction will not take place because the first reaction that would take place would be an acid-base reaction that would convert the ammonia to an ammonium ion. An ammonium ion, because it lacks an electron pair, is not nucleophilic. NH3 + + + NH 4 CH3OH 2 + CH3OH 6.26 The better yield will be obtained by using the secondary halide, 1-bromo-1-phenylethane, because the desired reaction is E2. Using the primary halide will result in substantial SN 2 reaction as well, producing the alcohol as well as the desired alkene. 6.27 Reaction (2) would give the better yield because the desired reaction is an SN 2 reaction, and the substrate is a methyl halide. Use of reaction (1) would, because the substrate is a secondary halide, result in considerable elimination by an E2 pathway. NaH Et2O (−H2) 6.28 (a) O − Na+ OH (−NaBr) O (b) SH NaH Et2O (−H2) Br S− Na+ Br (−NaBr) S IONIC REACTIONS NaH (c) OH CH3I O − Na+ (−H2) O (−NaI) O OH (d) Br O (−NaI) CN OH (−NaBr) + Na+ − CN O Ο (f ) O − Na+ Br + CH3CO2H O (−NaBr) H Br acetone (−NaBr) (g) Na+ OH− + (R)-2-Bromopentane (S )-2-Pentanol (S )-2-Chloro-4-methylpentane Br + Br (j) Br • C OH OH + −• H (R)-2-Iodo-4-methylpentane EtO− Na+ EtOH (−NaBr) (i) I acetone (−NaCl) H H HO Cl (h) Na+ I− (k) Na+ CH3 CH3I (−H2) O (e) O − Na+ NaH N •• − Na OH H2O/CH3OH (−NaBr) H OH (−NaBr) + H CN (S )-2-Bromobutane Cl (l) + Na+ I − acetone (−NaCl) 93 I CH3 94 IONIC REACTIONS General SN 1, SN 2, and Elimination O 6.29 (a) The major product would be (by an SN 2 mechanism) because the substrate is primary and the nucleophile-base is not hindered. Some would be produced by an E2 mechanism. (b) The major product would be (by an E2 mechanism), even though the substrate is primary, because the base is a hindered strong base. Some O would be produced by an SN 2 mechanism. (by an E2 mechanism) would be the only product (c) For all practical purposes, because the substrate is tertiary and the base is strong. (d) Same answer as (c) above. (e) t-Bu (formed by an SN 2 mechanism) would, for all practical purposes, be the only product. Iodide ion is a very weak base and a good nucleophile. I (f) Because the substrate is tertiary and the base weak, an SN 1 reaction (solvolysis) will occur, accompanied by elimination (E1). At 25◦ C, the SN 1 reaction would predominate. Cl t-Bu OCH3 MeOH, 25 °C t-Bu (g) + t-Bu OCH3 + t-Bu [also (Z )] (by an E2 mechanism) would be the major product because the substrate is secondary and the base/nucleophile is a strong base. Some of the ether OCH3 would be formed by an SN 2 pathway. Ο O (h) The major product would be ion is a weak base. Some pathway. (by an SN 2 mechanism) because the acetate and might be formed by an E2 HO (i) [also (Z )] and (by E2) would be major products, and [(S ) isomer] (by SN 2) would be the minor product. H IONIC REACTIONS OCH3 (j) (by SN 1) would be the major product. [also (Z)], and (k) H 95 [also (Z )], (by E1) would be minor products. I (by SN 2) would be the only product. 6.30 (a), (b), and (c) are all SN 2 reactions and, therefore, proceed with inversion of configuration. The products are H H H (b) (a) D D I H (c) I D H H I (d) is an SN 1 reaction. The carbocation that forms can react with either nucleophile (H2 O or CH3 OH) from either the top or bottom side of the molecule. Four substitution products (below) would be obtained. (Considerable elimination by an E1 path would also occur.) OH H CH3 and CH3 H D OH D OCH3 H CH3 and CH3 H D OCH3 D 6.31 Isobutyl bromide is more sterically hindered than ethyl bromide because of the methyl groups on the β carbon atom. H3C H β α C CH2 H Br CH3 Isobutyl bromide H C CH2 Br H Ethyl bromide This steric hindrance causes isobutyl bromide to react more slowly in SN 2 reactions and to give relatively more elimination (by an E2 path) when a strong base is used. 96 IONIC REACTIONS 6.32 (a) SN 2 because the substrate is a l◦ halide. (b) Rate = k [CH3 CH2 Cl][I− ] = 5 × 10−5 L mol−1 s−1 × 0.1 mol L−1 × 0.1 mol L−1 Rate = 5 × 10−7 mol L−1 s−1 (c) 1 × 10−6 mol L−1 s−1 (d) 1 × 10−6 mol L−1 s−1 (e) 2 × 10−6 mol L−1 s−1 − 6.33 (a) CH3 N H because it is the stronger base. (b) CH3 O− because it is the stronger base. (c) CH3 SH because sulfur atoms are larger and more polarizable than oxygen atoms. (d) (C6 H5 )3 P because phosphorus atoms are larger and more polarizable than nitrogen atoms. (e) H2 O because it is the stronger base. (f) NH3 because it is the stronger base. (g) HS− because it is the stronger base. (h) HO− because it is the stronger base. Br 6.34 (a) HO + O HO− − O Br− + Br − (b) N H 6.35 N OH Br N+ H C − H + CH3CH2 Br Br − N H H N CH3 δ− C C + Br− + H2O N CCH2CH3 + Br ‡ δ− Br − H H 6.36 Iodide ion is a good nucleophile and a good leaving group; it can rapidly convert an alkyl chloride or alkyl bromide into an alkyl iodide, and the alkyl iodide can then react rapidly with another nucleophile. With methyl bromide in water, for example, the following reaction can take place: H2O alone (slower) + CH3OH 2 + Br − CH3Br H2O containing I − (faster) CH3I H2O (faster) + CH3OH2 + I− IONIC REACTIONS 97 6.37 tert-Butyl alcohol and tert-butyl methyl ether are formed via an SN 1 mechanism. The rate of the reaction is independent of the concentration of methoxide ion (from sodium methoxide). This, however, is only one reaction that causes tert-butyl bromide to disappear. A competing reaction that also causes tert-butyl bromide to disappear is an E2 reaction in which methoxide ion reacts with tert-butyl bromide. This reaction is dependent on the concentration of methoxide ion; therefore, increasing the methoxide ion concentration causes an increase in the rate of disappearance of tert-butyl bromide. 6.38 (a) You should use a strong base, such as RO− , at a higher temperature to bring about an E2 reaction. (b) Here we want an SN 1 reaction. We use ethanol as the solvent and as the nucleophile, and we carry out the reaction at a low temperature so that elimination will be minimized. 6.39 1-Bromobicyclo[2.2.1]heptane is unreactive in an SN 2 reaction because it is a tertiary halide and its ring structure makes the backside of the carbon bearing the leaving group completely inaccessible to attack by a nucleophile. Br Nu •• − 1-Bromobicyclo[2.2.1]heptane is unreactive in an SN 1 reaction because the ring structure makes it impossible for the carbocation that must be formed to assume the required trigonal planar geometry around the positively charged carbon. Any carbocation formed from 1-bromobicyclo[2.2.1]heptane would have a trigonal pyramidal arrangement of the –CH2 – groups attached to the positively charged carbon (make a model). Such a structure does not allow stabilization of the carbocation by overlap of sp 3 orbitals from the alkyl groups (see Fig. 6.7). 6.40 The cyanide ion has two nucleophilic atoms; it is what is called an ambident nucleophile. − C N It can react with a substrate using either atom, although the carbon atom is more nucleophilic. Br CH2CH3 − N •• C 6.41 (a) I F H H + + −• C N •• CH3CH2 C N •• CH3CH2 Br CH3CH2 N C • (Formation of this product depends on the fact that bromide ion is a much better leaving group than fluoride ion.) 98 IONIC REACTIONS (b) (Formation of this product depends on the greater reactivity of 1◦ substrates in SN 2 reactions.) Cl I (c) S (Here two SN 2 reactions produce a cyclic molecule.) S Cl + OH (d) Cl (e) + NaH − H2 Et2O − NaNH2 O Na+ − O liq. NH3 Na+ CH3 I (−NaI) 6.42 The rate-determining step in the SN 1 reaction of tert-butyl bromide is the following: (CH3)3C slow Br + Br − (CH3)3C + H2O + (CH3)3COH2 (CH3 )3 C+ is so unstable that it reacts almost immediately with one of the surrounding water molecules, and, for all practical purposes, no reverse reaction with Br− takes place. Adding a common ion (Br− from NaBr), therefore, has no effect on the rate. Because the (C6 H5 )2 CH+ cation is more stable, a reversible first step occurs and adding a common ion (Br− ) slows the overall reaction by increasing the rate at which (C6 H5 )2 CH+ is converted back to (C6 H5 )2 CHBr. + (C6H5)2CH (C6H5)2CHBr H2O + Br − (C6H5)2CHOH 2+ 6.43 Two different mechanisms are involved. (CH3 )3 CBr reacts by an SN 1 mechanism, and apparently this reaction takes place faster. The other three alkyl halides react by an SN 2 mechanism, and their reactions are slower because the nucleophile (H2 O) is weak. The reaction rates of CH3 Br, CH3 CH2 Br, and (CH3 )2 CHBr are affected by the steric hindrance, and thus their order of reactivity is CH3 Br > CH3 CH2 Br > (CH3 )2 CHBr. 6.44 The nitrite ion is an ambident nucleophile; that is, it is an ion with two nucleophilic sites. The equivalent oxygen atoms and the nitrogen atom are nucleophilic. Nucleophilic site O N O − Nucleophilic site IONIC REACTIONS 99 6.45 (a) The transition state has the form: δ+ Nu δ− L R in which charges are developing. The more polar the solvent, the better it can solvate the transition state, thus lowering the free energy of activation and increasing the reaction rate. (b) The transition state has the form: δ+ R δ+ L in which the charge is becoming dispersed. A polar solvent is less able to solvate this transition state than it is to solvate the reactant. The free energy of activation, therefore, will become somewhat larger as the solvent polarity increases, and the rate will be slower. 6.46 (a) (b) Cl HO I Cl + some alkene 6.47 (a) In an SN 1 reaction the carbocation intermediate reacts rapidly with any nucleophile it encounters in a Lewis acid-Lewis base reaction. In the case of the SN 2 reaction, the leaving group departs only when “pushed out” by the attacking nucleophile and some nucleophiles are better than others. (b) CN− is a much better nucleophile than ethanol and hence the nitrile is formed in the SN 2 Cl . In the case of reaction of Cl , the tert-butyl cation reacts chiefly with the nucleophile present in higher concentration, here the ethanol solvent. Challenge Problems 6.48 (a) The entropy term is slightly favorable. (The enthalpy term is highly unfavorable.) (b) G ◦ = H ◦ − T S ◦ = 26.6 kJ mol−1 − (298)(0.00481 kJ mol−1 ) = 25.2 kJ mol−1 The hydrolysis process will not occur to any significant extent. (c) log K eq = = −G ◦ 2.303RT −25.2 kJ mol−1 (2.303)(0.008314 kJ mol−1 K−1 )(298 K) = −4.4165 K eq = 10−4.4165 = 3.85 × 10−5 100 IONIC REACTIONS (d) The equilibrium is very much more favorable in aqueous solution because solvation of the products (ethanol, hydronium ions, and chloride ions) takes place and thereby stabilizes them. 6.49 The mechanism for the reaction, in which a low concentration of OH− is used in the presence of Ag2 O, involves the participation of the carboxylate group. In step 1 (see following reaction) an oxygen of the carboxylate group attacks the chirality center from the back side and displaces bromide ion. (Silver ion aids in this process in much the same way that protonation assists the ionization of an alcohol.) The configuration of the chirality center inverts in step 1, and a cyclic ester called an α-lactone forms. − O O O δ− C O C C Step 1 C H CH3 O Br Ag + C O C δ− Br CH3 Ag + H H CH3 An α-lactone + AgBr The highly strained three-membered ring of the α-lactone opens when it is attacked by a water molecule in step 2. This step also takes place with an inversion of configuration. O O δ− C Step 2 O C C δ+ OH2 O H2O C H H CH3 CH3 O C O OH2 H − C CH3 OH + H3O + The net result of two inversions (in steps 1 and 2) is an overall retention of configuration. Ag2O 6.50 (a) and (b) H2O retention H HO OH O H Cl O HO KOH inversion OH O (S)-(−)-Chlorosuccinic acid PCl5 inversion HO H HO OH O OH O O (S)-(−)-Malic acid KOH inversion PCl5 inversion (R)-(+)-Malic acid Cl HO H O Ag2O OH O (R)-(+)-Chlorosuccinic acid (c) The reaction takes place with retention of configuration. H2O retention IONIC REACTIONS (d) H Cl O HO SOCl2 H HO OH KOH O (S )-(−)-Chlorosuccinic acid OH O 101 HO O H HO OH OH O (R)-(+)-Malic acid O (S)-(−)-Malic acid H O Cl KOH HO SOCl2 OH O (R)-(+)-Chlorosuccinic acid 6.51 (a) H N3− CH3 CH3O Cl H CH3 CH3O A N3 B (b) No change of configuration occurs, just a change in the relative priority of a group at the chirality center. 6.52 Comparison of the molecular formulas of starting material and product indicates a loss of HC1. The absence of IR bands in the 1620–1680 cm−1 region rules out the presence of the alkene function. A nucleophilic substitution agrees with the evidence: HO − H Cl S (−H2O) − Cl S S − S Cl + Cl − 102 IONIC REACTIONS 6.53 The IR evidence indicates that C possesses both an alkene function and a hydroxyl group. An E2 reaction on this substrate produces enantiomeric unsaturated alcohols. OH Br Br OH H tert-BuO H (a) − (b) C (racemic) OH OH + (b) (a) H OH (R) H OH (S ) 6.54 Regarding the SN 2 reaction, there is extreme steric hindrance for attack by the nucleophile from the back side with respect to the leaving group due to atoms on the other side of the rigid ring structure, as the following model shows. For the SN 1 reaction, formation of a carbocation would require that the bridgehead carbon approach trigonal planar geometry, which would lead to a carbocation of extremely high energy due to the geometric constraints of the bicyclic ring. 6.55 The lobe of the LUMO that would accept electron density from the nucleophile is buried within the bicyclic ring structure of 1-bromobicyclo[2.2.1]heptane (the large blue lobe), effectively making it inaccessible for approach by the nucleophile. 6.56 (a) The LUMO in an SN 1 reaction is the orbital that includes the vacant p orbital in our simplified molecular orbital diagrams of carbocations. (b) The large lobes above and below the trigonal planar carbon atom of the isopropyl group are the ones that would interact with a nucleophile. These are the lobes associated with stylized p orbitals we IONIC REACTIONS 103 draw in simplified diagrams of carbocations. (c) The HOMO for this carbocation shows the contribution of electron density from a nonbonding electron pair of the ether oxygen that is adjacent to the carbocation, This is evident by the lobes that extend over these two atoms and encompass the bond between them. In effect, this orbital model represents the resonance hybrid we can draw where a nonbonding electron pair from oxygen is shifted to the bonding region between the carbon and oxygen. (c) (b) The HOMO of this carbocation shows contribution of electron density from the ether oxygen to the adjacent carbon. This is evident by the lobes that encompass both atoms and extend over the bond between them. These lobes are indicated by the arrows. A nucleophile could contribute electron density to either lobe of the carbocation p-orbital indicated by these arrows. QUIZ 6.1 Which set of conditions would you use to obtain the best yield in the reaction shown? ? Br ONa, (a) H2 O, heat (b) (c) Heat alone (d) H2 SO4 OH, heat (e) None of the above 6.2 Which of the following reactions would give the best yield? (a) CH3ONa + (b) + ONa (c) CH3OH CH3 Br CH3 CH3Br heat + CH3 Br O O O 6.3 A kinetic study yielded the following reaction rate data: Experiment Number Initial Concentrations [HO− ] [R Br] 1 2 3 0.50 0.50 0.25 Initial Rate of Disappearance of R Br and Formation of R OH 0.50 0.25 0.25 1.00 0.50 0.25 Which of the following statements best describe this reaction? (a) The reaction is second order. (b) The reaction is first order. (c) The reaction is SN 1. (d) Increasing the concentration of HO− has no effect on the rate. (e) More than one of the above. 104 IONIC REACTIONS 6.4 There are four compounds with the formula C4 H9 Br. List them in order of decreasing reactivity in an SN 2 reaction. > > > 6.5 Supply the missing reactants, reagents, intermediates, or products. O H ONa O O A (C4H9Br) O OH O − K+ + O Br OH B (Major product) CH3O − Na+ H Cl + CH3OH C Br Na + CN − 25°C + D Major product Minor product 6.6 Which SN 2 reaction will occur most rapidly. (Assume the concentrations and temperatures are all the same.) (a) CH3O− + F O (b) CH3O− + I O (c) CH3O− + Cl O (d) CH3O− + Br O CH3 CH3 CH3 CH3 + F− + I− + Cl − + Br − IONIC REACTIONS 105 6.7 Provide three-dimensional structures for the missing boxed structures and formulas for missing reagents. Na+ (S ) −A (C5H11Br) − ? (S ) −B (C7H12) (S ) −C (C7H16) 7 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS. ELIMINATION REACTIONS OF ALKYL HALIDES SOLUTIONS TO PROBLEMS 7.1 (a) (E )-1-Bromo-1-chloro-1-pentene or (E )-1-Bromo-1-chloropent-1-ene (b) (E )-2-Bromo-1-chloro-1-iodo-1-butene or (E )-2-Bromo-1-chloro-1-iodobut-1-ene (c) (Z )-3,5-Dimethyl-2-hexene or (Z )-3,5-Dimethylhex-2-ene (d) (Z )-1-Chloro-1-iodo-2-methyl-1-butene or (Z )-1-Chloro-1-iodo-2-methylbut-1-ene (e) (Z,4S )-3,4-Dimethyl-2-hexene or (Z,4S )-3,4-Dimethylhex-2-ene (f) (Z,3S )-1-Bromo-2-chloro-3-methyl-1-hexene or (Z,3S )-1-Bromo-2-chloro-3-methylhex-1-ene 7.2 < 7.3 (a), (b) < H2 2-Methyl-1-butene (disubstituted) Pt pressure H2 3-Methyl-1-butene (monosubstituted) Pt pressure H2 Pt pressure Order of increasing stability ∆ H° = − 119 kJ mol −1 ∆ H° = − 127 kJ mol −1 ∆ H° = − 113 kJ mol −1 2-Methyl-2-butene (trisubstituted) (c) Yes, because hydrogenation converts each alkene into the same product. 106 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS H > (d) H (trisubstituted) H > H (disubstituted) 107 H (monosubstituted) Notice that this predicted order of stability is confirmed by the heats of hydrogenation. 2-Methyl-2-butene evolves the least heat; therefore, it is the most stable. 3-Methyl-1-butene evolves the most heat; therefore, it is the least stable. (e) H H H H 1-Pentene H H cis-2-Pentene H trans-2-Pentene (f) Order of stability: trans-2-pentene > cis-2-pentene >1-pentene 7.4 (a) 2,3-Dimethyl-2-butene would be the more stable because the double bond is tetrasubstituted. 2-Methyl-2-pentene has a trisubstituted double bond. (b) trans-3-Hexene would be the more stable because alkenes with trans double bonds are more stable than those with cis double bonds. (c) cis-3-Hexene would be more stable because its double bond is disubstituted. The double bond of 1-hexene is monosubstituted. (d) 2-Methyl-2-pentene would be the more stable because its double bond is trisubstituted. The double bond of trans-2-hexene is disubstituted. Four stereoisomers. However only two fractions are found as enantiomer pairs are indistinguishible by distillation. 7.5 enantiomers 7.6 enantiomers Br NaOEt EtOH, 55 °C most less least 108 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS Br OK 7.7 (a) + OH heat (trisubstituted, more stable) (monosubstituted, less stable) Major product Minor product Br OK (b) + OH heat (tetrasubstituted, more stable) Major product (disubstituted, less stable) Minor product 7.8 t-BuOK in t-BuOH 7.9 An anti coplanar transition state allows the molecule to assume the more stable staggered conformation, H H H H H Br whereas a syn coplanar transition state requires the molecule to assume the less stable eclipsed conformation. H H H H H Br 7.10 cis-1-Bromo-4-tert-butylcyclohexane can assume an anti coplanar transition state in which the bulky tert-butyl group is equatorial. Br H H H H B •• The conformation (above), because it is relatively stable, is assumed by most of the molecules present, and, therefore, the reaction is rapid. ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 109 On the other hand, for trans-1-bromo-4-tert-butylcyclohexane to assume an anti coplanar transition state, the molecule must assume a conformation in which the large tert-butyl group is axial: H Br Br H H H H H B •• Such a conformation is of high energy; therefore, very few molecules assume this conformation. The reaction, consequently, is very slow. 7.11 (a) Anti coplanar elimination can occur in two ways with the cis isomer. H (a) H H (a) Br CH3 H CH3 B •• or (b) cis-1-Bromo-2-methylcyclohexane CH3 (b) (major product) (b) Anti coplanar elimination can occur in only one way with the trans isomer. H Br H H CH3 H CH3 B •• trans-1-Bromo-2-methylcyclohexane O 7.12 (a) (1) CH3 CH OH H CH3 O S O O O H CH3 CH O O H CH S O H O H2 O CH3 CH3 (3) CH3 CH CH3 H CH O H CH3 (2) CH3 H CH2 H OSO3H CH3 CH CH2 HOSO3H 110 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS (b) By donating a proton to the OH group of the alcohol in step (1), the acid allows the loss of a relatively stable, weakly basic, leaving group (H2 O) in step (2). In the absence of an acid, the leaving group would have to be the strongly basic HO− ion, and such steps almost never occur. OH > 7.13 > OH OH 1° 2° 3° Order of increasing case of dehydration CH3 CH3 + 7.14 (1) CH3CCH2OH + H A CH3CCH2 OH2 CH3 A− + CH3 CH3 CH3 + (2) CH3CCH2 + CH3CCH2 OH2 CH3 + H2O CH3 1° Carbocation + + CH3 CH3 CH3 + (3) CH3CCH2 CH3C CH3C CH2 + + CH2 CH3 CH3 CH3 Transition state 1° Carbocation 3° Carbocation [Steps (2) and (3), ionization and rearrangement, may occur simultaneously.] CH3 (4) CH3 C + CH3 CH H C CH3 A− H CH3 + C CH3 2-Methyl-2-butene HA ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS •• OH •• 7.15 CH3CH2CHCH2 •• H + O + H (−H2O) CH3CH2CHCH2 (+H2O) H CH3 + OH 2 •• 111 (−H2O) (+H2O) CH3 2-Methyl-1-butanol H H CH3CH2 1,2-hydride + C CH2 shift* CH3 C CH3 CH3 3° Cation CH3 1° Cation CH3CH OH2 + CH CH3 + H3O + C CH3 2-Methyl-2-butene •• OH •• CH3CHCH2CH2 •• H + O + H (−H2O) CH3CHCH2CH2 (+H2O) H CH3 + OH2 (−H2O) (+H2O) CH3 3-Methyl-1-butanol H H CH3CH 1,2-hydride + CH CH2 CH3 shift* + CH OH2 CH3 CH3 CH3 CH3C C CH CH3 + H3O+ * The hydride shift may occur simultaneously with the preceding step. CH3 2-Methyl-2-butene CH3 CH3 7.16 HO = CH3 HO CH3 H3O+ (−H2O) + CH3 CH3 Isoborneol H CH3 CH3 CH3 = + + CH2 CH2 CH3 CH3 OH2 CH2 CH2 CH3 + H O+ 3 CH3 CH2 Camphene 112 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS CH2 + NaNH2 7.17 (a) CH3CH No reaction (b) CH3C C H + Na+ − NH2 Stronger Stronger acid base (c) CH3CH2CH3 + NaNH2 (d) CH3C C Stronger base (e) CH3C C CH3C C Weaker base + H OCH2CH3 Stronger acid − + H NH3 − CH3C CH + Weaker acid + Weaker acid Cl OCH2CH3 Weaker base CH + NH3 CH3C Stronger acid O Weaker base Cl (1) 3 equiv. NaNH2 PCl5 0 °C 7.18 Na+ + NH3 Weaker acid No reaction − Stronger base − mineral oil, heat (2) HA O 7.19 (a) CH3CCH3 PCl5 0 °C (b) CH3CH2CHBr2 (c) CH3CHBrCH2Br (d) CH3CH CH2 CH3CCl2CH3 (1) 3 NaNH2 mineral oil, heat (2) NH4 + [same as (b)] Br2 (1) 3 NaNH2 mineral oil, heat (2) NH4 + CH3C CH3C CH3CHCH2Br CH CH3C CH CH [same as (b)] CH CH3C Br CH3 7.20 CH3 C CH3 C C − H + Na+ NH2 (−NH3) CH3 CH3 (Starting the synthesis with 1-propyne and attempting to alkylate with a tert-butyl substrate would not work because elimination would occur instead of substitution.) C C C − Na+ CH3 CH3 CH3 CH3 C CH3 C C CH3 I ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 113 O 7.21 Compound A (1) Li, C2H5NH2, −78 °C (2) NH4Cl 7.22 (E)-2-Nonene 7.23 Route 1 CH3 CH3 HC CCH2CHCH3 NaNH2 Na+ − C (−NH3) CCH2CHCH3 CH3 (−NaBr) CH3 CH3 C CH3 CCH2CHCH3 H2 Pd, Pt, or Ni pressure Route 2 HC Br CH3CH2CH2CH2CHCH3 CH3 NaNH2 CH (−NH3) HC C − Na+ Br CH2CH2CHCH3 (−NaBr) CH3 CH3 HC C CH2CH2CHCH3 H2 Pd, Pt, or Ni pressure CH3CH2CH2CH2CHCH3 Route 3 CH3 CH3 HC CCHCH3 NaNH2 (−NH3) Na+ − C CCHCH3 C CCHCH3 (−NaBr) CH3 CH3 CH3CH2 CH3CH2Br H2 Pd, Pt, or Ni CH3CH2CH2CH2CHCH3 114 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS Etc. (using the other alkyne and alkyl halide homologue pairs) 7.24 (a) Undecane CH3C C − + X(CH2)7CH3 C − + X(CH2)8CH3 HC CH3CH2C C + X(CH2)6CH3 − CH3CH2CH2C C + X(CH2)5CH3 − 2-Methylheptadecane (after hydrogenation of the alkyne from one of the possible retrosynthetic disconnections) X + (or homologous pairs) − X + (Note that − + X − is not a good choice because the alkyl halide is branched at the carbon adjacent to the one which bears the halogen. + work because the alkyl halide is X − secondary. Both of these routes would lead to elimination instead of substitution.) Neither would (b) For any pair of reactants above that is a feasible retrosynthetic disconnection, the steps for the synthesis would be RC C H (a terminal alkyne; R = alkyl, H) NaNH2 R′ X R (R′ is (an alkynide primary and anion) unbranched at the second carbon) R (−NH3) C C − C C H2 Pd, Pt, or Ni pressure CH2CH2 R R′ R′ 7.25 (a) We designate the position of the double bond by using the lower number of the two numbers of the doubly bonded carbon atoms, and the chain is numbered from the end nearer the double bond. The correct name is trans-2-pentene. 5 4 3 2 1 not 1 2 3 4 5 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 115 (b) We must choose the longest chain for the base name. The correct name is 2-methylpropene. 3 1 2 (c) We use the lower number of the two doubly bonded carbon atoms to designate the position of the double bond. The correct name is 1-methylcyclohexene. 1 2 (d) We must number the ring starting with the double bond in the direction that gives the substituent the lower number. The correct name is 3-methylcyclobutene. 1 4 3 2 4 1 not 3 2 (e) We number in the way that gives the double bond and the substituent the lower number. The correct name is (Z )-2-chloro-2-butene or (Z )-2-chlorobut-2-ene. 3 2 1 2 not 4 3 4 1 Cl Cl (f) We number the ring starting with the double bond so as to give the substituents the lower numbers. The correct name is 3,4-dichlorocyclohexene. Cl 3 2 Cl 6 Cl 5 1 1 2 not Cl 7.26 (a) 4 (b) 3 4 (c) Br (f ) (e) (d) Br Br (g) (h) H ( j) Cl (i) H (k) H Cl (l) 116 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 7.27 (a) H H Br (2Z,4S)-4-Bromo-2-hexene or (2Z,4S)-4-Bromohex-2-ene H Br (2E,4R)-4-Bromo-2-hexene or (2E,4R)-4-Bromohex-2-ene H Br (2E,4S)-4-Bromo-2-hexene or (2E,4S)-4-Bromohex-2-ene Cl H (3R,4Z )-3-Chloro-1,4-hexadiene or (3R,4Z )-3-Chlorohexa-1,4-diene H Cl (3S,4Z )-3-Chloro-1,4-hexadiene or (3S,4Z )-3-Chlorohexa-1,4-diene Cl H (3R,4E )-3-Chloro-1,4-hexadiene or (3R,4E )-3-Chlorohexa-1,4-diene H Cl (3S,4E )-3-Chloro-1,4-hexadiene or (3S,4E )-3-Chlorohexa-1,4-diene Br (2Z,4R)-4-Bromo-2-hexene or (2Z,4R)-4-Bromohex-2-ene (b) (c) Cl Cl Cl H (2E,4R)-2,4-Dichloro-2-pentene or (2E,4R)-2,4-Dichloropent-2-ene Cl H (2Z,4R)-2,4-Dichloro-2-pentene or (2Z,4R)-2,4-Dichloropent-2-ene Cl Cl H Cl (2E,4S )-2,4-Dichloro-2-pentene or (2E,4S )-2,4-Dichloropent-2-ene H Cl (2Z,4S )-2,4-Dichloro-2-pentene or (2Z,4S )-2,4-Dichloropent-2-ene ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS (d) Br H Cl (3R,4Z )-5-Bromo-3-chloro-4hexen-1-yne or (3R,4Z )-5-Bromo-3-chlorohex4-en-1-yne Br H Cl (3R,4E )-5-Bromo-3-chloro-4hexen-1-yne or (3R,4E )-5-Bromo-3-chlorohex4-en-1-yne 117 Br Cl H (3S,4Z )-5-Bromo-3-chloro-4hexen-1-yne or (3S,4Z )-5-Bromo-3-chlorohex4-en-1-yne Br Cl H (3S,4E )-5-Bromo-3-chloro-4hexen-1-yne or (3S,4E )-5-Bromo-3-chlorohex4-en-1-yne An IUPAC* rule covers those cases in which a double bond and a triple bond occur in the same molecule: Numbers as low as possible are given to double and triple bonds as a set, even though this may at times give “-yne” a lower number than “-ene.” If a choice remains, preference for low locants is given to the double bonds. *International Union of Pure and Applied Chemistry, www.iupac.org. 7.28 (a) (E )-3,5-Dimethyl-2-hexene or (E )-3,5-dimethylhex-2-ene (b) 4-Chloro-3-methylcyclopentene (c) 6-Methyl-3-heptyne or 6-methylhept-3-yne (d) 1-sec-Butyl-2-methylcyclohexene or 1-methyl-2-(1-methylpropyl)cyclohexene (e) (4Z,3R)-3-Chloro-4-hepten-1-yne or (4Z,3R)-3-chlorohept-4-en-1-yne (f) 2-Pentyl-1-heptene or 2-pentylhept-1-ene 7.29 l-Pentanol > l-pentyne > l-pentene > pentane (See Section 3.8 for the explanation.) 118 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS Synthesis OK Cl 7.30 (a) OH Cl ONa (b) OH HA, heat OH (c) OH HA, heat (d) Br (e) Br H2 (1 equiv.) (2) HA Ni2B (P-2) H2 (f) 7.31 (a) (1) NaNH2 (3 equiv.) Ni2B (P-2) Br ONa OH heat (b) OH HA, heat 7.32 (a) NaNH2 liq. NH3 − Na+ CH3 (−NaI) I ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS NaNH2 (b) − (c) NaNH2 [from (a)] − liq. NH3 CH3 Na+ I (−NaI) H2 (d) [from (c)] Ni2B (P-2) NH2 (1) Li, (e) NH4Cl (2) [from (c)] (f) Br (−NaBr) Na+ liq. NH3 119 − Br (−NaBr) Na+ [from (a)] (g) NaNH2 − liq. NH3 [from (f )] CH3 I (−NaI) H2 (h) Ni2B (P-2) [from (g)] NH2 (1) Li, (i) (2) [from (g)] (j) Na+ − Na+ NH4Cl NaNH2 Br (−NaBr) liq. NH3 [from (a)] − Na+ Br − (k) Na+ D2O D [from ( j)] D2 (l) Ni2B (P-2) [from (c)] D D 7.33 We notice that the deuterium atoms are cis to each other, and we conclude, therefore, that we need to choose a method that will cause a syn addition of deuterium. One way would be to use D2 and a metal catalyst (Section 7.14) CH3 D D2 CH3 D Pt H 120 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 7.34 Br − Br Na+ 3NaNH2 (a) NH4Cl mineral oil, heat − Br (b) Br Na+ 3NaNH2 mineral oil, heat NH4Cl Br Br2 (c) − Na+ − Na+ 3NaNH2 Br mineral oil, heat NH4Cl Cl O Cl PCl5 (d) 3NaNH2 mineral oil, heat NH4Cl Dehydrohalogenation and Dehydration + + EtO δ − 7.35 EtO − H H C CH3 CH3 C CH3 Br H C CH3 C CH3 CH3 Br δ− CH3 H H CH3 C C CH3 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 121 7.36 Dehydration of trans-2-methylcyclohexanol proceeds through the formation of a carbocation (through an E1 reaction of the protonated alcohol) and leads preferentially to the more stable alkene. 1-Methylcyclohexene (below) is more stable than 3-methylcyclohexene (the minor product of the dehydration) because its double bond is more highly substituted. CH3 CH3 CH3 H OH A− −HA + HA −H2O CH3 + (major) (minor) Trisubstituted Disubstituted double bond double bond Dehydrohalogenation of trans-1-bromo-2-methylcyclohexane is an E2 reaction and must proceed through an anti coplanar transition state. Such a transition state is possible only for the elimination leading to 3-methylcyclohexene (cf. Review Problem 7.11). Br H H B 7.37 (a) − CH3 H CH3 3-Methylcyclohexene + major (d) + minor major minor (e) (b) + major only product minor + (c) (f ) major (+ stereoisomer) minor only product + (c) + 7.38 (a) major major minor (+ stereoisomer) minor + (d) (b) major only product minor (+ stereoisomer) 122 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS (e) + major minor Br 7.39 (a) OK [+(Z)] + major minor OH (b) ONa only product Br OH (c) ONa only product Br OH Br (d) ONa only product OH Br (e) ONa + OH major CH3 CH3 7.40 CH3CCH2CH3 OH 7.41 (a) 3° > CH3CHCHCH3 OH OH OH OH > CH3CH2CH2CH2CH2OH 1° 2° HA heat (−H2O) HA heat (−H2O) (b) (c) minor HA heat (−H2O) + + major minor + major (⫹ stereoisomer) minor + minor minor ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS HA heat (−H2O) rearrangement OH (d) HA heat (−H2O) OH (e) 123 + minor major + minor major 7.42 The alkene cannot be formed because the double bond in the product is too highly strained. Recall that the atoms at each carbon of a double bond prefer to be in the same plane. 7.43 Only the deuterium atom can assume the anti coplanar orientation necessary for an E2 reaction to occur. H Br H CH3 H H H3C D − OEt 7.44 (a) A hydride shift occurs. OH •• H2O •• HA OH2 (−H2O) + + H H hydride shift (may be concerted with departure of the leaving group) + + H3O + major product •• (b) A methanide shift occurs. H +OH 2 OH HA OH 2 •• + (−H2O) methanide shift + + + H3O major product 124 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS (c) A methanide shift occurs. I I Ag+ AgNO3 (−AgI) + methanide shift + H •• HOEt •• + (−EtOH2) major product (d) The required anti coplanar transition state leads only to (Z ) alkene: − H Ph Br O Ph H ONa heat OH + Br − + Na+ Ph Ph (Z ) only 7.45 (a) Three (b) One Index of Hydrogen Deficiency 7.46 (a) Caryophyllene has the same molecular formula as zingiberene (Review Problem 4.21); thus it, too, has an index of hydrogen deficiency equal to 4. That 1 mol of caryophyllene absorbs 2 mol of hydrogen on catalytic hydrogenation indicates the presence of two double bonds per molecule. (b) Caryophyllene molecules must also have two rings. (See Review Problem 23.2 for the structure of caryophyllene.) 7.47 (a) C30 H62 = formula of alkane C30 H50 = formula of squalene H12 = difference = 6 pairs of hydrogen atoms Index of hydrogen deficiency = 6 (b) Molecules of squalene contain six double bonds. (c) Squalene molecules contain no rings. (See Review Problem 23.2 for the structural formula of squalene.) Structure Elucidation 7.48 That I and J rotate plane-polarized light in the same direction tells us that I and J are not enantiomers of each other. Thus, the following are possible structures for I, J, and K. ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 125 (The enantiomers of I, J, and K would form another set of structures, and other answers are possible as well.) CH3 CH 3 CH3CH H2 H C CH Pt CH2 CH3 CH 3 I Optically active CH3 CH CH2 C CH3CH C H CH2CH3 3 H C H2 CH2CH3 Pt J Optically active 7.49 The following are possible structures: CH3 CH3 C C H CHCH3 L H2 Pt pressure CH3 CH3 CH3CH2CHCH(CH3)2 CH3 CHCH3 CH3 C H N Optically inactive but resolvable H2 C CH3 Pt pressure M (other answers are possible as well) Challenge Problems H 7.50 H2 Pt CH3 E Optically active (the enantiomeric form is an equally valid answer) H CH3 F Optically inactive and nonresolvable K Optically active ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS CH3 H C C C H2 CH3CH2CH2CH2CH2CH3 Pt pressure H G Optically active (the enantiomeric form is an equally valid answer) CH3CH2 H Optically inactive and nonresolvable 7.51 (a) We are given (Section 7.3A) the following heats of hydrogenation: Pt cis-2-Butene + H2 butane ∆H° = − 120 kJ mol−1 trans-2-Butene + H2 Pt butane ∆H° = − 115 kJ mol−1 Thus, for trans-2-butene cis-2-Butene ∆H ° = − 5.0 kJ mol−1 (b) Converting cis-2-butene into trans-2-butene involves breaking the π bond. Therefore, we would expect the energy of activation to be at least as large as the π -bond strength, that is, at least 264 kJ mol−1 . (c) + _ 264 kJ mol−1 G+> Free Energy 126 cis-2-Butene ∆ G° = − 5.0 kJ mol−1 trans-2-Butene Reaction coordinate 7.52 (a) With either the (1R,2R)- or the (1S,2S )-1,2-dibromo-1,2-diphenylethane, only one conformation will allow an anti coplanar arrangement of the H- and Br-. In either case, the elimination leads only to (Z )-1-bromo-1,2-diphenylethene: B H H − Ph Ph Br Br (1R,2R)-1,2-Dibromo-1,2-diphenylethane (anti coplanar orientation of H- and -Br) B − H Br Ph Ph H Br Ph (Z )-1-Bromo-1,2-diphenylethene Ph H Br Ph Br Ph H (1S,2S )-1,2-Dibromo-1,2-diphenylethane (anti coplanar orientation of H- and -Br) (Z )-1-Bromo-1,2-diphenylethene ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 127 (b) With (1R,2S )-1,2-dibromo-1,2-diphenylethane, only one conformation will allow an anti coplanar arrangement of the H- and Br-. In either case, the elimination leads only to (E )-1-bromo-1,2-diphenylethene: B H − Br Ph H Br H Ph Ph Ph Br (1R,2S )-1,2-Dibromo-1,2-diphenylethane (anti coplanar orientation of H and Br) (E )-1-Bromo-1,2-diphenylethene (c) With (1R,2S )-1,2-dibromo-1,2-diphenylethane, only one conformation will allow an anti coplanar arrangement of both bromine atoms. In this case, the elimination leads only to (E )-1,2-diphenylethene: I: Br − Ph H H Ph Br Ph H H Ph (1R,2S )-1,2-Dibromo-1,2-diphenylethane (E )-1,2-Diphenylethene (anti coplanar orientation of both -Br atoms) H2, Ni2B(P-2) 7.53 (a) HA or Na/NH3 + − HA A− + H (b) No, tetrasubstituted double bonds usually show no C infrared spectra. 7.54 OH OH A OH + A− − HA and its enantiomer H C (b) Six HA B + 7.55 (a) Three C stretching absorption in their O 128 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS QUIZ 7.1 Which conditions/reagents would you employ to obtain the best yields in the following reaction? ? Br (a) H2 O/heat (c) OK in (b) ONa in OH, heat (d) Reaction cannot occur as shown OH 7.2 Which of the following names is incorrect? (a) 1-Butene (b) trans-2-Butene (d) 1,1-Dimethylcyclopentene (c) (Z )-2-Chloro-2-pentene (e) Cyclohexene 7.3 Select the major product of the reaction ONa Br (a) OH (b) (c) (d) (e) O ? ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS 129 7.4 Supply the missing reagents. (a) trans-2-butene (b) 2-Butyne cis-2-butene (c) butane (d) Br (e) 7.5 Arrange the following alkenes in order of decreasing stability. 1-Pentene, cis-2-pentene, trans-2-pentene, 2-methyl-2-butene > > > Most stable Least stable 7.6 Complete the following synthesis. (a) 3 NaNH2 Br2 mineral oil 110–160 °C (c) (b) NaNH2 NH4Cl liq. NH3 (d) (e) 2-Pentyne 8 ALKENES AND ALKYNES II: ADDITION REACTIONS SOLUTIONS TO PROBLEMS CH3CHCH2I 8.1 Br 2-Bromo-1-iodopropane 8.2 (a) + δ+ δ− H Br Br − + Br Cl (b) + δ+ δ− I Cl 8.3 (a) + Cl I I δ+ δ− H I (c) I + − + δ+ δ− H Cl + 2° carbocation I − 1,2-hydride shift + 3° carbocation Cl− Cl 2-Chloro-2-methylbutane (from rearranged carbocation) Cl + Unrearranged 2° carbocation 130 Cl− 2-Chloro-3-methylbutane (from unrearranged carbocation) ALKENES AND ALKYNES II: ADDITION REACTIONS 131 Cl (b) + δ+ δ− H Cl Cl− + 1,2-methanide shift Cl Cl− 3-Chloro-2,2-dimethylbutane (from unrearranged carbocation) + 2-Chloro-2,3-dimethylbutane (from rearranged carbocation) + + 8.4 (a) H + O H (from dilute H2SO4) H + H O H H O H H H OH + + O H H + H3O+ (b) Use a high concentration of water because we want the carbocation produced to react with water. And use a strong acid whose conjugate base is a very weak nucleophile. (For this reason we would not use HI, HBr, or HCl.) An excellent method, therefore, is to use dilute sulfuric acid. (c) Use a low concentration of water (i.e., use concentrated H2 SO4 ) and use a higher temperature to encourage elimination. Distill cyclohexene from reaction mixture as it is formed, so as to draw the equilibrium toward product. (d) 1-Methylcyclohexanol would be the product because a 3◦ carbocation would be formed as the intermediate. + + + H O + O H H H H H O OH H + + O H H + H3O+ 8.5 CH2 CH2 + H2SO4 CH3CH2OSO3H H2O heat CH3CH2OH + H2SO4 ALKENES AND ALKYNES II: ADDITION REACTIONS H O+ H H + 8.6 H2O + methanide migration H H + Ο OH Ο Ο H H 132 H H + + H3O+ 8.7 The order reflects the relative ease with which these alkenes accept a proton and form a carbocation. (CH3 )2 C CH2 reacts faster because it leads to a tertiary cation, CH2 (CH3)2C CH3 H3O+ + CH3 C 3° Carbocation CH3 CH3 CH CH2 leads to a secondary cation, H CH3CH CH2 H3O+ + CH3 C 2° Carbocation CH3 and CH2 CH2 reacts most slowly because it leads to a primary carbocation. H CH2 CH2 H3O+ CH3 + C 1° Carbocation H Recall that formation of the carbocation is the rate-determining step in acid-catalyzed hydration and that the order of stabilities of carbocations is the following: 3◦ > 2◦ > 1◦ > + CH3 8.8 H OSO3H Me + Ο H Me +Ο H Me Ο (−H2SO4) − OSO3H ALKENES AND ALKYNES II: ADDITION REACTIONS OH (1) Hg(OAc)2/ THF−Η2O 8.9 (a) (2) NaBH4, 133 HO− OH (1) Hg(OAc)2/THF−Η2O (b) (2) NaBH4, HO− (1) Hg(OAc)2/THF−Η2O (c) (2) NaBH4, HO− OH can also be used. R O 8.10 (a) C C + δ+ + HgOCCF 3 C R C H − +Ο O H O C C A O HgOCCF3 HgOCCF3 δ+ RO −HA C C O HgOCCF3 O CH3 (b) CH3 C O CH3 CH2 Hg(OCCF3)2 THF-CH3OH solvomercuration CH3 C − CH2HgOCCF3 NaBH4/HO demercuration OCH3 CH3 CH3CCH3 + Hg + CF3COO − OCH3 (c) The electron-withdrawing fluorine atoms in mercuric trifluoroacetate enhance the electrophilicity of the cation. Experiments have demonstrated that for the preparation of tertiary alcohols in satisfactory yields, the trifluoroacetate must be used rather than the acetate. 8.11 (a) 3 BH3:THF B 134 ALKENES AND ALKYNES II: ADDITION REACTIONS BH3:THF (b) 3 B (c) 3 (or cis isomer) BH3:THF B H BH3:THF (d) 3 + enantiomer syn addition anti Markovnikov B H CH3 CH3 8.12 2 CH3C CHCH3 BH3:THF CH3CH CH 2 BH CH3 Disiamylborane (1) BH3:THF 8.13 (a) OH (2) H2O2, HO− (1) BH3:THF (b) OH (2) H2O2, HO− OH (1) BH3:THF (c) (2) H2O2, HO− [or (Z ) isomer] (1) BH3:THF (d) (2) H2O2, HO− OH (e) CH3 CH3 (1) BH3:THF + (2) H2O2, HO− OH enantiomer ALKENES AND ALKYNES II: ADDITION REACTIONS (1) BH3:THF (f) OH (2) H2O2, HO− 8.14 (a) 3 CH3CHCH BH3:THF CH2 CH3CHCH2CH2 CH3 3 B CH3CO2D CH3 3 CH3CHCH2CH2D CH3 B (b) 3 CH3C BH3:THF CHCH3 CH3CH CH CH3 CH3CO2D CH3 CH3 3 CH3CHCHDCH3 CH3 (c) 3 BH3:THF CH3 CH3 H (+ enantiomer) CH3CO2D H B CH3 3 H (+ enantiomer) H D CH3 (d) 3 D BD3: THF B H D 3 CH3 T H CH3CO2T CH3 (+ enantiomer) (+ enantiomer) 135 136 ALKENES AND ALKYNES II: ADDITION REACTIONS (a) + 8.15 δ+ δ− Br Br OH2 (a) Br + or (b) (b) Bromonium ion + OH2 Br Br OH H2O + Br OH2 + from (b) from (a) Br + OH Racemic mixture Because paths (a) and (b) occur at equal rates, these enantiomers are formed at equal rates and in equal amounts. 8.16 + δ+ δ− Br Br + Br Br − + Nu Br + Nu = H2O Br or Br − Br or Cl − Br A HA + OH2 Br Br Cl Cl 8.17 (a) Cl t-BuOK + CHCl3 H t-BuOK (b) CHBr3 H Br Br CH2I2/Zn(Cu) (c) 8.18 Et2O CHBr3 Br t-BuOK Br enantiomer OH ALKENES AND ALKYNES II: ADDITION REACTIONS CH3CHI2 8.19 + Zn(Cu) Η 8.20 (a) Η OH (1) OsO4, pyridine, 25 °C (2) NaHSO3 OH OH (b) OH (1) OsO4, pyridine, 25 °C (racemic) (2) NaHSO3 (1) OsO4, pyridine, 25 °C (c) (2) NaHSO3 (racemic) OH H OH O O (1) O3 8.21 (a) + (2) Me2S H O (b) (1) O3 (2) Me2S O O O (c) 8.22 (a) O (1) O3 H H + (2) Me2S (2) Me2S [or (E) isomer] H O (1) O3 (b) O O 2 H O (1) O3 (2) Me2S + 137 ALKENES AND ALKYNES II: ADDITION REACTIONS O (1) O3 (c) + O (2) Me2S H H 8.23 Ordinary alkenes ar e more reactive toward electrophilic reagents than alkynes. But the alkenes obtained from the addition of an electrophilic reagent to an alkyne have at least one electronegative atom (Cl, Br, etc.) attached to a carbon atom of the double bond. X C HX C C C H or X C X2 C C C X These alkenes are less reactive than alkynes toward electrophilic addition because the electronegative group makes the double bond “electron poor.” 8.24 The molecular formula and the formation of octane from A and B indicate that both comO pounds are unbranched octynes. Since A yields only on ozonolysis, A must OH be the symmetrical octyne . The IR absorption for B shows the . presence of a terminal triple bond. Hence B is O OH Since C (C8 H12 ) gives HO on ozonolysis, C must be cy- O 138 clooctyne. This is supported by the molecular formula of C and the fact that it is converted to D, C8 H16 (cyclooctane), on catalytic hydrogenation. 8.25 By converting the 3-hexyne to cis-3-hexene using H2 /Ni2 B (P-2). H2 Ni2B(P−2) H H Then, addition of bromine to cis-3-hexene will yield (3R,4R), and (3S,4S)-3,4dibromohexane as a racemic form. Br2 H H anti addition Br Br Br Br + H H (3R, 4R) H (3S, 4S) Racemic 3,4-dibromohexane H ALKENES AND ALKYNES II: ADDITION REACTIONS 139 Alkenes and Alkynes Reaction Tool Kit 8.26 The answers to all parts except (b), (j), (l), and (n) are formed as racemic mixtures. OH I (b) (a) OH O Cl (j) (i) OH Br (g) (f) (h) Br O H + H O (l) (d) Br (e) OSO3H (c) Br OH (k) H OH OH OH + CO2 (m) (n) OH 8.27 The answers to (g), (h), (k), and (n) are formed as racemic mixtures. OH I (b) (a) (c) OH OSO3H (d) Br Br OH Br (e) (f) Cl (g) O O OH H (k) OH O (l) OH O (j) (i) Br (h) (m) OH OH (n) Br Br 8.28 (a) (b) OH + OH OH enantiomers H OH H + (c) H enantiomers OH H 140 ALKENES AND ALKYNES II: ADDITION REACTIONS Br Br + (d) Br Br O H (e) 2 enantiomers Br 8.29 (a) (b) (c) Br Br Br Br (d) (e) (g) (f ) [An E2 reaction would take place and when − Na+ is treated with ] Br Br 8.30 (a) (b) Br Br (c) Br Br Br Br (d) Cl (e) (f ) (h) (i) Br Br (g) ( j) O OH (2 molar equivalents) (k) O OH (2 molar equivalents) (l) No reaction ALKENES AND ALKYNES II: ADDITION REACTIONS 8.31 (a) (b) Cl Br Cl Br (c) (d) O OH O Br 3 NaNH2 Br2 8.32 (a) CCl4 Br − (b) Na+ (c) (d) Cl t-BuOK 8.33 (a) Then as in (a) t-BuOH, heat 2 NaNH2 − Na+ NH4Cl − Na+ NH4Cl mineral oil, heat 3 NaNH2 mineral oil, heat Cl (e) NaNH2 − liq. NH3 Na+ Br H3O+, H2O OH (b) (c) HBr (no peroxides) Br Cl2, H2O Cl OH (d) mineral oil, heat NH4Cl Cl Cl OH + (1) BH3:THF (2) H2O2, HO− OH 141 142 ALKENES AND ALKYNES II: ADDITION REACTIONS H H T CH3 8.34 (a) D CH3 (b) H D + enantiomer + + H + enantiomer HO Cl OH CH3 (c) D + enantiomer 8.35 H CH2CH3 Cl− Cl O H Cl− −HCl + O 8.36 The rate-determining step in each reaction is the formation of a carbocation when the alkene accepts a proton from HI. When 2-methylpropene reacts, it forms a 3◦ carbocation (the most stable); therefore, it reacts fastest. When ethene reacts, it forms a 1◦ carbocation (the least stable); therefore, it reacts the slowest. H I fastest I− + I 3° cation I H I + I − 2° cation H I slowest + 1° cation I− I ALKENES AND ALKYNES II: ADDITION REACTIONS +H OH OH + δ H δ Loss of H2O and 1,2-hydride shift H − I 8.37 I− I + OH 8.38 + δ H − + δ I OH2 (−H2O) 2° carbocation I− 1,2- + Methanide shift I + 3° carbocation 8.39 (a) OH OH (1) OsO4 + (2) NaHSO3, H2O enantiomer H H syn addition OH OH (b) (c) (1) OsO4 (2) NaHSO3, H2O H + enantiomer H syn addition Br Br Br2 + enantiomer H H anti addition (d) Br2 Br Br H H anti addition + enantiomer 143 144 ALKENES AND ALKYNES II: ADDITION REACTIONS 8.40 (a) (2S, 3R)- [the enantiomer is (2R, 3S)-] (b) (2S, 3S)- [the enantiomer is (2R, 3R)-] (c) (2S, 3R)- [the enantiomer is (2R, 3S)-] (d) (2S, 3S)- [the enantiomer is (2R, 3R)-] 8.41 Because of the electron-withdrawing nature of chlorine, the electron density at the double bond is greatly reduced and attack by the electrophilic bromine does not occur. 8.42 The bicyclic compound is a trans-decalin derivative. The fused nonhalogenated ring prevents the ring flip of the bromine-substituted ring necessary to give equatorial bromines. CH3 Br CH3 Br2 Br H H Diequatorial conformation 8.43 H−OSO3H2 cat. − OSO3H + + + I (major) II (minor) Though II is the product predicted by application of the Zaitsev rule, it actually is less stable than I due to crowding about the double bond. Hence I is the major product (by about a 4:1 ratio). 8.44 The terminal alkyne component of the equilibrium established in base is converted to a salt by NaNH2 , effectively shifting the equilibrium completely to the right. R R R C R − NaOH is too weak a base to form a salt with the terminal alkyne. Of the equilibrium components with NaOH, the internal alkyne is favored since it is the most stable of these structures. Very small amounts of the allene and the terminal alkyne are formed. ALKENES AND ALKYNES II: ADDITION REACTIONS 8.45 OH HCO3− O− 145 I2 O O O O O − O I I + + H2SO4 H2O 8.46 + + OH2 δ+ Br 8.47 δ− Br + Br Br − OH HSO4− Br Br H2O Br H2O OH2 + Cl− Br OH Br Cl Enantiomers of each also formed. 8.48 (a) C10 H22 (saturated alkane) C10 H16 (formula of myrcene) H6 = 3 pairs of hydrogen atoms Index of hydrogen deficiency (IHD) = 3 (b) Myrcene contains no rings because complete hydrogenation gives C10 H22 , which corresponds to an alkane. (c) That myrcene absorbs three molar equivalents of H2 on hydrogenation indicates that it contains three double bonds. 146 ALKENES AND ALKYNES II: ADDITION REACTIONS O (d) O O H H (e) Three structures are possible; however, only one gives 2,6-dimethyloctane on complete hydrogenation. Myrcene is therefore 8.49 (a) (b) Four 2,6,10-Trimethyldodecane H O 8.50 (1) O3 + (2) Me2S H O O O + H O O H 8.51 Limonene 8.52 The hydrogenation experiment discloses the carbon skeleton of the pheromone. C13H24O Codling moth pheromone 2 H2 Pt OH C13H28O 3-Ethyl-7-methyl-1-decanol ALKENES AND ALKYNES II: ADDITION REACTIONS 147 The ozonolysis experiment allows us to locate the position of the double bonds. + O (1) O3 OH (2) Me2S + O O H O 5 6 2 4 H (2Z) 3 1 (6E) OH Codling moth pheromone OH General Problems 8.53 Retrosynthetic analysis Markov- Br + HBr nikov addition Br Br Markov- HC nikov addition CH + Br + HBr Synthesis HC CH NaNH2 liq. NH3 HC HBr C − Br + Na H Br HBr Br Br 8.54 Syn hydrogenation of the triple bond is required. So use H2 and Ni2 B(P-2) or H2 and Lindlar’s catalyst. 8.55 (a) 1-Pentyne has IR absorption at about 3300 cm−1 due to the C-H stretching of its terminal triple bond. Pentane does not absorb in that region. (b) 1-Pentene absorbs in the 1620–1680 cm−1 region due to the alkene function. Pentane does not exhibit absorption in that region. 148 ALKENES AND ALKYNES II: ADDITION REACTIONS (c) See parts (a) and (b). (d) 1-Bromopentane shows C-Br absorption in the 515–690 cm−1 region while pentane does not. (e) For 1-pentyne, see (a). The interior triple bond of 2-pentyne gives relatively weak absorption in the 2100–2260 cm−1 region. (f) For 1-pentene, see (b). 1-Pentanol has a broad absorption band in the 3200–3550 cm−1 region. (g) See (a) and (f). (h) 1-Bromo-2-pentene has double bond absorption in the 1620–1680 cm−1 region which 1-bromopentane lacks. (i) 2-Penten-1-ol has double bond absorption in the 1620–1680 cm−1 region not found in 1-pentanol. 8.56 The index of hydrogen deficiency of A, B, and C is two. C6 H14 C6 H10 H4 = 2 pairs of hydrogen atoms This result suggests the presence of a triple bond, two double bonds, a double bond and a ring, or two rings. Br2 A, B, C all decolorize a solution of bromine cold concd H2SO4 all soluble The fact that A, B, and C all decolorize Br2 and dissolve in concd. H2 SO4 suggests they all have a carbon-carbon multiple bond. A, B excess H2 ; C Pt (C6H10) (C6H14) excess H2 C6H12 Pt (C6H10) A must be a terminal alkyne, because of IR absorption at about 3300 cm−1 . Since A gives hexane on catalytic hydrogenation, A must be 1-hexyne. (1) KMnO4, HO− , heat 2H2 Pt A (2) H3O+ O OH + CO2 ALKENES AND ALKYNES II: ADDITION REACTIONS 149 This is confirmed by the oxidation experiment O (1) KMnO4, HO− , heat B 2 (2) H3O+ OH Hydrogenation of B to hexane shows that its chain is unbranched, and the oxidation experiment shows that B is 3-hexyne, . Hydrogenation of C indicates a ring is present. Oxidation of C shows that it is cyclohexene. O (1) KMnO4, HO−, heat OH (2) H3O+ C HO O 8.57 (a) Four (b) CH3(CH2)5 (CH2)7CO2H + enantiomer CH2 C HO H CH3(CH2)5 C H H CH2 H C HO H C C H + enantiomer C (CH2)7CO2H 8.58 Hydroxylations by OsO4 are syn hydroxylations (cf. Section 8.16). Thus, maleic acid must be the cis-dicarboxylic acid: H C H C CO2H HO2C H C (1) OsO4, pyr (2) NaHSO3/H2O syn hydroxylation CO2H HO2C OH C OH H meso-Tartaric acid Maleic acid Fumaric acid must be the trans-dicarboxylic acid: H HO2C C C CO2H (1) OsO4, pyr (2) NaHSO3/H2O syn hydroxylation H Fumaric acid HO2C H C OH HO + C CO2H H C H H C HO OH HO2C CO2H (±)-Tartaric acid 150 ALKENES AND ALKYNES II: ADDITION REACTIONS 8.59 (a) The addition of bromine is an anti addition. Thus, fumaric acid yields a meso compound. H HO2C HO2C H C CO2H C + Br2 C anti addition HO2C H C Br Br = C H Br H HO2C H CO2H C Br A meso compound (b) Maleic acid adds bromine to yield a racemic form. (a) H AlCl3 8.60 + Cl (b) + Cl + − (a) −H + AlCl3 (b) Hydride shift + More Less −H + + Methanide shift + Most [+ (Z )] −H + + Less [+ (Z )] + Least Each carbocation can combine with chloride ion to form a racemic alkyl chloride. 8.61 The catalytic hydrogenation involves syn addition of hydrogen to the predominantly less hindered face of the cyclic system (the face lacking 1,3-diaxial interactions when the molecule is adsorbed on the catalyst surface). This leads to I, even though II is the more stable of the two isomers since both methyl groups are equatorial in II. ALKENES AND ALKYNES II: ADDITION REACTIONS 8.62 (+) −A HBr ROOR (C7 H11 Br) t-BuOK + B Opt.active C both C7 H12 Br2 1 eq. t-BuOK Opt.inactive +) −A (− t-BuOK D O (1) O3 2 (2) Me2S H + H O O (C7 H10) Br + HBr Br Br (+) A Br B (optically active) Br C (a meso compound) Br B t-BuOK = 1 molar eq. Br (+) A (+) A C t-BuOK + 1 molar eq. Br Br (−) A (+) A A (1) O3 t-BuOK 1 molar eq. (2) Me2S O O O D + H H C 8.63 HC C C C C C H C HC C C C C (CH CH)2CH2CO2H (CH CH)2CH2CO2H C H 2 H H 151 152 ALKENES AND ALKYNES II: ADDITION REACTIONS 8.64 H H2 Pt D E Optically active (the other enantiomer is an equally valid answer) Br2 H2O 8.65 (a) Optically inactive (nonresolvable) A NaOH/H2O l eq. MeOH B C cat. HA C5 H8 O C6 H12 O2 −1 (no 3590–3650 cm IR absorption) − (3590–3650 cm 1 IR absorption) IR data indicate B does not possess an −OH group, but C does. (b) H Br H H OH A (and enantiomer) H O OCH3 H S H + R OH S OH B H CH3O R H racemate C (c) C, in contrast to its cis isomer, would exhibit no intramolecular hydrogen-bonding. This would be proven by the absence of infrared absorption in the 3500- to 3600-cm−1 region when studied as a very dilute solution. C would only show free O H stretch at about 3625 cm−1 Challenge Problems B 8.66 −H2O H2SO4 OH H OH2 + + ALKENES AND ALKYNES II: ADDITION REACTIONS 8.67 H 153 H C H CH2 N CH3CH2 Cl −C2H4 − C + H CH3CH2 N Cl CH2 Cl −Cl − C Cl CH3CH2 CH3CH2 + N C Cl CH3CH2 E CH3 H2O D CH3CH2 + N H −Cl − C O CH3CH2 H CH3CH2 H H N C CH3CH2 O Cl H −H + CH3CH2 N CH3CH2 H C + H O Cl H −H+ CH3CH2 N CH3CH2 F H C O QUIZ 8.1 A hydrocarbon whose molecular formula is C7 H12 , on catalytic hydrogenation (excess H2 /Pt), yields C7 H16 . The original hydrocarbon adds bromine and also exhibits an IR absorption band at 3300 cm−1 . Which of the following is a plausible choice of structure for the original hydrocarbon? (c) (a) (b) (d) (e) 154 ALKENES AND ALKYNES II: ADDITION REACTIONS 8.2 Select the major product of the dehydration of the alcohol, OH (a) (b) (d) (c) (e) 8.3 Give the major product of the reaction of cis-2-pentene with bromine. (a) CH3 (b) CH3 (c) CH3 CH3 (d) H Br Br H H Br Br H H Br Br H Br H H Br CH2CH3 CH2CH3 (e) A racemic mixture of (c) and (d) CH2CH3 CH2CH3 8.4 The compound shown here is best prepared by which sequence of reactions? + (a) + (b) NaNH2 then NaNH2 then Br product Br + H2 (c) Br (d) NaOC2H5 C2H5OH Pt product product product ALKENES AND ALKYNES II: ADDITION REACTIONS 155 8.5 A compound whose formula is C6 H10 (Compound A) reacts with H2 /Pt in excess to give a product C6 H12 , which does not decolorize Br2 . Compound A does not show IR absorption in the 3200–3400 cm−1 region. O Ozonolysis of A gives 1 mol of H (a) H and 1 mol of (b) (d) (c) (e) O . Give the structure of A. 8.6 Compound B (C5 H10 ) does not dissolve in cold, concentrated H2 SO4 . What is B? (a) (b) (c) (d) 8.7 Which reaction sequence converts cyclohexene to cis-1,2-cyclohexanediol? That is, H ? OH H OH (a) H2 O2 (b) (1) O3 (2) Me2 S O H (c) (1) OsO4 (2) NaHSO3 /H2 O (e) More than one of these (d) (1) R O O (2) H3 O+ /H2 O 156 ALKENES AND ALKYNES II: ADDITION REACTIONS 8.8 Which of the following sequences leads to the best synthesis of the compound ? (Assume that the quantities of reagents are sufficient to carry out the desired reaction.) Br2 (a) Br2 (b) Br NaOH H2O (1) NaNH2 (3 equiv.), heat (2) NH4Cl H2SO4 (c) Br (d) (e) Br2 NaNH2 light O3 Me2S 9 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY: TOOLS FOR STRUCTURE DETERMINATION SOLUTIONS TO PROBLEMS O (a) δ 0.8−1.10 (b) δ 2.1−2.6 (c) δ 3.3−3.9 9.1 CH3C (b) OCH2CH3 (c) (a) 9.2 The presence of two signals in the 1 H NMR spectrum signifies two unique proton environments in the molecule. The chemical shift of the downfield signal, (a), is consistent with protons and a chlorine on the same carbon. Its triplet nature indicates two hydrogens on the adjacent carbon. The upfield quintet (b) indicates four adjacent and equivalent hydrogens, which requires two hydrogens on each of two carbons. Integration data indicate a ratio of (approximately) 2:1, which is actually 4 : 2 in this case. (b) Cl Cl Of the isomeric structures with the formula C3 H6 Cl2 , only fits the (a) (a) evidence. 9.3 (a) We see below that if we replace a methyl hydrogen by Cl, then rotation of the methyl group or turning the whole molecule end-for-end gives structures that represent the same compound. This means that all of the methyl hydrogens are equivalent. H H C H C H H H replace H with Cl H H C H H C H H H Cl turn C Cl H C H H rotate CH2Cl group 157 158 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY H H C H C H H H turn Cl C H H C H Cl H Cl H rotate CH2Cl group H Cl C C H H turn H C H H C H H H All of these are the same compound. Replacing a ring hydrogen by Cl gives another compound, but replacing each ring hydrogen in turn gives the same compound. This shows that all four ring hydrogens are equivalent and that 1,4-dimethylbenzene has only two sets of chemical shift equivalent protons. CH3 CH3 CH3 CH3 CH3 Cl replace H by Cl Cl Cl CH3 CH3 Cl CH3 CH3 CH3 All of these are the same compound. (b) As in (a) we replace a methyl hydrogen by Cl yielding the compound shown. Essentially H H free rotation of the −CH2 Cl group means that all orientations of that group give the same compound. C H H C H Cl Substitution into the other methyl group produces the same compound, as seen when we flip and rotate to form that structure. H3C CH3 CH2Cl flip ClCH2 CH2Cl rotate Thus the methyl hydrogens are chemical shift equivalent. CH3 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 159 Ring substitution produces two different compounds. CH3 CH3 CH3 CH3 CH3 CH3 Cl CH3 CH3 = = Cl Cl Cl Again, flipping and rotating members of each pair demonstrates their equivalence. There are then, three sets of chemical shift equivalent protons—one for the methyl hydrogens and one each for the two different ring hydrogens. CH3 CH2Cl (c) on replacement of a methyl hydrogen by Cl. CH3 CH3 Again, different orientations (conformations) do not result in different compounds. Substitution in the other methyl group yields the same compound. CH3 CH2Cl flip CH3 CH2Cl Replacement of ring hydrogens produces three compounds: Cl CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 = C1 C1 Cl Four sets of chemical shift equivalent hydrogens are present in this isomer: one for the methyl hydrogens and three for the chemically non-equivalent ring hydrogens. 9.4 (a) One (d) One (b) Two (e) Two (c) Two (f) Two 9.5 (a) CH3 CH3 HO C H H C H CH3 replacement by Q CH3 HO C H HO C H H C Q Q C H CH3 CH3 Diastereomers 160 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY (b) Six (c) Six signals (a) CH3 (b) H C OH (c) (d) H C H (e) CH3 (f) 9.6 (a) Two (b) (a) (d) (a) (e) Two (b) (a) (g) (c) OH (a) (b) (a) (b) Four Br (a) H H (b) (h) Three (c) (c) Three (b) (b) H (c) Two (f) (a) H (c) (d) Br H (d) (b) (b) (b) (a) (a) (a) (a) Five Cl (a) H H (b) Three (b) (k) H (c) (d) OH (e) (a) (b) Six (b) (i) (b) (a) (b) (a) (j) H (d) (a) Four (c) H Four (b) (d) H(f) (l) Six (b) H (a) (c) H (d) H (e) H (c) H (f) Five (c) (m) (e) (d) (b) (b) Four (p) (a) (d) (c) (b) (c) (b) (d) (a) (o) Three (c) (a) (b) (c) (a) (n) O H (e) (c) H (d) H (c) Three (c) O (b) (b) (a) (c) (b) H (a) (a) O NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 161 9.7 Splitting tree diagram for a quartet. The signal from the observed hydrogen (a H) is split into four peaks of 1:3:3:1 intensity by three equivalent hydrogens (b H). The same method of analysis we used for the triplet pattern applies here, wherein each successively lower level in the tree diagram represents splitting by another one of the coupled hydrogens. Again, because in this case Jab is the same for splitting of the a H the signal by all three b H hydrogens, internal legs of the diagram overlap, and intensities amplify each time this occurs. The result is a pattern of 1:3:3:1 intensities, as we would observe for a quartet in an actual spectrum. The possible magnetic orientations shown under the tree diagram for the three b H hydrogens indicate that all three of the adjacent hydrogens may be aligned with the applied field, or two may be aligned with the field and one against (in three equal energy combinations), or two against and one with the field (in three equal energy combinations), or all three may be aligned against the applied field. aH (split by three bH protons) Jab Jab Jab Jab Jab Jab Applied magnetic field, B0 bH bH aH C C Jab Jab Jab bH 9.8 The determining factors here are the number of chlorine atoms attached to the carbon atoms bearing protons and the deshielding that results from chlorine’s electronegativity. In 1,1,2trichloroethane the proton that gives rise to the triplet is on a carbon atom that bears two chlorines, and the signal from this proton is downfield. In 1,1,2,3,3-pentachloropropane, the proton that gives rise to the triplet is on a carbon atom that bears only one chlorine; the signal from this proton is upfield. 162 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY Cl (a) Cl (a) Cl Cl (b) (a) Cl Cl 9.9 (a) (b) 6 7 5 3 4 δH (ppm) 2 TMS 1 0 The signal from the three equivalent protons designated (a) should be split into a doublet by the proton (b). This doublet, because of the electronegativity of the two attached chlorines, should occur downfield (δ ∼ 5–6). It should have a relative area (integration value) three times that of (b). The signal for the proton designated (b) should be split into a quartet by the three equivalent protons (a). The quartet should occur upfield (δ ∼ 1–2). I 9.10 A C3H7I is (a) (b) (a) (a) doublet δ 1.9 (b) septet δ 4.35 Cl B C H Cl is 2 4 2 (a) (b) Cl (a) doublet δ 2.08 (b) quartet δ 5.9 C C3H6Cl2 is (b) Cl (a) Cl (a) (a) triplet δ 3.7 (b) quintet δ 2.2 Hb φ = 180° Jax,ax = 8-10 Hz 9.11 (a) Ha Hb (b) Hc φ = 60° Jax,eq = 2-3 Hz NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 163 9.12 A chair conformation with both the bromine and the chlorine in equatorial positions is consistent with J1,2 = 7.8 Hz, since the hydrogens would be axial. Hax Br Jax,ax = 7 8 Hz Cl Hax trans-1-Bromo-2-chlorocyclohexane 9.13 NMR coupling constants would be distinctive for the protons indicated. Compound A would have an equatorial-axial proton NMR coupling constant of 2-3 Hz between the indicated protons, whereas B would have an axial-axial coupling constant of 8-10 Hz. OH H OH H b b O HO HO HO Ha OCH3 HO HO OCH3 Ha B Ja,b = 8-10 Hz A Ja,b = 2-3 Hz 9.14 (a) Jab = 2Jbc O HO (b) Jab Jab Jab Jbc Result: (nine peaks) 164 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY (b) Jab = Jbc (b) Jbc Jab Jab Jbc Jbc Jbc Jbc Jbc Result Jab Jbc Jbc (six peaks) 9.15 A single unsplit signal, because the environment of the proton rapidly changes, reversibly, from axial to equatorial. 9.16 A is 1-bromo-3-methylbutane. The following are the signal assignments: (a) (d) Br (c) (b) (a) (a) δ 23 (b) δ 27 (c) δ 32 (d) δ 42 B is 2-bromo-2-methylbutane. The following are the signal assignments: (c) (d) (b) (a) (b) (a) δ 11 (b) δ 33 (c) δ 40 (d) δ 68 Br NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 165 C is 1-bromopentane. The following are the signal assignments. (e) Br (c) (d) (a) (b) (a) δ 14 (b) δ 23 (c) δ 30 (d) δ 33 (e) δ 34 + 9.17 A peak at M . −15 involves the loss of a methyl radical and the formation of a 1◦ or 2◦ carbocation. + + + CH3 (M + − 15) (M +) or + + (M +) + CH3 (M + − 15) + A peak at M . −29 arises from the loss of an ethyl radical and the formation of a 2◦ carbocation. + + (M +) + (M + − 29) Since a secondary carbocation is more stable than a methyl carbocation, and since there are two cleavages that form secondary carbocations by loss of an ethyl radical, the peak at + M . −29 is more intense. 9.18 After loss of a pi bonding electron through electron impact ionization, both peaks arise from allylic fragmentations: + + Allyl radical + + + m/z 57 + m/z 41 Allyl cation 9.19 The spectrum given in Fig. 9.40 is that of butyl isopropyl ether. The main clues are the peaks at m/z 101 and m/z 73 due to the following fragmentations. NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY + loss of CH3 O + O m/z 101 + loss of O O + m/z 73 Butyl propyl ether (Fig. 9.41) has no peak at m/z 101 but has a peak at m/z 87 instead. + loss of + O O 166 m/z 87 9.20 (a) First calculate the expected masses of the compound shown for C6 H4 + 79 Br + 79 Br (M) = 234 m/z, C6 H4 + 81 Br + 79 Br (M+2) = 236 m/z, and C6 H4 + 81 Br + 81 Br (M+4) = 238 m/z. The relative ratio of 79 Br and 81 Br is 1:1. Therefore if you have two bromines in the molecule we must simplify the expression of (1:1)(1:1) = 1:2:1. The ratio of the peaks will be M (234), M+2 (236), M+4 (238) = (1:2:1) (b) First calculate the expected masses of the compound shown for C6 H4 + 35 Cl + 35 Cl (M) =146 m/z, C6 H4 + 35 Cl + 37 Cl (M+2) = 148 m/z, and C6 H4 + 37 Cl + 37 Cl (M+4) = 150 m/z. The relative ratio of 35 Cl and 37 Cl is 3:1. Therefore if you have two chlorines in the molecule we must simplify the expression of (3:1)(3:1) = 9:6:1. The ratio of the peaks will be M (146), M+2 (148), M+4 (150) = (9:6:1). 9.21 (1) We hypothesize that the highest m/z peaks with significant intensity, 78 and 80, relate to a molecular ion with isotopic contributions, i.e, that 78 and 80 are M+ and M+2 peaks, respectively. (2) Given that the ratio of m/z 78 is about three times the intensity of m/z 80, we hypothesize that chlorine is the likely isotope, i.e., that 35 Cl and 37 Cl are part of the M+ and M+2 peaks, respectively. (3) The base peak with a m/z of 43 indicates a C3 H7 + fragment. (4) The m/z peak at 43 plus either 35 or 37 for the isotopes of chlorine accounts for the M+ and M+2 at 78 and 80 and suggests the formula C3 H7 Cl. (5) Based on the 1 H NMR, with a small septet (1H) and a large doublet (2x CH3 ’s) means that the data indicate the compound is 2-chloropropane. Cl 2-Chloropropane NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 167 NMR Spectroscopy 9.22 Each number represents another signal, with the highest number indicating the total number of proton NMR signals for the molecule. Repeated use of the same number indicates positions bearing homotopic hydrogens. 2 (a) (b) H3 2 (c) 1 1 1 1 H 3 4 2 2 2 2 2 1 1 (d) (e) 6 2 3 1 (f) 1 2 Br 1 5 10,11 4 8,9 5 O 4 6,7 1 (g) 1 2 3 7 3 2 (h) 3,4 5,6 4 2 3 3 1 2 3,4 1 (i) 1 5 OH 2 2 2 OH 2 3 3 1 4 4 1 ∗ Two numbers at the same carbon indicates diastereotopic protons. 9.23 Each number represents another signal, with the highest number indicating the total number of carbon NMR signals for the molecule. Repeated use of the same number indicates positions bearing homotopic carbons. 2 (a) H 1 2 1 1 (b) 3 H (c) 1 3 3 2 2 2 2 2 3 1 1 (d) 2 3 6 8 2 7 9 3 1 (e) (f) 2 4 1 5 Br 5 3 6 3 4 7 5 6 1 2 3 8 O 4 (g) 1 2 1 4 3 2 6 5 4 1 2 1 1 (h) 2 OH (i) 1 5 2 2 1 6 OH 3 6 3 4 168 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 9.24 Chemical Shift Splitting Integration Structure (a) 0.9 (b) 1.2 t s 3H 6H (c) 1.3 (d) 2.0 q s 2H 1H CH3 adjacent to CH2 Two equivalent CH3 groups adjacent to no hydrogens CH2 adjacent only to CH3 OH (d) (b) HO (c) (b) (a) 9.25 Compound G is 2-bromobutane. Assignments are as follows: Br (a) (b) (a) (b) (c) (d) (d) (c) triplet δ 1.05 multiplet δ 1.82 doublet δ 1.7 multiplet δ 4.1 Compound H is 2,3-dibromopropene. Assignments are as follows: (b) H (a) H (c) Br Br (a) δ 4.2 (b) δ 6.05 (c) δ 5.64 Without an evaluation of the alkene coupling constants, it would be impossible to specify whether the alkene has (E ), (Z ), or geminal substitution. We do know from the integral values and number of signals that the alkene is disubstituted. For the analysis of the alkene substitution pattern these typical alkene coupling constants are useful: H C C H (trans), J = 12–18 Hz; H C C H (cis), J = 6–12 Hz; C CH2 (geminal), J = 0–3 Hz. Since the alkene signal for compound H shows little or no coupling, the data support the designation of H as 2,3-dibromopropene. 9.26 Run the spectrum with the spectrometer operating at a different magnetic field strength (i.e., at 30 or at 100 MHz). If the peaks are two singlets the distance between them—when measured in hertz—will change because chemical shifts expressed in hertz are proportional to the strength of the applied field (Section 9.7A). If, however, the two peaks represent a doublet, then the distance that separates them, expressed in hertz, will not change because this distance represents the magnitude of the coupling constant and coupling constants are independent of the applied magnetic field. NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 169 9.27 Compound O is 1,4-cyclohexadiene and P is cyclohexane. (a) (b) H2 (2 equiv.) (a) δ 26.0 (b) δ 124.5 Ni (b) CH2 (DEPT) CH (DEPT) (a) O P 9.28 The molecular formula of Q (C7 H8 ) indicates an index of hydrogen deficiency (Section 4.17) of four. The hydrogenation experiment suggests that Q contains two double bonds (or one triple bond). Compound Q, therefore, must contain two rings. Bicyclo[2.2.1]hepta-2,5-diene. The following reasoning shows one way to arrive at this conclusion: There is only one signal (δ 144) in the region for a doubly bonded carbon. This fact indicates that the doubly bonded carbon atoms are all equivalent. That the signal at δ 144 is a CH group in the DEPT spectra indicates that each of the doubly bonded carbon atoms bears one hydrogen atom. Information from the DEPT spectra tells us that the signal C at δ 85 is a —CH2 — group and the signal at δ 50 is a H group. The molecular formula tells us that the compound must contain two and since only one signal occurs in the 13 C spectrum, these equivalent. Putting this all together, we get the following: (a) (c) (c) H2 (a) δ 50 (b) Ni (c) (c) (b) δ 85 (c) δ 144 (a) Q R O 9.29 (a) 1 H NMR C C H groups must be CH (DEPT) CH2 (DEPT) CH (DEPT) O O O A B Compound A = δ 0.9 (t, 3 H), δ 2.0 (s, 3 H), δ 4.1 (q, 2 H) Compound B = δ 0.9 (t, 3 H), δ 2.0 (q, 2 H), δ 4.1 (s, 3 H) 1 (b) 1 1 2 1 4 H NMR 3 3 5 3 3 4 1 1 1 1 2 C NMR 1 3 2 4 4 1 4 4 5 4 4 2 2 3 6 3 3 1 1 1 2 3 13 2 1 1 2 1 1 H groups, 170 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY (c) 13 C NMR 3 4 4 1 2 CH3 2 CH3 3 1 1 1 CH3 CH3 3 2 3 3 2 3 2 5 2 H3C 4 3 1 CH3 3 1 9.30 That S decolorizes bromine indicates that it is unsaturated. The molecular formula of S allows us to calculate an index of hydrogen deficiency equal to 1. Therefore, we can conclude that S has one double bond. The 13 C spectrum shows the doubly bonded carbon atoms at δ 130 and δ 135. In the DEPT spectra, one of these signals (δ 130) is a carbon that bears no hydrogen atoms; the other (δ 135) is a carbon that bears one hydrogen atom. We can now arrive at the following partial structure. C C C C C H The three most upfield signals (δ 19, δ 28, and δ 31) all arise from methyl groups. The signal at δ 32 is a carbon atom with no hydrogen atoms. Putting these facts together allows us to arrive at the following structure. (c) (a) (e)( f ) (d) (a) δ 19 (b) δ 28 (c) δ 31 (c) (c) ( b) (d) δ 32 (e) δ 130 (f) δ 135 S Although the structure just given is the actual compound, other reasonable structures suggested by these data are and Mass Spectrometry 9.31 The compound is butanal. The peak at m/z 44 arises from a McLafferty rearrangement. + H + O H O + H m/z 72 (M+) CH2 H m/z 44 (M+−28) CH2 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 171 The peak at m/z 29 arises from a fragmentation producing an acylium ion. + O O H H + + m/z 29 9.32 + + OH + m/z: 144 (100.0%), 145 (10.0%) H2O m/z: 126 (100.0%), 127 (9.9%) + m/z: 111 (100.0%), 112 (8.8%) OH + m/z: 87 (100.0%), 88 (5.6%) 9.33 + + OH O m/z: 100 (100.0%), 101 (6.7%) McLafferty rearrangement m/z: 58 (100.0%), 59 (3.4%) O + m/z: 85 (100.0%), 86 (5.5%) O + m/z: 43 (100.0%), 44 (2.2%) NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 9.34 (a) + + C3H7 O + m/z: 71 (100.0%), 72 (4.4%) O 172 + m/z: 114 (100.0%), 115 (7.8%) + O C3H7 m/z: 71 (100.0%), 72 (4.4%) (b) OH + + + m/z: 86 (100.0%), 87 (5.6%) (c) H2O m/z: 68 (100.0%), 69 (5.5%) + + OH O m/z: 142 (100.0%), 143 (10.0%) + m/z: 58 (100.0%), 59 (3.4%) + OH + 9.35 First calculate the expected masses of the compound shown for C6 H4 + 79 Br + 35 Cl (M) = 190 m/z, C6 H4 + 37 Cl + 79 Br (M+2) = 192 m/z or C6 H4 + 35 Cl + 81 Br (M+2) = 192 m/z, and C6 H4 + 37 Cl + 81 Br (M+4) = 194 m/z. The relative ratio of 79 Br and 81 Br is 1:1 and 35 Cl and 37 Cl is 3:1. Therefore if you have 1 bromine and 1 chlorine in the molecule we must simplify the expression of (3:1)(1:1) = 3:4:1. The ratio of the peaks will be M (190), M+2 (192), M+4 (194) = (3:4:1) or (77%: 100%: 24%). 9.36 Ethyl bromide will have a significant M+2 for the 81 Br isotope at 110. The ratio of the peak at m/z 108 and m/z 110 will be approximately 1:1. The spectrum will also show peaks at M+1 and M+3 for the presence of a 13 C isotope (2.2 %) Methoxybenzene will have a small peak, M+1, for the 13 C isotope (7.8 %) + 9.37 The ion, CH2 NH2 , produced by the following fragmentation. R CH2 + NH2 −R + CH2 NH2 m/z 30 + CH2 NH2 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 173 Integrated Structure Elucidation (a) (b) OH 9.38 (a) (a) (a) Singlet, δ 1.28 (9H) (b) Singlet, δ 1.35 (1H) (a) Br (a) Doublet, δ 1.71 (6H) (b) Septet, δ 4.32 (1H) (b) (b) (a) (a) O (a) (c) (b) (c) (a) OH (a) Triplet, δ 1.05 (3H) (b) Singlet, δ 2.13 (3H) (c) Quartet, δ 2.47 (2H) C O, near 1720 cm−1 (s) (a) Singlet, δ 2.43 (1H) O H, 3200–3550 cm−1 (br) (b) Singlet, δ 4.58 (2H) (c) Multiplet, δ 7.28 (5H) (d) (b) (c) (e) (a) (a) Doublet, δ 1.04 (6H) (b) Multiplet, δ 1.95 (1H) (c) Doublet, δ 3.35 (2H) Cl (b) (c) O (f) (a) (a) Singlet, δ 2.20 (3H) C O, near 1720 cm−1 (s) (b) Singlet, δ 5.08 (1H) (c) Multiplet, δ 7.25 (10H) (d) OH (a) Triplet, δ 1.08 (3H) C O (acid), 1715 cm−1 (s) (b) Multiplet, δ 2.07 (2H) O H, 2500–3500 cm−1 (br) (c) Triplet, δ 4.23 (1H) (d) Singlet, δ 10.97 (1H) C6H5 (b) C6H5 (c) (b) (g) O (c) (a) Br (a) (h) (b) (c) (a) Triplet, δ 1.25 (3H) (b) Quartet, δ 2.68 (2H) (c) Multiplet, δ 7.23 (5H) NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY O (i) (a) (a) Triplet, δ 1.27 (3H) C O (acid), 1715 cm−1 (s) (b) Quartet, δ 3.66 (2H) O H, 2500–3550 cm−1 (br) (c) Singlet, δ 4.13 (2H) (d) Singlet, δ 10.95 (1H) (d) OH O (c) (b) NO2 (j) (b) (a) (a) (k) CH 3 (a) Doublet, δ 1.55 (6H) (b) Septet, δ 4.67 (1H) (a) (b) O O (b) (a) Singlet, δ 3.25 (6H) (b) Singlet, δ 3.45 (4H) (a) CH3 O (l) (a) Doublet, δ 1.10 (6H) (b) Singlet, δ 2.10 (3H) (c) Septet, δ 2.50 (1H) (c) (b) (a) C O, near 1720 cm−1 (s) (a) (a) Doublet, δ 2.00 (3H) (b) Quartet, δ 5.15 (1H) (c) Multiplet, δ 7.35 (5H) (b) (m) Br (c) 9.39 Compound E is phenylacetylene, C6 H5 C CH. We can make the following assignments in the IR spectrum: 100 90 E, C8H6 80 Transmittance (%) 174 70 Ar 60 C H C 50 40 30 20 C 10 0 4000 H 3600 3200 C (Ar) C 2800 2400 2000 1800 Wavenumber 1600 1400 1200 1000 800 (cm−1) The IR spectrum of compound E (Problem 9.40). Compound F is C6 H5 CBr2 CHBr2 . 650 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 175 9.40 Compound J is cis-1,2-dichloroethene. H H Cl Cl We can make the following IR assignments: 3125 cm−1 , alkene C H stretching 1625 cm−1 , C C stretching 695 cm−1 , out-of-plane bending of cis C H bonds The 1 H NMR spectrum indicates the hydrogens are equivalent. 9.41 (a) Compound K is, O (b) (c) (a) (b) (c) (d) (a) OH (d) Singlet, δ 2.15 C O, near 1720 cm−1 (s) Quartet, δ 4.25 Doublet, δ 1.35 Singlet, δ 3.75 (b) When the compound is dissolved in D2 O, the —OH proton (d ) is replaced by a deuteron, and thus the 1 H NMR absorption peak disappears. O O + + D2O OH DHO OD 9.42 The IR absorption band at 1745 cm−1 indicates the presence of a C O group in a five-membered ring, and the signal at δ 220 can be assigned to the carbon of the carbonyl group. There are only two other signals in the 13 C spectrum; the DEPT spectra suggest two equivalent sets of two groups each. Putting these facts together, we arrive at CH2 cyclopentanone as the structure for T. O (c) (b) (a) (a) δ 23 (b) δ 38 (c) δ 220 T 9.43 (1) Molecular ion = 96 m/z Potential molecular formula = C7 H14 or C6 H8 O (2) The presence of a strong C=O absorption at 1685 cm−1 in the IR and the integrals of the 1 H NMR spectra totaling 8, and the appearance of 6 unique carbons in the 13 C spectra lead to C6 H8 O as the correct molecular formula. 176 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY (3) Degree of unsaturation for C6 H8 O = 3 (4) 1 H NMR Letter ppm (a) 1.9 (b) 2.2 (c) 3.0 (d) 6.5 Splitting s q t t O Integration 3H 2H 2H 1H (c) Conclusion CH3 (allylic) CH2 (allylic) CH2 (adjacent to ketone and CH2 ) CH (adjacent to CH2 ) (a) (b) (d) (5) 13 C NMR 1. 2. 3. 4. 5. 6. δ207 – (ketone) δ145 – (CH – alkene) δ139 – (C – alkene) δ37 – (CH2 alpha to ketone) δ25 – (CH2 allylic) δ16 – (CH3 – allylic) O 1 4 5 3 6 2 9.44 (1) Molecular ion = 148 m/z Potential molecular formula = C10 H12 O (2) IR: 3065 cm−1 (C-H sp2 aromatic), 2960 cm−1 (C-H sp3 ), 2760 cm−1 (C-H aldehyde), 1700 cm−1 (C O conjugate aldehyde), 1600 cm−1 (C C aromatic) (3) Degree of unsaturation for C10 H12 O = 5 (4) 1 H NMR Letter ppm (a) 1.3 (b) 3.1 (c) 7.2 (d) 7.8 (e) 9.8 Splitting d septet d d s (c) Integration 6H 1H 2H 2H 1H O (d) H (e) (a) (d) (b) (a) (c) Conclusion Two CH3 adjacent to C-H C-H adjacent to two CH3 Aromatic ring disub - para Aromatic ring disub - para Aldehyde NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 177 (5) 13 C NMR 1. δ191 – CH aldehyde 2. δ154 – C aromatic 3. δ134 – C aromatic 4. δ130 – CH aromatic 5. δ127 – CH aromatic 6. δ36 – CH aliphatic 7. δ23 – CH3 aliphatic O 4 2 5 7 H 4 3 6 1 5 7 9.45 (1) Molecular ion = 204 m/z Potential molecular formula = C15 H24 (2) IR: 3065 cm−1 (C aromatic) H sp2 aromatic), 2960 cm−1 (C H sp3 ) 1600 cm−1 (C (3) Degree of unsaturation for C15 H24 = 4 (4) 1 H NMR Letter ppm (a) 1.3 (b) 2.8 (c) 6.9 Splitting d septet s Integration 6H 1H 1H (a) (b) (a) (c) (a) Conclusion Two CH3 next to CH CH next to two CH3 CH aromatic (c) (a) (b) (b) (c) (a) (a) (5) 13 C NMR 1. δ148 – (C aromatic) 2. δ122 – (CH aromatic) 3. δ37 – (CH aliphatic) 4. δ23 – (CH3 aliphatic) 4 4 3 1 2 4 2 1 3 4 2 4 1 3 4 C 178 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY 9.46 (1) Molecular Formula = C10 H12 O3 (2) IR − 3065 cm−1 (C H sp2 aromatic), 2960 cm−1 (C ester), 1600 cm−1 (C C aromatic) (3) Degree of unsaturation for C10 H12 O3 = 5 (4) 1 H NMR Letter ppm (a) 3.5 (b) 3.7 (c) 3.8 (d) 6.7 (e) 6.9 Splitting s s s d d Integration 2H 3H 3H 2H 2H H sp3 ), 1740 cm−1 (C O Conclusion CH2 CH3 CH3 CH (aromatic para-sub) CH (aromatic para -sub) The low ppm of the aromatic protons indicates electron donating groups are attached to the ring. O (b) (a) O (e) (e) (d) (d) O (c) 13 (4) C NMR 1. δ171 – (C O ester) 2. δ160 – (C aromatic) 3. δ130 – (CH aromatic) 4. δ127 – (C aromatic) 5. δ115 – (CH aromatic) 6. δ56 – (CH3 aliphatic) 7. δ52 – (CH3 ester) 8. δ46 – (CH2 benzylic) O 7 1 O 8 4 3 3 5 5 2 O 6 Challenge Problems 9.47 The product of the protonation is a relatively stable allylic cation. The six methyl, four methylene, and single methine hydrogens account for the spectral features. NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY H2SO4 179 + + + 9.48 A McLafferty rearrangement accounts for this outcome, as shown below for butanoic acid. In the case of longer chain carboxylic acids the additional carbons are eliminated in the alkene formed. H + + O H H OH H H H O + OH m/z 60 9.49 As in the case of IR spectroscopy of alcohols (Section 2.16B), the degree of intermolecular association of the hydroxyl groups is the principal determining factor. If the alcohol is examined in the vapor phase or at very low concentration in a nonpolar solvent such as carbon tetrachloride, the OH hydrogen absorption occurs at about δ 0.5. Increasing concentrations of the alcohol up to the neat state (solvent is absent) leads to a progressive shift of the absorption toward the value of δ 5.4. H + δ R Hydrogen bonding, O H O , results in a decrease of electron density (deδ− R shielding) at the hydroxyl proton. Thus, the proton peak is shifted downfield. The magnitude of the shift, at some particular temperature, is a function of the alcohol concentration. 9.50 At the lower temperature a DMF molecule is effectively restricted to one conformation due to the resonance contribution of II, which increases the bond order of the C N bond. At ∼ 300 K there is insufficient energy to bring about significant rotation about that bond. CH3 O CH3 N CH3 O + − N H I CH3 H II The methyl groups clearly are nonequivalent since each has a unique relationship to the single hydrogen atom. Hence each set of methyl group hydrogens is represented by its own signal (though they are quite close). At a sufficiently high temperature (> 130◦ C) there is enough energy available to overcome the barrier to rotation due to the quasi-double bond, C N , and substantially free rotation 180 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY occurs. The differences between the two sets of methyl group hydrogens are no longer discernable and all six methyl hydrogens are represented by a single signal. 9.51 Formed by fragmentation of the molecular ion, C4 H +3 possesses the requisite mass. Possible structures are [CH2 CH C C]+ and [HC CH—C CH]+ , each of which is resonancestabilized. Br 9.52 Ha Hb O Hc Jab = 5.3 Hz Jac = 8.2 Hz Jbc = 10.7 Hz (a) Ha Hb h,i Br O 9 signals f,g Hc d,e Ha (b) Jac = 8.2 Hz Jab = 5.3 Hz QUIZ 9.1 Propose a structure that is consistent with each set of the following data. (a) C4 H9 Br 1 (b) C4 H7 Br3 1 H NMR spectrum Singlet δ 1.95 (3H) Singlet δ 3.9 (4H) (c) C8 H16 1 H NMR spectrum Singlet δ 1.7 H NMR spectrum Singlet δ 1.0 (9H) Singlet δ 1.75 (3H) Singlet δ 1.9 (2H) Singlet δ 4.6 (1H) Singlet δ 4.8 (1H) IR spectrum 3040, 2950, 1640 cm−1 and other peaks. NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY (d) C9 H10 O 1 H NMR spectrum Singlet δ 2.0 (3H) Singlet δ 3.75 (2H) Singlet δ 7.2 (5H) IR spectrum 3100, 3000, 1720, 740, 700 cm−1 and other peaks. (e) C5 H7 NO2 1 IR spectrum 2980, 2260, 1750 cm−1 and other peaks. This compound has a nitro group. H NMR spectrum Triplet δ 1.2 (3H) Singlet δ 3.5 (2H) Quartet δ 4.2 (2H) 181 9.2 How many 1 H NMR signals would the following compound give? Cl (a) One (b) Two (c) Three (d) Four (e) Five 9.3 How many 1 H NMR signals would 1,1-dichlorocyclopropane give? (a) One (b) Two (c) Three (d) Four (e) Five 9.4 Which of these C6 H14 isomers has the greatest number of 13 C NMR signals? (a) Hexane (d) 2,2-Dimethylbutane (b) 2-Methylpentane (e) 2,3-Dimethylbutane (c) 3-Methylpentane 9.5 How many 13 C NMR signals would be given by the following compound? O (a) 7 (b) 8 (c) 10 (d) 11 (e) 13 9.6 Which of these is a true statement concerning mass spectrometry? (There may be more than one.) (a) The M+ peak is always the most prominent (largest m/z). (b) Only liquids can be analyzed by MS. (c) Unlike IR and NMR, MS is a destructive method of analysis. (d) The molecular ion is assigned a value of 100 on the vertical scale. (e) The initial event in the determination of a mass spectrum in the EI mode is the formation of a radical cation. 9.7 What is the structure of a compound C5 H12 which exhibits a prominent MS peak at m/z 57? 10 RADICAL REACTIONS SOLUTIONS TO PROBLEMS 10.1 > 3° > > 2° > 1° > CH3 > Methyl 10.2 The compounds all have different boiling points. They could, therefore, be separated by careful fractional distillation. Or, because the compounds have different vapor pressures, they could easily be separated by gas chromatography. GC/MS (gas chromatography/mass spectrometry) could be used to separate the compounds as well as provide structural information from their mass spectra. 10.3 Their mass spectra would show contributions from the naturally occurring 35 Cl and 37 Cl isotopes. The natural abundance of 35 Cl is approximately 75% and that of 37 Cl is approximately 25%. Thus, for CH3 Cl, containing only one chlorine atom, there will be an M +. peak and an M +. +2 peak in roughly a 3 : 1 (0.75 : 0.25) ratio of intensities. For CH2 Cl2 there will be M +. , M +. +2, and M +. +4 peaks in roughly a 9 : 6 : 1 ratio, respectively. [The probability of a molecular ion M +. with both chlorine atoms as 35 Cl is (.75)(.75) = .56, the probability of an M +. +2 ion from one 35 Cl and one 37 Cl is 2(.75)(.25) = .38, and the probability of an M +. +4 ion peak from both chlorine atoms as 37 Cl is (.25)(.25) = 0.06; thus, their ratio is 9 : 6 : 1.] For CHCl3 there will be M +. , M +. +2, and M +. +4, and M +. +6 peaks in approximately a 27 : 27 : 9 : 1 ratio, respectively (based on a calculation by the same method). (This calculation does not take into account the contribution of 13 C, 2 H, and other isotopes, but these are much less abundant.) 10.4 The use of a large excess of chlorine allows all of the chlorinated methanes (CH3 Cl, CH2 Cl2 , and CHCl3 ) to react further with chlorine. Cl2 10.5 hv or heat 2 Cl H Step 2a Cl 182 + Cl H Cl + Cl RADICAL REACTIONS 183 Cl Step 3a Step 2b Step 3b Cl Cl + + Cl H Cl + Cl Cl 1,1-dichloroethane Cl + Cl Cl + H Cl Cl Cl Cl Cl + Cl 1,2-dichloroethane 10.6 (a) There is a total of eight hydrogen atoms in propane. There are six equivalent 1◦ hydrogen atoms, replacement of any one of which leads to propyl chloride, and there are two equivalent 2◦ hydrogen atoms, replacement of any one of which leads to isopropyl chloride. Cl Cl + Cl2 + If all the hydrogen atoms were equally reactive, we would expect to obtain 75% propyl chloride and 25% isopropyl chloride: % Propyl chloride = 6/8 × 100 = 75% % Isopropyl chloride = 2/8 × 100 = 25% (b) Reasoning in the same way as in part (a), we would expect 90% isobutyl chloride and 10% tert-butyl chloride, if the hydrogen atoms were equally reactive. + Cl2 Cl + Cl % Isobutyl chloride = 9/10 × 100 = 90% % tert-Butyl chloride = 1/10 × 100 = 10% (c) In the case of propane (see Section 10.6), we actually get more than twice as much isopropyl chloride (55%) than we would expect if the 1◦ and 2◦ hydrogen atoms were equally reactive (25%). Clearly, then, 2◦ hydrogen atoms are more than twice as reactive as 1◦ hydrogen atoms. In the case of isobutane, we get almost four times as much tert-butyl chloride (37%) as we would get (10%) if the 1◦ and 3◦ hydrogen atoms were equally reactive. The order of reactivity of the hydrogens then must be 3◦ > 2◦ > 1◦ 184 RADICAL REACTIONS 10.7 The hydrogen atoms of these molecules are all equivalent. Replacing any one of them yields the same product. Cl + hv Cl2 (+ more highly chlorinated products) Cl + hv Cl2 (+ more highly chlorinated products) We can minimize the amounts of more highly chlorinated products formed by using a large excess of the cyclopropane or cyclobutane. (And we can recover the unreacted cyclopropane or cyclobutane after the reaction is over.) 10.8 (a) (b) Cl Cl Cl2 10.9 (a) Cl Cl + light (S )-2-Chloropentane Cl (2S,4S )-2,4-Dichloro- (2R,4S )-2,4-Dichloropentane pentane (b) They are diastereomers. (They are stereoisomers, but they are not mirror images of each other.) (c) No, (2R,4S)-2,4-dichloropentane is achiral because it is a meso compound. (It has a plane of symmetry passing through C3.) (d) No, the achiral meso compound would not be optically active. (e) Yes, by fractional distillation or by gas chromatography. (Diastereomers have different physical properties. Therefore, the two isomers would have different vapor pressures.) (f and g) In addition to the (2S,4S)-2,4-dichloropentane and (2R,4S)-2,4-dichloropentane isomers described previously, we would also get (2S,3S)-2,3-dichloropentane, (2S,3R)2,3-dichloropentane and the following: Cl Cl Cl (optically active) Cl Cl + + (optically inactive) Cl (optically active) 10.10 (a) The only fractions that would contain chiral molecules (as enantiomers) would be those containing 1-chloro-2-methylbutane and the two diastereomers of 2-chloro-3methylbutane. These fractions would not show optical activity, however, because they would contain racemic forms of the enantiomers. RADICAL REACTIONS 185 (b) Yes, the fractions containing 1-chloro-2-methylbutane and the two containing the 2-chloro-3-methylbutane diastereomers. (c) Yes, each fraction from the distillation could be identified on the basis of 1 H NMR spectroscopy. The signals related to the carbons where the chlorine atom is bonded would be sufficient to distinguish them. The protons at C1 of 1-chloro-2-methylbutane would be a doublet due to splitting from the single hydrogen at C2. There would be no proton signal for C2 of 2-chloro-2-methylbutane since there are no hydrogens bonded at C2 in this compound; however there would be a strong singlet for the six hydrogens of the geminal methyl groups. The proton signal at C2 of 2-chloro-3-methylbutane would approximately be a quintet, due to combined splitting from the three hydrogens at C1 and the single hydrogen at C3. The protons at C1 of 1-chloro-3-methylbutane would be a triplet due to splitting by the two hydrogens at C2. Br Br 10.11 (a) (i) Br (ii) (iii) Br (ii) Br (b) Br Br Br (iii) Br d (c) (i) (ii) from from d d d d d (iii) from d 10.12 leads to 1-chloro-1-phenylpropane 10.13 I leads to 2-chloro-1-phenylpropane II leads to 1-chloro-3-phenylpropane III The major product is 1-chloro-1-phenylpropane because I is the most stable radical. It is a benzylic radical and therefore is stabilized by resonance. Cl RADICAL REACTIONS 10.14 (a) Retrosynthetic Analysis − Na+ Synthesis NaNH2 − liq. NH3 CH3I Na+ (b) Retrosynthetic Analysis − Na+ Synthesis (1) NaNH2/liq. NH3 (2) Br (c) Retrosynthetic Analysis H H Synthesis H2 H Ni2B(P-2) H [from (a)] (d) Retrosynthetic Analysis H H Synthesis H Li, EtNH2 [from (a)] H 186 10.15 Head-to-tail polymerization leads to a more stable radical on the growing polymer chain. In head-to-tail coupling, the radical is 2◦ (actually 2◦ benzylic, and as we shall see in Section 15.12A this makes it even more stable). In head-to-head coupling, the radical is 1◦ . RADICAL REACTIONS 10.16 (a) OCH3 R R + OCH3 OCH3 (from initiator) 187 Monomer OCH3 R OCH3 OCH3 ( OCH3 R OCH3 OCH3 OCH3 repetition (n OCH3 OCH3 OCH3 (b) Cl R Cl Cl R + Cl Cl (from initiator) Cl Monomer Cl Cl R Cl Cl Cl Cl ( Cl Cl R Cl Cl Cl Cl Cl Cl repetition (n Cl Cl Cl Cl Cl Cl 10.17 In the cationic polymerization of isobutylene (see text), the growing polymer chain has a stable 3◦ carbocation at the end. In the cationic polymerization of ethene, for example, the intermediates would be much less stable 1◦ cations. H A H H CH3CH2+ + H H H H H H + etc. 1° Carbocation With vinyl chloride and acrylonitrile, the cations at the end of the growing chain would be destabilized by electron-withdrawing groups. 188 RADICAL REACTIONS Cl A H + Cl + Cl + etc. Cl Cl CN A H + CN + CN + etc. CN CN Radical Mechanisms and Properties 10.18 (1) Br2 (2) Br hv 2 Br + + (3) + Br2 then 2,3 HBr Br + Br 2,3 etc. , the rate-determining step is: 10.19 For formation of Br ⫽ d Br + H d Br H A⫽ A Br, the rate-determining step is: For formation of ⫽ d Br H + H B⫽ d Br B RADICAL REACTIONS 189 B⫽ B + HBr A⫽ Eact A + HBr + Br P.E. Eact Reaction coordinate · · A is a 3◦ alkyl radical and more stable than B , a 1◦ alkyl radical, a difference anticipated by the relative energies of the transition states. As indicated by the potential energy diagram, the activation energy for the formation of A–| is less than that for the formation of B–| . The lower energy of A–| means that a greater fraction of bromine atom-alkane collisions will lead to A rather than to B . Note there is a statistical advantage to the formation of B (6:1) but this is outweighed by the inherently greater reactivity of a 3◦ hydrogen. · Cl 10.20 and · · I are the only monosubstitution products which can be formed from cyclopentane and ethane, respectively. (However, direct iodination of alkanes is not a feasible reaction, practically speaking.) Br Br would be formed in amounts much larger than the isomeric alterna- and tives due to the highly selective nature of bromine. Br Formation of would be acompanied by considerable and Cl amounts of isomeric byproducts in each case. 10.21 Chain-Initiating Step Cl2 heat, hv light 2 Cl Chain-Propagating Steps Cl2 Cl + Cl + HCl Cl Cl + HCl Cl2 + Cl 190 RADICAL REACTIONS 10.22 (a) Three Cl Cl2 H + Cl H Cl + I II III Enantiomers as a racemic form (b) Only two: one fraction containing I, and another fraction containing the enantiomers II and III as a racemic form. (The enantiomers, having the same boiling points, would distill in the same fraction.) (c) Both of them. (d) The fraction containing the enantiomers. (e) In the 1 H spectrum for 1-chlorobutane the signal furthest downfield would be that for CH2 Cl; it would be a triplet. The corresponding signal for in either C CH Cl enantiomer of 2-chlorobutane (also furthest downfield) would be an approximate sextet. The DEPT spectra for 1-chlorobutane would identify one CH3 group and three CH2 groups; for 2-chlorobutane, two (non-equivalent) CH3 groups, one CH2 group, and one CH group would be specified. (f) Molecular ions from both 1-chlorobutane and the 2-chlorobutane enantiomers would be present (but probably of different intensities). M +. +2 peaks would also be present. Both compounds would likely undergo C Cl bond cleavage to yield C4 H + 9 cations. The mass spectrum of 1-chlorobutane would probably show loss of a propyl radical by C C bond cleavage adjacent to the chlorine, resulting in an m/z 49 peak for CH2 Cl+ (and m/z 51 from 37 C1). Similar fragmentation in 2-chlorobutane would produce an m/z 63 peak for CH3 CHCl+ (and m/z 65). 10.23 (a) Five H Cl H Cl Cl2 (R)-2Chlorobutane Cl H Cl H Cl Cl Cl + A + B H Cl Cl H + C D H Cl + E (b) Five. None of the fractions would be a racemic form. (c) The fractions containing A, D, and E. The fractions containing B and C would be optically inactive. (B contains no chirality center and C is a meso compound.) 10.24 (a) Oxygen-oxygen bonds are especially weak, that is, HO CH3CH2O OH OCH3 DH ° = 214 kJ mol−1 DH ° = 184 kJ mol−1 Cl RADICAL REACTIONS 191 This means that a peroxide will dissociate into radicals at a relatively low temperature. RO 100-200 °C OR 2 RO Oxygen-hydrogen single bonds, on the other hand, are very strong. (For HO H, DH ◦ = 499 kJ mol−1 .) This means that reactions like the following will be highly exothermic. + R RO H O (b) Step 1 H + R RO heat O O 2 Chain Initiation + O Step 2 + Br Step 3 R + R Step 4 Br R R Br H H > 10.25 OH + R H Br + Br Br + R ∼ > (3°) (2°) Chain Propagation (1°) (1°) 10.26 Single-barbed arrows show conversion of the enediyne system to a 1,4-benzenoid diradical via the Bergman cycloaromatization. Each alkyne contributes one electron from a pi bond to the new sigma bond. The remaining electrons in each of the pi bonds become the unpaired electrons of the 1,4-benzenoid diradical. The diradical is a highly reactive intermediate that reacts further to abstract hydrogen atoms from the DNA sugar-phosphate backbone. The new radicals formed on the DNA lead to bond fragmentation along the backbone and to double-stranded cleavage of the DNA. OMe OO NH Bergman cycloaromatization HO S O Sugar Calicheamicin enediyne intermediate OO OO OMe NH HO S Hydrogen abstraction from DNA OMe NH H HO S O O Sugar Sugar 1,4-Benzenoid diradical H 192 RADICAL REACTIONS 10.27 CH2 = CH H (CH3 )2 CH H CH2 = CHCH2 DH◦ (kJ mol−1 ) 465 413 369 H (a) It is relatively difficult to effect radical halogenation at Ha because of the large DH◦ for dissociation of a bond to an alkenyl (vinylic) hydrogen. (b) Substitution of Hb occurs more readily than Hc because DH◦ for the generation of an allylic radical is significantly smaller than that for the formation of a simple 2◦ radical. O O O 10.28 (1) Br + O heat Br O O (2) + CO2 O O O (3) Br Br + O + then 2,3 2,3 etc. Synthesis 10.29 (a) CH3CH3 (b) Br2 Br NaI (SN2) I OH NaH (−H2) O− Na+ heat, light Br NaOH (SN2) Br [from part (a)] O (c) Br2 heat, light Br OK OH heat (E2) RADICAL REACTIONS Br (d) (e) Br2 ONa heat, light OH, heat (E2) Br (by antiMarkovnikov addition) Br2 CH4 CH3Br heat, light H 1) NaNH2 H (f ) H HBr, ROOR, heat, light liq. NH3 2) CH3Br 1) NaNH2 H 1) NaNH2 H H liq. NH3 2) Br [from part (a)] liq. NH3 2) CH3Br H2, Ni2B or H2, Lindlar’s catalyst HA, H2O OH (g) Br + Na+ −N + N N− SN2 [from part (a)] Br Br2 10.30 (a) hv N + N ONa H2O OH H2SO4 N− (aqueous solvolysis of 2-bromo-2-methylpropane would also yield the desired alcohol) HBr (b) ROOR Br [from (a)] (1) BH3 : THF (c) (2) H2O2, OH⫺ [from (a)] OH OH 193 194 RADICAL REACTIONS O F F O⫺ Hg F (1)* (d) 2 THF-CH3OH OCH3 HO⫺ (2) NaBH4, *[Hg(OAc)2 would also work in step (1)] [from (a)] Br Br Br2 (e) hv Br Br2 ONa OH ⫹ enantiomer Br Br2 (f ) ONa hv OH OH Br2 NaOH H2O O H2O Br ⫹ enantiomer 10.31 (a) retro: O Br (b) CH3ONa NBS hν synthesis: Br O OH retro: Br CN CN Br (c) Br2 H2O NaCN NBS hν synthesis: Br retro: CN CN Br O OH Br O Br OH O + Br synthesis: NBS hν Br 1) BH3 : THF OH (2) Br O Br2 H2O O Br (1) NaH (1) NaNH2 (3 equiv) (2) NH4Cl 2) H2O2, HO− O RADICAL REACTIONS 195 10.32 (a) Br CN retro: Br Br synthesis: Br Br Br NBS hν CN NaCN Br Br (b) Br retro: + Br NaNH2 NBS hν synthesis: (c) Br retro: Br EtONa EtOH, heat NBS hν synthesis: 10.33 (a) retro: Br OH O + Br synthesis: NBS hν Br H2O OH (1) NaH or HO− O (2) NBS hν Br 196 RADICAL REACTIONS (b) + Br retro: + Br Br NaNH2 NBS hν synthesis: (1) NaNH2 (2) NBS hν 10.34 (1) N N Br 60 °C N 2 N N + N2 AIBN (2) N N H + + H O (3) + O O2 O (4) O then 3,4 3,4 etc. OH + H O + 10.35 Besides direct H· abstraction from C5 there would be many H· abstractions from the three methyl groups, leading to: RADICAL REACTIONS 197 Any of these radicals could then, besides directly attacking chlorine, intramolecularly abstract H· from C5 (analogous to the “back biting” that explains branching during alkene radical polymerization). O O 10.36 (1) Ph O O heat O 2 Ph O Ph O (2) Ph + Ph CO2 O (3) Ph H + PhH + O O (4) + CO O (5) H + then 4,5 4,5 etc. + 10.37 HO + O HO • HO O + HOH dimerization HO OH X N N 10.38 (1) + Bu3SnH + Bu3Sn H (from AIBN) I (2) Bu3Sn (3) + Bu3SnI + then 2,3 2,3 etc. Bu3SnH + + Bu3Sn 198 RADICAL REACTIONS O 10.39 O O O Ph O 2 Ph Ph O + Bu 3Sn O Ph Ph Ph O + Bu3Sn Bu 3SnH + CO2 PhH Br + + Bu3SnBr + Bu3SnH + Bu3SnH + Bu3Sn + Bu3Sn 10.40 Unpaired electron density in the methyl radical is localized solely above and below the carbon atom, in the region corresponding to the p orbital of the approximately sp2 -hybridized carbon atom. The ethyl radical shows some unpaired electron density at the adjacent hydrogen atoms, especially the hydrogen atom that in the conformation shown has its H C sigma bond aligned parallel to the unpaired electron density of the p orbital of the radical. The larger size of the spin density lobe of the hydrogen with its H C bond parallel to the p orbital of the radical indicates hyperconjugation with the radical. This effect is even more pronounced in the tert-butyl radical, where three hydrogen atoms with H C sigma bonds parallel to the radical p orbital (two hydrogens above the carbon plane and one below in the conformation shown) have larger unpaired electron density volumes than the other hydrogen atoms. 10.41 The sequence of molecular orbitals in O2 is σ 1s (HOMO-7), σ ∗ 1s (HOMO-6), σ 2s (HOMO-5), σ ∗ 2s (HOMO-4), π 2 p y (HOMO-3), π 2 pz (HOMO-2), σ 2 px (HOMO-1), π ∗ 2 p y (HOMO), π ∗ 2 pz (LUMO). Therefore (a) HOMO-3 and HOMO-2 represent bonding pi molecular orbitals, (b) HOMO-1 is a bonding sigma molecular orbital comprised of overlap of the px orbitals on each oxygen, and (c) the HOMO and LUMO represent the antibonding pi molecular orbital counterparts to the bonding pi molecular orbitals RADICAL REACTIONS 199 represented by HOMO-3 and HOMO-2. Note that the s and p orbitals in O2 are not hybridized. A diagram of the orbitals and their respective energy levels is shown below. HOMO (π∗2py) LUMO (π∗2pz) HOMO-1 (σ 2px) HOMO-3 (π2py) HOMO-2 (π2pz) HOMO-4 (σ∗2s) HOMO-5 (σ2s) HOMO-6 (σ∗1s) HOMO-7 (σ1s) 200 QUIZ RADICAL REACTIONS Use the single-bond dissociation energies of Table 10.1 (page 202): 10.1 On the basis of Table 10.1, what is the order of decreasing stability of the radicals, CH2 CH > CH2 CH2 HC C (a) HC (b) CH2 CH (c) CH2 CHCH2 > HC (d) CH2 CHCH2 > CH2 (e) CH2 CH C > > ? CHCH2 CH > CH2 CHCH2 C > CH2 CHCH2 HC CH2 > C CH2 CH CH > HC C CHCH2 > HC C 10.2 In the radical chlorination of methane, one propagation step is shown as Cl + CH4 + HCl CH3 Why do we eliminate the possibility that this step goes as shown below? Cl + CH4 CH3Cl + H (a) Because in the next propagation step, H· would have to react with Cl2 to form Cl· and HCl; this reaction is not feasible. (b) Because this alternative step has a more endothermic H ◦ than the first. (c) Because free hydrogen atoms cannot exist. (d) Because this alternative step is not consistent with the high photochemical efficiency of this reaction. 10.3 Pure (S)-CH3 CH2 CHBrCH3 is subjected to monobromination to form several isomers of C4 H8 Br2 . Which of the following is not produced? (a) H Br Br (b) H (d) (e) Br Br (c) H Br H H Br H Br Br H Br Br 10.4 Using the data of Table 10.1, calculate the heat of reaction, H ◦ , of the reaction, CH3CH3 (a) 47 kJ mol−1 (d) −1275 kJ mol−1 + Br2 (b) −47 kJ mol−1 (e) −157 kJ mol−1 Br + HBr (c) 1275 kJ mol−1 RADICAL REACTIONS 201 Table 10.1 Single-bond homolytic dissociation energies DH◦ at 25◦ C Compound H H D D F F Cl Cl Br Br I I H F H Cl H Br H I CH3 H CH3 F CH3 Cl CH3 Br CH3 I CH3 OH CH3 OCH3 CH3 CH2 H CH3 CH2 F CH3 CH2 Cl CH3 CH2 Br CH3 CH2 I CH3 CH2 OH CH3 CH2 OCH3 CH3 CH2 CH2 H CH3 CH2 CH2 F CH3 CH2 CH2 Cl CH3 CH2 CH2 Br CH3 CH2 CH2 I CH3 CH2 CH2 OH CH3 CH2 CH2 OCH3 (CH3 )2 CH H (CH3 )2 CH F (CH3 )2 CH Cl A:B kJ mol−1 436 443 159 243 193 151 570 432 366 298 440 461 352 293 240 387 348 421 444 353 295 233 393 352 423 444 354 294 239 395 355 413 439 355 A + B Compound kJ mol−1 (CH3 )2 CH Br (CH3 )2 CH I (CH3 )2 CH OH (CH3 )2 CH OCH3 (CH3 )2 CHCH2 H (CH3 )3 C H (CH3 )3 C Cl (CH3 )3 C Br (CH3 )3 C I (CH3 )3 C OH (CH3 )3 C OCH3 C6 H5 CH2 H CH2 CHCH2 H CH2 CH H C6 H5 H HC C H CH3 CH3 CH3 CH2 CH3 CH3 CH2 CH2 CH3 CH3 CH2 CH2 CH3 (CH3 )2 CH CH3 (CH3 )3 C CH3 HO H HOO H HO OH (CH3 )3 CO OC(CH3 )3 O C6H5CO 298 222 402 359 422 400 349 292 227 400 348 375 369 465 474 547 378 371 374 343 371 363 499 356 214 157 O OCC6H5 CH3 CH2 O OCH3 CH3 CH2 O H 139 184 431 O CH3C–H 364 202 RADICAL REACTIONS 10.5 What is the most stable radical that would be formed in the following reaction? Cl + 10.6 The reaction of 2-methylbutane with chlorine would yield a total of monochloro products (including stereoisomers). + HCl different 11 ALCOHOLS AND ETHERS SOLUTIONS TO PROBLEMS Note: A mixture of bond-line and condensed structural formulas is used for solutions in this chapter so as to aid your facility in using both types. 11.1 These names mix two systems of nomenclature (functional class and substitutive; see Section 4.3F). The proper names are: isopropyl alcohol (functional class) or 2-propanol (substitutive), and tert-butyl alcohol (functional class) or 2-methyl-2-propanol (substitutive). Names with mixed systems of nomenclature should not be used. 11.2 (a) OH O OH 2-Propanol or propan-2-ol (Isopropyl alcohol) 1-Propanol or propan-1-ol (Propyl alcohol) Methoxyethane (Ethyl methyl ether) OH (b) OH OH 1-Butanol or butan-1-ol (Butyl alcohol) OH 2-Methyl-1-propanol or 2-methylpropan-1-ol (Isobutyl alcohol) O 2-Butanol 2-Methyl-2-propanol or or butan-2-ol 2-methylpropan-2-ol (sec-Butyl alcohol) (tert-Butyl alcohol) O 1-Methoxypropane (Methyl propyl ether) O Ethoxyethane (Diethyl ether) 2-Methoxypropane (Isopropyl methyl ether) OH 11.3 (a) (b) OH (c) OH 203 204 ALCOHOLS AND ETHERS 11.4 A rearrangement takes place. + + H (a) 1,2-methanide H O shift + H H +O OH2 + H OH OH2 2,3-Dimethyl-2-butanol (major product) (b) (1) Hg(OAc)2 /THF-H2 O; (2) NaBH4 , HO− (oxymercuration-demercuration) 11.5 (a) + OH Stronger base Stronger acid (b) O− Na+ NaNH2 + OH Stronger acid NH3 Weaker acid Weaker base − H + Na+ O− Na+ Stronger base + Weaker base Weaker acid (d) O− Na+ Weaker base + OH + Stronger base O− Na+ NaOH Weaker base Weaker acid 11.6 Ο O− Na+ + OH Stronger base OH2 Br + H2O Stronger acid −H2O + Br + Br − OH Stronger acid + OH H H Weaker acid Ο (c) H ALCOHOLS AND ETHERS 205 11.7 (a) Tertiary alcohols react faster than secondary alcohols because they form more stable carbocations; that is, 3◦ rather than 2◦ : + O + + H H2O X− H X (b) CH3 OH reacts faster than 1◦ alcohols because it offers less hindrance to SN 2 attack. (Recall that CH3 OH and 1◦ alcohols must react through an SN 2 mechanism.) 11.8 (a) H3C SO2Cl CH3OH H3C base (−HCl) SO2OCH3 OH (b) CH3SO2Cl OSO2Me base (−HCl) OH (c) CH3SO2Cl OSO2Me base (−HCl) OH 11.9 (a) OTs + TsCl (pyr.) retention (−HCl) X OTs HO − (b) OH cis-2-Methylcyclohexanol OH inversion SN2 + − OTs Y Cl− TsCl retention (pyr.) OTs A + inversion Cl B − OTs 206 ALCOHOLS AND ETHERS 11.10 Use an alcohol containing labeled oxygen. If all of the label appears in the sulfonate ester, then one can conclude that the alcohol C O bond does not break during the reaction: R 18 O + H R′ SO2Cl HA base −H2O + 11.11 HO R (−HCl) 18 O + A− (−HA) O H R R′ R OH (1° only) H2O + SO2 R O This reaction succeeds because a 3◦ carbocation is much more stable than a 1◦ carbocation. Consequently, mixing the 1◦ alcohol and H2 SO4 does not lead to formation of appreciable amounts of a 1◦ carbocation. However, when the 3◦ alcohol is added, it is rapidly converted to a 3◦ carbocation, which then reacts with the 1◦ alcohol that is present in the mixture. 11.12 (a) O CH3 − O + (1) CH3 + L− L (L = X, OSO2R, or OSO2OR) CH3 L O (2) CH3O − + L− (L = X, OSO2R, or OSO2OR) (b) Both methods involve SN 2 reactions. Therefore, method (1) is better because substitution takes place at an unhindered methyl carbon atom. In method (2), where substitution must take place at a relatively hindered secondary carbon atom, the reaction would be accompanied by considerable elimination. L CH3O − + H 11.13 (a) HO− + HO + CH3OH Cl H2O + − O + Cl L− O + Cl– ALCOHOLS AND ETHERS (b) The O − 207 group must displace the Cl− from the backside, Cl Cl = H OH Cl OH − H H H O − OH trans-2-Chlorocyclohexanol O SN2 O (c) Backside attack is not possible with the cis isomer (below); therefore, it does not form an epoxide. H Cl = H OH Cl OH cis-2-Chlorocyclohexanol 11.14 OH K° (−H2) A TsCl pyr OTs C Br O− K+ O B OH K2CO3 O D 208 ALCOHOLS AND ETHERS O H 11.15 (a) O H2O HSO4− + S O H O + Br + H O OH Br + H3O + O Br H O H O O (b) S + O O + HSO4− H O H2O H OH + + + 11.16 (a) H3O+ H Me Me O +H + O I I − OH MeI + SN 2 attack by I− occurs at the methyl carbon atom because it is less hindered; therefore, the bond between the sec-butyl group and the oxygen is not broken. ALCOHOLS AND ETHERS 209 (b) + OMe + O H OMe + I− I H MeOH + + I− I In this reaction the much more stable tert-butyl cation is produced. It then combines with I− to form tert-butyl iodide. H 11.17 H Br O O + − Br Br + H Br OH H H2O + O Br + − Br H 210 ALCOHOLS AND ETHERS 11.18 HCl Cl OCH3 H Cl Cl − OCH3 + Cl− + + + H + CH3OH H H 11.19 (a) O + HA O HO Me HO + O Me H HO A − OMe Methyl Cellosolve HO (b) An analogous reaction yields ethyl cellosolve, . O O I − −O (c) H2O HO I O NH3 I −O + OH− HO + NH3 (d) O − OMe −O (e) NH2 MeOH HO OMe OMe + CH3O− 11.20 The reaction is an SN 2 reaction, and thus nucleophilic attack takes place much more rapidly at the primary carbon atom than at the more hindered tertiary carbon atom. MeO − + O fast MeOH MeO OH Major product ALCOHOLS AND ETHERS slow MeOH MeO − + O 211 Minor OCH3 product HO 11.21 Ethoxide ion attacks the epoxide ring at the primary carbon because it is less hindered, and the following reactions take place. * Cl + − * Cl OEt OEt O O− * OEt O H2O H H OH + OH H 11.22 + H3O + O H H2O+ H H H2O OH −H3O+ OH H (plus enantiomer, by attack at the other epoxide carbon) D: 11.23 A: 2-Butyne O B: H2 , Ni2 B (P-2) C: mCPBA 11.24 (a) E: MeOH, cat. acid (b) O O O O O O O O O 15-Crown-5 12-Crown-4 Problems Nomenclature 11.25 (a) 3,3-Dimethyl-1-butanol or 3,3-dimethylbutan-1-ol (b) 4-Penten-2-ol or pent-4-en-2-ol (c) 2-Methyl-1,4-butanediol or 2-methylbutan-1,4-diol (d) 2-Phenylethanol (e) 1-Methylcyclopentanol (f) cis-3-Methylcyclohexanol H 212 ALCOHOLS AND ETHERS 11.26 HO OH (a) OH (b) HO H (c) HO OH H OH (d) H OH (e) (f ) O Cl O (g) (h) O O (i) OH ( j) O Reactions and Synthesis 11.27 (a) (c) (b) (d) 11.28 (a) (c) (b) (d) or or or BH3 : THF 11.29 (a) 3 (hydroboration) 3 B H2O2/HO− OH (oxidation) (b) Cl Cl (c) HBr ROOR HO− OH OK OH Br HO− OH (or by hydroboration-oxidation of 1-butene formed by the elimination reaction) ALCOHOLS AND ETHERS H2 (d) Ni2B (P-2) [as in (a)] OH OH 11.30 Br + (a) 3 (b) PBr3 + 3 OK PBr3 OH H3PO3 Br OH Br HBr (no peroxides) (c) See (b) above. (d) HBr Ni2B (P-2) (no peroxides) (1) BH3 : THF (2) H2O2, HO− 11.31 (a) (b) H2 OH (1) BH3 : THF T O (2) OT O R BD3 : THF (c) Br B OT R T D D OH Na (d) ONa Br O [from (a)] 11.32 (a) OH + Cl + SO2 SOCl2 Cl (b) + HCl + HCl 213 214 ALCOHOLS AND ETHERS Br HBr (c) (no peroxides) H (1) BH3 : THF (d) (2) H2O2, HO− H + enantiomer OH Br OK (1) BH3 : THF (2) H O , HO− (e) OH 2 2 OH 11.33 (a) CH3Br + Br (c) Br Br (d) Br Br (b) + O− Na+ 11.34 A: B: Br Br (2 molar equivalents) G: O O Si H: SO2CH3 C: O O CH3 D: O E: + CH3SO3− Na+ SO2 O F: 11.35 A: B: I + SO3− Na+ Si I: OH + F J: Br K: Cl L: Br O− Na+ D: O O E: O CH3 + CH3SO3− Na+ SO2 C: O SO2CH3 F: I + SO3− Na+ ALCOHOLS AND ETHERS G: O H: O + Br K: Cl L: Br Si I: OH Si + F Br2 11.36 (a) Br heat, hv (b) J: ONa Br OH Br HBr (c) peroxides Br (d) I KI acetone Br OH HO− (e) H2O or OH (1) BH3:THF (2) H2O2, HO− HBr (f ) Br (no peroxides) Br (g) CH3ONa O CH3 CH3OH O O Br (h) ONa O O OH Br (i) Br ( j) NaCN CH3SNa CN SCH3 215 216 ALCOHOLS AND ETHERS Br2 ( k) H2O NaHSO3 OH H2O ( m) HO O HO NH3 excess Br ( n) O HO OsO4, pyr ( l) HO− Br NH2 HO –OEt O EtOH HO Br 11.37 H H NaNH2 liq. NH3 − H + Na A (C9H16) − NaNH2 + Na Br 4 liq. NH3 B (C9H15Na) 4 C (C19H36) H2 4 Ni2B (P-2) H mCPBA H D (C19H38) 4 H O H Disparlure (C19H38O) 1. PBr3 2. NaOC(CH3)3 3. H2O, H2SO4 cat. 11.38 ( a) OH OH Br ( b) ( c) 1. NaOC(CH3)3 2. CH3C(O)OOH I O 1. NaOC(CH3)3 2. H2O, H2SO4 cat. OH ALCOHOLS AND ETHERS ( d) OH NaOH, H2O or H2O, H2SO4 cat. O OH mCPBA ( e) O 1. NaOC(CH3)3 2. Br2, H2O Cl ( f) 217 Br OH SOCl2 OH 11.39 (a) OH (b) Cl Br HBr Br NaNH2 OH (c) Cl (d) O− Na+ NH3 PBr3 OH Br 1) TsCl, pyridine (e) OH 2) NaSCH2CH3 O O I H2SO4 HI (excess) (b) S NaI OH (f) 11.40 (a) + I I + I I OH HI (excess) + I OH (c) H2SO4, H2O O + OH 218 ALCOHOLS AND ETHERS (d) O HO NaOCH3 OCH3 (e) O H3CO HOCH3, H2SO4 OH EtS HO (f) 1. EtSNa 2. H2O O OH SEt + SEt HO EtS OH (g) O HCl (1 equiv) O MeONa Cl + HO (h) (i) 1. EtONa 2. MeI O no rxn MeO OEt (j) 11.41 (a) O HI O 1. EtSNa 2. MeI HO I MeO MeO + SEt O (b) 1. Na+ 2. H2O − 3. I O HBr (excess) (c) O SEt 2 Br ALCOHOLS AND ETHERS 1. HI 2. NaH or H2SO4 (cat) O (d) O 219 O + OH 11.42 Cl Cl2 Cl Cl2 400 °C Cl H2O HO−, l eq (b) OH HO HO− excess (a) OH Glycerol Epichlorohydrin CH3 11.43 (a) A = O Cl CH3 B = + enantiomer + enantiomer OTs OH CH3 C = + enantiomer OH CH3 D = + enantiomer I A and C are diastereomers. H OMs OMs (b) E = CH3 H = H C C CH3 H F = CH3 H H C H H CH 220 ALCOHOLS AND ETHERS (c) G = H = O− I = Na+ OMe OMs J = OMe H and J are enantiomers. Mechanisms 11.44 + OH OH2 HA −H2O + + H H + H2O 3° Carbocation is more stable A− + H + HA + OH 11.45 OH O H O S O O H +O O H O − O S O O H ALCOHOLS AND ETHERS 11.46 Br Br + OH Br Br 221 OH − H +O Br OH O 11.47 Br O H3PO4 (cat), CH3CH2OH O O P H O OH OH OH O HO H OH H O+ OH O + O Cl 11.48 (a) H O O OH HCl enantiomer OH (b) The trans product because the Cl− attacks anti to the epoxide and an inversion of configuration occurs. H H O HCl H H Cl O H H Cl H OH enantiomer 222 ALCOHOLS AND ETHERS 11.49 −O O − O Cl Cl H OH SN2 H H H − 1 OH SN2 H O H HO −O OH OH H H 2 Two SN2 reactions yield retention of configuration. 11.50 Collapse of the α-haloalcohol by loss of a proton and expulsion of a halide ion leads to the thermodynamically-favored carbonyl double bond. Practically speaking, the position of the following equilibrium is completely to the right. H O R O X R R′ R′ + HX + 11.51 HO O OH HA A− −HA OH HO+ OH2 OH −H2O + HO + The reaction, known as the pinacol rearrangement, involves a 1,2-methanide shift to the positive center produced from the loss of the protonated —OH group. ALCOHOLS AND ETHERS 223 11.52 The angular methyl group impedes attack by the peroxy acid on the front of the molecule (as drawn in the problem). II results from epoxidation from the back of the molecule—the less hindered side. 11.53 For ethylene glycol, hydrogen bonding provides stabilization of the gauche conformer. This cannot occur in the case of gauche butane. H H H O H − O δ H + Hδ H only van der Waals repulsive forces H H H Challenge Problems 11.54 The reactions proceed through the formation of bromonium ions identical to those formed in the bromination of trans- and cis-2-butene (see Section 8.12A). Me H Br A H Me HBr Me H Br H Me OH + + Br− + H Me Br − H Me Br OH2 Me H Br Me H −H2O (Attack at the other carbon atom of the bromonium ion gives the same product.) Br meso-2,3-Dibromobutane H Me Br B H Me OH HBr H Me Br H Me −H2O + OH2 − + Br (a) Br + H Me (a) Br − Br H Me H Me (b) Br H Me 2,3-Dibromobutane (racemic) Br (b) H Me H Me Br 224 ALCOHOLS AND ETHERS x 11.55 R H SOCl2 H −HCl x R ⫹ H OH x H R Cl O H − S O −SO2 R H 11.56 OH R R HO OH HO S Cl S OH OH achiral A R HO O H Cl The anion and cation form a close S ion pair through O which chloride is transferred with x retention. H B s OH S OH R HO S OH OH r pseudoasymmetric D C A and B are enantiomers A, C, and D are all diastereomers B, C, and D are all diastereomers C is meso D is meso CH3 11.57 ≠ O CH3 O H3C H3C O CH3 H DMDO H (Z)-2-Butene CH3 O H3C H3C O CH3 H H Concerted transition state + O H3C H3C CH3 Acetone H H Epoxide 11.58 The interaction of DMDO with (Z)-2-butene could take place with “syn” geometry, as shown below. In this approach, the methyl groups of DMDO lie over the methyl groups of (Z)-2-butene. This approach would be expected to have higher energy than that shown in ALCOHOLS AND ETHERS 225 the solution to Problem 11.57, an “anti” approach geometry. Computations have been done that indicate these relative energies. (Jenson, C.; Liu, J.; Houk, K.; Jorgenson, W. J. Am. Chem. Soc. 1997, 199, 12982–12983.) H3C Acetone ≠ H3C O DMDO H3C O H3C H3C O H3C H3C H O H3C H3C H (Z)-2-Butene with “syn” approach of DMDO O H3C + O H3C H H3C H H Epoxide Concerted transition state Anti transition state Syn transition state QUIZ 11.1 Which set of reagents would effect the conversion, OH ? (a) BH3 :THF, then H2 O2 /HO− (c) H3 O+ , H2 O (b) Hg(OAc)2 , THF-H2 O, then NaBH4 /HO− (d) More than one of these (e) None of these 11.2 Which of the reagents in item 11.1 would effect the conversion, H ? + OH H H enantiomer 226 ALCOHOLS AND ETHERS 11.3 The following compounds have identical molecular weights. Which would have the lowest boiling point? (a) 1-Butanol (b) 2-Butanol (c) 2-Methyl-1-propanol (d) 1,1-Dimethylethanol (e) 1-Methoxypropane 11.4 Complete the following synthesis: (1) OH O NaH (2) H3O+ (−H2) A CH3SO2Cl base (−HCl) B C ONa (−CH3SO2ONa) D 12 ALCOHOLS FROM CARBONYL COMPOUNDS: OXIDATION-REDUCTION AND ORGANOMETALLIC COMPOUNDS SOLUTIONS TO PROBLEMS O 1 12.1 OH 2 1 2 O H 2 1 OH Bonds to C1 Bonds to C1 Bonds to C1 2 to H = −2 1 to H = −1 3 to O = +3 1 to O = +1 2 to O = +2 = Oxid. state of C1 Total = −1 Total = +1 Bonds to C2 = Oxid. state of C1 = Oxid. state of C1 3 to H = −3 Bonds to C2 Bonds to C2 = Oxid. state of C2 3 to H = −3 3 to H = −3 = Oxid. state of C2 = Oxid. state of C2 2 2 12.2 (a) 3 1 3 1 Bonds to C1 Bonds to C1 2 to H = −2 = Oxid. state of C1 3 to H = −3 = Oxid. state of C1 Bonds to C2 Bonds to C2 1 to H = −1 = Oxid. state of C2 2 to H = −2 = Oxid. state of C2 Bonds to C3 Bonds to C3 3 to H = −3 = Oxid. state of C3 3 to H = −3 = Oxid. state of C3 227 228 ALCOHOLS FROM CARBONYL COMPOUNDS The oxidation states of both C1 and C2 decrease as a result of the addition of hydrogen to the double bond. Thus, the reaction can be considered as both an addition reaction and as a reduction reaction. (b) The hydrogenation of acetaldehyde is not only an addition reaction, but it is also a reduction because the carbon atom of the C=O group goes from a + 1 to a − 1 oxidation state. The reverse reaction (the dehydr ogenation of ethanol) is not only an elimination reaction, but also an oxidation. Ion-Electron Half-Reaction Method for Balancing Organic Oxidation-Reduction Equations Only two simple rules are needed: Rule 1 Electrons (e− ) together with protons (H+ ) are arbitrarily considered the reducing agents in the half-reaction for the reduction of the oxidizing agent. Ion charges are balanced by adding electrons to the left-hand side. (If the reaction is run in neutral or basic solution, add an equal number of HO− ions to both sides of the balanced half-reaction to neutralize the H+ , and show the resulting H+ + HO− as H2 O.) Rule 2 Water (H2 O) is arbitrarily taken as the formal source of oxygen for the oxidation of the organic compound, producing product, protons, and electrons on the righthand side. (Again, use HO− to neutralize H+ in the balanced half-reaction in neutral or basic media.) EXAMPLE 1 Write a balanced equation for the oxidation of RCH2 OH to RCO2 H by Cr2 O7 2− in acid solution. Reduction half-reaction: Cr2 O7 2− + H+ + e− = 2Cr3+ + 7H2 O Balancing atoms and charges: Cr2 O7 2− + 14H+ + 6e− = 2Cr3+ + 7H2 O Oxidation half-reaction: RCH2 OH + H2 O = RCO2 H + 4H+ + 4e− The least common multiple of a 6-electron uptake in the reduction step and a 4-electron loss in the oxidation step is 12, so we multiply the first half-reaction by 2 and the second by 3, and add: 3RCH2 OH + 3H2 O + 2Cr2 O7 2− + 28H+ = 3RCO2 H + 12H+ + 4Cr3+ + 14H2 O ALCOHOLS FROM CARBONYL COMPOUNDS 229 Canceling common terms, we get: 3RCH2 OH + 2Cr2 O7 2− + 16H+ −→ 3RCO2 H + 4Cr3+ + 11H2 O This shows that the oxidation of 3 mol of a primary alcohol to a carboxylic acid requires 2 mol of dichromate. EXAMPLE 2 Write a balanced equation for the oxidation of styrene to benzoate ion and carbonate ion by MnO4 − in alkaline solution. Reduction half-reaction: MnO4 − + 4H+ + 3e− = MnO2 + 2H2 O (in acid) Since this reaction is carried out in basic solution, we must add 4 HO− to neutralize the 4H+ on the left side, and, of course, 4 HO− to the right side to maintain a balanced equation. MnO4 − + 4H+ + 4 HO− + 3e− = MnO2 + 2H2 O + 4 HO− or, MnO4 − + 2H2 O + 3e− = MnO2 + 4 HO− Oxidation half-reaction: CH2 + 5H2 O = ArCO2 − + CO3 2− + 13H+ + 10e− ArCH We add 13 HO− to each side to neutralize the H+ on the right side, CH2 + 5H2 O + 13HO− = ArCO2 − + CO3 2− + 13H2 O + 10e− ArCH The least common multiple is 30, so we multiply the reduction half-reaction by 10 and the oxidation half-reaction by 3 and add: 3ArCH CH2 + 39HO− + 10MnO4 − + 20H2 O = 3ArCO2 − + 3CO3 2− + 24H2 O + 10MnO2 + 40HO− Canceling: 3ArCH CH2 + 10MnO4 − −→ 3ArCO2 − + 3CO3 2− + 4H2 O + 10MnO2 + HO− SAMPLE PROBLEMS Using the ion-electron half-reaction method, write balanced equations for the following oxidation reactions. (hot) (a) Cyclohexene + MnO4 − + H+ −→ HO2 C(CH2 )4 CO2 H + Mn2+ + H2 O (cold) (b) Cyclopentene + MnO4 − + H2 O −→ cis-1,2-cyclopentanediol + MnO2 + HO− (hot) (c) Cyclopentanol + HNO3 −→ HO2 C(CH2 )3 CO2 H + NO2 + H2 O (cold) (d) 1,2,3-Cyclohexanetriol + HIO4 −→ OCH(CH2 )3 CHO + HCO2 H + HIO3 230 ALCOHOLS FROM CARBONYL COMPOUNDS SOLUTIONS TO SAMPLE PROBLEMS (a) Reduction: MnO4 − + 8 H+ + 5e− = Mn2+ + 4 H2 O Oxidation: H CO2H 4 H2O = + H The least common multiple is 40: + CO2H 8 H+ + 8 e− 8 MnO4− + 64 H+ + 40 e− = 8Mn2+ + 32 H2 O H + 20 H2O = 5 CO2H 5 CO2H H + 40 H + + 40 e− Adding and canceling: H + 8 MnO4− + 24 H+ 5 5 H CO2H CO2H + 8 Mn 2 + + 12 H2O (b) Reduction: MnO4 − + 2 H2 O + 3e− = MnO2 + 4 HO− Oxidation: OH + 2 HO− + 2 e− = OH The least common multiple is 6: 2MnO4 − + 4 H2 O + 6e− = 2 MnO2 + 8 HO− OH 3 + 6 HO− + 6 e− = 3 OH Adding and canceling: OH 3 + 2 MnO4 − + 4 H2O + 2 MnO2 + 2 HO− 3 OH ALCOHOLS FROM CARBONYL COMPOUNDS 231 (c) Reduction: HNO3 + H+ + e− = NO2 + H2 O Oxidation: OH + 3 H2O = CO2H + 8 H+ + 8 e− CO2H The least common multiple is 8: 8 HNO3 + 8 H+ + 8 e− = 8 NO2 + 8 H2 O OH + 3 H2O = CO2H + 8 H+ + 8 e− CO2H Adding and canceling: OH CO2H + 8 NO2 + 5 H2O CO2H + 8 HNO3 (d) Reduction: HIO4 + 2 H+ + 2e− = HIO3 + H2 O Oxidation: HO H OH H + H O = 2 OH CHO OH H + H O = 2 OH CHO O OH + 4 H + + 4 e − + HC CHO H The least common multiple is 4: 2 HIO4 + 4 H+ + 4 e− = 2 HIO3 + 2H2 O HO H H Adding and canceling: HO H OH H + 2 HIO = 4 OH H O OH + 4 H + + 4 e − + HC CHO CHO O + HC CHO OH + 2 HIO3 + H2O 232 ALCOHOLS FROM CARBONYL COMPOUNDS 12.3 (a) LiAlH4 (d) LiAlH4 (b) LiAlH4 (e) LiAlH4 (c) NaBH4 O O 12.4 (a) (b) H No reaction. Ethers can not be oxidized. O (c) O + N H CrO3Cl− (PCC)/CH2Cl2 or 12.5 (a) (1) DMSO, (COCl)2, −60 ºC (2) Et3N (b) KMnO4 , HO− , H2 O, heat; then H3 O+ [or conditions as in (c) below] (c) H2 CrO4 /acetone [or conditions as in (b) above], or Swern oxidation (d) (1) O3 (2) Me2 S (e) LiAlH4 12.6 (a) MgBr Stronger base + H OH pKa 15.7 Stronger acid + pKa ∼ 50 Weaker acid − + Stronger base (c) H OMe pKa 15.5 Stronger acid + H O + pKa 43-45 Weaker acid + Me Ο − pKa 18 Stronger acid pKa ∼ 50 Weaker acid + Br − + Mg 2+ + Br − Weaker base Ο − + MgBr Stronger base Mg 2+ Weaker base MgBr (b) + OH Weaker base Mg 2+ + Br − ALCOHOLS FROM CARBONYL COMPOUNDS 233 (d) Li + pKa ∼ 50 Weaker acid pKa 18 Stronger acid Stronger base Ο + H O Mg 12.7 (a) + Li + Weaker base δ+ δ − MgBr Br − D D OD Et2O (b) 12.8 Br2 Mg hv Et2O Br D δ− OD MgBr δ+ D + O O− MgBr Cl + BrMg Cl −MgBrCl + O O− MgBr BrMg NH4Cl/H2O OH O OH + 12.9 (a) (1) CH3MgI O OH + CH3MgI (1) ether (2) + NH4/H2O 234 ALCOHOLS FROM CARBONYL COMPOUNDS O OH + (2) MgBr O OH (1) ether + MgBr (2) +NH4/H2O O OH (3) + O O + O OH (1) ether 2 CH3MgI (2) +NH4/ H2O O OH (b) (1) + O + (2) +NH OH 4 /H2O O MgBr + (2) O MgBr + CH3MgI (1) ether CH3MgI OH OH (1) ether (2) +NH4 /H2O O OH (3) OCH3 O OCH3 2 CH3MgI + 2 MgBr + (1) ether (2) +NH4 /H2O MgBr OH ALCOHOLS FROM CARBONYL COMPOUNDS MgBr O (c) (1) + H OH MgBr O H (1) ether (2) H3O+ + OH O CH3MgI (2) + H OH O + CH3MgI H (1) ether (2) H3O+ OH O MgBr + (d) (1) OH O MgBr + (1) ether (2) +NH4 /H2O OH O + (2) CH3MgI OH O + CH3MgI (1) ether (2) +NH4/H2O OH 235 236 ALCOHOLS FROM CARBONYL COMPOUNDS O MgBr + (3) OH O MgBr + (1) ether (2) +NH4/H2O OH OH O O (e) (1) MgBr + 2 OH O O MgBr + 2 (1) ether (2) +NH4 OH O MgBr + (2) OH O MgBr + (1) ether (2) +NH4 ALCOHOLS FROM CARBONYL COMPOUNDS MgBr OH (f) (1) O + Br 237 H OH H MgBr O + MgBr O (1) OH , ether Br PBr3 (2) H3O+ OMgBr H3O+ (1) Mg, ether OH O (2) H H OH (2) MgBr O MgBr OH Br + O + H H O MgBr (1) H (2) H3O (1) Mg, ether (2) O OH PBr3 H , ether Br + OMgBr H3O+ OH 238 ALCOHOLS FROM CARBONYL COMPOUNDS OH MgBr O 12.10 + (a) OH H MgBr Swern Oxidation: (1) DMSO, (COCl)2, −60 ºC (2) Et3N OH O or PCC, CH2Cl2 ether H OMgBr OH + H 3O H2O O (b) MgBr OH H O MgBr (1) H H, ether OH (2) H3O+ Swern Oxidation: (1) DMSO, (COCl)2, −60 ºC (2) Et3N O H or PCC, CH2Cl2 OH O (c) MgBr + MeO MgBr O 2 OMgBr OH 2 +NH 4 MeO ether [from part (a)] OH (d) H2O O H HO Swern Oxidation: (1) DMSO, (COCl)2, −60 ºC (2) Et3N or PCC, CH2Cl2 MgBr + O HO MgBr H ether OMgBr H3O+ H2O OH ALCOHOLS FROM CARBONYL COMPOUNDS 239 Problems Reagents and Reactions 12.11 (a) CH3CH3 (b) (c) D OH Ph OH (d) Ph OH (e) Ph Ph Ph (g) CH CH 3 3 OH (f) + OH 12.12 (a) (b) OH OH OH (c) (e) 12.13 (a) (d) (g) (j) + + (d) OLi O D + OLi OCH3 (h) CH3OLi (k) (c) OH (e) + OH + N (b) + OH (f) (i) OH + H OH OH Li 240 ALCOHOLS FROM CARBONYL COMPOUNDS (d) (1) KMnO4 , HO− , heat (2) H3 O+ 12.14 (a) LiAlH4 (b) NaBH4 (or H2 CrO4 /acetone) (c) LiAlH4 (e) PCC/CH2 Cl2 or Swern oxidation (1) DMSO, (COCl)2, −60°C 12.15 (a) Myrtenol (2) Et3N O H O (1) DMSO, (COCl)2, −60°C (b) Chrysanthemyl alcohol OMgX O 12.16 (a) EtMgBr + EtO OEt EtO OMgX OEt Et Et + EtMgX OH NH4 Et Et Et Et Et H2O Et OMgX O −EtOMgX + H OEt H OEt Et Et OMgX O + EtMgX Et OEt Et OMgX O H −EtOMgX EtMgX Et (b) EtMgBr OEt Et O −EtOMgX Et H (2) Et3N OH H3O H Et Et H Et ALCOHOLS FROM CARBONYL COMPOUNDS OH H 12.17 (a) O O OH 1. NaBH4, EtOH 2. H2O OH 1. LiAlH4, THF 2. H2O/H2SO4 O (b) OH OH OH O 1. NaBH4, EtOH 2. H2O (c) O O O O O 12.18 (a) 1. KMnO4, NaOH OH OH 2. H3O+ OH O PCC (b) CH2Cl2 O OH (c) H (1) DMSO, (COCl)2, −60 ºC (2) Et3N OH H2CrO4 (d) O No Rxn 3° Alcohols do not oxidize O H2CrO4 (e) H OH 12.19 (a) OH Swern Oxidation (as in 12.18(c)) or OH PCC CH2Cl2 OH O 241 ALCOHOLS FROM CARBONYL COMPOUNDS O (b) H2CrO4 HO HO OH OH O O (c) Swern Oxidation or PCC O O O CH2Cl2 O O O HO (d) (1) LAH (2) aq. H2SO4 OH O OH O OH NaBH4 (e) (1) CH3MgBr (2) H3O+ 12.20 (a) H O 242 HO Ο (b) (1) HO MgBr (2) NH4Cl, H2O Ο OH MgBr (c) Ο OH (1) Ο Ο O (2) H3O+ + Ο Ο (d) Ο (1) CH3CH2Li (excess) (2) NH4Cl, H2O ΟΗ ΗΟ ALCOHOLS FROM CARBONYL COMPOUNDS 12.21 (a) (1) MeMgBr (excess) (2) NH4Cl, H2O O + MeOH OH O (1) Mg O OH (b) Br 243 (2) H H (3) H3O+ (c) (1) PBr3 (2) Mg (3) H3O+ OH (1) PCC or Swern oxidation (2) MgBr OH (d) OH O+ (3) H3 (1) EtMgBr (2) H3O+ (3) NaH (4) CH3Br (e) H O O (1) O O (f) H O MgBr (excess) O O O O O (2) NH4Cl, H2O (3) PCC O O 1. 12.22 O BrMg 2. H3O+ (1 equiv) MgBr OH + MeOH ALCOHOLS FROM CARBONYL COMPOUNDS Mechanisms O 12.23 (1) NaBD4 (2) H3O+ HO D (1) NaBH4 (2) D3O+ DO H (1) NaBD4 (2) D3O+ DO D O O BrMg + MgBr 12.24 MgBr O O O − O O− O H O H O O H +O H 12.25 H − O− H HO − O H H − − H O O− O H HO H +O OH H +O OH H 244 −O H O− H O+ H ALCOHOLS FROM CARBONYL COMPOUNDS 245 12.26 The three- and four-membered rings are strained, and so they open on reaction with RMgX or RLi. THF possesses an essentially unstrained ring and hence is far more resistant to attack by an organometallic compound. 12.27 R1 R2 R O C Mg X R Mg Mg C R2 X R O R1 + Mg X R X The alkylmagnesium group of the alkoxide can go on to react as a nucleophile with another carbonyl group. Synthesis 12.28 (a) CH4 + (b) HO Li HO (c) O (d) + (e) OH O OMs (f) O (g) (h) + Na+ −OSO2Me Na+ −OSO2Me OH + OH 246 ALCOHOLS FROM CARBONYL COMPOUNDS OH Br + 12.29 (a) 3 PBr3 Br + MgBr O H3PO3 MgBr ether Mg + 3 (1) ether + (2) H3O+ H OH MgBr O (1) ether + (b) H [from part (a)] MgBr (c) O (1) ether + (2) H3O+ [from part (a)] + (1) ether (2) H3O+ H [from part (a)] OH D MgBr + (e) OH SOCl2 Cl O MgBr (d) OH (2) H3O+ H D2O [from part (a)] PBr3 12.30 (a) OH Br OK (b) [from (a)] Br OH O OH (1) Hg 2 /THF-H2O O (c) [from (b)] − (2) NaBH4/HO ALCOHOLS FROM CARBONYL COMPOUNDS (d) [from (b)] H2/Pt, Pd, or Ni pressure OH (e) Br PBr3 [from (c)] O (1) H Mg (f ) [from (a)] Br MgBr Et2O H (2) H3O + OH (1) (g) [from (f )] (h) MgBr O OH (2) H3O+ (1) DMSO, (COCl)2, −60 ºC (2) Et3N OH [from (c)] H or PCC, CH2Cl2 OH (i) O O H2CrO4 acetone, H2O O (1) KMnO4, HO−, heat ( j) OH OH (2) H3O+ (k) (1) OH H2SO4 140 ºC 2O (2) OH NaH (−H2) ONa Br 2O [from (a)] Br2 (l) Br [from (b)] 3 NaNH2 heat Br − + Na H3O+ H HBr (1 equiv.) (m) H [from (l)] (no peroxides) Br 247 248 ALCOHOLS FROM CARBONYL COMPOUNDS (n) Br [from (a)] Li Et2O Li O OH [from (i)] (1) (o) MgBr [from (f )] (2) +NH4/H2O 12.31 (a) NaBH4 , − OH O (f) (1) PBr3 (b) 85% H3 PO4 , heat (2) Mg (4) HA (3) (g) (1) CH3 MgBr/Et2 O O (c) H2 /Ni2 B (P-2) (d) (1) BH3 :THF (2) H2 O2 ,− OH (2) (3) NH4 Cl/H2 O (e) (1) NaH (2) (CH3 )2 SO4 12.32 (a) Retrosynthetic Analysis OH O O H + MgBr Synthesis OH (1) DMSO, (COCl)2, −60 ºC (2) Et3N or PCC, CH2Cl2 OH O H Br MgBr H2CrO4 acetone Mg Et2O O (b) Retrosynthetic Analysis HO 2 MgBr + H3O+ O OCH3 ALCOHOLS FROM CARBONYL COMPOUNDS 249 Synthesis 2 PBr3 OH Br 2 Mg MgBr 2 Et2O O HO OCH3 (1) (2) +NH4/H2O (c) Retrosynthetic Analysis O + MgBr H OH Synthesis OH Br PBr3 MgBr Mg Et2O H OMgBr OH + H3O O [from (a)] (d) Retrosynthetic Analysis O H MgBr + OH O Synthesis O Br Mg Et2O MgBr (1) DMSO, (COCl)2, −60 ºC OH (2) Et3N or PCC, CH2Cl2 (1) (2) H3O+ O H (e) Retrosynthetic Analysis O OH OH MgBr + O 250 ALCOHOLS FROM CARBONYL COMPOUNDS Synthesis PBr3 OH MgBr O O − (1) (2) H3O Mg Et2O Br OH + (1) KMnO4, HO , heat OH (2) H3O+ (f ) Retrosynthetic Analysis O OH MgBr + Synthesis O OH (1) H2CrO4 MgBr OH [from (b)] acetone (2) +NH4/H2O (g) Retrosynthetic Analysis OH O O + BrMg H Synthesis (1) DMSO, (COCl)2, −60 ºC (2) Et3N OH OH O H or PCC, CH2Cl2 H2CrO4 acetone MgBr (1) [from (e)] (2) H3O+ O ALCOHOLS FROM CARBONYL COMPOUNDS (h) Retrosynthetic Analysis Br MgBr OH + OH O MgBr H + O Synthesis MgBr OH (1) DMSO, (COCl)2, −60 ºC (2) Et3N H O or PCC, CH2Cl2 (1) [from (a)] (2) H3O+ H2CrO4 acetone MgBr O OH OH Br HBr (1) PBr3 (2) Mg, Et2O OH 251 252 ALCOHOLS FROM CARBONYL COMPOUNDS 12.33 (a) H OH O MgBr MgBr O or + + H H O OH MgBr (1) Et2O + (2) H3O+ or OH MgBr O (1) Et2O + (2) H3O+ H (b) or OH O + O BrMg + BrMg + (1) Et2O BrMg OH (2) NH4Cl/H2O O BrMg or O + (1) Et2O (2) NH4Cl/H2O OH ALCOHOLS FROM CARBONYL COMPOUNDS (c) 253 O OH O + CH3CH2MgBr (excess) OH O OH O + MgBr excess (1) Et2O (2) NH4Cl/H2O OH 12.34 There are multiple strategies to approach this these problems. OH BrMg O (1) DMSO, (COCl)2, −60 ºC (2) Et3N (a) or PCC, CH2Cl2 HO H OH (1) (2) H3O+ BrMg (1) DMSO, (COCl)2, −60 ºC or PCC, CH2Cl2 (1) PBr3 (2) Et3N (2) Mg°, ether OH Br O (1) PBr3 (2) Mg°, ether MgBr (1) (2) H3O+ OH 254 ALCOHOLS FROM CARBONYL COMPOUNDS (b) OH (1) DMSO, (COCl)2, −60 ºC (2) Et3N O + H (2) H3O or PCC, CH2Cl2 CH3OH (1) PBr3 (2) Mg°, ether OH (1) CH3MgBr CH3MgBr (1) DMSO, (COCl)2, −60 ºC or PCC, CH 2 Cl2 (2) Et N 3 O (1) CH3MgBr (2) H3O+ OH MgBr (c) (1) DMSO, (COCl)2, −60 ºC OH O (2) Et3N (1) (2) H3O+ H or PCC, CH2Cl2 OH OH Br MgBr (1) PBr3 (1) Mg, ether O (2) Mg, ether (2) H H (3) H3O+ 12.35 Retrosynthetic Analysis OH + O MgBr + Mg O Br + Synthesis R OOH O R (1) OOH O (2) H3O+ MgBr OH ALCOHOLS FROM CARBONYL COMPOUNDS 12.36 Retrosynthetic Analysis O OH Na + − + Synthesis O OH (1) NaNH2 − Na + liq. NH3 12.37 255 (2) + NH4/H2O Retrosynthetic Analysis OH MgBr HO O + HO H Br HO Synthesis Br Si Cl imidazole HO Br (1) Mg, Et2O O (2) SiO H (3) H3O+ OH OH +− Bu4NF SiO HO Before converting the reactant to a Grignard reagent it is first necessary to mask the alcohol, such as by converting it to a tert-butyldimethylsilyl ether. After the Grignard reaction is over, the protecting group is removed. 256 ALCOHOLS FROM CARBONYL COMPOUNDS O H 12.38 H Br Br2, hv H H (1) BH3 (2) NaOH, H2O2 NaNH2 O H Br MgBr (1) HO Mg Et2O (2) NH4Cl, H2O NaH, EtI O Challenge Problems 12.39 2-Phenylethanol, 1,2-diphenylethanol, and 1,1-diphenylethanol are distinct from 2,2diphenylethanoic acid and benzyl 2-phenylethanoate in that they do not have carbonyl groups. IR spectroscopy can be used to segregate these five compounds into two groups according to those that do or do not exhibit carbonyl absorptions. 1 H NMR can differentiate among all of the compounds. In the case of the alcohols, in the 1 H NMR spectrum of 2-phenylethanol there will be two triplets of equal integral value, whereas for 1,2-diphenylethanol there will be a doublet and a triplet in a 2:1 area ratio. The triplet will be downfield of the doublet. 1,1-Diphenylethanol will exhibit a singlet for the unsplit methyl hydrogens. The broadband proton-decoupled 13 C NMR spectrum of 2-phenylethanol should show 6 signals (assuming no overlap), four of which are in the chemical shift region for aromatic carbons. 1,2-Diphenylethanol should exhibit 10 signals (assuming no overlap), 8 of which are in the aromatic region. 1,1-Diphenylethanol should show 6 signals (assuming no overlap), four of which would be in the aromatic region. The DEPT 13 C NMR spectra would give direct evidence as to the number of attached hydrogens on each carbon. Regarding the carbonyl compounds, both 2,2-diphenylethanoic acid and benzyl 2phenylethanoate will show carbonyl absorptions in the IR, but only the former will also have a hydroxyl absorption. The 1 H NMR spectrum of 2,2-diphenylethanoic acid should ALCOHOLS FROM CARBONYL COMPOUNDS 257 show a broad absorption for the carboxylic acid hydrogen and a sharp singlet for the unsplit hydrogen at C2. Their integral values should be the same, and approximately one-tenth the integral value of the signals in the aromatic region. Benzyl 2-phenylethanoate will exhibit two singlets, one for each of the unsplit CH2 groups. These signals will have an area ratio of 2 : 5 with respect to the signal for the 10 aromatic hydrogens. The broadband 1 H decoupled 13 C NMR spectrum for 2,2-diphenylethanoic acid should show four aromatic carbon signals, whereas that for benzyl 2-phenylethanoate (assuming no overlapping signals) would show 8 signals in the aromatic carbon region. Aside from the carbonyl and aromatic carbon signals, benzyl 2-phenylethanoate would show two additional signals, whereas 2,2-diphenylethanoic acid would show only one. DEPT 13 C NMR spectra for these two compounds would also distinguish them directly. 12.40 It makes it impossible to distinguish between aldehyde and ketone type sugars (aldoses and ketoses) that had been components of the saccharide. Also, because the R groups of these sugars contain chirality centers, reduction of the ketone carbonyl will be stereoselective. This will complicate the determination of the ratio of sugars differing in configuration at C2. 12.41 The IR indicates the presence of OH and absence of C C and C O. The MS indicates a molecular weight of 116 amu and confirms the presence of hydroxyl. The reaction data indicate X contains 2 protons per molecule that are acidic enough to react with a Grignard reagent, meaning two hydroxyl groups per molecule. (This analytical procedure, the Zerewitinoff determination, was routinely done before the advent of NMR.) Thus X has a partial structure such as: C6 H10 (OH)2 with one ring, or C5 H6 O(OH)2 with two rings, or (less likely) C4 H2 O2 (OH)2 with three rings. QUIZ 12.1 Which of the following could be employed to transform ethanol into (a) Ethanol + HBr, then Mg/diethyl ether, then H3 O+ O (b) Ethanol + HBr, then Mg/diethyl ether, then H (c) Ethanol + H2 SO4 /140 ◦ C O (d) Ethanol + Na, then H H , then H3 O+ (e) Ethanol + H2 SO4 /180 ◦ C, then O H , then H3 O+ OH ? 258 ALCOHOLS FROM CARBONYL COMPOUNDS 12.2 The principal product(s) formed when 1 mol of methylmagnesium iodide reacts with 1 mol of O OH (a) (d) O CH4 OH + OMgI (b) O OMgI OH (c) (e) None of the above O O CH3 12.3 Supply the missing reagents. (1) A O (1) B HO (2) NH4+/ H2O CH3 (2) C O 12.4 Supply the missing reagents and intermediates. O B A (1) MgBr + (2) H3O C (1) D OH + (2) H3O 12.5 Supply the missing starting compound. A CH3 (1) CH3MgBr (2) +NH 4/H2O OH The solutions in this section correspond with problems that can be found only in the WileyPLUS course for this text. WileyPLUS is an online teaching and learning solution. For more information about WileyPLUS, please visit http://www.wiley.com/college/wileyplus ANSWERS TO FIRST REVIEW PROBLEM SET 1. (a) + OH + OH2 H −H2O A A− +H2O 2° Carbocation H A− + methanide shift + HA 3° Carbocation (b) + Br + Br Br + − Br then, Br + Br − Cl Cl (c) The enantiomer of the product given would be formed in an equimolar amount via the following reaction: Cl − + Cl Br Br 259 260 ANSWERS TO FIRST REVIEW PROBLEM SET The trans-1,2-dibromocyclopentane would be formed as a racemic form via the reaction of the bromonium ion with a bromide ion: Br + + Br Br Br − + Br Br Racemic trans-1,2-dibromocyclopentane And, trans-2-bromocyclopentanol (the bromohydrin) would be formed (as a racemic form) via the reaction of the bromonium ion with water. + + OH2 + + Br H2O OH2 + Br OH A− Br OH + + HA Br Br Racemic trans-2-bromocyclopentanol (d) H + Br 1,2-Hydride Shift Br + H 2. (a) CHCl3 − (b) The cis isomer Br (c) CH3 Cl 3. This indicates that the bonds in BF3 are geometrically arranged so as to cancel each others’ polarities in contrast to the case of NF3 . This, together with other evidence, indicates that BF3 has trigonal planar structure and NF3 has trigonal pyramidal structure. (1) Br2, H2O (2) NaOH (3) NaOMe, MeOH 4. (a) OCH3 OH (1) BH3:THF (2) NaOH, H2O2 (3) NaH (4) Cl (b) (c) O D D OH (1) PBr3 (2) NaSCH3, HSCH3 SCH3 ANSWERS TO FIRST REVIEW PROBLEM SET (d) 261 (1) mCPBA (2) OH, H2SO4 OH OEt (e) (1) TsCl (2) NaN3 OH N3 (f) O concd. H2SO4, O OH OH (g) NaNH2 OH 5. (a) TS2 ONa + NH3 TS3 Ea3 II Free energy OH H2SO4 Ea2 ΔG3 ΔG2 I Δ G1 TS1 Ea1 + I II III III TS = transition state; Ea = activation energy Reaction coordinate TS2 (b) Free energy HBr OH Ea2 I Δ G1 TS1 Ea1 I Δ G2 Reaction coordinate II Br II 262 ANSWERS TO FIRST REVIEW PROBLEM SET 6. A= ( ) B= ( ) C= ( ) 11 − Na+ 11 ( ) 6 11 Muscalure = ( ) ( ) 11 6 C6H5 7. C6H5 (E )-2,3-Diphenyl-2-butene C6H5 C6H5 (Z )-2,3-Diphenyl-2-butene Because catalytic hydrogenation is a syn addition, catalytic hydrogenation of the (Z) isomer would yield a meso compound. C6H5 H2 Pd C6H5 C6H5 H H C6H5 C6H5 H H C6H5 (by addition at one face) A meso compound (Z ) (by addition at the other face) Syn addition of hydrogen to the (E) isomer would yield a racemic form: C6H5 H2 C6H5 (E) Pd C6H5 H H C6H5 + H C6H5 (by addition at one face) H C6H5 (by addition at the other face) Enantiomers - a racemic form ANSWERS TO FIRST REVIEW PROBLEM SET 263 8. From the molecular formula of A and of its hydrogenation product B, we can conclude that A has two rings and a double bond. (B has two rings.) From the product of strong oxidation with KMnO4 and its stereochemistry (i.e., compound C), we can deduce the structure of A. (1) KMnO4, HO−, heat HO2C CO2H = CO2H (2) H3O + A meso-1,3-Cyclopentanedicarboxylic acid Compound B is bicyclo[2.2.1]heptane and C is a glycol. H2 cat. A B OH OH H KMnO4, HO− cold, dilute A C H Notice that C is also a meso compound. (1) BH3:THF (2) NaOH, H2O2 9. (a) OH (1) Hg(OAc)2, H2O, THF (2) NaBH4, NaOH (b) OH mCPBA (c) O concd. H2SO4 (d) O OH (e) (1) Br2, H2O (2) NaOH O CO2H 264 ANSWERS TO FIRST REVIEW PROBLEM SET (f) (1) concd. H2SO4 (2) BH3:THF (3) NaOH, H2O2 OH OH OH Br PBr3 10. (a) O (b) O SOCl2 O O OH Cl Br (c) Br HCl OH Cl or SOCl2 Br OH (d) HBr (2 eq) or PBr3 (2 eq) HO Br H (e) OH H (1) TsCl, Pyr H (2) H OK (f) OH NaNH2 11. (a) liq. NH3 H2 (b) Ni2B (P-2) [from (a)] Li liq. NH3 (c) [from (a)] H2SO4 + CH3I ANSWERS TO FIRST REVIEW PROBLEM SET (d) (1) NaNH2 H2 (2) CH3Br Ni2B(P-2) or (1) Na, liq. NH3 (2) NH4Cl or HBr OK Br OK NBS (e) Br OH heat [from (d)] OH Br (f ) [from (d)] Br HBr ROOR (g) or HBr no peroxides Br racemic [from (b) or (c)] or [from (d)] HBr no peroxides Br Br2 (h) racemic Br H H [from (c)] (anti addition) Br2 (i) [from (b)] (anti addition) (cf. Section 8.11) Br (2R,3S) A meso compound Br H H Br Br Br H H (2R,3R) (2S,3S) A racemic form 265 266 ANSWERS TO FIRST REVIEW PROBLEM SET (1) OsO4 (j) [from (b)] H H (2) NaHSO3 OH (syn addition) (cf. Section 8.15) OH O or (1) R OOH H H OH (anti addition) [from (c)] Br HBr, Br− (k) (cf. Section 11.15) OH (cf. Section 8.18) O OH 12. Br Br2 (cf. Section 10.5) hv, heat (a) ONa Br OH heat OH (b) H2O [from (a)] (1) BH3 :THF − (c) [from (a)] OH HBr ROOR heat (d) [from (a)] OK Br OH Br2 3 NaNH2 Br heat Br − (e) [from (d)] Na HBr ROOR heat H Br ANSWERS TO FIRST REVIEW PROBLEM SET 267 HCl (f) [from (d)] Cl Cl HCl (g) [from (a)] (h) Br [from (e)] NaI acetone SN2 I O O (1) O3 (i) H (2) Me2S [from (a)] O (1) O3 ( j) O H (2) Me2S [from (d)] H H Cl 13. Cl2 + hv, heat + Cl Cl B C (racemic) D A B cannot undergo dehydrohalogenation because it has no β hydrogen; however, C and D can, as shown next. C Cl H2 ONa Pt OH E Cl D HCl + + H E + Cl − Cl + Cl − (1) Mg, ether (2) H3O+ F G A 268 ANSWERS TO FIRST REVIEW PROBLEM SET 14. , No IR absorption in 2200–2300 cm−1 region. H2, Pt A H2 Ni2B (P-2) HO (1) OsO4 (2) NaHSO3 (syn hydroxylation) B OH H H C (a meso compound) 15. The eliminations are anti eliminations, requiring an anti coplanar arrangement of the bromine atoms. H Br H KI Br − meso-2,3-Dibromobutane H H KI Br IBr + IBr H OH (2S,3S)-2,3-Dibromobutane cis-2-Butene Br H KI Br − + trans-2-Butene Br H − H OH H H (2R,3R)-2,3-Dibromobutane OH H H + IBr cis-2-Butene 16. The eliminations are anti eliminations, requiring an anti coplanar arrangement of the and Br. Br Br Br H C6H5 H C6H5 C6H5 C6H5 H H − EtO meso-1,2-Dibromo1,2-diphenylethane Br C6H5 H Br C6H5 H (E)-1-Bromo-1,2diphenylethene Br C6H5 C6H5 H − EtO (2R,3R)-1,2-Dibromo1,2-diphenylethane (Z)-1-Bromo-1,2diphenylethene (2S,3S)-1,2-Dibromo-1,2-diphenylethane will also give (Z)-1-bromo-1,2-diphenylethene in an anti elmination. ANSWERS TO FIRST REVIEW PROBLEM SET 269 17. In all the following structures, notice that the large tert-butyl group is equatorial. (a) Br H (bromine addition is anti; cf. Section 8.11) Br + enantiomer as a racemic form OH (b) (syn hydroxylation; cf. Section 8.15) OH H + enantiomer as a racemic form OH (c) (anti hydroxylation; cf. Section 11.15) H OH + enantiomer as a racemic form H (d) OH (syn and anti Markovnikov addition of H and OH; cf. Section 8.9) H + enantiomer as a racemic form OH (e) (Markovnikov addition of H and OH; cf. Section 8.4, 8.5) OH (f) H Br + enantiomer as a racemic form (anti addition of Br and OH, with Br and OH placement resulting from the more stable partial carbocation in the intermediate bromonium ion; cf. Section 8.13) 270 ANSWERS TO FIRST REVIEW PROBLEM SET Cl (g) (anti addition of I and Cl, following Markovnikov’s rule; cf. Section 8.11) H I + enantiomer as a racemic form O (h) H O D (i) (syn addition of deuterium; cf. Section 7.12) D H + enantiomer as a racemic form D (j) (syn, anti Markovnikov addition of T , with B H + enantiomer as a racemic form B D and being replaced by T where it stands; cf. Section 8.6) OH 18. A = BH B= C= 2 19. (a) The following products are diastereomers. They would have different boiling points and would be in separate fractions. Each fraction would be optically active. H H Br2 H + Br Br (R)-3-Methyl-1-pentene H Br (optically active) Br H (optically active) Diastereomers (b) Only one product is formed. It is achiral, and, therefore, it would not be optically active. H H H2 Pt (optically inactive) ANSWERS TO FIRST REVIEW PROBLEM SET 271 (c) Two diastereomeric products are formed. Two fractions would be obtained. Each fraction would be optically active. H H (1) OsO4 H + OH (2) NaHSO3 OH H OH (optically active) HO H (optically active) Diastereomers (d) One optically active compound is produced. H H (1) BH3: THF (2) H2O2, HO− OH (optically active) (e) Two diastereomeric products are formed. Two fractions would be obtained. Each fraction would be optically active. O H (1) O 2 Hg, THF–H2O (2) NaBH4, HO− H H + H OH (optically active) HO H (optically active) Diastereomers (f) Two diastereomeric products are formed. Two fractions would be obtained. Each fraction would be optically active. H (1) mCPBA H H OH + (2) H3O+, H2O H OH OH HO H ANSWERS TO FIRST REVIEW PROBLEM SET H 20. H (1) Mg, ether (2) NH4+, H2O Cl H + H H + enantiomer B + enantiomer A (a meso compound) C H2 ONa Pt OH D (1) O3 (2) Me2S O O 272 O 21. (a) (1) O Li OH (2) H3O+ (b) O O (c) HO (1) CH3MgBr (2) H3O+ CH3 (1) CH3MgBr (excess), ether (2) H3O+ + CH3OH OH 22. (1) Br2, hv (2) Mg, ether (3) O (4) H3O+ OH ANSWERS TO FIRST REVIEW PROBLEM SET 273 SOCl2 23. (a) pyridine OH Cl OH O (1) LiAlH4 (excess) (2) H3O+ O (b) O OH HO (c) (1) LiAlH4 (2) H3O+ O OH O OH (1) OsO4 (2) NaHSO3 (d) Br (e) OH NaH O OH O OH (f) O (1) O (g) OH H2CrO4 (2) H3O+ HO O MgBr (1 equiv.) O HO O O (h) + O CH3 O (1) NaBH4 (2) H3O+ HO CH3 O ANSWERS TO FIRST REVIEW PROBLEM SET 24. PBr3 Mg, ether HO Br MgBr A (1) A (2) H3O+ H PCC HO 274 or Swern oxidation O OH 25. OH O (1) PhMgBr (2) NH4Cl/H2O OH O (1) CH3MgBr (2) NH4Cl/H2O OH O MgBr (1) (2) NH4Cl/H2O 26. O O 1 O O Y O H NMR Singlet at δ 1.4 13 Presence of only one proton signal suggests that all 18 protons are equivalent. The chemical shift suggests that an electronegative atom is not bonded at the carbon bearing the protons. C NMR δ 87 δ 151 IR 1750–1800 cm−1 C C O O C O 2 carbonyls indicated by split peak ANSWERS TO FIRST REVIEW PROBLEM SET 27. m/z 120 = M +. H 105 = M +. −15(CH3 ·) = + C6H5 77 = M +. − 43(i-Pr·) = C6 H5 + 5 ring protons CH of isopropyl group equivalent CH3 s of isopropyl group δ7.2–7.6 2.95 1.29 28. C5 H10 O has IHD = 2 IR absorption indicates C O 13 C NMR spectrum for X is consistent with structure O 29. (a) H Cl H H Cl H Cl H Cl Cl H Cl Cl H Cl H Cl H H Cl 2 Cl Cl H H H H H H Cl Cl Cl H H H H H Cl 6 Cl meso Cl H Cl H Cl H 3 Cl Cl H H Cl meso Cl H Cl Cl H Cl meso 5 H Cl 1 meso Cl H Cl Cl H H Cl H H Cl Cl Cl meso 4 Cl H H H Cl Cl Cl H H Cl H H Cl Cl Cl H H Cl H H Cl H Cl Cl H H Cl Cl H H Cl Cl 9 H meso Cl 8 7 Enantiomers H H Cl meso Cl H 275 276 ANSWERS TO FIRST REVIEW PROBLEM SET (b) Isomer 9 is slow to react in an E2 reaction because in its more stable conformation (see following structure) all the chlorine atoms are equatorial and an anti coplanar transition state cannot be achieved. All other isomers 1–8 can have a Cl axial and thus achieve an anti coplanar transition state. H H Cl Cl H H Cl Cl Cl 9 Cl H H 30. (a) F F2 F H 1 H 2 Enantiomers (obtained in one fraction as an optically inactive racemic form) F F F H 4 3 (achiral and, therefore, optically inactive) F H 5 6 Enantiomers (obtained in one fraction as an optically inactive racemic form) (achiral and, therefore, optically inactive) (b) Four fractions. The enantiomeric pairs would not be separated by fractional distillation because enantiomers have the same boiling points. (c) All of the fractions would be optically inactive. (d) The fraction containing 1 and 2 and the fraction containing 4 and 5. 31. (a) F H F2 (R)-2-Fluorobutane F H F H F F F 1 2 (optically active) (achiral and, therefore, optically inactive) F F H H F F H 3 (optically active) F H 4 meso compound (optically inactive) 5 (optically active) ANSWERS TO FIRST REVIEW PROBLEM SET 277 (b) Five. Compounds 3 and 4 are diastereomers. All others are constitutional isomers of each other. (c) See above. 32. H H H (R) H (S) H (R) H meso H H (S) meso Each of the two structures just given has a plane of symmetry (indicated by the dashed line), and, therefore, each is a meso compound. The two structures are not superposable one on the other; therefore, they represent molecules of different compounds and are diastereomers. 33. Only a proton or deuteron anti to the bromine can be eliminated; that is, the two groups undergoing elimination (H and Br or D and Br) must lie in an anti coplanar arrangement. The two conformations of erythro-2-bromobutane-3-d in which a proton or deuteron is anti coplanar to the bromine are I and II. Br D D − HBr H ΕtO− I Br − DBr H D ΕtO− II Conformation I can undergo loss of HBr to yield cis-2-d-2-butene. Conformation II can undergo loss of DBr to yield trans-2-butene. To a minor extent, a proton of the methyl group can be eliminated with the bromine. Br − HBr H H H ΕtO− D D 13 CONJUGATED UNSATURATED SYSTEMS SOLUTIONS TO PROBLEMS 13.1 The following two allylic radicals are possible, differing only because of the isotopic label. Together they allow for four constitutional isomers with respect to the 13 C label. (In the absence of the isotopic label, only one constitutional isomer (as a racemic mixture) would be possible.) δ 夹 夹 δ δ δ Br 夹 夹 夹 + 夹 Br + Br (夹 = 13C labeled position) (b) + + 13.2 (a) Br + because it represents a 2º carbocation. Cl (c) Cl and (racemic) 13.3 (a) (b) + (c) 278 + + CONJUGATED UNSATURATED SYSTEMS 279 + (d) + (e) + + + H H O + H O O − (f ) + Br Br + + + (g) + O O (h) − (i) S + − + S + ( j) + O N O − + N O − O O 2+ N O − − (minor) 13.4 (a) because the positive charge is on a tertiary carbon atom rather than a + primary one (rule 8). (b) + because the positive charge is on a secondary carbon atom rather than a primary one (rule 8). (c) N + because all atoms have a complete octet (rule 8b), and there are more covalent bonds (rule 8a). O (d) OH because it has no charge separation (rule 8c). 280 CONJUGATED UNSATURATED SYSTEMS (e) one (rule 8). because the radical is on a secondary carbon atom rather than a primary N because it has no charge separation (rule 8c). (f ) NH2 13.5 In resonance structures, the positions of the nuclei must remain the same for all structures (rule 2). The keto and enol forms shown differ not only in the positions of their electrons, but also in the position of one of the hydrogen atoms. In the enol form, it is attached to an oxygen atom; in the keto form, it has been moved so that it is attached to a carbon atom. 13.6 (a) (3Z )-Penta-1,3-diene, (2E,4E )-2,4-hexadiene, (2Z,4E )-hexa-2,4-diene, and 1,3cyclohexadiene are conjugated dienes. (b) 1,4-Cyclohexadiene and 1,4-pentadiene are isolated dienes. (c) Pent-1-en-4-yne (1-penten-4-yne) is an isolated enyne. 13.7 The formula, C6 H8 , tells us that A and B have six hydrogen atoms less than an alkane. This unsaturation may be due to three double bonds, one triple bond and one double bond, or combinations of two double bonds and a ring, or one triple bond and a ring. Since both A and B react with 2 mol of H2 to yield cyclohexane, they are either cyclohexyne or cyclohexadienes. The absorption maximum of 256 nm for A tells us that it is conjugated. Compound B, with no absorption maximum beyond 200 nm, possesses isolated double bonds. We can rule out cyclohexyne because of ring strain caused by the requirement of linearity of the —C C— system. Therefore, A is 1,3-cyclohexadiene; B is 1,4-cyclohexadiene. A has three signals in its 13 C NMR spectrum. With its higher symmetry, B shows only two 13 C NMR signals. 13.8 All three compounds have an unbranched five-carbon chain, because the product of hydrogenation is unbranched pentane. The formula, C5 H6 , suggests that they have one double bond and one triple bond. Compounds D, E, and F must differ, therefore, in the way the multiple bonds are distributed in the chain. Compounds E and F have a terminal —C CH [IR absorption at ∼3300 cm−1 ]. The UV absorption maximum near 230 nm for D and E suggests that in these compounds, the multiple bonds are conjugated. Absence of UV absorption beyond 200 nm indicates that the unsaturation sites are isolated in F. The structures are or H H D E Cl 13.9 (a) Cl + stereoisomers + stereoisomers H F CONJUGATED UNSATURATED SYSTEMS 281 Cl Cl (b) D D + stereoisomers + stereoisomers D D Cl Cl + stereoisomers + stereoisomers 13.10 Addition of the proton gives the resonance hybrid. + + (a) I II The inductive effect of the methyl group in I stabilizes the positive charge on the adjacent carbon. Such stabilization of the positive charge does not occur in II. Because I contributes more heavily to the resonance hybrid than does II, C2 bears a greater positive charge and reacts faster with the bromide ion. (b) In the 1,4-addition product, the double bond is more highly substituted than in the 1,2-addition product; hence it is the more stable alkene. 13.11 Endo and exo product formation in the Diels-Alder reaction of cyclopentadiene and maleic anhydride. H H H H Endo approach O H H O H O O O (Major product) H O H H H H O O O Exo approach O H H (Minor product) O O H H CONJUGATED UNSATURATED SYSTEMS 13.12 (a) The dienophile can approach the diene in an endo fashion from above or below the diene. Approach from above leads to one enantiomer. Approach from the bottom leads to the other. (b) They are diastereomers. CH3 CH3 13.13 + OCH3 O CH3 OCH3 CH3 Minor-by exo addition (plus enantiomer) Major-by endo addition (plus enantiomer) O HO H + (a) O OMe H (b) H OMe O O O H O (c) + HO O H H O 13.14 O 282 (major product) (minor product) O O (d) O H H + enantiomer H (e) O H CONJUGATED UNSATURATED SYSTEMS 283 13.15 (a) Use the trans diester because the stereochemistry is retained in the adduct. O H OMe + MeO H O (b) Here, the cis relationship of the acetyl groups requires the use of the cis dienophile. O H O + H O O OMe O OMe + O O OH (dienophile underneath the plane of the diene) OH O 13.17 O O O OH (dienophile above the plane of the diene) OH O 13.16 O (Or, in each case, the other face of the dienophile could present itself to the diene, resulting in the respective enantiomer.) 284 CONJUGATED UNSATURATED SYSTEMS Problems Conjugated Systems OK (2 equiv.) Br 13.18 (a) Br OH, heat concd H2SO4 OH (b) HO heat concd H2SO4 OH (c) heat OK Cl (d) OH, heat OK (e) Cl OH, heat concd H2SO4 heat (f ) OH (g) + H2 Ni2B (P-2) 13.19 Cl 13.20 (a) Cl + Cl Cl (E) + (Z) (racemic) Cl Br (b) Cl Cl Cl (3 stereoisomers) (d) (c) Br Br Br (3 stereoisomers) CONJUGATED UNSATURATED SYSTEMS OH + (e) Cl Cl (E) + (Z) OH (racemic) Cl + Cl + Cl Cl (E) + (Z) (racemic) (f ) 4 CO2 (Note: KMnO4 oxidizes HO2C CO2H to 2 CO2.) OH + (g) 285 (E) + (Z ) OH (racemic) H2, Pd/C 13.21 (a) (1) O3 (b) O (2) Me2S O Br2, hv (c) Br HBr, heat (d) Br O 13.22 (a) + N Br ROOR ∆ Br + Br (racemic) (E) + (Z ) O (NBS) OK OH, heat Note: In the second step, both allylic halides undergo elimination of HBr to yield 1,3butadiene; therefore, separating the mixture produced in the first step in unnecesary. The BrCH2 CH=CHCH3 undergoes a 1,4-elimination (the opposite of a 1,4-addition). 286 CONJUGATED UNSATURATED SYSTEMS (b) + ROOR ∆ NBS Br (racemic) Br OK + (E) + (Z ) OH, heat Here again both products undergo elimination of HBr to yield 1,3-pentadiene. concd H2SO4 (c) OH [as in (a)] + heat + Br2 Br Br heat (E) + (Z ) Br + (d) NBS ROOR ∆ Br (E) + (Z) + (racemic) Br NBS, ROOR OK light heat + Br2 (e) ∆ (excess) Br (racemic) OH, heat Br OK (f) same as OH, heat 13.23 R O O R R O + H Br + HBr heat or light 2R O R O H + Br Br Br H Br + H Br Br [ + (Z) isomer] + Br CONJUGATED UNSATURATED SYSTEMS 13.24 (a) 13 C NMR UV-Vis IR 2 signals 217 nm (s) [conj. system] ∼1600 cm−1 (s) [conj. system] UV-Vis 1 H NMR 217 nm (s) [conj. system] δ ∼ 5.0 [ CH2 ] δ ∼ 6.5 [—CH ] δ ∼ 117 [ CH2 ] δ ∼ 137 [ CH—] 4 signals 185 nm (w) ∼3300 cm−1 (s) [ C—H] 2100-2260 cm−1 (w) [C C] (b) 13 C NMR transparent in UV-Vis δ ∼1.0 (t) [CH3 ] δ ∼1.4 (q) [CH2 ] δ ∼13 [CH3 ] δ ∼25 [CH2 ] (c) OH IR 2800–3300 cm [C—H] 13 2 signals m/z 58 (M+. ) C NMR MS −1 (s, sharp) 3200–3550 cm−1 (s, broad) [O—H] ∼900 cm−1 (s) [ CH H] ∼1000 cm−1 (s) [ C H] 4 signals m/z 72 (M+.), 54 (M+. − 18) (d) Br MS 13 C NMR UV-Vis + m/z 54 (M .) 2 signals 217 nm (s) [conj. system] + m/z 134 (M . ), 136 (M+. +2) 4 signals <200 nm Br (e) Br 1 H NMR IR Br 2 signals 960–980 cm−1 [trans C—H] s = strong; w = weak; q = quartet; t = triplet Br 1 signal no alkene C—H 287 288 CONJUGATED UNSATURATED SYSTEMS 13.25 HCl + Ι + ΙΙ + + The resonance hybrid, I, has the positive charge, in part, on the tertiary carbon atom; in II, the positive charge is on primary and secondary carbon atoms only. Therefore, hybrid I is more stable and will be the intermediate carbocation. A 1,4-addition to I gives Cl Br (racemic) and 13.26 The products are Br [also (Z )] . They are formed from an allylic radical in the following way: Br2 Br 2 Br (from NBS) + + HBr [(Z) and (E)] δ Br δ + Br2 (racemic) + Br + Br [also (Z )] 13.27 (a) Because a highly resonance-stabilized radical is formed: (b) Because the carbanion is more stable: − − − − − That is, for the carbanion derived from the diene we can write more resonance structures of nearly equal energies. CONJUGATED UNSATURATED SYSTEMS O O 13.28 N hv Br + N Br O O + HBr H Br Br Br Br Br Br Br Where did the Br2 come from? It was formed from NBS, as follows. O O N + Br N HBr + H Br + A Br2 O O 13.29 H + Br− B Br− Br Br Kinetic product (1,2-Addition) Thermodynamic product (1,4-Addition) C D 289 CONJUGATED UNSATURATED SYSTEMS C is the kinetic product because it has a smaller Eact from B than D. B Free Energy 290 A C D is the thermodynamic product because it is more stable. D Reaction Coordinate Br HBr −15 °C 13.30 (a) Br HBr 40 °C (b) + Br Major (c) Br NBS, hv heat + Br Major 13.31 Protonation of the alcohol and loss of water leads to an allylic cation that can react with a chloride ion at either C1 or C3. H3O + H −H2O +O OH H + + Cl Cl− + Cl [also (Z )] (racemic) CONJUGATED UNSATURATED SYSTEMS + 13.32 (1) + Cl Cl2 + δ+ Cl δ+ δ+ Cl − + Cl (2) Cl 291 δ+ OMe MeOH Cl Cl + HA (racemic) OMe [also (Z )] 13.33 (a) The same carbocation (a resonance hybrid) is produced in the dissociation step: Cl Ag+ Ag+ + + I + AgCl II Cl H2O + OH (85%) OH (15%) (b) Structure I contributes more than II to the resonance hybrid of the carbocation (rule 8). Therefore, the hybrid carbocation has a larger partial positive charge on the tertiary carbon atom than on the primary carbon atom. Reaction of the carbocation with water will therefore occur more frequently at the tertiary carbon atom. 292 CONJUGATED UNSATURATED SYSTEMS 13.34 A six-membered ring cannot accommodate a triple bond because of the strain that would be introduced. Too highly strained X Br 2 RO− Na+ Br 13.35 (a) Propyne. (b) Base (:B− ) removes a proton, leaving the anion whose resonance structures are shown: H H − − − C C C H C C C CH2 C CH2 + Β Η Β + H H H I II Reaction with H:B may then occur at the CH2 carbanion (II). The overall reaction is CH2 C − CH2 + Β − [CH2 C − CH2 CH CH3 C C CH] + Η Β − CH + Β 13.36 The product formed when 1-bromobutane undergoes elimination is 1-butene, a simple monosubstituted alkene. When 4-bromo-1-butene undergoes elimination, the product is 1,3-butadiene, a conjugated diene, and therefore, a more stable product. The transition states leading to the products reflect the relative stabilities of the products. Since the transition state leading to 1,3-butadiene has the lower free energy of activation of the two, the elimination reaction of 4-bromo-1-butene will occur more rapidly. Diels-Alder Reactions 13.37 LUMO HOMO Energy CONJUGATED UNSATURATED SYSTEMS 293 13.38 The diene portion of the molecule is locked into an s-trans conformation. It cannot, therefore, achieve the s-cis conformation necessary for a Diels-Alder reaction. 13.39 s-cis conformation is disfavored (steric strain) No steric strain in s-cis conformation 4 Locked s-cis conformation 2 3 1 Cannot achieve s-cis conformation for Diels-Alder reaction. O O OMe 13.40 (a) + enantiomer OMe (b) CN CN (c) 13.41 + enantiomer + enantiomer (d) + enantiomer + O CF3 OMe 13.42 (a) (b) OMe CF3 O O 13.43 (a) + MeO O Me + ∆ MeO O Major O 294 CONJUGATED UNSATURATED SYSTEMS (b) + ∆ CN CN O + (c) ∆ O O (d) O OMe OMe O + ∆ CO2Me CO2Me Major O O + 2 ∆ O O O (f ) CO2Me CO2Me + O (e) O + H (1) ∆ (2) NaBH4 (3) H2O OH + H OH OH O OH Major 2 MeO (g) CO2Me CO2Me OMe O O ∆ MeO2C CO2Me CN (h) + + ∆ CN CN Major CN CN CN CONJUGATED UNSATURATED SYSTEMS O Ο 13.44 (a) 295 + (b) Η OCH3 + OCH3 O O (c) O OCH3 + (d) O + CH3O O O CN (e) + (f ) + OCH3 O Ο (g) + Η 13.45 The endo adduct is less stable than the exo, but is produced at a faster rate at 25◦ C. At 90◦ C the Diels-Alder reaction becomes reversible; an equilibrium is established, and the more stable exo adduct predominates. 13.46 Cl Cl Cl Cl Cl Cl H + Cl Cl Cl Cl Cl Aldrin Cl Cl Cl O H Cl Cl OOH H Cl Dieldrin H H O H Cl CONJUGATED UNSATURATED SYSTEMS H + 13.47 (a) H Norbornadiene Cl Cl + (b) ONa OH heat Cl Cl Cl Cl 13.48 Cl Cl (dark) Cl Cl Cl Cl Cl Cl Cl2 + Cl Cl Cl Cl Cl Cl Chlordan Cl Cl Note: The other double bond is less reactive because of the presence of the two chlorine substituents. Cl Cl Cl allylic Cl chlorination Cl Cl Cl Cl Cl Cl Cl Cl Cl Heptachlor Cl Cl Cl 13.49 Cl Cl Cl Cl Cl Cl Cl Cl Cl Isodrin O O O 13.50 (a) ∆ O 296 CONJUGATED UNSATURATED SYSTEMS 297 CN NC (b) CN CN ∆ O O Br (1) pyridine (a weak base) (c) (2) ,∆ O Br (d) (1) KO-t-Bu (2) HOOC HO COOH ,∆ HO O O (e) H ,∆ (1) O O (2) O3 (3) Me2S H O Challenge Problems Diene 13.51 Dienophile O OCH3 H + O− OCH3 − + H + + O O H − H − OCH3 OCH3 O H + + CH3O Good polarity interactions lead to major product. Major Minor 298 CONJUGATED UNSATURATED SYSTEMS 13.52 The first crystalline solid is the Diels-Alder adduct below, mp 125◦ C, O H H O O O On melting, this adduct undergoes a reverse Diels-Alder reaction, yielding furan (which vaporizes) and maleic anhydride, mp 56◦ C, O Ο H H heat O O + O O Furan O Ο Maleic anhydride (mp 56° C) H 13.53 MeO2C H H H OSi(tert-Bu)Ph2 H H 13.54 + + + δ+ δ+ δ+ The map of electrostatic potential for the pentadienyl carbocation shows greater electron density or less positive charge (less blue color) in the vicinity of the C1—C2 and C4—C5 bonds, suggesting that the most important contributing resonance structure is the one with the most positive charge near C3, a secondary carbon. The other contributing resonance structures have a positive charge at primary carbon atoms. CONJUGATED UNSATURATED SYSTEMS 299 QUIZ 13.1 Which is the 1,4-addition product from the following reaction: + (a) ? HCl (b) Cl (c) Cl (d) Cl Cl Cl (e) Cl 13.2 Which diene and dienophile could be used to synthesize the following compound? CN H H CN CN H (a) + CN H H NC + CN CN 13.3 Which reagent(s) could be used to carry out the following reaction? Br O (a) NBS, ROOR, ∆ NBS = NBr O CN + CN (e) H H (c) NC CN + CN + (b) H (d) H (b) NBS, ROOR, , then Br2 /hv H 300 CONJUGATED UNSATURATED SYSTEMS ΟΚ (c) Br2 , hv, then (d) ΟΚ Ο Η , , then NBS, ROOR,  Ο Η , then NBS, ROOR,  13.4 Which of the following structures does not contribute to the hybrid for the carbocation formed when 4-chloro-2-pentene ionizes in an SN 1 reaction? + + (a) + (b) (c) (d) All of these contribute to the resonance hybrid. 13.5 Which of the following resonance structures accounts, at least in part, for the lack of SN 2 reactivity of vinyl chloride? Cl (a) Cl (b) − + (c) Neither (d) Both SUMMARY OF REACTIONS BY TYPE CHAPTERS 1–13 I. SUBSTITUTION REACTIONS Type Stereochemical Result Favoring Conditions inversion 1◦ , 2◦ , benzylic (1◦ or 2◦ ), or allylic (1◦ or 2◦ ) leaving group (e.g., halide, tosylate, mesylate); strong nucleophile; polar aprotic solvent SN 1 Chapter 6 racemization (via carbocation) 3◦ , benzylic, or allylic leaving group Radical substitution racemization (via radical) 3◦ , benzylic, or allylic hydrogen; peroxide, heat/light SN 2 Chapter 6 See Section 6.14 for a summary of nucleophiles mentioned in early chapters Chapter 10 Section 13.2 II. ELIMINATION REACTIONS Type E2 (dehydrohalogenation) Section 7.6 Stereochemical/Regiochemical Result elimination to form the most substituted alkene (Zaitsev elimination) with small bases formation of the less substituted alkene with use of a bulky base (e.g., tert-BuOK/ tert-BuOH) E1 (dehydrohalogenation) Sections 6.17 and 7.7A Dehydration Section 7.7 Favoring Conditions strong base (e.g., NaOEt/EtOH, KOH/EtOH, tert-BuOK/tert-BuOH); 2◦ or 3◦ leaving group (e.g., halide, tosylate, mesylate, etc.); heat formation of most substituted alkene; may occur with carbocation rearrangement 3◦ leaving group; weak base; heat formation of most substituted alkene; may occur with carbocation rearrangement catalytic acid (H+ , e.g., concd H2 SO4 , H3 PO4 ); heat 301 302 SUMMARY OF REACTIONS BY TYPE III. MECHANISTIC SUMMARY OF ALKENE AND ALKYNE ADDITION REACTIONS Reactant Alkenes Stereochemical Result Regiochemical Result syn H2 /Pt, Pd or Ni anti *X2 Markovnikov *H2 O, HA (hydration) anti-Markovnikov (i) BH3 :THF (ii) HO− , H2 O2 (Sections 4.17A, 7.12, 7.13) (Section 8.11) (Section 8.4) (Section 8.6) (i) OsO4 , (ii) NaHSO3 * X2 ,H2 O (Section 8.15) (Section 8.13) * (i) Hg(OAc)2 , H2 O, THF (ii) NaBH4 , HO− (Section 8.5) RCO3 H (e.g., mCPBA) (i) RCO3 H, (ii) H3 O+ or HO− (Section 11.14) (Section 11.13) * (i) BH3 :THF (ii) H2 O2 , HO− (Sections 8.6 and 8.7) * X2 /ROH (Section 8.13) * HX (no peroxides) HBr (w/peroxides) (Section 8.2) (Section 10.10) * additon of other alkenes (Section 10.11, Special Topic B) * X2 , Nucleophile (e.g., RO− , RMgX) (Section 8.13) C Addition of carbenes (Section 8.14) * Shares mechanistic themes with other reactions denoted by the same symbol. Alkynes H2 , Pd on CaCO3 w/quinoline (Lindlar’s cat.) (i) Li, EtNH2 , −78 ◦ C (ii) NH4 + Cl− HX (one or two molar equivalents) (Section 7.14A) (Section 7.14B) (Section 8.18) H2 , Ni2 B (P-2) X2 (2 equiv.) (Section 7.14A) (Section 8.17) H2 (2 equiv.), Pt or Ni (complete hydrogenation) (Sections 4.16A, 7.14) SUMMARY OF REACTIONS BY TYPE 303 IV. ALKENE AND ALKYNE CLEAVAGE WITH OXIDATION Conditions Reactant Product(s) Alkenes (i) KMnO4 /HO− , heat; (ii) H3 O+ (Section 8.16A) (i) O3 , (ii) Me2S (Section 8.16B) tetrasubstituted alkene trisubstituted alkene monosubstituted alkene −→ −→ −→ two ketones one ketone, one carboxylic acid one carboxylic acid and CO2 tetrasubstituted alkene trisubstituted alkene monosubstituted alkene −→ −→ −→ two ketones one ketone, one aldehyde one aldehyde and formaldehyde terminal alkyne internal alkyne −→ −→ a carboxylic acid and formic acid two carboxylic acids terminal alkyne internal alkyne −→ −→ a carboxylic acid and formic acid two carboxylic acids Alkynes (i) KMnO4 /HO− , heat; (ii) H3 O+ (Section 8.19) (i) O3 , (ii) HOAc (Section 8.19) V. CARBON—CARBON BOND-FORMING REACTIONS (a) Alkylation of alkynide anions (with 1◦ alkyl halides, epoxides, and aldehydes or ketones) (Sections 7.11, 8.20 and 12.8D) (b) Grignard reaction (with aldehydes and ketones, or epoxides) (Sections 12.7B, C, and 12.8) (c) Carbocation addition to alkenes (e.g., polymerization) (Special Topic B) (d) Diels-Alder reaction (Section 13.10) (e) Addition of a carbene to an alkene (Section 8.14) (f) Heck, Suzuki, Stille, Sonogashira, Grubbs, and other transition metal organometallic reactions (Special Topic G) VI. REDUCTIONS/OXIDATIONS (NOT INCLUDING ALKENES/ALKYNES) (a) 2 R—X (w/Zn/H3 O+ ) −→ 2 R—H + ZnX2 (Section 5.15A) (b) R X Mg ether R MgX H3O+ R H (Section 12.7A) (c) Lithium aluminum hydride (LiAlH4 ) reduction of alkyl halides to hydrocarbons (Section 12.3D) (d) Lithium aluminum hydride (LiAlH4 ) and sodium borohydride (NaBH4 ) reduction of carbonyl compounds (Section 12.3A, B, C) 304 SUMMARY OF REACTIONS BY TYPE (e) Oxidation of alcohols by the Swern oxidation (Section 12.4B), with chromic acid (Section 12.4C), pyridinium chlorochromate (PCC, Section 12.4D), or hot potassium permanganate (Section 12.4E) VII. MISCELLANEOUS (a) R—CO—R′ + PCl5 −→ R—CCl2 —R′ −→ alkynes (Section 7.10) (b) R—COOH + SOCl2 or PCl5 −→ R—COCl−→ acyl chlorides for Friedel-Crafts reactions, esters, amides, etc. (Sections 15.7 and 17.5) (c) Terminal alkynes + NaNH2 in liq. NH3 −→ alkynide anions (Sections 8.20, 12.8D) (d) ROH + TsCl, MsCl, or TfCl (with pyridine) −→ R—OTs, R—OMs, or R—OTf (Section 11.10) (e) R—OH + Na (or NaH) −→ R—O− Na+ + H2 (Section 6.15B, 11.11B) VIII. CHEMICAL TESTS (a) Alkenes/Alkynes: Br2 (Section 8.11) (b) Rings/Unsaturation/etc.: Index of Hydrogen Deficiency (Section 4.17) (c) Position of Unsaturation: KMnO4 (Section 8.16A); O3 (Section 8.16B) METHODS FOR FUNCTIONAL GROUP PREPARATION CHAPTERS 1–13 I. ALKYL HALIDES (a) HX addition to alkenes [Markovnikov (Section 8.2) and HBr anti-Markovnikov (Section 10.10)] (b) X2 addition to alkenes (Section 8.11) (c) Radical halogenation (i) X2 /light/heat for alkanes (Sections 10.3, 10.4, 10.5) (ii) N-Bromosuccinimide (NBS)/heat/light for allylic and benzylic substitution (Sections 10.8B, 10.9) (d) R—OH + SOCl2 (with pyridine) −→ R—Cl + SO2 + pyridinium hydrochloride (Section 11.9) (e) 3 R—OH + PBr3 −→ 3 R—Br + P(OH)3 (Section 11.9) (f ) R—OH + HX −→ R—X + H2 O (Sections 11.7, 11.8) II. ALCOHOLS (a) Markovnikov addition of H2 O with catalytic HA (Section 8.4) (b) Markovnikov addition of H2 O via (i) Hg(OAc)2 , H2 O, THF; (ii) NaBH4 , HO− (Section 8.5) (c) Anti-Markovnikov addition of H2 O via (i) BH3 :THF; (ii) HO− , H2 O2 (Section 8.6) (d) OsO4 addition to alkenes (syn) (Section 8.15) (e) (i) RCO3 H (peroxycarboxylic acids, e.g., mCPBA); (ii) H3 O+ (Section 11.15) (f ) Cleavage of ethers with HX (Section 11.12A) (g) Opening of epoxides by a nucleophile (Sections 11.14, 11.15) (h) Lithium aluminum hydride or sodium borohydride reduction of carbonyl compounds (Section 12.3) (i) Grignard reaction with aldehydes, ketones, and epoxides (Sections 12.7B, C, 12.8) ( j) Cleavage of silyl ethers (Section 11.11E) III. ALKENES (a) E2 Dehydrohalogenation (preferred over E1 for alkene synthesis) (Section 7.6) (b) Dehydration of alcohols (Section 7.7) 305 306 METHODS FOR FUNCTIONAL GROUP PREPARATION (c) Hydrogenation of alkynes: (Z ) by catalytic hydrogenation, (E ) by dissolving metal reduction (Sections 7.14A, B) (d) Diels-Alder reaction (forms one new double bond with ring formation) (Section 13.11) (e) Heck, Suzuki, Stille, Sonogashira cross coupling reactions (Special Topic G) (f) Grubbs olefin metathesis (Special Topic G) IV. ALKYNES (a) Alkylation of alkynide anions (Sections 7.11 and 8.20) (b) Double dehydrohalogenation of vicinal or geminal dihalides (Section 7.10) (c) Sonogashira cross coupling (Special Topic G) V. CARBON—CARBON BONDS (a) Alkylation of alkynide anions (Section 7.11 and 8.20) (b) Organometallic addition to carbonyl compounds and epoxides (e.g., Grignard or RLi reactions) (Sections 12.7B, C, 12.8) (c) Diels-Alder reaction (Section 13.11) (d) Addition of alkenes to other alkenes (e.g., polymerization) (Section 10.11 and Special Topic B) (e) Addition of a carbene to an alkene (Section 8.14) (f) Heck, Suzuki, Stille, Sonogashira, Grubbs, and other transition metal organometallic reactions (Special Topic G) VI. ALDEHYDES (a) (i) O3 ; (ii) Me2S with appropriate alkenes (Section 8.16B) (b) Swern oxidation of 1◦ alcohols (Section 12.4B) (c) Pyridinium chlorochromate (PCC) oxidation of 1◦ alcohols (Section 12.4D) VII. KETONES (a) (i) O3 ; (ii) Me2S with appropriate alkenes (Section 8.16B) (b) KMnO4 /HO− cleavage of appropriate alkenes (Section 8.16A) (c) Swern oxidation of 2◦ alcohols (Section 12.4B) (d) H2 CrO4 oxidation of 2◦ alcohols (Section 12.4C) (e) Stille carbonylation by cross coupling with carbon monoxide (Special Topic G) METHODS FOR FUNCTIONAL GROUP PREPARATION 307 VIII. CARBOXYLIC ACIDS (a) (i) KMnO4 /HO− ; (ii) H3 O+ with 1◦ alcohols (Section 12.4E) (b) (i) O3 ; (ii) HOAc with alkynes (Section 8.20) (c) (i) KMnO4 /HO− , heat; (ii) H3 O+ with alkynes and alkenes (Sections 8.20 and 8.17A) (d) H2 CrO4 oxidation of 1◦ alcohols (Section 12.4C) IX. ETHERS (INCLUDING EPOXIDES) (a) RO− + R′ X −→ ROR’ + X− (Williamson synthesis) (Section 11.11B) (b) 2 ROH (+cat. H2 SO4 , heat) −→ ROR + HOH (for symmetrical ethers only) (Section 11.11A) (c) Alkene + RCO3 H (a peroxycarboxylic acid, e.g., mCPBA) −→ an epoxide (Section 11.13) (d) An epoxide + RO− −→ an α-hydroxy ether (also acid catalyzed) (Section 11.14) (e) ROH + ClSiR′ 3 −→ ROSiR′ 3 (silyl ether protecting groups, e.g., TBDMS ethers) (Section 11.11E) 14 AROMATIC COMPOUNDS SOLUTIONS TO PROBLEMS 14.1 (a) 4-Bromobenzoic acid (or p-bromobenzoic acid) (b) 2-Benzyl-1.3-cyclohexadiene (c) 2-Chloro-2-phenylpentane (d) Phenyl propyl ether 14.2 Compounds (a) and (b) would yield only one monosubstitution product. 14.3 Resonance structures are defined as being structures that differ only in the positions of the electrons. In the two 1,3,5-cyclohexatrienes shown, the carbon atoms are in different positions; therefore, they cannot be resonance structures. 14.4 The cyclopentadienyl cation would be a diradical. We would not expect it to be aromatic. 14.5 (a) No, the cycloheptatrienyl anion (below) would be a diradical. (b) The cycloheptatrienyl cation (below) would be aromatic because it would have a closed bonding shell of delocalized π electrons. 14.6 If the 1,3,5-cycloheptatrienyl anion were aromatic, we would expect it to be unusually stable. This would mean that 1,3,5-cycloheptatriene should be unusually acidic. The fact that 1,3,5cycloheptatriene is not unusually acidic (it is less acidic than 1,3,5-heptatriene) confirms the prediction made in the previous problem, that the 1,3,5-cycloheptatrienyl anion should not be aromatic. 308 AROMATIC COMPOUNDS Br 309 Br 14.7 (a) heat −HBr (b) Br + Br Br− Tropylium bromide These results suggest that the bonding in tropylium bromide is ionic; that is, it consists of a positive tropylium ion and a negative bromide ion. 14.8 The fact that the cyclopentadienyl cation is antiaromatic means that the following hypothetical transformation would occur with an increase in π -electron energy. π-electron energy increases HC+ + H2 + 14.9 (a) The cyclopropenyl cation (below). + or SbCl6− + (b) Only one 13 C NMR signal is predicted for this ion. 14.10 (a) 3 (b) 4 (c) 7 (d) 5 14.11 The upfield signal arises from the six methyl protons of trans-15,16-dimethyldihydropyrene, which by virtue of their location are strongly shielded by the magnetic field created by the aromatic ring current (see Figure 14.8). 14.12 Major contributors to the hybrid must be ones that involve separated charges. Contributors like the following would have separated charges, and would have aromatic five- and sevenmembered rings. − − + etc. + − + SH 14.13 (a) OH N N (b) N N H N N N N H 310 AROMATIC COMPOUNDS 14.14 Because of their symmetries, p-dibromobenzene would give two 13 C signals, o-dibromobenzene would give three, and m-dibromobenzene would give four. Br (a) (b) (b) (b) (a) Br Two signals (c) (b) (b) Br (a) (a) Br Br (b) (c) (d) (b) (a) (b) Br (c) (c) Three signals Four signals Br 14.15 Br Br Br A B C D A. Strong absorption at 740 cm−1 is characteristic of ortho substitution. B. A very strong absorption peak at 800 cm−1 is characteristic of para substitution. C. Strong absorption peaks at 680 and 760 cm−1 are characteristic of meta substitution. D. Very strong absorption peaks at 693 and 765 cm−1 are characteristic of a monosubstituted benzene ring. Problems Nomenclature O OH Br Br 14.16 (a) (c) (b) NO2 Br O NO2 OH OH (d) (e) NO2 (f ) O2N NO2 NO2 AROMATIC COMPOUNDS O SO3H (g) 311 SO2OCH3 (h) (i) Cl Cl Br NH2 ( j) (l) (k) NO2 O OH (m) (o) (n) Br OH OCH3 HO (p) (r) (q) C6H5 C6H5 Cl Br Br Br Br Br 14.17 (a) Br 1,2,3-Tribromobenzene Br Br 1,2,4-Tribromobenzene Br 1,3,5-Tribromobenzene 312 AROMATIC COMPOUNDS OH OH Cl OH Cl Cl (b) Cl 2,3-Dichlorophenol OH Cl Cl Cl 2,4-Dichlorophenol 2,5-Dichlorophenol OH OH Cl Cl 2,6-Dichlorophenol NH2 Cl 3,4-Dichlorophenol Cl Cl 3,5-Dichlorophenol NH2 NH2 NO2 (c) NO2 NO2 4-Nitroaniline (p-nitroaniline) SO3H 3-Nitroaniline (m-nitroaniline) SO3H 2-Nitroaniline (o-nitroaniline) SO3H 4-Methylbenzenesulfonic acid (p-toluenesulfonic acid) 3-Methylbenzenesulfonic acid (m-toluenesulfonic acid) 2-Methylbenzenesulfonic acid (o-toluenesulfonic acid) Isobutylbenzene sec-Butylbenzene (d) (e) Butylbenzene tert-Butylbenzene AROMATIC COMPOUNDS Aromaticity + + 14.18 (a) Antiaromatic (b) Aromatic (d) Aromatic (f ) Antiaromatic (h) Aromatic ( j) Aromatic (l) Aromatic O (c) N Antiaromatic + N + (e) N Aromatic N − (g) − Antiaromatic + (i) Nonaromatic + − (k) S Antibonding 14.19 (a) Energy + Bonding Antibonding (b) Energy − Bonding Aromatic 313 314 AROMATIC COMPOUNDS H N +N HCl (1 equiv.) 14.20 (a) N Cl− N CH3 CH3 H H N N HCl (1 equiv.) (b) Cl− N+ H N 14.21 H B or − − A H B or − − B The conjugate base of A is a substituted cyclopentadienyl anion and both rings are aromatic. In B, the five-membered ring does not contribute to anion stability. The additional stabilization provided by the cyclopentyldienyl moiety is the reason for the greater acidity of A. AROMATIC COMPOUNDS 315 14.22 + The major resonance form leaves the electrons in the cyclopentadiene ring to make both rings aromatic. Therefore, the central bond has more single bond character and rotation around that bond is easily achieved. − 14.23 Hückel’s rule should apply to both pentalene and heptalene. Pentalene’s antiaromaticity can be attributed to its having 8 π electrons. Heptalene’s lack of aromaticity can be attributed to its having 12 π electrons. Neither 8 nor 12 is a Hückel number. 14.24 (a) The extra two electrons go into the two partly filled (nonbonding) molecular orbitals (Fig. 14.7), causing them to become filled. The dianion, therefore, is not a diradical. Moreover, the cyclooctatetraene dianion has 10 π electrons (a Hückel number), and this apparently gives it the stability of an aromatic compound. (The highest occupied molecular orbitals may become slightly lower in energy and become bonding molecular orbitals.) The stability gained by becoming aromatic is apparently large enough to overcome the extra strain involved in having the ring of the dianion become planar. (b) The strong base (butyllithium) removes two protons from the compound on the left. This acid-base reaction leads to the formation of the 10 π electron pentalene dianion, an aromatic dianion. 2− + 2 Li + 2 BuLi − = − Pentalene dianion 14.25 The bridging CH2 group causes the 10 π electron ring system (below) to become planar. This allows the ring to become aromatic. CH2 14.26 (a) Resonance contributions that involve the carbonyl group of I resemble the aromatic cycloheptatrienyl cation and thus stabilize I. Similar contributors to the hybrid of II resemble the antiaromatic cyclopentadienyl cation (see Problem 14.8) and thus destabilize II. 316 AROMATIC COMPOUNDS O O − Contributors like IA are exceptionally stable because they resemble an aromatic compound. They therefore make large stabilizing contributions to the hybrid. + (a) I IA O O − + II (b) Contributors like IIA are exceptionally unstable because they resemble an antiaromatic compound. Any contribution they make to the hybrid is destabilizing. IIA O O 14.27 Ionization of 5-chloro-1,3-cyclopentadiene would produce a cyclopentadienyl cation, and the cyclopentadienyl cation (see Problem 14.8) would be highly unstable because it would be antiaromatic. Cl SN1 + + Cl − Antiaromatic ion (highly unstable) 14.28 (a) The cyclononatetraenyl anion with 10 π electrons obeys Hückel’s rule. _ 2 Cyclononatetraenyl anion 10 π electrons Aromatic _ Cyclohexadecaoctaenyl dianion 18 π electrons Aromatic (b) By adding 2 π electrons, [16] annulene becomes an 18 π electron system and therefore obeys Hückel’s rule. 14.29 As noted in Problem 13.42, furan can serve as the diene component of Diels-Alder reactions, readily losing all aromatic character in the process. Benzene, on the other hand, is so AROMATIC COMPOUNDS 317 unreactive in a Diels-Alder reaction that it can be used as a nonreactive solvent for DielsAlder reactions. Spectroscopy and Structure Elucidation O O O 14.30 (a) O 4 total signals aromatic region, 2 doublets 6 total signals aromatic region, 2 doublets and 2 doublets of doublets 3 O O 1 O O (b) 2 Signal 1 (~4 ppm, singlet) Signal 2 (~2 ppm, singlet) 4 Signal 3 (~2 ppm, singlet) Signal 4 (~2 ppm, singlet) Br (c) 5 total signals aromatic region: 1 singlet, 2 doublets, and 1 doublet of doublets 14.31 A (c) B (d ) Br 3 total signals aromatic region: 2 doublets CH3 (a) H (b) CH3 (a) (a) doublet δ 1.25 (b) septet δ 2.9 (c) multiplet δ 7.3 CH3 (a) H (b) NH2 (c) (a) doublet δ 1.35 (b) quartet δ 4.1 (c) singlet δ 1.7 (d ) multiplet δ 7.3 318 AROMATIC COMPOUNDS H C (a) H H (c) (a) triplet δ 2.9 (b) quintet δ 2.05 (c) multiplet δ 7.1 (b) H H H (a) 14.32 A 1 H NMR signal this far upfield indicates that cyclooctatetraene is a cyclic polyene and is not aromatic; its π electrons are not fully delocalized. 14.33 Compound F is p-isopropyltoluene. 1 H NMR assignments are shown in the following spectrum. The IR absorptions from 3020 to 2870 cm−1 indicate sp 2 and sp 3 C-H stretching vibrations, as would be present in an alkyl subsituted benzene ring. The absorptions at 1517 and 1463 cm−1 range are characteristic of benzene ring stretching vibrations. The absorption at 818 cm−1 suggests para substitution of the benzene ring. F, C10H14 (d) (c) CH (b) CH3 (a) CH3 (a) (a) CH3 (d) (b) (c) TMS 3.0 8 7 6 2.8 1.4 1.2 5 4 δH (ppm) 3 We can make the following 1 H NMR assignments: (b) CH3 (c) CH (d) (a) CH3 CH3 (a) (a) doublet δ 1.25 (c) septet δ 2.85 (b) singlet δ 2.3 (d) multiplet δ 7.1 2 1 0 AROMATIC COMPOUNDS 319 14.34 Compound L is allylbenzene, H (a) (c) H C C CH2 (d) H (b) (d) (a) (a) (c) (e) Doublet, δ 3.1 (2H) or (b) Multiplet, δ 4.8 or (b) Multiplet, δ 5.1 Multiplet, δ 5.8 Multiplet, δ 7.1 (5H) (e) The following IR assignments can be made. 3035 cm−1 , C H stretching of benzene ring 3020 cm−1 , C H stretching of CH CH2 group 2925 cm−1 and 2853 cm−1 , C H stretching of CH2 group 1640 cm−1 , C C stretching 990 cm−1 and 915 cm−1 , C H bendings of CH CH2 group 740 cm−1 and 695 cm−1 , C H bendings of C6 H5 group The UV absorbance maximum at 255 nm is indicative of a benzene ring that is not conjugated with a double bond. 14.35 Compound M is m-ethyltoluene. We can make the following assignments in the spectrum. CH3 (c) (d) CH2 (b) CH3 (a) (a) triplet δ 1.4 (b) quartet δ 2.6 (c) singlet δ 2.4 (d ) multiplet δ 7.05 Meta substitution is indicated by the very strong peaks at 690 and 780 cm−1 in the IR spectrum. 14.36 Compound N is C6 H5 CH CHOCH3 . The absence of absorption peaks due to O H or C O stretching in the IR spectrum of N suggests that the oxygen atom is present as part of an ether linkage. The (5H) 1 H NMR multiplet between δ 7.1–7.6 strongly suggests the presence of a monosubstituted benzene ring; this is confirmed by the strong peaks at ∼690 and ∼770 cm−1 in the IR spectrum. 320 AROMATIC COMPOUNDS We can make the following assignments in the 1 H NMR spectrum: (a) C6H5 (b) CH (c) CH (d) OCH3 (a) Multiplet δ 7.1–7.6 (b) Doublet δ 6.1 (c) Doublet δ 5.2 (d ) Singlet δ 3.7 14.37 Compound X is m-xylene. The upfield signal at δ 2.3 arises from the two equivalent methyl groups. The downfield signals at δ 6.9 and 7.1 arise from the protons of the benzene ring. Meta substitution is indicated by the strong IR peak at 680 cm−1 and very strong IR peak at 760 cm−1 . 14.38 The broad IR peak at 3400 cm−1 indicates a hydroxyl group, and the two bands at 720 and 770 cm−1 suggest a monosubstituted benzene ring. The presence of these groups is also indicated by the peaks at δ 4.4 and δ 7.2 in the 1 H NMR spectrum. The 1 H NMR spectrum also shows a triplet at δ 0.85 indicating a CH3 group coupled with an adjacent CH2 group. There is a complex multiplet at δ 1.7 and there is also a triplet at δ 4.5 (1H). Putting these pieces together in the only way possible gives us the following structure for Y. OH Y, C9H12O (c) CH (b) CH2 (a) CH3 (a) OH (d) (e) (b) (e) (c) (d) TMS 4.6 8 7 4.4 6 1.8 5 1.6 0.9 0.8 4 δH (ppm) 3 2 1 0 AROMATIC COMPOUNDS 321 14.39 (a) Four unsplit signals. (d) O H3C N N O (a) CH3 N (b) H N CH3 (c) (b) Absorptions arising from: , C CH3 , and C O groups. H Challenge Problems 14.40 The vinylic protons of p-chlorostyrene should give a spectrum approximately like the following: Jbc (c) (a) (b) Jac Jac Jbc Jab Jab Jbc 7.0 6.0 5.0 δH (ppm) Ph 14.41 Ph O Ph Ph A t-Bu + Ph B Ph Br− 322 AROMATIC COMPOUNDS 14.42 − Na+ Fe (a “sandwich compound”) C D Ph 14.43 Ph E 14.44 Third unoccupied π MO (five nodal planes) Second unoccupied π MO (four nodal planes) First unoccupied π MO (three nodal planes) Highest energy occupied π MO (two nodal planes) Second highest energy occupied π MO (one nodal plane) Lowest energy occupied π MO (no nodal planes) AROMATIC COMPOUNDS 323 QUIZ 14.1 Which of the following reactions of benzene is inconsistent with the assertion that benzene is aromatic? no reaction (a) Br2/25 °C no reaction (b) H 2 /Pt/25 °C (c) Br2/FeBr3 C6H5Br + HBr no reaction (d) KMnO4 /H 2O/25 °C (e) None of the above 14.2 Which is the correct name of the compound shown? NO2 Cl (a) 3-Chloro-5-nitrotoluene (b) m-Chloro-m-nitrotoluene (c) l-Chloro-3-nitro-5-toluene (d) m-Chloromethylnitrobenzene (e) More than one of these 14.3 Which is the correct name of the compound shown? OH F (a) 2-Fluoro-1-hydroxyphenylbenzene (b) 2-Fluoro-4-phenylphenol (c) m-Fluoro-p-hydroxybiphenyl (d) o-Fluoro-p-phenylphenol (e) More than one of these 324 AROMATIC COMPOUNDS 14.4 Which of the following molecules or ions is not aromatic according to Hückel’s rule? (a) (b) (c) (d) + (e) All are aromatic. 14.5 Give the structure of a compound with the formula C7 H7 Cl that is capable of undergoing both SN 1 and SN 2 reactions. 14.6 Write the name of an aromatic compound that is isomeric with naphthalene. 15 REACTIONS OF AROMATIC COMPOUNDS SOLUTIONS TO PROBLEMS E + H 15.1 E E H + A− H + A− −HA −HA E A− −HA E E 15.2 The rate is dependent on the concentration of NO2 + ion formed from protonated nitric acid. H + O NO2 + NO2+ HA H3O+ + + A− H (where HA = HNO3 or HOSO3H) Because H2 SO4 (HOSO3 H) is a stronger acid, a mixture of it and HNO3 will contain a higher concentration of protonated nitric acid than will nitric acid alone. That is, the reaction, H O NO2 + H HOSO3H + O NO2 + HSO4− H Protonated nitric acid produces more protonated nitric acid than the reaction, H O NO2 + H O NO2 H + O NO2 + NO3− H 325 326 REACTIONS OF AROMATIC COMPOUNDS + Step 1 15.3 H + F + δ+ Step 2 + − H + δ+ δ+ H δ+ Step 3 δ+ δ+ F F − + HF H O+ OH + BF3 15.4 BF3− Rearrangement to the secondary carbocation occurs as the leaving group departs. H H O+ hydride shift BF3− + + HOBF2 + F− Both carbocations are then attacked by the ring. H A− + + HA + H H + − A + HA + O 15.5 (a) O Cl AlCl3 heat O (b) NH2NH2, KOH O Cl AlCl3 Zn(Hg) HCl, reflux REACTIONS OF AROMATIC COMPOUNDS O C6H5 (c) 327 O Cl NH2NH2, KOH AlCl3 heat O O O O NH2NH2, KOH O (d) O OH AlCl3 O OH SOCl2 heat Cl O Zn(Hg) AlCl3 HCl reflux 15.6 If the methyl group had no directive effect on the incoming electrophile, we would expect to obtain the products in purely statistical amounts. Since there are two ortho hydrogen atoms, two meta hydrogen atoms, and one para hydrogen, we would expect to get 40% ortho (2/5), 40% meta (2/5), and 20% para (1/5). Thus, we would expect that only 60% of the mixture of mononitrotoluenes would have the nitro group in the ortho or para position. And we would expect to obtain 40% of m-nitrotoluene. In actuality, we get 96% of combined o- and p-nitrotoluenes and only 4% m-nitrotoluene. This shows the ortho-para directive effect of the methyl group. CH3 CH3 SO3H SO3 H2SO4 15.7 (a) CH3 minor O (b) OH O SO3H OH HNO3 H2SO4 major + NO2 (c) NO2 NO2 Br2 FeBr3 Br 328 REACTIONS OF AROMATIC COMPOUNDS O O Cl (d) + AlCl3 minor O major 15.8 As the following structures show, attack at the ortho and para positions of phenol leads to arenium ions that are more stable (than the one resulting from meta attack) because they are hybrids of four resonance structures, one of which is relatively stable. Only three resonance structures are possible for the meta arenium ion, and none is relatively stable. Ortho attack OH OH OH Br H Br H Br+ + OH + OH Br + H + Br H Relatively stable Meta attack OH OH + + Br+ OH OH Br Br Br + H H H Para attack OH OH OH + OH OH + + + Br+ H Br H Br H Br Relatively stable H Br REACTIONS OF AROMATIC COMPOUNDS 329 15.9 (a) The atom (an oxygen atom) attached to the benzene ring has an unshared electron pair that it can donate to the arenium ions formed from ortho and para attack, stabilizing them. (The arenium ions are analogous to the previous answer with a COCH3 group replacing the H of the phenolic hydroxyl). (b) Structures such as the following compete with the benzene ring for the oxygen electrons, making them less available to the benzene ring. O O − + O O This effect makes the benzene ring of phenyl acetate less electron rich and, therefore, less reactive. (c) Because the acetamido group has an unshared electron pair on the nitrogen atom that it can donate to the benzene ring, it is an ortho-para director. (d) Structures such as the following compete with the benzene ring for the nitrogen electrons, making them less available to the benzene ring. O O − + N H N H 15.10 The electron-withdrawing inductive effect of the chlorine of chloroethene makes its double bond less electron rich than that of ethene. This causes the rate of reaction of chloroethene with an electrophile (i.e., a proton) to be slower than the corresponding reaction of ethene. When chloroethene adds a proton, the orientation is governed by a resonance effect. In theory, two carbocations can form: Cl Cl + HCl + I (less stable) Cl− Cl Cl + Cl + Cl II (more stable) Carbocation II is more stable than I because of the resonance contribution of the extra structure just shown in which the chlorine atom donates an electron pair (see Section 15.11D). 330 REACTIONS OF AROMATIC COMPOUNDS 15.11 Ortho attack Br H Br+ Br H + Br H + + Relatively stable Para attack + + Br+ H + Br H Br Relatively stable H Br 15.12 The phenyl group, as the following resonance structures show, can act as an electronreleasing group and can stabilize the arenium ions formed from ortho and para attack. Ortho attack NO2+ H NO2 NO2 H + + H + NO2 H NO2 H NO2 + + H NO2 NO2 + A− −HA In the case of the arenium ions above and following, the unsubstituted ring can also be shown in the alternative Kekule structure. REACTIONS OF AROMATIC COMPOUNDS 331 Para attack + H H + NO2 O2N + O2N + + + H H H O2N O2N O2N A− −HA + H O2N O2N In the case of both ortho and para substitution, the phenyl group functions as an activating group, resulting in a faster reaction then in the case of benzene. 15.13 The addition of hydrogen bromide to 1-phenylpropene proceeds through a benzylic radical in the presence of peroxides, and through a benzylic cation in their absence (cf., a and b as follows). (a) Hydrogen bromide addition in the presence of peroxides. Chain Initiation Step 1 R O O R 2R O O Step 2 RO + H Step 3 Br + Br C6H5 R H + Br C6H5 Br A benzylic radical 332 REACTIONS OF AROMATIC COMPOUNDS Chain Propagation Step 4 + H C6H5 Br + Br C6H5 Br Br 2-Bromo-1-phenylpropane The mechanism for the addition of hydrogen bromide to 1-phenylpropene in the presence of peroxides is a chain mechanism analogous to the one we discussed when we described anti-Markovnikov addition in Section 10.9. The step that determines the orientation of the reaction is the first chain-propagating step. Bromine attacks the second carbon atom of the chain because by doing so the reaction produces a more stable benzylic radical. Had the bromine atom attacked the double bond in the opposite way, a less stable secondary radical would have been formed. Br + C6H5 Br C6H5 A secondary radical (b) Hydrogen bromide addition in the absence of peroxides. + + HBr C6H5 + Br− C6H5 A benzylic cation Br C6H5 1-Bromo-1-phenylpropane In the absence of peroxides, hydrogen bromide adds through an ionic mechanism. The step that determines the orientation in the ionic mechanism is the first, where the proton attacks the double bond to give the more stable benzylic cation. Had the proton attacked the double bond in the opposite way, a less stable secondary cation would have been formed. + HBr C6H5 15.14 (a) + + C6H5 A secondary cation Br− because the more stable carbocation intermediate is the benzylic Cl carbocation, , which then reacts with a chloride ion. + REACTIONS OF AROMATIC COMPOUNDS 333 (b) OH because the more stable intermediate is a mercurinium ion in which a partial positive charge resides on the benzylic carbon, which then reacts with H2 O. 15.15 (a) The first method would fail because introducing the chlorine substituent first would introduce an ortho-para directing group. Consequently, the subsequent Friedel-Crafts reaction would not then take place at the desired meta position. The second method would fail for essentially the same reasons. Introducing the ethyl group first would introduce an ortho-para director, and subsequent ring chlorination would not take place at the desired meta position. (b) If we introduce an acetyl group first, which we later convert to an ethyl group, we install a meta director. This allows us to put the chlorine atom in the desired position. Conversion of the acetyl group to an ethyl group is then carried out using the Clemmensen (or Wolff-Kishner) reduction. O O O Cl AlCl3 Cl2 FeCl3 Zn(Hg) HCl Cl CN OH Cl OCH3 OCH3 NO2 O2N (b) 15.16 (a) O2N (c) + SO3H NO2 CF3 NO2 NO2 15.17 (a) In concentrated base and ethanol (a relatively nonpolar solvent), the SN 2 reaction is favored. Thus, the rate depends on the concentration of both the alkyl halide and Since no carbocation is formed, the only product is ONa . EtO (b) When the concentration of O− ion is small or zero, the reaction occurs through the SN 1 mechanism. The carbocation that is produced in the first step of the SN 1 mechanism is a resonance hybrid. + Cl + + Cl − 334 REACTIONS OF AROMATIC COMPOUNDS O− or This ion reacts with the nucleophile ( OH ) to produce two isomeric ethers OEt EtO and 15.18 (a) The carbocation that is produced in the SN 1 reaction is exceptionally stable because one resonance contributor is not only allylic but also tertiary. SN1 Cl + + A 3° allylic carbocation OH (b) OH + 15.19 Compounds that undergo reactions by an SN 1 path must be capable of forming relatively stable carbocations. Primary halides of the type ROCH2 X form carbocations that are stabilized by resonance: O O + CH2 R R + CH2 15.20 The relative rates are in the order of the relative stabilities of the carbocations: Ph + Ph + + < Ph < Ph Ph < Ph + Ph The solvolysis reaction involves a carbocation intermediate. 15.21 REACTIONS OF AROMATIC COMPOUNDS 335 Mechanisms 15.22 (a) 1. Generate the electrophile O H O O + S O HO + H N O− O N N+ O O + + H2O O O O− H H 2. Attack the electrophile with aromatic ring to form the sigma complex. O + N+ + + O H H NO2 H NO2 NO2 3. Elimination to regain aromaticity + + H NO2 H3O+ NO2 O H H (b) 1. Generate the electrophile Br Br + Br+ FeBr3 + FeBr4− 2. The aromatic ring attacks the electrophile to generate a resonance-stabilized sigma complex. + Br+ + FeBr4 − Br H Br + H Br + H 336 REACTIONS OF AROMATIC COMPOUNDS 3. Elimination to regain aromaticity Br + Br H Br− (c) 1. Generate the electrophile + AlBr3 + Br Br + AlBr4− − AlBr3 2. Attack the electrophile with the aromatic ring to form the sigma complex. + + + H 3. Elimination to regain aromaticity + H Br − + H H REACTIONS OF AROMATIC COMPOUNDS 15.23 O H O O S O H +O O + H + + OH H OH H2O H O+ OH H O H O S H O + + H2O + H2O H O OH 337 338 REACTIONS OF AROMATIC COMPOUNDS 15.24 (a) Electrophilic aromatic substitution will take place as follows: E O EA O O + O HA + E O + O HA (b) The ring directly attached to the oxygen atom is activated toward electrophilic attack because the oxygen atom can donate an unshared electron pair to it and stabilize the intermediate arenium ion when attack occurs at the ortho or para position. 15.25 H Br + O + OH H OH − H H + OH OH2 H − Br Br H H + −HBr Br −H2O REACTIONS OF AROMATIC COMPOUNDS −H2O HA 15.26 (a) + OH OH2 rearrangement −HA + H H + H A− + HA (b) + δ+ δ+ −HA δ+ H −A + 15.27 (a) C6H5 C6H5 + H X + C6H5 δ+ δ+ C6H5 δ+ X δ+ C6H5 (b) 1,2-Addition. X − C6H5 339 340 REACTIONS OF AROMATIC COMPOUNDS (c) Yes. The carbocation given in (a) is a hybrid of secondary allylic and benzylic contributors and is therefore more stable than any other possibility; for example, C6H5 C6H5 + H + A hybrid of allylic contributors only X + C6H5 (d) Since the reaction produces only the more stable isomer—that is, the one in which the double bond is conjugated with the benzene ring—the reaction is likely to be under equilibrium control: Cl δ+ C6H5 More stable isomer δ+ C6H5 Actual product Cl + Cl − Not formed C6H5 Less stable isomer Reactions and Synthesis OCH3 Cl Cl + 15.28 (a) OCH3 + (b) Cl F Cl O F OH Cl + (c) (d) Cl Cl Cl Cl NO2 Cl (e) (f ) + Cl Cl Cl (g) + Cl REACTIONS OF AROMATIC COMPOUNDS Cl Cl O + (h) O O O O O HN HN O NO2 NO2 + 15.29 (a) O + (b) NO2 NO2 (mainly) (mainly) Cl Cl Cl O2N NO2 (c) + (d) OH OH O NO2 O O OH O NO2 (e) Br O 15.30 (a) O + Br Br O (b) Br + N H O Br (c) O O N H 341 342 REACTIONS OF AROMATIC COMPOUNDS Cl 15.31 (a) (c) (b) OH (d) (e) Br (f ) O OH (g) Cl 15.32 (a) + (b) + AlCl3 AlCl3 Cl O O (c) Cl + (Note: The use of Cl product, isopropylbenzene.) O (d) NH2NH2, KOH heat AlCl3 in a Friedel-Crafts synthesis gives mainly the rearranged O AlCl3 + Cl Zn(Hg) or NH2NH2,KOH HCl, heat reflux + (e) [from (b)] Cl2 FeCl3 dark Cl REACTIONS OF AROMATIC COMPOUNDS Cl Cl Cl2 AlCl3 + (f ) 343 hv ONa OH heat Η H (g) [from (f )] ΟΗ (1) BH3:THF Η (2) H2O2, HO− (syn addition) + NO2 NO2 (h) + HNO3 HNO3 H2SO4 H2SO4 NO2 (i) enantiomer NO2 NO2 + Br2 FeBr3 Br Br ( j) + Br2 NO2 HNO3 FeBr3 + H2SO4 ortho isomer Br Cl (k) + Cl2 FeCl3 Cl Cl SO3H SO3 + H2SO4 SO3H H2O/H2SO4, heat (separate) 344 REACTIONS OF AROMATIC COMPOUNDS Cl (l) Cl SO3 HNO3 H2SO4 H2SO4 [from (k)] Cl Cl NO2 heat SO3H SO3H (+ ortho isomer) NO2 NO2 SO3 (m) H2SO4 SO3H [from (h)] Cl 15.33 (a) C6H5 (b) C6H5 Cl2 H2 Ni pressure Cl C6H5 C6H5 OH (c) C6H5 1) OsO4, pyridine OH C6H5 2) NaHSO3/H2O O − (d) C6H5 (1) KMnO4, HO , heat C6H5 (2) H3O+ OH OH (e) C6H5 H2O H2SO4 C6H5 Br (f ) C6H5 (g) C6H5 HBr no peroxides C6H5 (1) BH3:THF − (2) H2O2, HO C6H5 H2O/H2SO4 OH NO2 REACTIONS OF AROMATIC COMPOUNDS D (1) BH3:THF (h) C6H5 (2) 345 C6H5 O OD HBr peroxides (i) C6H5 (j) C6H5 [from (i)] I Br + NaI + CN − Br (k) C6H5 Br C6H5 C6H5 CN C6H5 [from (i)] D (l) C6H5 C6H5 Ni pressure C6H5 (m) D D2 + (n) C6H5 [from (g)] C6H5 heat C6H5 H2 Ni pressure OH Na C6H5 ONa CH3I C6H5 O CH3 346 REACTIONS OF AROMATIC COMPOUNDS 15.34 O CH3 KMnO4, HO− (a) OH OH O H3O+ Cl2 FeCl3 heat Cl CH3 CH3 O (b) + AlCl3 + Cl ortho isomer O CH3 (c) CH3 CH3 HNO3 Br2 H2SO4 FeBr3 NO2 NO2 (+ ortho isomer) CH3 (d) Br OH O CH3 Br2 KMnO4, HO− FeBr3 heat Br (+ ortho isomer) H3O + Br REACTIONS OF AROMATIC COMPOUNDS CCl3 CCl3 CH3 (e) 347 Cl2 (excess) Cl2 light FeCl3 Cl CH3 CH3 Cl (f ) AlCl3 + ortho isomer + CH3 Cl (g) AlCl3 + H3C + ortho isomer CH3 CH3 O2N HNO3 (excess) H2SO4 (h) NO2 NO2 CH3 CH3 SO3 (i) CH3 H2SO4 SO3H (+ ortho isomer) OH O NO2 (1) KMnO4, HO−, heat H2O/H2SO4 heat SO3H CH3 CH3 NO2 (2) H3O+ NO2 Cl2 FeCl3 Cl Cl CH3 CH3 O ( j) NO2 HNO3 H2SO4 + Cl AlCl3 CH3 NH2NH2, KOH heat O (+ ortho isomer) 348 REACTIONS OF AROMATIC COMPOUNDS O NH2 O HN O HN Br2 (1) H3O+ FeBr3 H2O (2) HO− Cl 15.35 (a) NH2 excess Br Br (+ ortho isomer) O O O HN HN HN HN SO3H SO3 (b) O + H2SO4 [from (a)] (minor product) Br Br2 FeBr3 SO3H (major product) SO3H NH2 Br (1) H2O/H2SO4 heat (2) HO− O NH2 (c) HN O Br O HN Br Cl NH2 Br HNO3 H2SO4 excess [from (b)] O HN (d) SO3H [from (b)] H2O (2) HO− NO2 NO2 O O HN H2SO4 NO2 Br2 FeBr3 SO3H NH2 HN NO2 HNO3 Br (1) H3O+ NO2 (1) H3O+ H2O, heat (2) HO− Br Br REACTIONS OF AROMATIC COMPOUNDS NH2 349 NH2 Br2 (e) Br Br H2O Br 15.36 (a) Step (2) will fail because a Friedel-Crafts reaction will not take place on a ring that bears an NO2 group (or any meta director). NO2 O HNO3 no Friedel-Crafts reaction Cl AlCl3 H2SO4 (b) The last step will first brominate the double bond. Br Br Br2 ONa NBS light FeBr3 OH heat Br Br Br Br Br Br + + Br 15.37 This problem serves as another illustration of the use of a sulfonic acid group as a blocking group in a synthetic sequence. Here we are able to bring about nitration between two meta substituents. OH OH concd H2SO4 HO3S concd HNO3 60–65°C concd H2SO4 OH OH SO3H G OH OH NO2 HO3S NO2 H3O +, H2O heat OH OH SO3H H I 350 REACTIONS OF AROMATIC COMPOUNDS OH 15.38 OH OH H2SO4 Cl2 100 °C Fe OH Cl Cl SO3H O O AlCl3 O + Cl heat SO3H 15.39 dilute H2SO4 NH2NH2, KOH heat O Succinic anhydride Toluene HO O A (C11H12O3) SOCl2 AlCl3 Cl HO O O B (C11H14O2) C (C11H13ClO) NaBH4 H2SO4 heat O OH D (C11H12O) E (C11H14O) Br F (C11H12) 2-Methylnaphthalene NBS ONa light OH heat G (C11H11Br) Cl REACTIONS OF AROMATIC COMPOUNDS NH2NH2, KOH 15.40 (a) heat O NaBH4 (b) HO− O OH (1) CH3MgI (c) (2) NH4+ OH O (1) C6H5MgBr (d) HA heat (2) +NH4 C6H5 O OH C6H5 H2/Ni pressure C6H5 O 15.41 (a) Cl Cl B A C Br (c) H C6H5 (b) E H CH3 Br Br (d) H C6H5 F G General Problems 15.42 (a) C6H5 Br (b) C6H5 Br NaCN DMF (−NaBr) CH3ONa CH3OH (−NaBr) D C6H5 CN C6H5 OCH3 CH3 H Br 351 352 REACTIONS OF AROMATIC COMPOUNDS O (c) C6H5 O ONa Br C6H5 O O OH (−NaBr) (d) C6H5 (e) (f ) Br Br NaI acetone (−NaBr) C6H5 NaN3 I N3 acetone (−NaBr) ONa Br O OH (−NaBr) Br 15.43 A = B = CH3 CH3 NO2 HNO3 15.44 H2SO4 CH3 (only possible mononitro product) CH3 15.45 (a) Large ortho substituents prevent the two rings from becoming coplanar and prevent rotation about the single bond that connects them. If the correct substitution patterns are present, the molecule as a whole will be chiral. Thus, enantiomeric forms are possible even though the molecules do not have a chirality center. The compound with 2-NO2 , 6-CO2 H, 2′ -NO2 , 6′ -CO2 H is an example. HO O2N O NO2 O and O2N O NO2 O OH HO OH REACTIONS OF AROMATIC COMPOUNDS 353 The molecules are atropisomers (see Section 5.18). Furthermore, they are nonsuperposable mirror images and, thus, are enantiomers. OH HO O O (b) Yes and Br O Br O OH HO These molecules are enantiomeric atropisomers (Section 5.18). (c) This molecule has a plane of symmetry; hence, its mirror image forms are equivalent. Br NO2 NO2 O OH The plane of the page is a plane of symmetry. O 15.46 Cl O AlCl3 + AlCl4− + + O O − + AlCl4 + + O H O A− + HA + Cl − Cl C8H13OCl 15.47 (a and b) The tert-butyl group is easily introduced by any of the variations of the FriedelCrafts alkylation reaction, and, because of the stability of the tert-butyl cation, it is easily removed under acidic conditions. (c) In contrast to the −SO3 H group often used as a blocking group, −C(CH3 )3 activates the ring to further electrophilic substitution. 15.48 At the lower temperature, the reaction is kinetically controlled, and the usual o,p-directive effects of the −CH3 group are observed. At higher temperatures, the reaction is thermodynamically controlled. At reaction times long enough for equilibrium to be reached, the most stable isomer, m-toluenesulfonic acid, is the principal product. 354 REACTIONS OF AROMATIC COMPOUNDS 15.49 The evidence indicates that the mechanistic step in which the C H bond is broken is not rate determining. (In the case cited, it makes no difference kinetically if a C H or C D bond is broken in electrophilic aromatic substitution.) This evidence is consistent with the two-step mechanism given in Section 15.2. The step in which the aromatic compound reacts with the electrophile (NO2 + ) is the slow rate-determining step. Proton (or deuteron) loss from the arenium ion to return to an aromatic system is a rapid step and has no effect on the overall rate. C6H5 15.50 C6H5 OH C6H5 C6H5 HA + C6H5 OH2 C6H5 − :A C6H5 + C6H5 O C6H5 H +H2O C6H5 + C6H5 Very stable carbocation + OH − OH C6H5 :A− HA C6H5 −H2O C6H5 O C6H5 Br 15.51 (a) would be the most reactive in an SN 2 reaction because it is a 1◦ allylic halide. There would, therefore, be less steric hindrance to the attacking nucleophile. would be the most reactive in an SN 1 reaction because it is a 3◦ allylic halide. Br The carbocation formed in the rate-determining step, being both 3◦ and allylic, would be the most stable. (b) Challenge Problems 15.52 Resonance structures for adding the E+ to the 2 position E O E E O O+ + + Resonance structures for adding the E+ to the 3 position + E O+ O E There are more resonance structures for substitution at the 2 position of furan. The pathway for EAS should be lower in energy for substitution at the 2 position over the 3 position. REACTIONS OF AROMATIC COMPOUNDS 355 15.53 The final product is o-nitroaniline. (The reactions are given in Section 15.14A.) The presence of six signals in the 13 C NMR spectrum confirms that the substitution in the final product is ortho and not para. A final product with para substitution (i.e., p-nitroaniline) would have given only four signals in the 13 C NMR spectrum. 15.54 15.55 OH HO (Glycerol). After sodium borohydride reduction of the aldehyde, ozonolOH ysis oxidatively degrades the aromatic ring, leaving only the polyhydroxy side chain. Water is an alternative to HOAc sometimes used to work up ozonolysis reactions. O OH HO H H D (Threonic acid) OH OH Ozonolysis oxidatively degrades the aromatic rings, leaving only the carboxyl carbon as a remnant of the alkyl-substituted benzene ring. Water is an alternative to HOAc used to work up the ozonolysis reaction. QUIZ 15.1 Which of the following compounds would be most reactive toward ring bromination? O HN OH (a) CH3 O (b) (c) (d) (e) 15.2 Which of the following is not a meta-directing substituent when present on a benzene ring? (a) C6H5 (b) + NO2 (c) N(CH3)3 (d) C N (e) CO2H Cl Br2 15.3 The major product(s), C, of the reaction, Cl C , would be FeBr3 Cl Cl Cl (a) Br (b) (c) Cl Cl (d) Equal amounts of (a) and (b) Br Br (e) Equal amounts of (a) and (c) Cl 356 REACTIONS OF AROMATIC COMPOUNDS 15.4 Complete the following syntheses. CH3 B CH3 A HNO3 (a) H2SO4 SO3H D CH3 C NO2 (1) KMnO4, HO−, heat (2) H3O+ O O A OH (b) O B Cl C D Br 16 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP SOLUTIONS TO PROBLEMS O 16.1 (a) O H Pentanal O H H 2-Methylbutanal O 3-Methylbutanal O O H 2,2-Dimethylpropanal O 2-Pentanone 3-Pentanone 3-Methyl-2-butanone O O CH3 O (b) H H Acetophenone (methyl phenyl ketone) O H Phenylethanal (phenylacetaldehyde) O 2-Methylbenzaldehyde (o-tolualdehyde) H CH3 3-Methylbenzaldehyde (m-tolualdehyde) CH3 4-Methylbenzaldehyde (p-tolualdehyde) 357 358 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 16.2 (a) 1-Pentanol, because its molecules form hydrogen bonds to each other. (b) 2-Pentanol, because its molecules form hydrogen bonds to each other. (c) Pentanal, because its molecules are more polar. (d) 2-Phenylethanol, because its molecules form hydrogen bonds to each other. (e) Benzyl alcohol because its molecules form hydrogen bonds to each other. O (1) DMSO, (COCl)2, −60 °C OH 16.3 (a) O (b) H (2) Et3N O (1) LiAlH SOCl2 OH Cl ( ( O O 3 H Et2O, −78 °C (2) H2O O 16.4 (a) Br2 Br Fe Mg MgBr Et2O OH (1) H H + (2) H3O O (1) DMSO, (COCl)2, –60 ºC (2) Et3N (b) CH3 H3O + Cl OH (1) LiAlH O O Cl AlCl3 ( O Et2O, −78 °C (2) H2O KMnO4 (c) SOCl2 OH O (d) O (1) KMnO4, HO−, heat (2) H O ( 3 O H ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Br CN (e) MgBr CN − 359 diethyl ether NMgBr O H3O+ O O (f ) OCH3 (1) i-Bu2AlH, hexane, −78 °C H (2) H3O+ 16.5 (a) The nucleophile is the negatively charged carbon of the Grignard reagent acting as a carbanion. (b) The magnesium portion of the Grignard reagent acts as a Lewis acid and accepts an electron pair of the carbonyl oxygen. This acid-base interaction makes the carbonyl carbon even more positive and, therefore, even more susceptible to nucleophilic attack. δ+ δ− C O MgX C δ+ Mg O R X R δ− (c) The product that forms initially (above) is a magnesium alkoxide salt. (d) On addition of water, the organic product that forms is an alcohol. 16.6 The nucleophile is a hydride ion. H H 16.7 O + O H H H O H O − + H H OH H OH H 16.8 Acid-Catalyzed Reaction O + HA A − OH OH +H218O −H218O +OH 2 +H2O 18 OH HA A− HA H218O+ −H2O OH A− 18 OH 18 + OH A− HA 18 O 360 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Base-Catalyzed Reaction OH − H218O + OH − 18 + H2O O− O + OH H2O OH − 18 OH − 18 OH OH Acetal group OH HO OH O OH − Acetal group OH O HO O HO O + OH − OH OH O HO 18 − OH − 18 16.9 H2O 18 OH O OH − HO OH O OH O HO OH OH Sucrose Sucrose H H 16.10 H HA A− O + OH O A− CH3 CH3 −HO CH3 O CH3 + A− H OH2 + CH3 OH H O HA H OH (hemiacetal) HA +HO −H2O +H2O + O CH3 H H + O H CH3 +HO CH3 − HO CH3 O H + O CH3 CH3 A− HA O H O (acetal) CH3 CH3 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP A + OH A− O OH OH − O H OH + 16.11 H OH +O OH 361 H H A− O HA H O OH + O A− OH O + HA H −H2O H OH O + O A− O O HA O +H2O 16.12 HO OH O OH O OH OEt 16.13 (a) O O OEt O HA A OH OMgI O 2 CH3MgI NH4+ O H2O O C (b) Addition would take place at the ketone group as well as at the ester group. The product (after hydrolysis) would be OH HO H A 16.14 (a) + O O + HOR O O H R O A− O R 362 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP (b) Tetrahydropyranyl ethers are acetals; thus, they are stable in aqueous base and hydrolyze readily in aqueous acid. O O H2O −ROH HA R + O O R O + H O A O + O − HO H O OH H 5-Hydroxypentanal H Cl (c) HO O HA O Cl O Mg Et2O O O MgCl O OMgCl O O OH O + NH 4 SH + H2 O CH3CH3 SH Raney Ni CH3 H2 OH O HCN H + CH3CH3 OH HCl, H2O CN reflux OH O Lactic acid (b) A racemic form NiS S H 16.16 (a) + S BF3 + HS H S Raney Ni (b) HO S BF3 O + HS 16.15 (a) + HO H2O + NiS ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP O 16.17 (a) CH3I (1) (C6H5)3P (2) RLi − + H2C P(C6H5)3 O (b) Br + (1) (C6H5)3P P(C6H5)3 (2) RLi − O − + (c) H2C P(C6H5)3 [from part (a)] O − + (d) H2C P(C6H5)3 [from part (a)] O (e) Br + (1) (C6H5)3P (2) RLi P(C6H5)3 − O H (f) Br (1) (C6H5)3P (2) RLi − + P(C6H5)3 O − (g) Br (1) (C6H5)3P (2) RLi + P(C6H5)3 H 363 364 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP O + 16.18 O− C6H5 H P(C6H5)3 H (C6H5)3P+ C6H5 H C6H5 H H (C6H5)3P + O − (C6H5)3P O + (C6H5)3PO H + O O H H O O O 16.19 H H R O + O O H O H O H H H O O O O 16.21 O O + R R O − H O H O O 16.20 O O R R O − O O H O O R O O O H O H O O O R OH R ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Problems O 16.22 (a) Methanal H (b) H O Ethanal H O (c) C6H5 (d) (e) (f ) Phenylethanal H O Propanone O Butanone O 1-Phenylethanone or methyl phenyl ketone C6H5 O (g) C6H5 Diphenylmethanone or diphenyl ketone C6H5 O H (h) 2-Hydroxybenzaldehyde or o-hydroxybenzaldehyde OH O CH3O (i) H 4-Hydroxy-3-methoxybenzaldehyde HO (j) O 3-Pentanone O (k) 2-Methyl-3-pentanone O (l) 2,4-Dimethyl-3-pentanone 365 366 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP O (m) 5-Nonanone O (n) 4-Heptanone O (o) (E)-3-Phenylpropenal C6H5 H 16.23 (a) OH (h) O OH (i) (b) O − NH4 + + Ag C6H5 OH ( j) (c) N OH H O (d) N (k) N O− C6H5 O (l) (e) O− S (f) OH (m) S O (g) (n) + O O (o) OH CH3CH3 + NiS ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP OH 16.24 (a) (h) OH (b) (i) No reaction C6H5 OH (c) OH (j) N C6H5 N (d) No reaction (k) (e) (l) No reaction (f) (g) O OH (m) S (n) O H N S + CH3CH3 O 16.25 (a) NNHC6H5 (b) NO2 OH (d) (c) OH (e) (f) O (g) O + NiS 367 368 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP H O 16.26 (a) O O H2SO4 (cat.) O (b) H OH, HO H2SO4 (cat.) H O OH OH O (c) , H2SO4 (cat.) OH O OH O (1) H2SO4 , H2O O O O O (2) H H2SO4 (cat.) (d) O H OH O (e) O H2SO4 (cat.) O O HO OH O (f) OH H2O, H2SO4 (cat.) O O + H CH3 O (g) O O H2O, H2SO4 (cat.) −CH3OH O O OH OH O O ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP H H CH3NH2, cat. HA 16.27 (a) O N CH3 N O N H , cat. HA (b) N O NH2 cat. HA (c) O O O PPh3 (d) (1) HS SH (e) (2) Raney Ni, H2 O O OH − CH2 (f) O O (g) O O (h) (i) HO H2O, HA OH + PPh3 (excess) O H OH 369 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP (1) O H O O 16.28 (a) O O MgBr (excess) O (2) H3O+ (3) H2CrO4 HO H2SO4 (cat.) O O (b) OH NH2 (c) HA (cat.) O N OH O (d) (1) HCN H H2N (2) LiAlH4 (3) H3O+ O (e) OPh PCC 16.29 (a) or Swern oxidation HO O H OH O (b) H (1) CH3MgBr (2) H3O+ (3) PCC or Swern oxidation (4) CH3CH2MgBr (5) H3O+ O 370 O ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP (1) OsO4 O (2) HA (cat.) H O (c) O O (1) O3 (2) CH3SCH3 (d) O (3) HA (cat.) OH HO OH (e) HA (cat.) O O OH O SH (f) (g) SH , HA (cat.) S S NH2NH2, KOH heat (h) mCPBA 16.30 (a) O O H + O O S O + O HO H O H OH O O O 371 372 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP H (b) CH3O + H CH3OH H H +O O O OH OH + O H CH3O + H CH3OH H O+ H3C + O H O + O CH3OH H3C O H H O O O + + CH3OH2 O H ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 373 H N N+ N H + O H O H O H H H + H H O H N + H H N O O H H + H H O H + O NH + H N + O H H O (c) O 374 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP O + O (d) + O H O O H OH O+ H H O H OH H OH OH H + O H O+ O H OH + H O H O + OH O H H H O + (1) Swern oxidation or PCC MgBr (3) H3O+ H O HO H OH + H3O+ OH OH (2) H H O HO 16.31 H (1) Swern oxidation or PCC (2) MgBr (3) H3O+ ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Synthesis O O AlCl3 + 16.32 (a) Cl or O O AlCl3 + 2 O NH2NH2, KOH O heat (b) S HS S SH HA (Other methods are possible, e.g., NaBH4 reduction of the starting ketone followed by dehydration and hydrogenation.) Raney Ni H2 O H 16.33 (a) OH NaBH 4 O O H (b) Ag(NH3)2+ H3O+ OH NH3/H2O O O OH (c) [from (b)] SOCl2 Cl 375 376 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP OMgBr O MgBr H (d) diethyl ether + H OH H3O+ O H2CrO4 H acetone or O O Cl (1) AlCl3 + (2) H3O+ [from (c)] OH O H (e) (1) CH3MgI H (2) H3O+ + enantiomer OH O H (f ) MgBr (1) H (2) H3O+ OH (g) + enantiomer Br PBr3 [from (a)] CH3 Br LiAIH4 (h) or Zn, HOAc [from (g)] or S O HS H SH BF3 S Raney Ni (H2) CH3 or Wolff-Kishner reduction of benzaldehyde ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP OCH3 O H (i) 377 OCH3 CH3OH HA 18 O H (j) O H218O H H318O+ D O H (k) (1) NaBD4 (2) H3O H + OH + enantiomer OH O H (l) (See Problem 16.8 for the mechanism) HCN H CN + enantiomer (a cyanohydrin) O H (m) OH NH2OH N HA (an oxime) + stereoisomer H O H (n) H + H N N H3O+ N (a phenylhydrazone) O H (o) + + (C6H5)3P − (a Wittig reagent) O O 16.34 (a) + Cl AlCl3 C6H5 + stereoisomer C6H5 H N 378 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP O N (2) H3O+ + (b) Li (or ethyl magnesium bromide) OH O H (c) H2CrO4 (2) H3O+ + acetone MgBr (or ethyl lithium) O O H OH Swern oxidation or PCC 16.35 (a) O O Cl OH SOCl2 (b) LiAlH ( O ( O H 3 Et2O, −78 °C O (1) KMnO4, HO− (c) (2) H3O O H [as in (b)] OH + O (1) KMnO4, HO− (d) (2) H3O (e) (1) ( ( 2 O AlH H , −78 °C (2) H2O N (f) OH [as in (b)] + O OCH3 O (1) ( ( (2) H2O 2 O AlH , −78 °C H H ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP OH 16.36 O H2CrO4 acetone (1) CH3MgI (2) +NH OH 4/H2O A 379 HA heat B O (1) O3 (1) Ag2O, HO− H (2) HA (2) Me2S C O D O HO O E O H O H O H3O+ 16.37 + heat OH O H H OH O The compound C7 H6 O3 is 3,4-dihydroxybenzaldehyde. The reaction involves hydrolysis of the acetal of formaldehyde. O MgBr (1) Br Mg 16.38 (a) Et2O MgBr (1) (b) H (2) H3O racemic + OH Swern oxidation O OH or PCC (2) H3O+ O [from (a)] H − (c) Br (1) (C6H5)3P O + P(C6H5)3 (2) C6H5Li (a Wittig reagent) C6H5 H 380 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP O MgBr (d) + (2) H3O+ H H2CrO4 acetone HO [from (a)] racemic O O 16.39 Br OH HO H Ο , HA Br Mg Ο Et2O A O (1) O BrMg OH H O + (2) H3O , H2O O H C B (1) CH3OH O O (2) HA OH OCH3 (a hemiacetal) D (an acetal) O 16.40 (7S) O (3S) (7R) (3S) 3,7-Diethyl-9-phenylnonan-2-one (dianeackerone) stereoisomers O 16.41 OH PCC CH2Cl2 OCH3 KMnO4, HO− cold, dilute A HO OCH3 OH C OCH3 CH3OH H HA OCH3 B O H3O+ H2O HO H OH Glyceraldehyde The product would be racemic since no chiral reagents were used. ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP H OH O 16.42 NaBH4 H HO H + H (R)-3-Phenyl-2-pentanone 381 (S) H (R) (R) (R) Diastereomers 16.43 Br ( ) 7 Br + (1) 2 (C6H5)3P (2) 2 RLi − − ( ) (C6H5)3P 7 + P(C6H5)3 A O 2 ( ) 11 ( ) ( ) 11 H2 ( ) 7 11 B ( ) Pt pressure 11 ( ) 9 ( ) 11 C 16.44 (a) Retrosynthetic Analysis N OH O H HO O O OH O O + NH Br HO MgBr + O OH O O 382 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Synthesis Br H HA (cat.), Br O H HO (3) H3O+ O OH (1) Mg O (2) O OH (1) H2O, HA (cat.) (2) PCC or Swern oxidation (3) HA (cat.) O O N HO NH OH (b) Retrosynthetic Analysis O OH OH + MgBr OH O H O + MgBr ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 383 Synthesis − OH (1) H2C PPh3 (excess) (2) Swern oxidation or PCC O H O (1) CH3CH2MgBr (2) H3O+ OH OH (1) Swern oxidation or PCC (2) CH2CHMgBr (3) NH4Cl/H2O OH 16.45 O H O + OH O S O HO H O+ + OH OH OH OH O H HO OH HO H OH + O + H3O OH OH O+ O H H OH 16.46 The two nitrogen atoms of semicarbazide that are adjacent to the C O group bear partial positive charges because of resonance contributions made by the second and third structures below. H2N N O− O− O NH2 H2N H This nitrogen is the most nucleophilic. + N H NH2 H2N + N H NH2 384 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 16.47 Hydrolysis of the acetal linkage of multistriatin produces the ketodiol below. O HO 16.48 O OH O O O NaH O + (CH3O)2P AcO H O O H AcO H 16.49 Compound W is O singlet δ 3.4 CH2 C multiplet δ 7.3 IR peak near 1715 cm−1 O CH2 (1) KMnO4, HO−, heat O Compound X is (2) H3O+ O COH COH C multiplet δ 7.5 CH2 CH2 triplet δ 2.5 O Phthalic acid triplet δ 3.1 16.50 Each 1 H NMR spectrum (Figs. 16.4 and 16.5) has a five-hydrogen peak near δ 7.2, suggesting the Y and Z each has a C6 H5 group. The IR spectrum of each compound shows a strong peak near 1710 cm−1 . This absorption indicates that each compound has a C O group not adjacent to the phenyl group. We have, therefore, the following pieces, O and If we subtract the atoms of these pieces from the molecular formula, C10H12O − C7H5O (C6H5 + C C3H7 we are left with O) In the 1 H NMR spectrum of Y, we see an ethyl group [triplet, δ 1.0 (3H) and quartet, δ 2.45 (2H)] and an unsplit CH2 group signal [singlet, δ 3.7 (2H)]. This means that Y must be O 1-Phenyl-2-butanone ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 385 In the 1 H NMR spectrum of Z, we see an unsplit CH3 group signal [singlet, δ 2.1 (3H)] and two triplets at δ 2.7 and 2.9. This means Z must be O 4-Phenyl-2-butanone 16.51 That compound A forms a phenylhydrazone, gives a negative Tollens’ test, and gives an IR band near 1710 cm−1 indicates that A is a ketone. The 13 C spectrum of A contains only four signals indicating that A has a high degree of symmetry. The information from the DEPT 13 C NMR spectra enables us to conclude that A is diisobutyl ketone: O (a) (b) (c) (d ) Assignments: (a) δ 22.6 (b) δ 24.4 (c) δ 52.3 (d) δ 210.0 16.52 That the 13 C spectrum of B contains only three signals indicates that B has a highly symmetrical structure. The information from DEPT spectra indicates the presence of equivalent methyl groups (CH3 at δ 19), equivalent C groups (at δ 71), and equivalent C groups (at δ 216). These features allow only one possible structure for B: O (b) (c) (a) O Assignments: (a) δ 19 (b) δ 71 (c) δ 216 O 386 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Challenge Problems 16.53 (a) O H stretch at about 3300 cm−1 ; C O stretch at about 1710 cm−1 OH O (Intramolecular hemiacetal from C) (b) OH 16.54 H3CO H3CO E QUIZ 16.1 Which Wittig reagent could be used to synthesize ? (Assume any other needed reagents are available.) − P(C6H5)3 (a) + (d) More than one of these − P(C6H5)3 (b) + (e) None of these − (c) P(C6H5)3 + 16.2 Which compound is an acetal? OH OCH3 (a) (b) (d) More than one of these (e) None of these O O (c) O OCH3 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 387 O 16.3 Which reaction sequence could be used to convert ( ) ( ) to 4 4 ? O (a) O3 , then Me2 S, then AlCl3 , then OH O ( ) (b) (1) H2 , P-2 cat.; (2) Hg O oxidation or H2 CrO4 2 /THF-H2 O; (3) NaBH4 , HO− ; (4) PCC or Swern O OH (c) HCl, then (d) O3 , then Me2 S, then H2 SO4 , HgSO4 , H2 O, heat O OH , then H2 O2 , HO− /H2 O (e) 16.4 Complete the following syntheses. If more than one step is required for a transformation, list them as (1), (2), (3), and so on. B A (a) CH3 NBS hv N THF O C H OH A O (b) Cl Cl H B C O Cl O ClMg O O 388 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP B (c) C6H5Li A Br − C P(C6H5)3 + A O B O (d) H OH OH C NH2 N (racemic) (racemic) 16.5 An industrial synthesis of benzaldehyde makes use of toluene and molecular chlorine Cl as starting materials to produce Cl . This compound is then converted to benzaldehyde. Suggest what steps are involved in the process. 16.6 In the case of aldehydes and unsymmetrical ketones, two isomeric oximes are possible. What is the origin of this isomerism? 17 CARBOXYLIC ACIDS AND THEIR DERIVATIVES: NUCLEOPHILIC ADDITIONELIMINATION AT THE ACYL CARBON SOLUTIONS TO PROBLEMS 17.1 (a) 2-Methylbutanoic acid (b) (Z )-3-Pentenoic acid or (Z )-pent-3-enoic acid (c) Sodium 4-bromobutanoate (d) 5-Phenylpentanoic acid (e) (E)-3-Ethyl-3-pentenoic acid or (E)-3-Ethylpent-3-enoic acid 17.2 Acetic acid, in the absence of solvating molecules, exists as a dimer owing to the formation of two intermolecular hydrogen bonds: O O H O O H At temperatures much above the boiling point, the dimer dissociates into the individual molecules. O 17.3 (a) F (F is more electronegative than H−) (F is more electronegative than Cl−) OH O (b) F OH O (c) OH (F is closer to CO2 H) F 389 390 CARBOXYLIC ACIDS AND THEIR DERIVATIVES O + + (d) Me3N [Me3 N is more electronegative than H ] OH O (e) CF3 (CF3 is more electronegative than CH3 ) OH O O 17.4 (a) O (b) OCH3 O O2N O O CH3 (c) CH3O OCH3 N (d) CH3 O (e) OCH3 OCH3 (f ) CN O O O (h) O O (g) CH3 H N CH3 O O O (i) Br O O ( j) Br O O 17.5 (a) (1) KMnO4, HO−, heat (2) H3O+ OH + CO2 CARBOXYLIC ACIDS AND THEIR DERIVATIVES 391 O MgBr Br Mg (b) OMgBr H3O+ CO2 Et2O O OH O O OH (1) Cl2/NaOH (c) (2) H3O+ + CHCl3 O (1) KMnO4, HO−, heat (d) OH (2) H3O+ O (1) KMnO4, HO−, heat OH (e) OH (2) H3O+ O O (1) KMnO4, HO− H (f ) OH (2) H3O+ 17.6 These syntheses are easy to see if we work backward using a retrosynthetic analysis before writing the synthesis. (a) Retrosynthetic analysis OH MgBr Br O O + C O Synthesis Br Mg Et2O MgBr OH (1) CO2 (2) H3O+ O 392 CARBOXYLIC ACIDS AND THEIR DERIVATIVES (b) Retrosynthetic analysis O MgBr Br OH + C O O O Synthesis MgBr Br Mg (1) CO2 OH (2) H3O+ Et2O (c) Retrosynthetic analysis O MgBr OH Br + O C O O Synthesis Br MgBr Mg (1) CO2 OH (2) H3O+ Et2O (d) Retrosynthetic analysis O MgBr OH Br + O C O Synthesis O MgBr Br Mg (1) CO2 Et2O (2) H3O+ OH (e) Retrosynthetic analysis O MgBr OH O Synthesis + C Br O O Br Mg Et2O MgBr (1) CO2 (2) H3O+ OH CARBOXYLIC ACIDS AND THEIR DERIVATIVES 17.7 (a) 393 OH O Retrosynthetic analysis OH Br N O Synthesis − Br CN N OH H3O+ heat O O OH Retrosynthetic analysis O N Br OH Synthesis Br O (1) −CN (2) H3O+, ∆ OH O OH Retrosynthetic analysis O N Br OH Synthesis Br O (1) −CN (2) H3O+, heat OH (b) A nitrile synthesis. Preparation of a Grignard reagent from HO not be possible because of the presence of the acidic hydroxyl group. 17.8 Since maleic acid is a cis dicarboxylic acid, dehydration occurs readily: O O OH OH O Maleic acid 200 °C O + H2O O Maleic anhydride Br would 394 CARBOXYLIC ACIDS AND THEIR DERIVATIVES Being a trans dicarboxylic acid, fumaric acid must undergo isomerization to maleic acid first. This isomerization requires a higher temperature. O O HO O −H2O OH OH heat OH O O O O Fumaric acid 17.9 17.10 The labeled oxygen atom should appear in the carboxyl group of the acid. (Follow the reverse steps of the mechanism in Section 17.7A of the text using H2 18 O.) + O H H O H A OCH3 O HO OCH3 OCH3 O H + Intermolecular proton transfer H + H O O A O O − O + HOCH3 CH3 + O O H 17.11 (a) (1) + C6H5SO2Cl OH OSO2C6H5 A OH HO−, heat (inversion) + C6H5SO3− B O (2) + C6H5 Cl OH O C6H5 C O HO−, heat (retention) + C6H5 OH D O O− CARBOXYLIC ACIDS AND THEIR DERIVATIVES O H (3) O − Na+ Br + ( ) O 4 H (inversion) ( ) O E 4 + NaBr O H HO−, heat (retention) 395 + ( ) HO O − Na+ 4 F H (4) Br ( ) 4 H HO−, heat (inversion) + ( ) HO Br− 4 F (b) Method (3) should give a higher yield of F than method (4). Since the hydroxide ion is a strong base and since the alkyl halide is secondary, method (4) is likely to be accompanied by considerable elimination. Method (3), on the other hand, employs a weaker base, acetate ion, in the SN 2 step and is less likely to be complicated by elimination. Hydrolysis of the ester E that results should also proceed in high yield. 17.12 (a) Steric hindrance presented by the di-ortho methyl groups of methyl mesitoate prevents formation of the tetrahedral intermediate that must accompany attack at the acyl carbon. (b) Carry out hydrolysis with labeled H18 O− in labeled H2 18 O. The label should appear in the methanol. O − HO H2O O 17.13 (a) C6H5 O− + C6H5 N Η Ν O H3O+ H2O Η C6H5 OH + O − O (b) HO H2O −O NH2 O N H H3O+ H2O + HO NH3 + Ν Η 396 CARBOXYLIC ACIDS AND THEIR DERIVATIVES O O − HO H2O O HO O NH2 O + NH2 −O NH2 N (c) − C6H5 O H H3O+ C6H5 H2O O + NH3 HO + NH3 + HO C6H5 17.14 (a) OH SOCl2 NH3 Cl O NH2 O O P4O10 heat CN (b) An elimination reaction would take place because − CN is a strong base, and the substrate is a 3◦ alkyl halide. − H NC + HCN Br + − Br O CH2OH 17.15 (a) O C N CH2 C N O + H O O + 4 CH3NH2 C (b) Cl C Cl CH3N H + − NCH3 + 2 CH3NH3 + 2 Cl H O CH2 (c) C O Cl + HO− + H3NCH2CO2− O CH2 CH2CO2− C O N H + Cl − CARBOXYLIC ACIDS AND THEIR DERIVATIVES 397 O CH2CO2− C CH2 O (d) N + H2 H3NCH2CO2− Pd + H CH3 CO2 + O CH2CO2− C CH2 O (e) N + HBr CH3CO2H H H3NCH2CO2H + CH2Br CO2 + O HO−, H2O C (f ) H2N heat NH2 2 NH3 + − CO32 17.16 (a) By decarboxylation of a β-keto acid: O O O 100−150 °C + OH CO2 (b) By decarboxylation of a β-keto acid: O O O OH HO 100−150 °C OH + CO2 (c) By decarboxylation of a β-keto acid: O O O OH 100−150 °C + CO2 (d) By decarboxylation of a β-keto acid: O HO O OH 100−150 °C O OH + CO2 398 CARBOXYLIC ACIDS AND THEIR DERIVATIVES 17.17 (a) The oxygen-oxygen bond of the diacyl peroxide has a low homolytic bond dissociation energy (DH◦ ∼ 139 kJ mol−1 ). This allows the following reaction to occur at a moderate temperature. O R O O O O 2 R R 139 kJ mol−1 ∆H ° O (b) By decarboxylation of the carboxyl radical produced in part (a). O R O R + CO2 (c) Chain Initiation O Step 1 O O heat R O O 2 R R O O Step 2 R R O + CO2 Chain Propagation Step 3 R Step 4 R + R + R Steps 3, 4, 3, 4, and so on. Problems Structure and Nomenclature O O 17.18 (a) OH (b) NH2 O O (c) N (d) OH [ the (Z) isomer ] CARBOXYLIC ACIDS AND THEIR DERIVATIVES O O (e) HO OH OH (f ) OH O O OH O O (h) ( g) O O O OH O O (i) O ( j) O O (k) (l) Cl O CN O O (m) O 17.19 (a) Benzoic acid (b) Benzoyl chloride (c) Acetamide or ethanamide (d) Acetic anhydride or ethanoic anhydride (e) Benzyl benzoate (f ) Phenyl propanoate or phenyl propionate (g) Isopropyl acetate or 1-methylethyl ethanoate (h) Acetonitrile or ethanenitrile (n) O O O 399 400 CARBOXYLIC ACIDS AND THEIR DERIVATIVES 17.20 Alkyl groups are electron releasing; they help disperse the positive charge of an alkylammonium salt and thereby help to stabilize it. RNH2 + H3O + R NH3+ Stabilized by electron-releasing alkyl group + H2O Consequently, alkylamines are somewhat stronger bases than ammonia. O Amides, on the other hand, have acyl groups, R , attached to nitrogen, and acyl groups are electron withdrawing. They are especially electron withdrawing because of resonance contributions of the kind shown here, O O R NH2 R − + NH2 This kind of resonance also stabilizes the amide. The tendency of the acyl group to be electron withdrawing, however, destabilizes the conjugate acid of an amide, and reactions such as the following do not take place to an appreciable extent. O + O H3O+ + + R NH3 Destabilized by electron-withdrawing acyl group NH2 R Stabilized by resonance H2O 17.21 (a) The conjugate base of an amide is stabilized by resonance. O R O NH2 + B− NH− R O R + BH − NH This structure is especially stable because the negative charge is on oxygen. There is no resonance stabilization for the conjugate base of an amine, RNH− . CARBOXYLIC ACIDS AND THEIR DERIVATIVES 401 (b) The conjugate base of an imide is stabilized by additional resonance structures, O R O O + NH R An imide O − OH − R O O R N R + H2O N − O R − O N R R Functional Group Transformations OH O 17.22 (a) (b) OH + HCl O O (c) (d) NH2 O O O O (f ) + (e) CH3 H H3C O O (g) (h) ONa N CH3 H3C + H H O ( i) CH3 N CH3 O + H3C Cl− N+ H3C O H ( j) O H O (k) O O (l) O H Cl− N+ O H 402 CARBOXYLIC ACIDS AND THEIR DERIVATIVES + 17.23 (a) CH3 CONH2 + CH3 CO2 − NH4 (b) 2 CH3 CO2 H (c) CH3 CO2 CH2 CH2 CH3 + CH3 CO2 H (d) C6 H5 COCH3 + CH3 CO2 H + (e) CH3 CONHCH2 CH3 + CH3 CO2 − CH3 CH2 NH3 + (f) CH3 CON(CH2 CH3 )2 + CH3 CO2 − (CH3 CH2 )2 NH2 O 17.24 (a) O− O + N H4 (b) H2N OH HO O O O O OH (c) OH (d) O O O O O− (e) O + N (f) O H N O O 17.25 (a) OH + OH + OH O (b) − O O (c) + O O (d) CH3 OH + N H (e) OH OH + OH (f ) C6H5 C6H5 + + O − H2N H3N OH OH CARBOXYLIC ACIDS AND THEIR DERIVATIVES O 17.26 (a) + OH + NH4 O− + NH3 O (b) CN (c) O 17.27 (a) O OH (b) O [(Z) + (E )] O O O (d) (c) O O O O O O (e) (f) NH NH General Problems O O OH 17.28 (a) (b) OH O Cl SOCl2 OH excess O O H2SO4 (cat.) N (c) H2O, H2SO4 (cat.) O OH O (d) O H H2CrO4 OH 403 404 CARBOXYLIC ACIDS AND THEIR DERIVATIVES (e) (1) Mg° (2) CO2 Br O (3) H3O+ O (f) OH OH (1) NaOH, H2O O O + (2) H3O+ OH H (g) H2O, H2SO4 (cat.) N OH + O O O (1) H2CrO4 (2) SOCl2 OH (h) Cl O (1) H2CrO4 (2) CH3OH (excess), H2SO4 (cat.) OH 17.29 (a) O (b) OCH3 O (1) NaOH, H2O O OH (2) H3O+ OH (1) H2O, H2SO4 (cat.) (2) SOCl2 (c) O N Cl O LiAlH (d) ( ( O O 3 H Cl O O OCH3 (e) LiAlH ( O O H 3 O NH (f) ( HCl, H2O OH + + NH3 Cl− + NH3 CARBOXYLIC ACIDS AND THEIR DERIVATIVES O O Cl 17.30 (a) S SH O O NH2 (b) Cl HN N O O OH O O O O HO (c) O O HO O H2SO4 (cat.) (d) OH O O O O O AlCl3 (e) + O Br OH OH (1) KCN (2) H2O, H2SO4 (cat.) 17.31 (a) OH CN O N (b) (1) DIBAL-H (2) H3O+ O N (c) H O (1) CH3MgBr (2) H3O+ 405 406 CARBOXYLIC ACIDS AND THEIR DERIVATIVES O O H2O, H2SO4 (cat.) NH (d) + HO NH3 HO O (1) MgBr (excess) (2) H3O+ OCH3 (e) Mechanisms 17.32 See the mechanisms in Section 17.8F, where R = CH2 CH3 for propanamide. O 17.33 (a) O S H O H − O H O O O S O H O + O OH H + O OH OH OH H3C O OH HO +O HO CH3 OH O H O − O S O S H CH3 O H O H O O O H H + O H O OH O + − O OCH3 CH3 + H2O S O H O O OCH3 CARBOXYLIC ACIDS AND THEIR DERIVATIVES −O O (b) O Cl +N Cl 407 O H +N H NH H H NH2 + H2N + NH3 O (c) O S O H H O H H O H O + O O N H H O O H S H O O + O O NH2 NH2 Tautomerization N S H H N H O O+ N O − O H H H H Intermolecular OH proton transfer H +O H NH2 HO OH OH NH3 + + H OH O+ O O −O S H O O 17.34 cis-4-Hydroxycyclohexanecarboxylic acid can assume a boat conformation that permits lactone formation. O OH OH O OH O O − H2O HO Neither of the chair conformations nor the boat form of trans-4-hydroxycyclohexanecarboxylic acid places the OH group and the CO2 H group close enough together to permit lactonization. CARBOXYLIC ACIDS AND THEIR DERIVATIVES Synthesis O CH3 (1) KMnO4, HO−, heat 17.35 (a) (2) H3O+ Cl OH Cl O O OH (b) (1) SOCl2 (2) LiAlH Cl [from (a)] H ( ( O 3 HCN Cl Et2O, −78 °C OH OH CN OH H3O+, H2O O 408 heat Cl Cl CH3 Br NBS hv or ROOR (c) Cl NaCN Cl O CN OH H3O+, H2O heat Cl Cl O OH O (1) SOCl2 (d) ( ( [from (c)] HO CN HCN (2) LiAlH O Cl H Cl 3 Cl Et2O, −78 °C O HO H3O+, H2O O OH OH Cl heat Cl [(E) + (Z)] CARBOXYLIC ACIDS AND THEIR DERIVATIVES O (1) KMnO4, HO−, heat OH 17.36 (a) HO O O (1) KMnO4, HO−, heat (b) OH HO (2) H3O+ OH 409 HO (2) H3O+ OH O 17.37 (a) O (1) KMnO4, HO−, heat OH OH (2) H3O+ O (b) H3O+ (1) Mg, Et2O Br OMgBr (2) CO2 O OH CN− Br H3O+, H2O CN heat O OH (c) O (d) H OH O (1) Ag(NH3)2OH (2) H3O+ OH O O OH CH3 2 (2) H3O+ [(E) or (Z )] 17.38 O (1) KMnO4, HO−, heat O Cl + SOCl2 CH3 2 NH N CH3 410 CARBOXYLIC ACIDS AND THEIR DERIVATIVES O 17.39 O SOCl2 (1) O , AlCl3 AlCl3 O O (2) Wolff-Kishner reduction O Cl OH Br (1) C6H5MgBr HA (2) NH4Cl/H2O heat C6H5 NBS hv or ROOR OH C6H5 C6H5 ONa C6H5 H CO2H 17.40 (a) + ∆ CO2H H H CO2H (b) CO2H + CO2H CO2H ∆ HO2C H CO2H CO2H (c) + CO2H ∆ CO2H CO2H − TsCl, pyridine 17.41 (a) OH (R)-(−)-2-butanol (retention) A OTs CN (inversion) B CN (1) LiAlH4 H2SO4, H2O (retention) (+)-C CO2H (2) H2O (retention) (−)-D OH CARBOXYLIC ACIDS AND THEIR DERIVATIVES PBr3 (b) − pyridine (inversion) OH CN (inversion) Br (R)-(−)-2-butanol CN E F (1) LiAlH4 H2SO4, H2O CO2H (retention) (2) H2O (retention) OH (−)-C (+)-D O (c) O− (inversion) OTs O HO− (retention) O A + G (S)-(+)-2-butanol Mg PBr3 (d) O− OH (+)-H O diethyl ether (retention) (retention) OH Br J (−)-D (1) CO2 (2) H3O+ (retention) MgBr CO2H K L O OH OH (e) HO HCN H CN HO OH (R)-(+)-Glyceraldehyde + HO CN OH OH CN CN HO HO HO M HO OH H2SO4, H2O heat O OH HO HO OH P OH N O (f ) M 411 [O] HNO3 O OH HO OH meso-Tartaric acid HO 412 CARBOXYLIC ACIDS AND THEIR DERIVATIVES O H2SO4, H2O (g) N HO OH H 17.42 [O] OH H OH O Br2, H2O OH HO HNO3 HO OH (−)-Tartaric acid Q O O [O] OH HO heat O O PBr3 OH OH H OH Br H OH (R)-(−)Glyceric acid (R)-(+)Glyceraldehyde O OH O OH (S )-(−)-3-Bromo2-hydroxypropanoic acid O H3O+ OH NaCN CN heat OH OH (R)-(+)-Malic acid H H H 17.43 (a) H OH HO H HO O (R)-(+)Glyceraldehyde N H2SO4 H2O H OH HO HCN OH (R)-(C4H5NO3) CN + HO H CN H M H OH O OH H [cf. Problem 17.41(e)] OH N [O] HO OH HNO3 H OH HO OH O OH O H OH (−)-Tartaric acid [cf. Problem 17.41(g)] O HO OH O H OH (S)-(−)-Malic acid Zn H3O+ HO Br H PBr3 O OH O H OH CARBOXYLIC ACIDS AND THEIR DERIVATIVES OH by a reaction that proceeds with inversion (b) Replacement of either alcoholic produces the same stereoisomer. OH O H HO 2 1 O H Br PBr3 OH (inversion OH 413 H O HO OH O H at C2) OH O H HO OH OH O H Br PBr3 (inversion at C1) OH O H HO 1 2 O H OH OH (c) Two. The stereoisomer given in (b) and the one given next, below. H HO OH 2 O 1 O H H PBr3 HO OH (retention OH Br O at C2) H OH O H OH O HO OH OH O H Br PBr3 (retention at C1) OH O H HO 2 O H 1 OH OH (d) It would have made no difference because treating either isomer (or both together) with zinc and acid produces ( )-malic acid. Br H O HO Zn OH H3O+ O H OH O HO OH O H OH (−)-Malic acid H Zn HO + H3O O H Br O OH OH 414 CARBOXYLIC ACIDS AND THEIR DERIVATIVES 17.44 (a) CH3O2C C CO2CH3 . This is a Diels-Alder reaction. C (b) H2 , Pd. The disubstituted double bond is less hindered than the tetrasubstituted double bond and hence is more reactive. (c) CH2 CH CH2 . Another Diels-Alder reaction. CH (d) LiAlH4 O (e) CH3 S Cl and pyridine O (f) CH3 CH2 S− (g) OsO4 , then NaHSO3 (h) Raney Ni (i) Base. This is an aldol condensation (see Section 19.4). (j) C6 H5 Li (or C6 H5 MgBr) followed by H3 O+ (k) H3 O+ . This is an acid-catalyzed rearrangement of an allylic alcohol. O (l) CH3CCl, pyridine , (m) O3 , followed by oxidation (n) Heat Spectroscopy O O 17.45 O HO−, H2O N H reflux O NH2 Phenetidine Phenacetin An interpretation of the 1 H NMR spectral data for phenacetin is as follows: O (c) (a) O N H (e) (d) (a) triplet δ 1.4 (b) singlet δ 2.1 (c) quartet δ 3.95 (d ) doublet δ 6.8; doublet, δ 7.4 (e) broad singlet δ 9.0 (b) + O− CARBOXYLIC ACIDS AND THEIR DERIVATIVES O (c) 17.46 (a) (a) (b) O (b) (a) O O (c) Interpretation: O (a) Triplet δ 1.2 (6H) , 1740 cm−1 (ester carbonyl) (b) Singlet δ 2.5 (4H) (c) Quartet δ 4.1 (4H) O (c) O (b) O (d ) H (b) (a) (a) Interpretation: O , 1720 cm−1 (ester carbonyl) (a) Doublet δ 1.0 (6H) O (b) Multiplet δ 2.1 (1H) (c) Doublet δ 4.1 (2H) (d ) Multiplet δ 7.8 (5H) (b) (c) (a) O O (c) (d ) Interpretation: O (a) Triplet δ 1.2 (3H) (b) Singlet δ 3.5 (2H) (c) Quartet δ 4.1 (2H) (d ) Multiplet δ 7.3 (5H) (b) H (d) Cl Cl , 1740 cm−1 (ester carbonyl) O O OH (a) Interpretation: (a) Singlet δ 6.0 (b) Singlet δ 11.70 OH, 2500–2700 cm−1 O , 1705 cm−1 (carboxylic acid carbonyl) OH 415 416 CARBOXYLIC ACIDS AND THEIR DERIVATIVES Cl O (e) (b) HH (b) O (c) (a) Interpretation: O (a) Triplet δ 1.3 , 1745 cm−1 (ester carbonyl) (b) Singlet δ 4.0 (c) Quartet δ 4.2 O 17.47 That compound X does not dissolve in aqueous sodium bicarbonate indicates that X is not a carboxylic acid. That X has an IR absorption peak at 1740 cm−1 indicates the presence of a carbonyl group, probably that of an ester (Figure 17.2). That the molecular formula of X (C7 H12 O4 ) contains four oxygen atoms suggests that X is a diester. The 13 C spectrum shows only four signals indicating a high degree of symmetry for X. The single signal at δ 166.7 is that of an ester carbonyl carbon, indicating that both ester groups of X are equivalent. Putting these observations together with the information gathered from DEPT 13 C spectra and the molecular formula leads us to the conclusion that X is diethyl malonate. The assignments are (c) (a) O O (d) (a) δ 14.2 O (b) (c) (d) O (a) (b) δ 41.6 (c) δ 61.3 (d) δ 166.7 17.48 The very low hydrogen content of the molecular formula of Y (C8 H4 O3 ) indicates that Y is highly unsaturated. That Y dissolves slowly in warm aqueous NaHCO3 suggests that Y is a carboxylic acid anhydride that hydrolyzes and dissolves because it forms a carboxylate salt: O O R R O NaHCO3 H2O, heat O− Na+ R + O O (insoluble) R O− Na+ (soluble) The infrared absorption peaks at 1779 and 1854 cm−1 are consistent with those of an aromatic carboxylic anhydride (Figure 17.2). That only four signals appear in the 13 C spectrum of Y indicates a high degree of symmetry for Y. Three of the signals occur in the aromatic region (δ 120-140) and one signal is downfield (δ 162) CARBOXYLIC ACIDS AND THEIR DERIVATIVES 417 These signals and the information from the DEPT 13 C NMR spectra lead us to conclude that Y is phthalic anhydride. The assignments are (a) (c) (b) O (d) O O O (a) δ 125 (b) δ 130 O (c) δ 136 (d) δ 162 Y OH O OH OH O AA O Z Z is phthalic acid and AA is ethyl hydrogen phthalate. Challenge Problems 17.49 (a) Ethyl acetate (b) Acetic anhydride (c) N-Ethylacetamide. 17.50 In the first instance, nucleophilic attack by the amine occurs preferentially at the less hindered carbon of the formyl group. (Recall that aldehydes are more reactive than ketones toward O is a better leaving group nucleophiles for the same reason.) In the second case, F O− O F F since the former is the conjugate base of the stronger acid. than H O− H H O O CH3 OH O Cl O N H 17.51 CH3NH2 COCl2 O 17.52 C6H6 + O Cl AlCl3 Clemmensen reduction Cl O CN KCN H H HCl, ZnCl2 O CN NaNH2 H3O+ CH3I heat OH 418 CARBOXYLIC ACIDS AND THEIR DERIVATIVES O O 17.53 A = C = ΟΗ B = CN D = (racemic) In the last step, HI/red P accomplishes both the reduction of of the nitrile function. OH to H and the hydrolysis 17.54 (a) The signal at δ 193.8 is consistent with the carbonyl carbon of an aldehyde and shows that the PCC reaction produced cinnamaldehyde. (b) The signal at δ 164.5 is consistent with the carbonyl carbon of a carboxylic acid, and suggests that the oxidation with K2 Cr2 O7 in sulfuric acid produced cinnamic acid. QUIZ 17.1 Which of the following would be the strongest acid? (a) Benzoic acid (b) 4-Nitrobenzoic acid (d) 4-Methoxybenzoic acid (e) 4-Ethylbenzoic acid (c) 4-Methylbenzoic acid 17.2 Which of the following would yield (S )-2-butanol? O (a) (R)-2-Bromobutane (b) (R)-2-Bromobutane (c) (S)-2-Butyl acetate O− Na+ + product HO−, H2O heat HO−, H2O heat HO−, H2O heat (d) All of the above (e) None of the above 17.3 Which reagent would serve as the basis for a simple chemical test to distinguish between hexanoic acid and hexanamide? (a) Cold dilute NaOH (b) Cold dilute NaHCO3 (c) Cold concd H2 SO4 (d) More than one of these (e) None of these CARBOXYLIC ACIDS AND THEIR DERIVATIVES 419 17.4 Give an acceptable name for: OCH3 O Cl NHCH3 O OH A NO2 B C 17.5 Complete the following synthesis. B A O (a) CH3 OH C O (CH3)2NH + Cl (CH3)2NH2+ Cl − D E O O + C6H5 OH O OH 2 F O NH2 + C6H5NH3+ C6H5 A O (b) + CH3 18 + HCl OH Cl B C + NaOH, H2O heat O− 420 CARBOXYLIC ACIDS AND THEIR DERIVATIVES A O (c) Cl NH3(excess) + NH4+ Cl− P4O10, heat C B (1) DIBAL-H hexane, −78 °C (2) H2O 17.6 Which of these acids would undergo decarboxylation most readily? O (a) O OH (b) O (c) O OH O OH OH (d) O O (e) OH 18 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS: ENOLS AND ENOLATES SOLUTIONS TO PROBLEMS OH O 18.1 Cyclohexa-2,4-dien-1-one (keto form) Phenol (enol form) The enol form is aromatic, and it is therefore stabilized by the resonance energy of the benzene ring. O does not have a hydrogen attached to its α-carbon atom 18.2 No. (which is a chirality center) and thus enol formation involving the chirality center is not O possible. With the chirality center is a β carbon and thus enol formation does not affect it. 421 422 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS 18.3 EtO− H O EtO H O− H O + EtO− H cis-Decalone O H H Protonation on the bottom face H trans-Decalone O H H A large group is axial in cis-decalone. H No bulky groups are axial in trans-decalone. 18.4 The reaction is said to be “base promoted” because base is consumed as the reaction takes place. A catalyst is, by definition, not consumed. 18.5 (a) The slow step in base-catalyzed racemization is the same as that in base-promoted halogenation—the formation of an enolate anion. (Formation of an enolate anion from 2-methyl-1-phenylbutan-1-one leads to racemization because the enolate anion is achiral. When it accepts a proton, it yields a racemic form.) The slow step in acid-catalyzed racemization is also the same as that in acid-catalyzed halogenation—the formation of an enol. (The enol, like the enolate anion, is achiral and tautomerizes to yield a racemic form of the ketone.) (b) According to the mechanism given, the slow step for acid-catalyzed iodination (formation of the enol) is the same as that for acid-catalyzed bromination. Thus, we would expect both reactions to occur at the same rate. (c) Again, the slow step for both reactions (formation of the enolate anion) is the same, and consequently, both reactions take place at the same rate. REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS LDA, THF −78 °C O 18.6 (a) O CH3I CH3 O I O−Li+ O O O (b) OCH3 423 O O− Li+ O− Li+ 2 eq. LDA, THF Br OCH3 (CH3)3Si O O O H3O+ OCH3 (CH3)3Si OCH3 (CH3)3Si 18.7 Working backward, (a) O O O heat −CO2 O O OEt (1) OEt O ONa O Br (2) C6H5 O Br (1) O (2) H3O+ OK (2) (b) (1) dil. NaOH, heat OH O O heat − CO2 OEt O (1) HO−, heat OH (2) H3O+ C6H5 O O (1) OEt C6H5 O ONa (2) C6H5 Br O− Li+ O OEt 424 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS 18.8 (a) Reactivity is the same as with any SN 2 reaction. With primary halides substitution is highly favored, with secondary halides elimination competes with substitution, and with tertiary halides elimination is the exclusive course of reaction. (b) Acetoacetic ester and 2-methylpropene (isobutylene) (c) Bromobenzene is unreactive to nucleophilic substitution because the halogen is bonded to an sp2 carbon. 18.9 The carboxyl group that is lost more readily is the one that is β to the keto group (cf. Section 17.10 of the text). 18.10 Working backward O O O (1) HO−, H2O, heat heat −CO2 C6H5 OH (2) H3O+ O O C6H5 O O O (1) O ONa OEt O OEt O Br (2) C6H5 C6H5 18.11 Working backward O O O C6H5 O heat −CO2 O C6H5 O O (1) NaH OEt (2) C6H5 O (2) H3O+ OH C6H5 O (1) HO−, H2O, heat O OEt Cl O 18.12 Working backward, Ο O (a) OH (1) HO−, H2O, heat ΟΗ ΟΗ heat −CO2 (2) H3O+ Ο EtO Ο Ο OEt (1) OEt (2) O ONa Br O EtO REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS HO O O heat (b) −CO2 OH (1) HO−, H2O, heat (2) H3O+ O HO EtO EtO O O MeI O O OK EtO EtO EtO (1) O ONa Br (2) O EtO O (c) HO heat −CO2 OH (1) HO−, H2O, heat O (2) H3O+ O HO EtO EtO O (1) O ONa (2) O Br EtO O EtO CN 18.13 2 Br OEt + base −2 HBr O HO−, heat −CO2, −EtOH CN OEt O CN (1) HO−, heat − NH3 (2) H3O+ O OH Valproic acid 425 426 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS 18.14 These syntheses are easier to see if we work backward. O + O N Cl− O O H2O Cl (a) 3N O H N N HA, −H2O O (b) Br− + Η2Ο N O Br H N N HA, −H2O O O O (1) Br (c) N (2) H2O H N HA, −H2O O O OEt O (d) O (1) Br N OEt (2) H2O H N HA, −H2O O REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O 18.15 (a) H C 3 O O CH2 H3C O H O CH3 O O O H3C CH3 H H 1 2 4 3 O O O O O 427 (b) H H H H H 4 1 OCH3 O H H 3 O 2 H H (c) H H H H O O 4 3 H3C CH3 H3C N 1 2 18.16 Abstraction of an α hydrogen at the ring junction yields an enolate anion that can then accept a proton to form either trans-1-decalone or cis-1-decalone. Since trans-1-decalone is more stable, it predominates at equilibrium. H O− O O A − − (−H + ) H H H H O HA H O + H H (95%) trans-1-Decalone (more stable) (5%) cis-1-Decalone (less stable) 18.17 In a polar solvent, such as water, the keto form is stabilized by solvation. When the interaction with the solvent becomes minimal, the enol form achieves stability by internal hydrogen bonding. (1) LDA (2) I O 18.18 (a) O O O 428 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O O O (b) Br2, Br OH NaH (c) Cl N N Br Br (d) O O Br2 (excess), NaOH H (e) O (2) Cl O O (1) LDA O H O O (3) H2O O O OH (1) I2, NaOH (f ) (2) H3O+ O H O O +O H H O − OH O 18.19 (a) Br Br − Br Br HBr + H O+ O Br REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O O H − O − (b) I OH O 429 I H I − OH − O I I H − I OH I − O I I I I O − O I I I I I I OH − OH O O O− Na+ H + HCI3 O − CI3 I 430 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS H O+ H O H (c) H3C O O H3C O H3C O + H O O O+ H H H O O H O + H3C H O H + CH3OH O H O H +O O H O O H H3C O O H H H O + H H H REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS H3C O O (d) S O H 3C OH +O O O H O O O H +O H CH3 +O O O +O CH3 H O H O + O O OCH3 + CH3OH2 OCH3 OCH3 O H OH CH3 H O OH H +O CH3 H O H 3C O H O+ OH H H O CH3 H 3C H OCH3 OH H O H H O O + 431 O − OH 18.20 (a) H − OH + − OH 432 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS DO − D O (b) DO OD − D O H H D − OD O O − D DO O O D D D D D D H DO DO − H O D D D O D OD D D H D − DO O O D D D − OD H D D D D D − D D D − O D D D − DO − D D O O D D D D D D Repeat last two steps × 2 D D D D D D D D OD REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O (c) O O H − OH H − OH H N O − H OH HO (d) O O Li + H − I − CH3 I O SUMMARY OF ACETOACETIC ESTER AND MALONIC ESTER SYNTHESES A. Acetoacetic Ester Synthesis O (1) OEt ONa O O (2) RX (1) OEt (2) R'X OH heat −CO2 OK R H O O (1) HO −, H2O OEt R R' O O (2) H3O+ R R' CH3 O + O 433 O R' R 434 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS B. Malonic Ester Synthesis O O (1) (2) RX OEt EtO O O ONa EtO OK (1) OEt (2) R'X OH heat −CO2 R H O O EtO (2) H3O+ OEt O O (1) HO−, H2O HO R R' O R' HO R R' R Acetoacetic Ester and Malonic Ester Syntheses O O O 18.21 (a) O O OEt (2) MeI O O OH Br (1) ONa (2) Br O PBr3 (1) HO−, H2O, heat OEt (2) H3O+ O OH O (c) O heat −CO2 (1) OEt O (2) H2O O OEt O OH (1) Mg (2) H3O+ O (b) O O O O O O OEt (1) LiAlH4 OEt HA O O OK (1) OEt (2) MeI HO O ONa (1) (2) O ONa O O OEt Br O O O OH heat −CO2 O O (1) HO−, H2O, heat (2) H3O+ REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O O (d) O ONa (1) OEt O (1) HO−, H2O, heat (2) H3O+ OEt O (2) Br 435 OEt OEt O O O OH O OH heat OH −CO2 NaBH4 OH OH O O O O O (e) O ONa (1) OEt (2) O O OH NaBH4 OH C6H5 O O OEt (1) ONa OEt (1) OEt (2) Br OEt (2) MeI (a) O OEt OK OEt O O O HO−, H2O, heat O+ (2) H3 O OH heat OH −CO2 OH O O (1) LiAlH4 OH (b) (2) H3O+ OH [from Problem 18.12 (c)] O (c) OH C6H5 O (1) OH (1) LiAlH4 + OEt (2) H3O Br (f ) Continuing from problem 18.11 18.22 O OEt (1) LiAlH4 OEt (2) H3O+ O [from (a) above] OH OH 436 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O O OEt (1) (d) (2) Br aq, HCl HO heat OH OEt OEt O O O OEt EtO O OEt O ONa OH O O O heat − CO2 OH HO (1) LiAlH4 OH HO (2) H3O+ O O O (1) NaOCH3, CH3OH (2) Br 18.23 (a) (3) HCl, H2O, heat O O CH3 O O O OH (b) heat O HO O HO 18.24 The following reaction took place: O O O OEt + Br ONa Br EtO Br O EtO ONa O− EtO O Br OH O (1) HO−, H2O, heat O Perkin’s ester HO O + (2) H3O O Perkin’s acid REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O OEt 18.25 (a) Br ONa + + Br OEt O O O OEt Br OK OEt Br − OEt OH O O O O OEt (1) HO−, H2O, heat OEt (2) H3O+ (b) 2 EtO O OEt − Na+ + Br Br O O EtO OEt EtO OEt O O A O O (1) 2 OEt EtO ONa EtO OEt O OO ONa Br (2) Br2 EtO OH (3) heat, −CO2 O O O HO OEt (1) HO−, H2O OH OO heat −2 CO2 (2) H3O+ O O OEt EtO B OEt O O OH HO C 437 438 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS HO O O OH + O OH O O OH O OH E Meso compound HO D Racemic form O Na+ (c) Br O OEt OEt − OEt O Br Br OEt O O O OEt OK OEt OH (1) HO−, H2O (2) H3O+ (3) heat, −CO2 O O O O 18.26 (a) O O (1) O ONa (2) Cl O O Cl (1) NaOH + (2) H3O+ or H3O O O OH (ester hydrolysis under either basic or acidic conditions) O heat HO OH REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O O O (b) O (1) O O O ONa (2) Cl 439 O Cl (1) NaOH + (2) H3O+ or H3O O LiAlH4 OH O (c) O heat OH Et2O O O (1) O HO O ONa O (2) Br (1) NaOH + (2) H3O+ or H3O O O heat O OH O OH 440 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O (d) O O O (1) O O O ONa (2) Cl O Cl (1) NaOH + (2) H3O+ or H3O O O OCH3 MeOH, cat.H2SO4 O heat OH HO (1) LDA (2) CH3l O CH3 O O O O (e) (1) O O ONa O (2) Br (1) (2) Br O O O OH heat O ONa Ph (1) NaOH + (2) H3O+ or H3O O O O OH REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O O O (f) (1) ONa (2) Br O 441 O O (1) ONa (2) BrCH3 O O O heat O (1) NaOH + (2) H3O+ or H3O OH O O General Problems O O O 18.27 (a) O (1) (2) ONa Br (3) H3O+, heat (1) O (b) O O ONa (2) O (3) O Br OH ONa O (4) O (5) (1) (c) O O O H3O+, OH ONa O (2) Br (3) ONa Br (4) O heat (5) H3O+, heat (1) LDA (2) CH3l O (d) O O O (1) Br (e) Br O O CH3 O NaOCH3 (excess) (2) H3O+, heat O 442 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS (1) HA (cat.) (2) O (f) + NH O Br (3) H3O+ (1) LDA (2) O O O O Cl (g) (3) NaOCH3 (2) CH3Br O Na+ 18.28 + HBr − O OEt O Br OEt F O G OH O − (1) (1) dil. NaOH + (2) H3O (3) heat (2) H3O Li + + I H H2 linalool Lindlar's catalyst O 18.29 EtO EtO Na+ EtO − ONa Br EtO EtO EtO H O O O Br O O (C10H16O4) Br (C10H17BrO4) O OEt OEt (1) LiAlH4 (2) H2O OH OH (C6H12O2) HBr Br Br O 2 ONa O OEt OEt O REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS O O − OH (1) HO , H2O (2) H3O + heat OH + CO2 OH J O (C9H12O4) O + Br 18.30 (a) CH3 443 (C8H12O2) O (2) RLi (C6H5)3P CH3 P(C6H5)3 (b) Hydrolysis of the ether yields a hemiacetal that then goes on to form an aldehyde. + OCH3 A− CH3O O HA [also (Z)] CH3 CH3O OCH3 +H2O + −H2O OH2 CH3O OCH3 A− OH HA CH3O (hemiacetal) O −CH3OH +CH3OH H CH3O O O (c) + CH3 OCH3 P(C6H5)3 A− + OH2 + OCH3 +H2O OCH3 −H2O OH OCH3 +CH3OH A− HA O −CH3OH HA H 444 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS 18.31 (a) The hydrogen atom that is added to the aldehyde carbon atom in the reduction must come from the other aldehyde rather than from the solvent. It must be transferred as a hydride ion and directly from molecule to molecule, since if it were ever a free species it would react immediately with the solvent. A possible mechanism is the following: O− O H + OH − − O O− Then O O− H H + − OH O− H + O− O O H H O− H + H DO− D2O OD H H (b) Although an aldol reaction occurs initially, the aldol reaction is reversible. The Cannizzaro reaction, though slower, is irreversible. Eventually, all the product is in the form of the alcohol and the carboxylate ion. O O HO LiAlH4 18.32 TsO TsCl base HO A base B O R OOH Lewis Acid O C O D O O O Multistriatin REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS 445 Spectroscopy 18.33 (a) Compound U is ethyl phenyl ketone. (b) Compound V is benzyl methyl ketone: δ 2.0 O O 1690 cm−1 δ 3.5 δ 3.0, δ 1.2 δ 7.7 1705 cm−1 δ 7.1 U V OCH3 O δ 3.2 18.34 A is OCH3 δ 2.1 δ 4.7 δ 2.6 I2 OCH3 O CHI3 ↓ NaOH OCH3 Ag(NH3)2+ OH − no reaction H3O+ O O Ag(NH3)2+ OH − H O O Ag ↓ + O− Challenge Problem 18.35 TfO Si Si O OH H NR3 + O O H H Proton removal and hydride migration Si O + O H + R3NH H 446 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS QUIZ 18.1 What would be the major product of the following reaction? O + Br2 + HO− O O Br O Br (a) Br OH (c) (b) Br (d) (e) None of these 18.2 Supply formulas for the missing reagents and intermediates in the following synthesis. A O HA + B N (−H2O) C O H2O + + Br− N H H 18.3 What alkyl halide would be used in the malonic ester synthesis of 4-methylpentanoic acid? (a) Br (b) (c) Br (d) (e) Br Br Br REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS 447 18.4 Which is the preferred base for the alkylation of a monoalkylacetoacetic ester? (a) NaOH (b) KOH (c) OK (e) OK ONa (d) 18.5 To which of these can the haloform reaction be applied for the synthesis of a carboxylic acid? O O (b) (a) O O (c) (d) O (e) 19 CONDENSATION AND CONJUGATE ADDITION REACTIONS OF CARBONYL COMPOUNDS: MORE CHEMISTRY OF ENOLATES SOLUTIONS TO PROBLEMS O O 19.1 (a) Step 1 − OEt + O OEt + HO − H O − OEt O O O Step 2 + OEt − − OEt O OEt EtO O O − + O OEt − O O O Step 3 OEt + − O OEt H (b) O O OH OEt 448 O O + OEt + OH CONDENSATION AND CONJUGATE ADDITION REACTIONS O O OEt O ONa (1) 19.2 2 449 (2) H3O+ OEt O O O (1) NaOH, H2O, heat heat −CO2 OH (2) H3O+ O O OEt 19.3 (a) (b) To undergo a Dieckmann condensation, diethyl 1,5-pentanedioate would have to form a highly strained four-membered ring. O O 19.4 C6H5 OEt O− + O OEt O O− + EtO O O− + OEt OH + O OEt EtO EtO C6H5 OEt O EtO OEt − O C6H5 − C6H5 O EtO OEt − C6H5 Resonance stabilized C6H5 O H3O+ OH + O EtO OEt C6H5 OEt + 19.5 (a) O O O EtO OEt (1) (2) H3O O ONa OEt OEt + O O O O (1) + (b) OEt H OEt O ONa (2) H3O+ H O OEt 450 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O O 19.6 (a) + H ONa (1) OEt O (2) H3O H + O O O OEt O ONa (1) (b) (2) H3O + O O O + (c) EtO OEt EtO OEt ONa (1) (2) H3O + O OEt O O O ONa (1) OEt O EtO + (2) H3O O O 19.7 + OH + + HCl OH + OEt O O HO OEt O Cl− OH + OH + + OH2 O O + Intermolecular proton transfer H HA 2,6-Dimethylhepta-2,5dien-4-one 19.8 Drawing the molecules as they will appear in the final product helps to visualize the necessary steps: CH3 H3C CH3 Mesitylene CONDENSATION AND CONJUGATE ADDITION REACTIONS 451 The two molecules that lead to mesitylene are shown as follows: CH3 O H3C (4-Methylpent-3-en-2-one is formed by an acid-catalyzed condensation between two molecules of acetone as shown in the text.) O CH3 H3 C CH3 The mechanism is the following. CH3 CH3 O + +O HA CH3 CH3 H CH3 + O CH3 CH3 HO HA A − HO HA H3 C CH3 CH3 H CH3 CH3 CH3 + O + HO H3C CH3 − A HO H CH3 HO CH3 HO CH3 H3C + A CH3 CH3 CH3 + −H2O HO CH3 H CH3 + HO − H2O H CH3 H3C CH3 CH3 CH3 −H2O HA + A− H3C CH3 CH3 H3C CH3 HA 452 CONDENSATION AND CONJUGATE ADDITION REACTIONS H 19.9 (a) H HO − O O H H O − H O H O H2O O O− H H OH O OH HO − O O (b) To form , a hydroxide ion would have to remove a β proton H OH in the first step. This does not happen because the anion that would be produced, H , cannot be stabilized by resonance. O O (c) H [both (E) and (Z )] OH O 19.10 (a) 2 O HO− H2O H H O (b) Product of (a) HA H (1) LiAlH4, Et2O (2) H2O OH (c) Product of (b) (d) Product of (a) OH H2 Pt pressure NaBH4 OH OH CONDENSATION AND CONJUGATE ADDITION REACTIONS O O OH H H PCC CH2Cl2 19.11 HO− C11H14O O O H H H2 Pd-C C14H18O Lily aldehyde (C14H20O) 19.12 Three successive aldol additions occur. First Aldol Addition O O HO− + H O H − + H O H + H2O O − −O H H O O −O H Second Aldol Addition HO + H2O H O H + HO − HO + HO− HO H2O + HO− O H − + H − O H + O O H HO HO H O− O HO O H O− + H2O HO H OH 453 454 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O Third Aldol HO Addition H − HO HO− + OH O H H H H + O− + HO BF3 + HO− H HO H2O OH H + O OH O− HO O O 19.13 (a) 2 H HO O HO H2O O − OH O HO + OH HO + H O OBF3− + OH O O + + H + O OH Pseudoionone O O H + O O α-Ionone OH + O + OH β-Ionone (b) In β-ionone both double bonds and the carbonyl group are conjugated; thus it is more stable. (c) β-Ionone, because it is a fully conjugated unsaturated system. CONDENSATION AND CONJUGATE ADDITION REACTIONS 19.14 (a) (b) − N C O− N − − N N N C O O H + OH − + OH N OH N N heat −H2O + (Z) O O− Li+ (1) LDA 19.15 (a) − − N O 455 O OH H (2) H2O Kinetic enolate O O− Li+ O (1) C6H5 OH O H LDA (b) (2) H2O C6H5 Kinetic enolate O− Li+ (1) O (c) O O H LDA (2) H2O Kinetic enolate OH 456 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O H 19.16 HO− O− H − O O O OH −H2O H2O O O O O H − − HO O O O− O OH O H H2O Ο 19.17 (a) H O H −H2O H H O (b) (c) O O O O Notice that starting compounds are drawn so as to indicate which atoms are involved in the cyclization reaction. 19.18 It is necessary for conditions to favor the intramolecular reaction rather than the intermolecular one. One way to create these conditions is to use very dilute solutions when we carry out the reaction. When the concentration of the compound to be cyclized is very low (i.e., when we use what we call a “high dilution technique”), the probability is greater that one end of a molecule will react with the other end of that same molecule rather than with a different molecule. CONDENSATION AND CONJUGATE ADDITION REACTIONS O O HO− 19.19 (a) HA C6H5 C6H5 − C6H5 − + C6H5 C6H5 O− O O O C6H5 C6H5 C6H5 HO− HA C6H5 O O C6H5 H H − A− (b) C6H5 HA O − + C6H5 O− C6H5 C6H5 C6H5 A− HA O C6H5 C6H5 O O 19.20 H2N NH2 + conjugate H addition O− H N NH2 N H H N + H H OH N N H −H2O H N N H 457 458 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O 19.21 (a) H − HO− + O O O O− O O HA − + O A− O O O O O A− HA O O O − O O− O O HA −H2O A− O O OH (b) 2-Methylcyclohexane-1,3-dione is more acidic because its enolate ion is stabilized by an additional resonance structure. O O− O A− O − H O O H H O O O− O A− H − H O− CONDENSATION AND CONJUGATE ADDITION REACTIONS 459 Me H 19.22 (a) O + CH2 Me2NH + HA H + H2O + A − N Me − A H Me O O + CH2 O N + Me NMe2 + HA H (b) O + H + HA N N+ CH2 H H2O + A − + − A H O O CH2 N + Ο Ν Me H (c) O CH2 + Me2NH + HA + + H2O + A − N H Me + Η O Η O Me + CH2 −: A Η N+ NMe2 Me CH3 CH3 OH OH NMe2 CH3 repetition of similar steps NMe2 + HA Me2N CH3 + HA 460 CONDENSATION AND CONJUGATE ADDITION REACTIONS Problems Claisen Condensation Reactions O 19.23 (a) O ONa O O OH O O (b) O + O O ONa O O OH O O O (c) O ONa + O O OH O (d) O O + O O ONa O O OH O O O + (e) O O O ONa OH O O O (f) + ONa OH Cl O O O O CONDENSATION AND CONJUGATE ADDITION REACTIONS 19.24 Working backward in each case: O O O heat (a) (1) HO−, H2O, heat OH (2) H3O+ − CO2 O O O OEt ONa (1) (2) H3O O (b) O HA OEt OEt 2 + OH OH O O O heat −CO2 O OH (1) HO−, H2O, heat O (2) H3O+ OH O O OEt O (1) EtO OEt O O O , ONa OEt (2) H3O+ OEt O O H (c) (1) H OEt C6H5 O (2) H3O+ OEt, ONa OEt C6H5 O 461 462 CONDENSATION AND CONJUGATE ADDITION REACTIONS 19.25 (a) O O (b) O (c) H OCH3 O CO2Et O O O O Cl O CH3 + O + CO2Et O + O H or (or other pairings of reactants) O O + Cl OCH3 O (or other pairings of reactants) O 19.26 O O O ONa, OH O O O (major) O + O O (minor) (The minor product has no alpha hydrogen, so a retro-Dieckmann reaction takes place, favoring the starting material instead.) CONDENSATION AND CONJUGATE ADDITION REACTIONS 463 Claisen and Dieckmann condensations are reversible. The driving force is the deprotonation of the alpha hydrogen in the product between the two carbonyls. This deprotonation prevents the reaction from reversing, pulling the product out of the equilbrium. − O O O O O H − O O Molecule can not undergo retroDieckmann in deprotonated form. 19.27 Intramolecular cyclization (which would give a product of formula C6 H8 O3 ) is not favored because of ring strain. The formula of the product actually obtained suggests a 1:1 intermolecular reaction: O OEt O 2 EtO O ONa OEt OEt EtO O O O O ONa O OEt EtO O O C12H16O6 19.28 The synthesis actually involves two sequential Claisen condensations, with ethyl acetate serving as the source of the carbanionic species. O O OEt OEt + O OEt O NaH OEt OEt O not isolated O O NaH O 464 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O − 19.29 OEt O − O OH OEt HO R R O O O + − O O R + OEt − OH OEt R HO− O + R EtOH O− 19.30 A gamma hydrogen is removed by base (as is an alpha hydrogen in the usual Claisen) to give a resonance-stabilized species: O − O OEt − OEt Ethyl crotonate differs from ethyl acetate by −CH CH−, a vinyl group. The transmission of the stabilizing effect of the −COOC2 H5 group is an example of vinylogous reactivity (reactivity through conjugation with a double bond). − O O EtO H − OEt + EtOH OEt O OEt EtO O O OEt EtO − O O OEt EtO − O H O OEt EtO O OEt EtO O O O− O OEt EtO O O + EtOH CONDENSATION AND CONJUGATE ADDITION REACTIONS O O 19.31 − OEt + − 465 OEt O O OEt OEt O − O OEt OEt O OEt H O − OEt O OEt OEt O O − O O O OEt OEt The two-stage reaction involves a reverse Dieckmann condensation followed by a forward Dieckmann using a different enolate. The original enolate cannot lead back to the starting material since the crucial α-hydrogen is lacking. O 19.32 EtO O O H O OEt OEt OEt ONa H NH2 H2N ONa O HN O N H O thymine Aldol Reactions NaOH H 19.33 (a) 2 H2O O H (b) H + H O OH O NaOH H H2O O H OH O O O (c) O H + NaOH H H2O H − 466 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O (d) H + H NaOH H2O O O O (e) O NaOH H + H2O O O H OH H 19.34 OH OH H H OH O 1 2 3 O O dil HO− + O O H (b) 4 O H 19.35 (a) O dil HO− + O O (c) H + NO2 dil HO− NO2 CONDENSATION AND CONJUGATE ADDITION REACTIONS dil HO− (d) 467 (1) LiAlH4, Et2O (2) H2O O O O OH O CN H (e) + base CH3CN CH3O CH3O O O H dil HO− 5° C (f) H heat OH O H OH (1) LiAlH4, Et2O (2) H2O O O O dil HO− + (g) H HS S S Raney Ni SH BF3 H2 O O O 19.36 (a) O (c) H + (b) O + H O O H (d) H O H 468 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O (e) H H + N O (f) O + O O H (g) (h) O O H (1) O3 19.37 (a) base (2) Me2S H (aldol condensation) O O H O (1) O3 (b) O (2) Me2S (2) NaHSO3 H (aldol condensation) OH (1) OsO4 (c) base O O (Section 22.6D) OH O H base (aldol condensation) HIO4 H + O H O CONDENSATION AND CONJUGATE ADDITION REACTIONS 469 + N O base + (d) N N (aldol condensation) O − O N 19.38 (a) The conjugate base is a hybrid of the following structures (cf. Rev. Prob. 19.30): O − O− O H H H − This structure is especially stable because the negative charge is on the oxygen atom. O (b) O A− H − H HA O O H O− − + H O OΗ H O HA H A− O − H2O H 470 CONDENSATION AND CONJUGATE ADDITION REACTIONS O O + 19.39 H (1) OEt (2) O ONa O O H OH O O O OH O H O O− 2 H O aldol −H2O O O O O O O HO H H O − O H O O O O O −Η2O OH O O 19.40 This difference in behavior indicates that, for acetaldehyde, the capture of a proton from the solvent (the reverse of the reaction by which the enolate anion is formed) occurs much more slowly than the attack by the enolate anion on another molecule. When acetone is used, the equilibrium for the formation of the enolate anion is unfavorable, but more importantly, enolate attack on another acetone molecule is disfavored due to steric hindrance. Here proton capture (actually deuteron capture) competes very well with the aldol reaction. CONDENSATION AND CONJUGATE ADDITION REACTIONS 471 Conjugate Addition Reactions O O OEt OEt − + 19.41 (a) O + − OH OEt OEt O O O O O OEt + OEt C6H5 OEt C6H5 O− EtO O OEt O C6H5 O CH3 OMe + (b) CH3NH2 OEt O C 6H 5 OEt OEt O EtO HA OEt O− EtO − O O O OEt O− H + N OMe H O O O CH3 OMe N OMe OMe CH3N OMe H O O base CH3−N O OMe − OMe OMe CH3−N O OMe O− O OMe CH3−N O 472 CONDENSATION AND CONJUGATE ADDITION REACTIONS O− O (c) EtO OEt O O− OEt + EtO O O O− EtO OEt O OEt OEt O OEt + OH O O OEt + EtO O OEt − O The Michael reaction is reversible, and the reaction just given is an example of a reverse Michael reaction. 19.42 Two reactions take place. The first is a reaction called the Knoevenagel condensation, initiated by attack of the conjugate base of the dicarbonyl compound on the ketone, O O O R R' R base −H2O + R' O O Then the α,β-unsaturated diketone reacts with a second mole of the active methylene compound in a Michael addition. O O O R O base + R' O O O − O 19.43 R R' O O − HO C6H5 C6H5 O O O C6H5 C6H5 − O OH C6H5 − H HO H O O− O − C6H5 O O OH H OH C6H5 H O − OH C6H5 −H2O O CONDENSATION AND CONJUGATE ADDITION REACTIONS 473 19.44 There are two stages to the reaction. The initial Michael addition is followed by the reaction of an enolate with an ester carbonyl to form a diketone (in equilibrium with an enol form). O H ONa 1 + 2 O (1) O ONa O (2) HCl OEt O EtO O O O EtO O OH O EtO O Another (less likely) possibility is the formation of a seven-membered ring. O O OH O O OH O OEt O O OEt O O O OEt General Problems O O O O O Br 19.45 (a) O H3O+ NaOH O O NaOH, H2O O (b) O + O NaOH, H2O O 474 CONDENSATION AND CONJUGATE ADDITION REACTIONS − O O O 19.46 H − O O OH − H − O O O − H OH OH OH OH Repeat aldol reaction on other side; same steps as above HO H − O O O − H − OH O (1) O 19.47 (a) O O ONa, O (2) O OH O O O + (3) H3O (workup) O CONDENSATION AND CONJUGATE ADDITION REACTIONS O H (b) H (1) LDA O (2) O O O Cl (1) O ONa, OH + (2) H3O (workup) (3) NaOH O O (c) O O Br (4) O ONa, O (1) O (2) O O (d) OH O O (3) H3O+ O O O (1) I2 (excess), NaOH, H2O (2) H3O+ 19.48 (a) O O H H (b) 2 OH KOH, H2O, + O O OH O O O O (c) KOH, H2O, OH O O (d) (1) LDA (1.1 equiv) (2) O H (3) H3O+ O OH 475 476 CONDENSATION AND CONJUGATE ADDITION REACTIONS O OH (1) CH3MgI 19.49 (1) O3 HA heat (2) + NH4/H2O (2) Me2S B A O O OH OH O H O O O pyridine OH Queen substance C O O H2 (1) I2/NaOH OH Pd (2) H3O+ D O O HO OH E O 19.50 (a) (b) KMnO4 , HO− , then H3 O+ OMe (d) CH3 ONa, then H3 O+ (c) CH3 OH, HA MeO O MeO O O (e) and (f ) OMe and O O OMe O − (g) HO , H2 O, then H3 O+ (h) heat (−CO2 ) (i) CH3 OH, HA (j) MeO O O (k) H2 , Pt (m) 2 NaNH2 + 2 CH3 I OMe (l) CH3 ONa, then H3 O+ CONDENSATION AND CONJUGATE ADDITION REACTIONS O O O Mannich reaction H + HN(CH3)2 + H 19.51 C6H5 477 (−H2O) CH3 C6H5 N CH3 O MgBr OH (1) C6H5 C6H5 O CH3 2 N (2) H3O+ C6H5 O CH3 O C6H5 CH3 N C6H5 O Darvon O − 19.52 EtO EtO EtO O O − + EtO O O O O O OEt EtO O OEt EtO − O O − O− O O EtO O O OEt O − EtO O (1) HO− O (2) H3O+ (hydrolysis and decarboxylation) O − O OH EtO O OH O CH3 478 CONDENSATION AND CONJUGATE ADDITION REACTIONS OEt OEt + HO HO O − H2SO4 19.53 OEt H O + O A O (1) Aromatization by proton removal using A − (2) HA (Protonation of the ether.) H OH A− O H − OH O O O OEt O− + Cl OEt − O O R R' Et + OH O 19.54 (a) Cl O O + Cl OH R O R OEt − + Cl R' OEt O− R' O OEt O (b) Decarboxylation of the epoxy acid gives an enolate anion which, on protonation, gives an aldehyde. O R' O R' R −CO2 + C5H5N H R O H − O O R R' + C5H5 N H H O O (c) Cl O OEt O OEt ONa β-Ionone O (1) HO−, H2O OH O + (2) H3O H O − C5H5N, heat −CO2 H + + C5H5NH O CONDENSATION AND CONJUGATE ADDITION REACTIONS O O H + 19.55 (a) O O O O− K+ O 479 OH + O OH O H (b) O O O + O OH O − K+ O Cl Cl + O OH Spectroscopy 19.56 (a) In simple addition, the carbonyl peak (1665–1780-cm−1 region) does not appear in the product; in conjugate addition, it does. (b) As the reaction takes place, the long-wavelength absorption arising from the conjugated system should disappear, One could follow the rate of the reaction by following the rate at which this absorption peak disappears. O Formed by the aldol condensation of acetone with two molecules of benzaldehyde. 19.57 X 480 CONDENSATION AND CONJUGATE ADDITION REACTIONS O − H 19.58 (a) O O O O H O O O O O O O − O O O − O O O O O O O O O O O O H O O− O O O O O O − O O O O O −O O O H H O O − O O − O H H H O H O O O O O + − O − O+ H H O O O Acid is added after formation of the anion is complete at each b -dicarbonyl group. O O O O O O CONDENSATION AND CONJUGATE ADDITION REACTIONS O (b) O O − O O − O O − O O O O 481 O O O O O H H O − − O O O O O O O H − O − O O O O O O O H O − H +O H O O O 482 CONDENSATION AND CONJUGATE ADDITION REACTIONS OH O O + 19.59 (a,b) N N N HA H O O O O N + N N −C4H9N H O O O O + O Ο O Η O O O A Br Br 19.60 HO COOH Mannich N reaction Br COOH HO CH3 Br CH3 OH C OH B COOH HO Hydrogenolysis CH3 OH D QUIZ 19.1 Which hydrogen atoms in the following ester are most acidic? O O a CH3 (a) a b CH2 C (b) b c CH2 C d OCH2 (c) c e CH3 (d) d (e) e CONDENSATION AND CONJUGATE ADDITION REACTIONS 483 19.2 What would be the product of the following reaction? O (1) 2 OEt ONa (2) H3O ? + O O (a) OEt O (b) O O O O (c) (d) OEt OEt O O O (e) OEt 19.3 What starting materials could be used in a crossed Claisen condensation to prepare the following compound? O O EtO OEt O O O (a) O and OEt O O (d) EtO and EtO OEt O O O (b) and OEt OEt (e) More than one of the above EtO O O O (c) and OEt H OEt H OEt 484 CONDENSATION AND CONJUGATE ADDITION REACTIONS 19.4 Supply the missing reagents, intermediates, and products. O O EtO OEt ONa (1) + (a) OEt + (2) H3O A (− OH ) B OK OH C (1) HO−, H2O I (−KI) (2) H3O+ D E heat −CO2 A O H N O + (b) EtO O O OEt OEt EtO OEt O OH B OEt ONa, OH O O C (1) HO−, H2O (2) H3O+ heat OH −2 CO2 OH O CONDENSATION AND CONJUGATE ADDITION REACTIONS B A O 485 LDA (c) (−LiI) THF O O O A O HA + (d) N (−H2O) B C O H2O + + Br− N H H 19.5 Supply formulas for the missing reagents and intermediates in the following synthesis. O (b) (a) HO− O −H2O heat H (e) CH3 CH3 + enantiomer and simple addition product (d) (c) 486 CONDENSATION AND CONJUGATE ADDITION REACTIONS 19.6 Which would be formed in the following reaction? O O + H OH H HO− 25 °C ? OH O (a) OH O (b) H OH O (c) H H O (d) (e) All of these will be formed. H 19.7 Supply formulas for the missing reagents and intermediates in the following synthesis. O (a) H CN (b) + HO− O (c) CONDENSATION AND CONJUGATE ADDITION REACTIONS 487 19.8 Supply formulas for the missing reagents and intermediates in the following synthesis. (a) O (b) HO− 0-10 °C H NaBH4 HA heat (e) (d) (c) OH [also (Z )] (f ) OH (g) H2 Ni pressure OCH3 OCH3 [also (Z )] H3O+ H2O (i) (h) OH O O 20 AMINES PREPARATION AND REACTIONS OF AMINES A. Preparation 1. Preparation via nucleophilic substitution reactions. NaN3, OH R Ο N3 ΝΚ R Na/ OH or LiAlH4 O Ο Br N R NH3 NH2 + R NH2 N N NH2NH2 H 2NH + R 2. Preparation through reduction of nitro compounds. NO2 H2, cat. or HNO3 (1) Fe, HCl (2) H O− H2SO4 3. Preparation via reductive amination. R O R' NH3 R' H [H] R NH2 R' H R N Aldehyde or ketone R O R' O 488 [H] R" H R R' R"NH2 R"R'"NH [H] R' H R N R" R'" O H O R R NH2 O +R NH2 3N (poor method) AMINES 489 4. Preparation of amines through reduction of amides, oximes, and nitriles. NH2 H2 CN − R Br R R N Ni R O R NH2OH Na/ethanol NOH R' R' O R NH2 + O Cl H R NH2 R (1) LiAlH4 R R' R' N R' (2) H2O N R' H H O RNHR' + O (1) LiAlH4 R R" Cl N R" R (2) H2O 5. Preparation through the Hofmann rearrangement of amides. O O O SOCl2 NH3 Br2 /NaOH R OH R" R' R' R N Cl R NH2 (NaOBr) RNH2 + CO22− B. Reaction of Amines 1. As a base or a nucleophile As a base N + H + O H N + H + H2O H As a nucleophile in alkylation N + X N + + X− R As a nucleophile in acylation R O N + O Cl (−HCl) R N R H 2. With nitrous acid R NH2 1° aliphatic HONO HX R N2+ X − (unstable) −N2 R+ alkenes, alcohols, and alkyl halides 490 AMINES Ar NH2 HONO HX (0–5 °C) N2+ X− Ar 1° aromatic CuCl ArCl + N2 CuBr ArBr + N2 CuCN ArCN + N2 KI ArI + N2 HBF4 ArN2+ BF4− Cu2O ArOH + N2 Cu2+, H2O H3PO2  ArF + N2 + BF3 ArH + N2 OH Ar N N OH Ar N N NR2 NR2 HONO R2NH R2N N O 2° aliphatic HONO ArNHR N O ArN R 2° aromatic R3N NaNO2 HX + R3NH+ X − + R3N O X− N 3° aliphatic HONO R2N R2N N O 3° aromatic 3. With sulfonyl chlorides R NH2 + ArSO2Cl (−HCl) RNHSO2Ar 1° amine R2NH + ArSO2Cl 2° amine (−HCl) R2NSO2Ar HO− H3O+ [RNSO2Ar]− + H2O AMINES 491 4. The Hofmann elimination + HO− + N(CH3)3 heat + (CH3)3N + H2O H SOLUTIONS TO PROBLEMS 20.1 Dissolve both compounds in diethyl ether and extract with aqueous HCl. This procedure gives an ether layer that contains cyclohexane and an aqueous layer that contains hexylaminium chloride. Cyclohexane may then be recovered from the ether layer by distillation. Hexylamine may be recovered from the aqueous layer by adding aqueous NaOH (to convert hexylaminium chloride to hexylamine) and then by ether extraction and distillation. + NH2 (in diethyl ether) H3O+Cl−/H2O ether layer aqueous layer + NH3 Cl − (evaporate ether and distill) HO− NH2 (extract into ether and distill) 20.2 We begin by dissolving the mixture in a water-immiscible organic solvent such as CH2 Cl2 or diethyl ether. Then, extractions with aqueous acids and bases allow us to separate the components. [We separate 4-methylphenol ( p-cresol) from benzoic acid by taking advantage of benzoic acid’s solubility in the more weakly basic aqueous NaHCO3 , whereas p-cresol requires the more strongly basic, aqueous NaOH.] 492 AMINES OH O NH2 OH , , , CH3 (in CH2Cl2) aqueous layer NaHCO3/H2O CH2Cl2 layer OH O NH2 O− Na+ , H3O+ , CH3 NaOH, H2O O aqueous layer CH2Cl2 layer OH O− Na+ NH2 Separate and recrystallize , H3O+Cl−/H2O CH3 aqueous layer H3O+ CH2Cl2 layer + NH3 Cl − OH Isolate by distillation HO− CH3 NH2 Extract into CH2Cl2 and distill Extract into CH2Cl2 and distill AMINES 493 20.3 (a) Neglecting Kekulé forms of the ring, we can write the following resonance structures for the phthalimide anion. O − O − N O O N N O O− (b) Phthalimide is more acidic than benzamide because its anion is stabilized by resonance to a greater extent than the anion of benzamide. (Benzamide has only one carbonyl group attached to the nitrogen atom, and thus fewer resonance contributors are possible.) (c) NH2NH2 + O− O H NH2NH2 O NHNH2 NH2NH2 N R N O R N O O O O NHNH2 NHNH2 N O Then, − R NH H O H2NNH2 + O O NHNH2 NH NH NH2+ R − O O O O NHR O H H − N N N N H NH2R + R + RNH2 H O R + NH2NH2 494 AMINES O O KOH NH 20.4 C6H5 N − K+ O Br (−KBr) O O O H C6H5 N N NH2NH2 NH2 + C6H5 OH reflux N Benzylamine H O O O + NH3 20.5 (a) H H2 O + C6H5NH2 (b) H N H CH3OH LiBH3CN + (CH3)2NH N C6H5 CH3OH H CH3 O O SOCl2 20.6 (a) C6H5 LiBH3CN CH3 O (c) C6H5 NH2 Ni pressure OH NH2 Cl C6H5 O C6H5 LiAlH4 N H N H N NaCN Br C6H5 (b) NH2 LiAlH4 O H N O SOCl2 (c) Cl OH O N LiAlH4 N HO O (d) NH2OH N Na/ OH NH2 C6H5 AMINES 20.7 (a) Fe HCl H2SO4 CH3O NH2 NO2 HNO3 495 CH3O CH3O (+ ortho; separate) NH2 O O NH3, H2 Cl (b) Ni AlCl3 CH3O CH3O CH3O CH3 CH3 Cl2 hv (c) + (CH3)3N Cl Cl− N CH3 CH3 (excess) O CH3 SOCl2 (1) KMnO4, HO− (d) OH (2) H3O+ O2N O2N O O NH2 Cl NH 3 NH2 O2N O2N CH3 (e) ROOR + NBS Br Br2 HO− KCN hv NH2 N (1) LialH4, Et2O (2) H2O 20.8 An amine acting as a base. NH2 + H + O + NH3 + H H H An amine acting as a nucleophile in an alkylation reaction. CH3 N + CH3 I N + I − O H O2N 496 AMINES An amine acting as a nucleophile in an acylation reaction. O O + (CH3)2NH + (excess) Cl + (CH3)2NH2 Cl− (CH3)2N An amino group acting as an activating group and as an ortho-para director in electrophilic aromatic substitution. NH2 NH2 Br Br Br2, H2O room temp. Br 20.9 (a, b) −O O + H3O+ N HO N HO N + O + H3O+ HO O + H2O O + H2O N H + HO N + H2O + N O O H + N(CH3)2 N(CH3)2 N(CH3)2 + + N A− O H + NO HA NO + (c) The NO ion is a weak electrophile. For it to react with an aromatic ring, the ring must have a powerful activating group such as −OH or −NR2 . NO2 20.10 (a) NO2 NH2 HNO3 Cl2 (1) Fe, HCl, heat H2SO4 Fe (2) HO− Cl NO2 (b) [from part (a)] NO2 Cl NH2 Br2 (1) Fe, HCl, heat FeBr3 (2) HO− Br Br AMINES 497 O H NO2 O NH2 2O (1) Fe, HCl, heat (c) N concd H2SO4 − (2) HO [as in (a)] O O H H N N NH2 NO2 HNO3 SO3H NO2 (1) dil. H2SO4, heat (2) HO− SO3H O O H H N N (d) NH2 HNO3 (1) H3O+ H2SO4 (2) HO− [from part (c)] NO2 NO2 (plus a trace of ortho) CH3 20.11 (a) Toluene HNO3 H2SO4 p-Nitrotoluene (+ o-nitrotoluene) O 2O (1) Fe, HCl (2) HO − NH2 CH3 CH3 CH3 (1) H3O+ (2) HO− HNO3 H2SO4 HONO (0−5°C) NO2 NO2 N NH2 N H H O O CH3 CH3 CH3 H3PO2 (1) Fe, HCl (2) HO− (−N2) N2+ NO2 NO2 NH2 m-toluidine 498 AMINES CH3 CuCl (b) Cl CH3 CuBr (c) CH3 CH3 Br CH3 HONO (0−5 °C) N2+ NH2 KI (d) I CH3 CuCN (e) H3O+, H2O N CH3 (f ) OH O NH2 N2+ NH2 Br 20.12 Br Br NO2 NO2 Br NO2 Br Br Br NO2 CuBr (0−5 °C) FeBr3 Br Br H2SO4/NaNO2 Br2 Br Br Br H2 H2SO4 /NaNO2 Pt (0−5 °C) Br N2+ NH2 Br Br H3PO2 − N2 Br AMINES 20.13 HO3S NH2 H2SO4 /NaNO2 (0−5 °C) 499 N2+ HO3S SO3Na N OH N OH NaOH pH 8-10 Orange II NO2 HNO3 H2SO4 20.14 + NH2 N2 H2SO4 /NaNO2 (0−5 °C) (1) Fe, HCl (2) HO− N(CH3)2 N(CH3)2 N N pH 5−7 Butter yellow OH + NH2 N2 H2SO4 /NaNO2 20.15 HO− (0−5 °C) O O OH N NaOH Br N O A O SnCl2 N N O B O NH2 2 2 O H N 2O 2 O O C Phenacetin 500 AMINES 20.16 (1) That A reacts with benzenesulfonyl chloride in aqueous KOH to give a clear solution, which on acidification yields a precipitate, shows that A is a primary amine. (2) That diazotization of A followed by treatment with 2-naphthol gives an intensely colored precipitate shows that A is a primary aromatic amine; that is, A is a substituted aniline. (3) Consideration of the molecular formula of A leads us to conclude that A is a methylaniline (i.e., a toluidine). C7H9N −C6H6N = CH3 CH3 NH2 But is A, 2-methylaniline, 3-methylaniline, or 4-methylaniline? (4) This question is answered by the IR data. A single absorption peak in the 680–840 cm−1 region at 815 cm−1 is indicative of a para-substituted benzene. Thus, A is 4-methylaniline ( p-toluidine). CH3 A NH2 20.17 First convert the sulfonamide to its anion, then alkylate the anion with an alkyl halide, then remove the SO2 C6 H5 group by hydrolysis. For example, H N SO2C6H5 R − N HO− R' R' N R SO2C6H5 X SO2C6H5 R R' H3O+ heat HO− + N R H C6H5SO3− AMINES O O H O NH2 H N N HOSO2Cl 2O 20.18 (a) Aniline SO2Cl O H NH2 N N H2N S (1) dil. HCl, heat N (2) HO− SO2 S N H S N H Sulfathiazole O H OH N Ο (b) O Ο N SO2 Ο S N H Succinylsulfathiazole CH3 N H 20.19 (a) (b) 3 N CH3 N CH3 (d) (c) NH2 (e) N H CH3 N SO2 (f ) N 501 502 AMINES O (g) OH (h) +N Br− N H N ( j) (i) O N H H (k) CH3 N N + H Cl− (l) CH3 N CH3 H (m) H2N (n) OH 4N + Cl− CH3 N CH3 (p) (o) H3C N H NH2 (r) (CH3)4N + OH − (q) CH3O O H N OH (s) H2N (t) CH3 AMINES 20.20 (a) Propylamine (or 1-propanamine) (b) N -Methylaniline 503 (h) Benzylaminium chloride (i) N ,N -Dipropylaniline (j) Benzenesulfonamide (k) Methylaminium acetate (l) 3-Amino-1-propanol (or 3-aminopropan-1-o1) (m) Purine (n) N -Methylpyrrole (c) Isopropyltrimethylammonium iodide (d) 2-Methylaniline (o-toluidine) (e) 2-Methoxyaniline (or o-methoxyaniline) (f) Pyrazole (g) 2-Aminopyrimidine Amine Synthesis and Reactivity HN 20.21 NH2 N H2N N sp3 hybridization stabilizes the corresponding conjugate acid. N Alkyl groups are electron donating, stabilizing the corresponding conjugate acid. Lone pair is not involved in aromaticity. N (2) H2O LiAlH4 + 20.22 (a) NH2 O (b) NH2 (c) Br + + LiAlH4 (2) H2O NH2 NH3 (excess) NH2 O O Br + N NK O O O NH2NH2 NH NH2 + NH O 504 AMINES OTs (d) + NH2 NH3 (excess) O H (e) NH2 H2 Ni + NH3 pressure NO2 (f ) + 3 H2 NH2 (g) O NH2 Pt pressure NH2 + CO32− Br2 HO− NO2 20.23 (a) HNO3 (1) Fe, HCl, heat H2SO4 (2) HO− Br O MgBr Mg (b) NH2 (1) CO2 Et2O (2) H3O OH O SOCl2 Cl NH3 + (excess) O NH2 NH2 Br2 NaOH O (c) NH2 Br2 HO− Hofmann rearrangement NH2 AMINES 505 O NK PBr3 OH 20.24 (a) O Br O NH2NH2 NH2 N O O H N + N O Br (b) N (1) LiAlH Et O 4, 2 NaCN NH2 (2) H2O [from part (a)] O (1) KMnO4, HO− OH (c) H (2) H3O+ OH O (1) SOCl2 Br2 (2) NH3 HO− NH2 NH2 O OH (d) CH3NH2 PCC CH2Cl2 H H2, Ni pressure O H NH2 2O 20.25 (a) NH2 (b) N O O O + O O heat N O H N CH3 506 AMINES O O H H N NH2 N (1) H3O+, H2O HNO3 H2SO4 (c) [from part (a)] (2) OH − NO2 (+ trace of ortho) O NO2 O H H N NH2 N (1) NH3 HOSO2Cl (d) (2) H3O+, heat (3) HO− [from part (a)] SO2Cl CH3 SO2NH2 CH3 NH2 N 2 CH3Ι (e) base N2+ X− NH2 N2 + BF4− HONO (f) HBF4 F heat (0–5 °C) N2+ X − N2+ X − Cl Br CuBr CuCl (g) (h) [from part (f )] [from part (f )] N N2+ X − + X− N2 I KI CuCN (i) ( j) [from part (f )] N [from part (f )] O (1) H3O+, H2O (k) (2) HO− [from part (j)] OH AMINES N2+ X − N2+ X − OH H3PO2 Cu2O (l) (m) 2+ Cu , Η2O [from part (f )] [from part (f )] OH N2+ X − OH N HO− pΗ 8-10 + (n) Η2O [from part (f )] N [from part (l)] CH3 N2+ X − N H3O+ pΗ 5-7 + (o) N N(CH3)2 [from part (f )] N [from part (e)] (1) cat. HA (2) NaBH3CN NH2 20.26 (a) N H O (1) NaCN (2) LiAlH4 (3) H3O+ Br (b) NH2 (1) NaN3 (2) LiAlH4 (3) H3O+ Br (c) NH2 O (1) NH2OH, cat. HA (2) NaBH3CN (d) O NH2 (e) NH2 (1) O (2) LiAlH4 (3) H3O+ O HN CH3 507 508 AMINES 20.27 (a) NH2 HONO NaNO2/HCl −N2 N2+ 0-5 °C + hydride + Cl− shift Cl− −H+ −H+ H2O H2O OH Cl Cl OH (b) HONO NaNO2/HCl 0−5 °C NH 2 N N O 2 H (c) N (d) N (e) 20.28 (a) N HONO NaNO2/HCl 0−5 °C HONO NaNO2/HCl 0−5 °C NH2 NH2 + C6H5SO2Cl O N O N N HONO, 0−5 °C NaNO2/HCl KOH H2O N2+ Cl− N − K+ SO2C6H5 Clear solution O+ H3 H N SO2C6H5 Precipitate (b) NH + C6H5SO2Cl KOH H2O N SO2C6H5 Precipitate H3O+ No reaction (precipitate remains) AMINES SO2C6H5 H (c) KOH + C6H5SO2Cl H O 2 N N Precipitate O+ H3 (d) No reaction (precipitate remains) No reaction (3° amine is insoluble) KOH + C6H5SO2Cl H O 2 N + N H3O+ H 3° Amine dissolves KOH (e) NH2 + C6H5SO2Cl H O 2 N− K+ SO C H 2 6 5 Clear solution H3O+ H N SO2C6H5 Precipitate HONO 20.29 (a) N H (b) N H + C6H5SO2Cl N NaNO2/ HCl KOH H2O N O N SO2C6H5 509 510 AMINES O 20.30 (a) 2 NH2 + C6H5 O N H Cl C6H5 + + O O O (b) 2 CH3NH2 + O + CH3NH3 O− O O (c) + CH3N H 2O NH3 Cl− NHCH3 + 2 CH3NH2 + O− O H3NCH3 O O heat (d) [product of (c)] N + CH3 H2O + CH3NH2 O O O N O O + (e) N O OH H O O + (f ) N + N 2O O H O O (g) 2 OH NH2 + Cl + N H NH3+ Cl− AMINES (h) HO− N+ H heat H H + H2O H CH3 CH3 + (i) N + 511 Br2 (excess) H2O Br Br NH2 NH2 NO2 HNO3 20.31 (a) CH3 CH3 CH3 + H2SO4 Separated CH3 NO2 CH3 CH3 (1) Fe, HCl HONO (2) HO− 0−5 °C NO2 HBF4 heat N2+ X − NH2 CH3 CH3 F CH3 NO2 (b) NH2 (1) Fe, HCl (2) HO− HONO 0−5 °C [from part (a)] CH3 CH3 +X− I N2 KI CH3 CH3 (c) CH3 (1) Fe, HCl, heat (1) H2SO4/NaNO2, 0−5 °C (2) HO− (2) Cu2O, Cu2+, H2O NO2 [from part (a)] NH2 OH 512 AMINES N2+ X − NH2 Cl HCl/NaNO2 (d) CuCl 0−5 °C Cl Cl [from Problem 20.10(a)] Cl NH2 NO2 fuming HNO3 (e) Fe, HCl, heat HCl/NaNO2 0−5 °C H2SO4, heat N + N2 NH2 NO2 X− CuCN N2+ X − NO2 N NH2 NO2 (1) Fe, HCl, heat Br2 (f ) (2) HO− FeBr3 Br Br [from Problem 20.10(a)] N (1) H2SO4 /NaNO2, 0−5 °C (2) CuCN Br NO2 NO2 Br2 (g) H2SO4 / NaNO2 FeBr3 0−5 °C Br NH2 [from Problem 20.10(d)] Br NH2 NO2 NO2 H3PO2 Br Br Br + N2 NO2 NH2 (1) Fe, HCl, heat (h) (2) HO− Br Br [from part (g)] Br Br Br AMINES NO2 NO2 (1) HBr/ NaNO2, 0−5 °C (i) (1) Fe, HCl, heat (2) HO− (2) CuBr Br Br Br Br Br NH2 [from part (g)] OH NH2 (1) H2SO4 /NaNO2, 0−5 °C (2) Cu2O, Cu2+, H2O Br Br Br Br Br Br N NH2 (1) H2SO4/ NaNO2, 0−5 °C (j) (2) CuCN Br Br Br Br Br Br [from part (i)] NO2 NO2 H3O+ (1) H2SO4/NaNO2, 0−5 °C (k) heat (2) CuCN Br Br Br Br NH2 [from part (g)] N NO2 H2 Pt Br Br O NH2 OH (1) H2SO4/ NaNO2, 0−5 °C (2) H3PO2 Br Br O 513 OH Br Br O OH pressure 514 AMINES NO2 NO2 (l) Br (1) H2SO4/ NaNO2, 0−5 °C (1) Fe, HCl, heat (2) KI (2) HO− Br Br NH2 [from part (g)] Br I NH2 (1) H2SO4/ NaNO2, 0−5 °C (2) H3PO2 Br Br Br I CH3 (m) Br I CH3 CH3 Br2 HNO3 FeBr3 H2SO4 NO2 (1) Fe, HCl, heat (2) H O− Br Br CH3 CH3 NH2 (1) H2SO4/ NaNO2, 0−5 °C (2) CuCN Br NH2 (n) Br N2+ X − H2SO4/NaNO2, 0−5 °C H3C H3C [from part (c)] OH , pH 8-10 OH N N H3C N AMINES 515 OH N2+ X − (o) HO H3C [from part (c)] N pH 8-10 H3C [from part (n)] N CH3 H3C 20.32 (a) Benzylamine dissolves in dilute HCl at room temperature, + NH2 + H3O+ + Cl − NH3 Cl − 25 °C benzamide does not dissolve: Ο + H3O+ + Cl− NH2 25 °C No reaction (b) Allylamine reacts with (and decolorizes) bromine instantly, Br ΝΗ2 + Br Br2 NH2 propylamine does not: ΝΗ2 No reaction if the mixture is not heated or irradiated + Br2 (c) The Hinsberg test: H3C NH2 + C6H5SO2Cl KOH H2O K−+ NSO2C6H5 H3C Soluble H3O+ H3C NHSO2C6H5 Precipitate NHCH3 + C6H5SO2Cl KOH H2O NSO2C6H5 CH3 Precipitate H3O+ Precipitate remains 516 AMINES (d) The Hinsberg test: NH2 + C6H5SO2Cl K+− NSO2C6H5 KOH H2O H3O+ Soluble NHSO2C6H5 Precipitate H + C6H5SO2Cl N KOH SO2C6H5 N H2O H3O+ Precipitate remains Precipitate (e) Pyridine dissolves in dilute HCl, N + H3O+ + Cl− + H Cl− N benzene does not: + H3O+ + Cl− No reaction (f ) Aniline reacts with nitrous acid at 0–5 ◦ C to give a stable diazonium salt that couples with 2-naphthol, yielding an intensely colored azo compound. NH2 + H2SO4 / NaNO2 N2 2-naphthol 0−5 °C HO N N Cyclohexylamine reacts with nitrous acid at 0–5 ◦ C to yield a highly unstable diazonium salt—one that decomposes so rapidly that the addition of 2-naphthol gives no azo compound. NH2 H2SO4 /NaNO2 0−5 °C + N2 alkenes, alcohols, and alkyl halides − N2 2-naphthol + No reaction AMINES 517 (g) The Hinsberg test: + C6H5SO2Cl 3N KOH H2O No reaction H3O+ + 3NH Soluble + C6H5SO2Cl 2NH KOH 2NSO2C6H5 H2O H3O+ Precipitate remains Precipitate (h) Tripropylaminium chloride reacts with aqueous NaOH to give a water insoluble tertiary amine. + 3NH Cl− NaOH 3N H2O Water soluble Water insoluble Tetrapropylammonium chloride does not react with aqueous NaOH (at room temperature), and the tetrapropylammonium ion remains in solution. + 4N + NaOH H2O Cl− Water soluble 4N [Cl − or HO− ] Water soluble (i) Tetrapropylammonium chloride dissolves in water to give a neutral solution. Tetrapropylammonium hydroxide dissolves in water to give a strongly basic solution. 20.33 Follow the procedure outlined in the answer to Problem 20.2. Toluene will show the same solubility behavior as benzene. Mechanisms O CN 20.34 H2 O NH2 −H2O (several steps) Pd H N N H2 Pd 518 AMINES O N 20.35 (a) O O H− − N ΟH H Br Br N Br H H − OH O H H − C O OH O N − OH N − Br N O H H N O N H O − H − O NH2 H OH + CO2 O O Cl P (b) Cl Cl O Cl NH2 O − P Cl + N H O + H N H H Cl − + N N Cl − H AMINES 519 20.36 Carry out the Hofmann reaction using a mixture of 15 N labeled benzamide, C6 H5 CO∗ NH2 , and p-chlorobenzamide. If the process is intramolecular, only labeled aniline, C6 H5 ∗ NH2 , and p-chloroaniline, will be produced. If the process is one in which the migrating moiety truly separates from the remainder of the molecule, then, in addition to the two products mentioned above, there should be produced both the unlabeled aniline and labeled p-chloroaniline, p-ClC6 H4 ∗ NH2 . Note: When this experiment is actually carried out, analysis of the reaction mixture by mass spectrometry shows that the process is intramolecular. General Synthesis O O + O 20.37 NH3 OH (2) H3O+ H2N O O O Br2, HO− O− H2N 2− − CO3 ( )8 20.38 (a) HO O H3O+ 2 PBr3 OH + O− H3N Br ( )8 + (CH3)3N 2 (CH3)3N Br + ( )8 N(CH3)3 2 Br − O (b) OH HO + 2 Br OH O HA (−2 H2O) O Br O O Br 2 (CH3)3N O O + (CH3)3N O O O + N(CH3)3 2 Br − 520 AMINES 20.39 H N 2 O NH2 N + N Br H2N −H2O Br NH2 Br N N NH2 OH Br Br N Br OH H H2N −HBr N Br N N HCO3− −HBr Br N OH OH O H2N N N N O OH N H Br + N −HBr folic acid O H2N OH Ag2O + CH3I 20.40 N I− N + H3C CH3 N CH3 H3C + CH3 S Ag2O (CH3)3N I− H2O + H2O N(CH3)3 + V + T U heat + (CH3)3N W O CH3 20.41 (CH3)2NH + N O CH3 N CH3 CH3 OH Cl base (−HCl) CH3 O O CH3I −H2O N CH3 R CH3I CH3 heat HO− H2O I− CH3 N O + CH3 Acetylcholine iodide O HO− AMINES 521 O O NH3 20.42 HO NH2 H N HO OH O O O Br Br2 O Cl 20.43 AlCl3 OH O H N N Diethylpropion O 20.44 Cl O SOCl2 OH Cl O NH2 Cl Cl OH N H O O N base H Spectroscopy 20.45 The results of the Hinsberg test indicate that compound W is a tertiary amine. The 1 H NMR provides evidence for the following: (1) The signals at δ 7.2 and 6.7 suggest hydrogens bonded to a benzene ring. Their integrals relative to those for other signals suggests two C6 H5 groups. (2) A CH3 CH2 group (the quartet at δ 3.5 and the triplet at δ 1.2). (3) An unsplit CH2 group (the singlet at δ 4.5). There is only one reasonable way to pull all of this together. N Thus W is N -benzyl-N -ethylaniline. 522 AMINES 20.46 Compound X is benzyl bromide, C6 H5 CH2 Br. This is the only structure consistent with the 1 H NMR and IR data. (The monosubstituted benzene ring is strongly indicated by the 5H, δ 7.3 1 H NMR absorption and is confirmed by the peaks at 690 and 770 cm−1 in the IR spectrum.) Compound Y, therefore, must be phenylacetonitrile, C6 H5 CH2 CN, and Z must be 2-phenylethylamine, C6 H5 CH2 CH2 NH2 . Br NaCN X C7H7Br NH2 LiAlH4 N Y C8H7N Z C8H11N Interpretations of the IR and 1H NMR spectra of Z are as follows. (c) Z, C8H11N H (d) [5H] H (a) (c) [2H] NH2 (d) H (b) [2H] H (b) (a) [2H] TMS 3.0 8 7 6 5 2.5 4 δ H (ppm) 3 2 1 0 (c) H H (a) NH2 (d) H H (b) (a) (b) (c) (d) singlet δ 1.0 triplet δ 2.7 triplet δ 2.9 multiplet δ 7.25 20.47 That A contains nitrogen and is soluble in dilute HCl suggests that A is an amine. The two IR absorption bands in the 3300–3500-cm−1 region suggest that A is a primary amine. The 13 C spectrum shows only two signals in the upfield aliphatic region. There are four signals downfield in the aromatic region. The information from the DEPT spectra suggests AMINES 523 an ethyl group or two equivalent ethyl groups. Assuming the latter, and assuming that A is a primary amine, we can conclude from the molecular formula that A is 2,6-diethylaniline. The assignments are NH2 (b) ( f ) (a) (e) (d) (c) (a) (b) (c) (d) (e) (f) δ 12.9 δ 24.2 δ 118.1 δ 125.9 δ 127.4 δ 141.5 (An equally plausible answer would be that A is 3,5-diethylaniline.) 20.48 That B dissolves in dilute HCl suggests that B is an amine. That the IR spectrum of B lacks bands in the 3300–3500-cm−1 region suggests that B is a tertiary amine. The upfield signals in the 13 C spectrum and the DEPT information suggest two equivalent ethyl groups (as was also true of A in the preceding problem). The DEPT information for the downfield peaks (in the aromatic region) is consistent with a monosubstituted benzene ring. Putting all of these observations together with the molecular formula leads us to conclude that B is N ,N -diethylaniline. The assignments are (b) (a) N (f) (c) (e) (d ) (a) (b) (c) (d) (e) (f) δ 12.5 δ 44.2 δ 112.0 δ 115.5 δ 128.1 δ 147.8 20.49 That C gives a positive Tollens’ test indicates the presence of an aldehyde group; the solubility of C in aqueous HCl suggests that C is also an amine. The absence of bands in the 3300–3500-cm−1 region of the IR spectrum of C suggests that C is a tertiary amine. The signal at δ 189.7 in the 13 C spectrum can be assigned to the aldehyde group. The signal at δ 39.7 is the only one in the aliphatic region and is consistent with a methyl group or with two equivalent methyl groups. The remaining signals are in the aromatic region. If we O N(CH3)2 group, assume that C has a benzene ring containing a H group and a then the aromatic signals and their DEPT spectra are consistent with C being p-(N ,N dimethylamino)benzaldehyde. The assignments are O (f) H (c) (d) (b) (e) N CH3 CH3 (a) (a) (b) (c) (d) (e) (f) δ 39.7 δ 110.8 δ 124.9 δ 131.6 δ 154.1 δ 189.7 524 AMINES Challenge Problems CH3 20.50 (a) C6H5N H I H (b) I −e− (during electron impact mass spectrometry) C6H5 CH2 + N m/z 107 H −H C6H5 + CH2 N m/z 106 H H H N N 20.51 (a,b) CH3 H H N N + O O O A S O CH3 + +O N S Cl O − N+ H H N C N N N B Diphenylcarbodiimide H O S O O− + O N+ + CH3 − CH3 S O O N AMINES O O OH H 20.52 O H + N 525 O −O +N O− O O O O O O O O O O H − N + O N O O O− O O O O + O− O O O O O + N O− O O H − + HO + N O− O O 20.53 An abbreviated version of one of several possible routes: O O O H O −H2O O −H +; +H + OH repeated sequence O O O NH3 N H C "triacetoneamine" NH2 (with two proton transfers) 526 AMINES 20.54 O-protonated acetamide cation N-protonated acetamide cation The N -protonated acetamide cation shows significant localization of positive charge near its NH3 + group, as indicated by the strong blue color at that location in its electrostatic potential map. The N -protonated acetamide model also shows noticeable negative charge (or less positive charge) near its oxygen atom, as indicated by yellow in that region. These observations suggest a lack of delocalization of the positive charge. On the other hand, the O-protonated acetamide cation has some blue as well as green mapped to both its NH2 group and O H (protonated carbonyl) group, suggesting delocalization of the positive charge between these two locations. An attempt to draw resonance structures for these two protonated forms shows that two resonance contributors are possible for O-protonated acetamide, whereas only one proper Lewis structure is possible for N -protonated acetamide. The resonance contributors for O-protonated acetamide distribute the formal positive charge over both the oxygen and nitrogen atoms. H + H O O O versus + + NH2 NH3 NH2 O-protonated acetamide cation N-protonated acetamide cation QUIZ 20.1 Which of the following would be soluble in dilute aqueous HCl? NH2 (a) (b) NH2 AMINES 527 O NH2 (c) (d) Two of the above (e) All of the above 20.2 Which would yield propylamine? (a) (1) NaCN Br (2) LiAlH4 O NH3 (b) H H2/Ni O Br2 (c) NH2 HO− (d) Two of the above (e) All of the above 20.3 Select the reagent from the list below that could be the basis of a simple chemical test that would distinguish between each of the following: O (a) (b) NH2 NH2 H N and NH2 (c) NH2 and CH3 NH2 and 1. Cold dilute NaHCO3 2. Cold dilute HCl 3. NaNO2 , HCl, 5 ◦ C, then 2-naphthol 4. C6 H5 SO2 Cl, HO− , then HCl 5. Cold dilute NaOH 528 AMINES 20.4 Complete the following syntheses: A CH3 B CH3 (1) Fe, HCl (a) (2) HO− H2N + isomer C CH3 CH3 E CH3 D H2N + N2 N N(CH3)2 Cl− F N N CH3 O A B Cl (b) N C heat H2O Br D C (1) HONO, 0−5 °C Br2 (xs) H2O (2) Br Br E + CO2 20.5 Select the stronger base from each pair (in aqueous solution): NH2 (a) NH2 or H3C (1) (2) O NH2 (b) NH2 or (2) (1) NH2 (c) NH2 or H3C (1) O2N (2) O AMINES or (d) N H N CH3 (2) (1) H N (e) H N CH3 or (2) (1) or (f ) N N (1) CH3 (2) 529 21 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION SOLUTIONS TO PROBLEMS 21.1 The inductive effect of an electron-releasing group (i.e., CH3 ) changes the charge distribution in the molecule so as to increase the electron density of the ring and oxygen, causing the proton to be held more strongly; it also destabilizes the phenoxide anion by intensifying its negative charge. These effects make the substituted phenol less acidic than phenol itself. O− OH + H2O + CH3 CH3 H3O+ δ+ Electron-releasing CH3 destabilizes the anion more than the acid. pKa is larger than for phenol. 21.2 The inductive effect of an electron-withdrawing group such as chlorine changes the charge distribution in the molecule so as to decrease the electron density of the ring and oxygen, causing the proton to be held less strongly; it also can stabilize the phenoxide ion by dispersing its negative charge. These effects make the substituted phenol more acidic than phenol itself. O− OH + H3O+ + H2O Cl − Cl δ Electron-withdrawing chlorine stabilizes the anion by dispersing the negative charge. pKa is smaller than for phenol. Nitro groups are very powerful electron-withdrawing groups by their inductive and resonance effects. Resonance structures (B–D) below place a positive charge on the hydroxyl 530 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 531 oxygen. This effect makes the hydroxyl oxygen dramatically more positive, causing the proton to be held much less strongly. These contributions explain why 2,4,6-trinitrophenol (picric acid) is so exceptionally acidic. H O −O H O− O O N+ +N + −O O N+ N O− + N+ −O A + −O H O O O+ + N + N O −O − +N −O O B H O N −O O O− O+ − + N N O− −O O− O O+ +N O D C 21.3 Dissolve the mixture in a solvent such as CH2 Cl2 (one that is immiscible with water). Using a separatory funnel, extract this solution with an aqueous solution of sodium bicarbonate. This extraction will remove the benzoic acid from the CH2 Cl2 solution and transfer it (as sodium benzoate) to the aqueous bicarbonate solution. Acidification of this aqueous extract will cause benzoic acid to precipitate; it can then be separated by filtration and purified by recrystallization. The CH2 Cl2 solution can now be extracted with an aqueous solution of sodium hydroxide. This will remove the 4-methylphenol (as its sodium salt). Acidification of the aqueous extract will cause the formation of 4-methylphenol as a water-insoluble layer. The 4-methylphenol can then be extracted into ether, the ether removed, and the 4-methylphenol purified by distillation. The CH2 Cl2 solution will now contain only toluene (and CH2 Cl2 ). These can be separated easily by fractional distillation. O O CH3 21.4 (a) O CH3 (b) NO2 (c) Br OH + HO (d) H3C 532 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 21.5 If the mechanism involved dissociation into an allyl cation and a phenoxide ion, then recombination would lead to two products: one in which the labeled carbon atom is bonded to the ring and one in which an unlabeled carbon atom is bonded to the ring. O 14CH O− 2 14 dissociation H2C + δ+ CH2 (recombination) δ+ OH OH 14CH2 14CH2 + The fact that all of the product has the labeled carbon atom bonded to the ring eliminates this mechanism from consideration. O OH 21.6 A B O O 21.7 (a) (b) O 21.8 O O O O O O O + OH 2 NaBH4 OH NH2 O NO2 NO2 (b) 21.9 (a) NO2 NHC6H5 NO2 (c) NO2 S NO2 (d) NO2 NO2 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 533 21.10 That o-chlorotoluene leads to the formation of two products (o-cresol and m-cresol) when submitted to the conditions used in the Dow process suggests that an elimination-addition mechanism takes place. CH3 CH3 CH3 CH3 Cl H −OH − 350 °C HO− OH OH HO− CH3 CH3 H − HO− H −OH OH OH Inasmuch as chlorobenzene and o-chlorotoluene should have similar reactivities under these conditions, it is reasonable to assume that chlorobenzene also reacts by an eliminationaddition mechanism in the Dow process. 21.11 2-Bromo-1,3-dimethylbenzene, because it has no α-hydrogen, cannot undergo an elimination. Its lack of reactivity toward sodium amide in liquid ammonia suggests that those compounds (e.g., bromobenzene) that do react do so by a mechanism that begins with an elimination. 21.12 (a) One molar equivalent of NaNH2 converts acetoacetic ester to its anion, O O O OEt + H − NH 2 O liq. NH3 − OEt + NH3 and one molar equivalent of NaNH2 converts bromobenzene to benzyne (cf. Section 21.11B): H2N − H Br + NH3 + Br− 534 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION Then the anion of acetoacetic ester adds to the benzyne as it forms in the mixture. OEt O − O OEt O − O OEt O − OEt O Na+ O then NH4Cl O This is the end product of the addition. (b) 1-phenyl-2-propanone, as follows: O OEt OH O (1) H O−, H2O, heat O (2) H3O + heat − CO2 O O (c) By treating bromobenzene with diethyl malonate and two molar equivalents of NaNH2 to form diethyl phenylmalonate. OEt O O Br OEt OEt 2 NaNH 2 + liq. NH3 O OEt O [The mechanism for this reaction is analogous to that given in part (a).] Then hydrolysis and decarboxylation will convert diethyl phenylmalonate to 2-phenylethanoic acid. O OEt OEt O O OH OH (1) H O−, H2O, heat (2) H3O+ O OH heat + O CO2 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 535 Problems Physical Properties OH O OH OH OH OH O 21.13 5 = most acidic OCH3 2 NO2 1 CF3 3 CF3 5 4 21.14 (a) 4-Fluorophenol because fluorine is an electron withdrawing group and a methyl is an electron donating group. (b) 4-Nitrophenol because a nitro group is electron withdrawing and a methyl is an electron donating group. (c) 4-Nitrophenol because the nitro group can exert an electron-withdrawing effect through resonance. + OH − O OH N+ − O N+ O O− In 3-nitrophenol these resonance structures are not possible. (d) 4-Methylphenol because it is a phenol, not an alcohol. (e) 4-Fluorophenol because fluorine is more electronegative than bromine. 21.15 (a) ONa + OH (b) ONa + H2O O ONa (c) OH + NaCl (d) OH 21.16 (a) 4-Chlorophenol will dissolve in aqueous NaOH; 4-chloro-1-methylbenzene will not. (b) 4-Methylbenzoic acid will dissolve in aqueous NaHCO3 ; 4-methylphenol will not. (c) 2,4,6-Trinitrophenol, because it is so exceptionally acidic (pK a = 0.38), will dissolve in aqueous NaHCO3 ; 4-methylphenol (pK a = 10.17) will not. (d) 4-Ethylphenol will dissolve in aqueous NaOH; ethyl phenyl ether will not. 536 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION General Reactions OH OH OH SO3H 21.17 (a) + (b) Br (major) SO3H (major) O OH OH O SO3H + (c) S (d) O CH3 CH3 SO3H (major) OH Br O Br (e) OC6H5 (f ) OH O Br OH C6H5 O Br Br (h) (g) O CH3 ONa (i) Same as (h) + O C6H5 O (k) ( j) CH3 + CH3OSO2O − O− O (l) C6H5OCH3 C6H5 (m) 21.18 Predict the product of the following reactions. O (a) HBr (excess) OH + Br PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 537 O OH (1) NaH (b) Br (2) O O O O OH O (c) + O O O CH3 CH3 OH Br OH (d) H2O H3C OH Br2 (excess) O H3C Br O O OH O Excess (e) O O O OH O H3C (1) NaH, (f) Br H3C (2) heat OH OH (g) HNO3, H2SO4 OH NO2 OH + OH 538 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION Br NH2 H2N NaNH2, NH3 (h) NH2 + CH3 + CH3 CH3 CH3 (1) NaOH (i) (2) OH O OH O Mechanisms and Synthesis OH OH Cl O + 21.19 Cl O Epichlorohydrin CH3 CH3 OH O HO− O N O H2N CH3 CH3 H Toliprolol 21.20 One can draw a resonance structure for the 2,6-di-tert-butylphenoxide ion which shows the carbon para to the oxygen as a nucleophilic center; that is, the species is an ambident nucleophile. O − − O Given the steric hindrance about the oxygen, the nucleophilic character of the para carbon is dominant and an SN Ar reaction occurs at that position to produce this biphenyl derivative: O2N OH PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 539 21.21 Both o- and m-toluidine are formed in the following way: CH3 CH3 CH3 2 H−NH2 − H − Cl NH2 NH2 − NH CH3 NH2 CH3 350 ° C − CH3 − NH2 H−NH2 NH2 NH2 Both m- and p-toluidine are formed from another benzyne-type intermediate. CH3 − CH3 NH2 Cl − H−NH2 − CH3 350 ° C H CH3 − NH2 NH2 NH2 CH3 CH3 NH2 NH2 NH2 − H − NH2 OH OH H2SO4 21.22 OH SO3H Cl SO3H 2 Cl2 25 °C H3O+ H2O (steam) Cl − OH O Na+ Cl Cl O NaOH O + Cl ONa Cl Cl OH Cl ONa OH O Cl O O Cl H3O+ Cl O Cl 2,4-D 540 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION O O H − 21.23 O − O O OH Cl H H O O− O Cl Cl O H O O O H O O H H − O OH O O O O Cl − O H Cl etc. O O O O O O O Dibenzo-18-crown-6 + H H O H H O H O+ H O H O 21.24 OH H + H H + O H H H O H 21.25 O + OH OH OH + PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION OH OH 21.26 (a) OH HA + OCH3 541 + OCH3 OCH3 BHA OH OH + (b) HA 2 CH3 CH3 BHT Notice that both reactions are Friedel-Crafts alkylations. − O O O O NO2 O S O N+ S 21.27 O H − NO2 O NO2 − OCH3 − O O − SO2 NO2 O O − N+ S O O NO2 NO2 21.28 The phenoxide ion has nucleophilic character both at the oxygen and at the carbon para to it; it is ambident. O− − O The benzyne produced by the elimination phase of the reaction can then undergo addition by attack by either nucleophilic center: reaction at oxygen producing 1 (diphenyl ether), and reaction at carbon producing 2 (4-hydroxybiphenyl). 542 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION − CN Cl 21.29 (a) CN 2 eq. KNH2 (2) + NH4 liq. NH3 −33 ° C CN 2 NH CH3 (b) Cl 2 eq. NaNH2 liq. NH3 −33 °C − N CH3 2 N CH3 + N H3C 21.30 The proximity of the two OH groups results in the two naphthalene nuclei being noncoplanar. As a result, the two enantiomeric forms are nonequivalent and can be separated by a resolution technique. OH HO HO ≠ OH O 21.31 (a) H + O R O + H O R (b) Because the phenolic radical is highly stabilized by resonance (see the following structures), it is relatively unreactive. O O O O PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 543 Spectroscopy 21.32 X is a phenol because it dissolves in aqueous NaOH but not in aqueous NaHCO3 . It gives a dibromo derivative and must therefore be substituted in the ortho or para position. The broad IR peak at 3250 cm−1 also suggests a phenol. The peak at 830 cm−1 indicates para substitution. The 1 H NMR singlet at δ 1.3 (9H) suggests nine equivalent methyl hydrogen atoms, which must be a tert-butyl group. The structure of X is OH 21.33 The broad IR peak at 3200–3600 cm−1 suggests a hydroxyl group. The two 1 H NMR peaks at δ 1.67 and δ 1.74 are not a doublet because their separation is not equal to other splittings; therefore, these peaks are singlets. Reaction with Br2 suggests an alkene. If we put these bits of information together, we conclude that Z is 3-methyl-2-buten-1-ol (3-methylbut-2en-1-ol). (a) CH3 H (c) C C (b) CH3 (a) (c) (d) (e) (d) (e) CH2OH and (b) singlets δ 1.67 and δ 1.74 multiplet δ 5.40 doublet δ 4.10 singlet δ 2.3 Challenge Problems 21.34 Initially, rearrangement to the ortho position does occur, but subsequent aromatization cannot take place. A second rearrangement occurs to the para position (with the labeled carbon again being at the end of the allyl group); now tautomerization can result in reestablishing an aromatic system. * O CH3 O CH3 CH3 O CH3 CH3 heat * (Claisen rearr.) (Cope rearr.) Not isolated CH3 * H Not isolated OH CH3 CH3 Tautomerization * 544 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 21.35 There are alternative representations for I, namely: O − − O A B Thus, I is an ambident nucleophile. O-alkylation via the nucleophile represented by A is anticipated since the negative charge resides primarily on the more electronegative atom. Use of DMF as solvent provides for a nucleophile unencumbered with solvent. When protic solvents are used, they strongly hydrogen bond to the oxygen atom and so reduce the nucleophilicity of the oxygen; C-alkylation then is favored. 21.36 In those cases where the SN Ar mechanism holds, the first step is rate-determining; the C Nu bond forms, but there is no C F bond breaking. The strongly electronegative fluorine facilitates this step by helping to disperse the negative charge on the carbanionic intermediate. The other halogens, being less electronegative, are less effective in this. In SN 1 reactions the considerable strength of the C F bond results in very slow dissociation of the fluoro compound. In SN 2 reactions, since F− is the most basic halide ion it is the poorest leaving group. 21.37 Examination of the resonance structures that can be drawn for the SN Ar intermediate in the first case reveals that the five-membered ring has cyclopentadienyl anion character. This is a stabilizing feature due to aromaticity. − Nu − X X Nu X X − etc. − Nu Nu In the latter case, electron-delocalization into the other (seven-membered) ring results in eight electrons in that ring, which does not provide aromatic stabilization. X Nu − X Nu X Nu etc. − − PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 21.38 CH3 + CH3 N CH3 X− − CH3 − CH3 + N CH3 CH3 N X− CH2 CH2 − H NH2 CH2 CH3 CH3 CH3 N N Tautomeric shift CH3 H CH3 Cl Cl excess Cl2 21.39 + X− 545 Cl Cl + Fe Cl Cl Cl Cl Cl Cl Cl Cl H O + HA _ H OH Hexachlorophene 2,4,5-TCP HA 1. HO 2. H3O+ H Cl HO OH + OH Cl H Cl Cl Cl Cl 2,4,5-Trichlorophenol (2,4,5-TCP) 21.40 (a) Carry out the reaction in the presence of another aromatic compound, ideally another phenol. If the reaction is intermolecular, acylation of the other phenol would be expected. If that acylation does not occur, the reaction apparently is intramolecular. (In principle, this 546 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION approach should provide the answer to the question. However, the situation is more complex since other research supports an intermolecular process.) (b) The temperature effects suggest that the para product is kinetically favored with the higher temperatures leading to the thermodynamically favored (more stable) ortho product. The ortho product is stabilized by hydrogen bonding or complex formation with AlCl3 . O S 21.41 S S N NaBH4 W H H S S N X N O S Cl C6H5 O Y Raney Ni N Z HO OH HO OH 21.42 (a, b) OH BH4− O HO HO O H H H3O+ (or BH3) HO OH HO O −H2O O HO H (or −HOBH3− ) A H +O H H(or BH3− ) 21.43 (a) HOMO-1 best represents the region where electrons of the additional π bond would be found because it has a large lobe extending toward the periphery of the carbon ring, signifying the high probability of finding the additional π electrons in this region. HOMO-2 also has a small lobe inside the ring, corresponding to the second lobe expected for a typical π bond. Overall, the molecular orbital bears less resemblance to that of a simple alkene π bond, however, because of the complexity of the molecule. (b) Only a vacant orbital can accept an electron pair from a Lewis base, hence it is the LUMO+1 (next to the lowest unoccupied molecular orbital) that is involved in the nucleophilic addition step. PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 547 (c) The HOMO, HOMO-2 and HOMO-3 are the orbitals associated with the six electrons of the aromatic π system. These orbitals approximate the shape of and have the same symmetry as the three bonding aromatic π molecular orbitals of benzene. (b) LUMO+1 (a) HOMO-1 (c) HOMO HOMO-2 HOMO-3 QUIZ 21.1 Which of the following would be the strongest acid? (a) O2N OH (d) OH (b) CH3 OH (e) (c) OH OH 21.2 What products would you expect from the following reaction? OCH3 Cl OCH3 NaNH2 liq. NH3 OCH3 OCH3 NH2 (a) (b) (only) (d) Two of the above (c) (only) NH2 (e) All of the above NH2 (only) 548 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 21.3 Which of the reagents listed here would serve as the basis for a simple chemical test to distinguish between OH CH3 OH? and (a) Ag(NH3 )2 OH (b) NaOH/H2 O (c) Dilute HCl (d) Cold concd H2 SO4 (e) None of the above 21.4 Indicate the correct product, if any, of the following reaction. OH + HBr CH3 ? Br Br (a) CH3 (b) CH3 OH OH (c) CH3 Br Br OH (d) CH3 (e) None of these. Br 21.5 Complete the following synthesis: A O B NaOH I Br O O + NH2 NH2 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 21.6 Give the products: Br CH3O concd HBr heat 21.7 Select the stronger acid. O or (a) OH (1) (b) O2N OH + CH3 OH (2) or CH3 OH (1) (2) Cl (c) Cl OH (1) or OH (2) 549 The solutions in this section correspond with problems that can be found only in the WileyPLUS course for this text. WileyPLUS is an online teaching and learning solution. For more information about WileyPLUS, please visit http://www.wiley.com/college/wileyplus ANSWERS TO SECOND REVIEW PROBLEM SET These problems review concepts from Chapters 13–21. Increasing acidity 1. O O (a) < OH < O O MeO < OMe OH O H (b) OH < OH < < OH O O (c) OH O OH O OH O < (d) < OH OH OH Cl Cl Cl O NH2 O NH2 < (e) O + < (CH3)3N < < NH O 2. Increasing basicity O (a) (b) < NH2 NH2 (c) O2N NH3 < < NH2 CH3 NH2 < < NH2 < NH2 NH2 CH3 H (d) 550 < O CH3 CH3 < N CH3 CH3 NH2 ANSWERS TO SECOND REVIEW PROBLEM SET O 3. HO (1) HCN (2) LAH (a) 551 NH2 (3) H2O O (b) (1) CH3NH2, HA cat. (2) NaCNBH3 H H3C H N O NH2 N cat. HA (c) O N N (d) H cat. HA O (c) H (1) LiAlH4 (2) H3O+ O 4. O (b) (1) (2) H3O+ Li (excess) N O O O (3) 2 O (d) (1) SOCl2 (2) NH3 (excess) O OH (3) Br2, ONa, OH (1) SOCl2 (2) PhMgBr (excess) (3) H3O+ OH, H2SO4 O NH, DCC OH O (a) O N (f ) (e) ANSWERS TO SECOND REVIEW PROBLEM SET 5. (a) O OH O O O+ + + H O H O H O S O H O O + OH O H H H O O O H H O 552 − O O OH O H OH O − − (b) O O O O O O O O O O − H − −O O OH H O O HO H O H − OH HO O − O O OH ANSWERS TO SECOND REVIEW PROBLEM SET 6. (a) OH PBr3 553 Br (or HBr) O NK O Br (b) O N [from part (a)] O O H H2NNH2 heat N N + NH2 H O Br (c) NaCN LiAlH4 CN Et2O [from part (a)] NH2 O OH (d) (1) KMnO4, HO−, heat OH (2) H3O+ O CN (e) + H3O , H2O OH heat [from part (c)] O (f ) OH [from part (d)] Cl [from part (f )] SOCl2 Cl O NH3 NH2 O (h) NH4+ O O (g) + O OH Cl [from part (f )] O base O (i) (1) Br2, HO− NH2 (2) H3O+ [from part (g)] NH2 + CO32− 554 ANSWERS TO SECOND REVIEW PROBLEM SET O O ( j) Cl [from part (f )] O O (k) Zn(Hg), HCl or Wolff-Kishner reduction AlCl3 O ONa Cl [from part (f )] 2O O OEt OEt Na+ − Br (l) O − OEt (1) HO , H2O, heat OEt (2) H3O+ O [from part (a)] O O O OH OH heat −CO2 OH O Br Br2 7. (a) Et2O FeBr3 CH3 (1) CH3 (separate from ortho isomer) O (2) H3O+ MgBr Mg CH3 OH H PCC or Swern oxidation CH3 O CH3 CH3 HO− (aldol condensation) O H [(E) or (Z)] CH3 ANSWERS TO SECOND REVIEW PROBLEM SET 555 Br NBS, hv (b) HF OH (1) BH3 : THF base (2) H2O2, HO− O (1) NaH (2) Br H O NH2 N Cl2, FeCl3 2O (c) O (separate from ortho isomer) + O H OH N O Cl Br HO− H2O heat NH2 Cl NH2 Br2 H2O Cl Br (1) HONO, 0–5 °C (2) H3PO2 Br Cl Br 556 ANSWERS TO SECOND REVIEW PROBLEM SET CH3 (d) CH3 HNO3, H2SO4 (separate from ortho isomer) (1) KMnO4, HO−, heat (2) H3O+ O2N O O OH Cl SOCl2 O2N LiAlH O 3 diethyl ether −78 °C O2N O O O H HO− O2N O2N CH3 (e) Br NBS, hv (1) H2, Lindlar’s catalyst O (2) (i) Hg O (ii) NaBH4 Na [(E) or (Z )] + − HCN O 2 CN H3O+ OH heat , THF-H2O (3) H2CrO4 O O OH OH OH −H2O [(E) or (Z )] ANSWERS TO SECOND REVIEW PROBLEM SET 8. O O COEt EtO + Diels-Alder reaction OEt COEt A O Diethyl fumarate 2-Methyl-1,3butadiene 557 O + enantiomer OH Br (1) LiAlH4, Et2O PBr3 (2) H3O+ OH B Br C + enantiomer + enantiomer (1) Mg (2) H3O+ D + enantiomer 9. (a) A is , C is BrMgO OH MgBr (b) A is an allylic alcohol and thus forms a carbocation readily. B is a conjugated enyne and is therefore more stable than A. HA OH A + −H2O + H2O −H3O+ HO B 558 ANSWERS TO SECOND REVIEW PROBLEM SET 10. H O + Et2O BrMgO MgBr (2) H3O+ H2 OH Ni2B (P-2) OH D O OH 2 OH O E OH −H2O O O F O O Vitamin A acetate + HCl + 11. + HO H OH OH + OH O OH A− −HA HO OH HA H O H + −H2O OH ANSWERS TO SECOND REVIEW PROBLEM SET + OH + 559 OH A− −HA H OH HO OH HO Bisphenol A O CH3 12. (1) KMnO4, HO−, heat (2) H3 O2N O2N HO O A O N N O Cl O2N SOCl2 OH O+ C O2N B O O H2 cat. H2N O 13. Procaine O OH (1) − : Na+ N Cl O Cl Cl O (2) NH4Cl/H2O A B O NH3 NH2 O Ethinamate 560 ANSWERS TO SECOND REVIEW PROBLEM SET 14. (a) O C6H5 OH (1) C6H5MgBr (2) H3O+ H Br PBr3 C6H5 C6H5 A (CH3)2N B C6H5 OH −HBr C6H5 C6H5 C6H5 N(CH3)2 O Diphenhydramine (b) The last step probably takes place by an SN 1 mechanism. Diphenylmethyl bromide, B, ionizes readily because it forms the resonance-stabilized benzylic carbocation. H + C6H5 C6H5 Br 15. (a) For this synthesis we need to prepare the benzylic halide, Br and then allow it to react with (CH3 )2 NCH2 CH2 OH as in Problem 14. This benzylic halide can be made as follows: OH O H (1) C6H5MgBr (2) H3O Br , PBr3 + Br Br Br (b) For this synthesis we can prepare the requisite benzylic halide in two ways: CH3 O CH3 Br CH3 OH H (1) C6H5MgBr (2) H3O PBr3 + or CH3 MgBr O H (1) (2) H3O CH3 CH3 OH Br PBr3 + We shall then allow the benzylic halide to react with (CH3 )2 NCH2 CH2 OH as in Problem 14. ANSWERS TO SECOND REVIEW PROBLEM SET O O OEt 16. O O EtO EtO OEt O− CN, OEt EtO O− Br O (Michael addition) O A O O O EtO OH , HA (converts C CO2Et OEt NC O EtO N to OEt EtO O O EtO B O EtO C O O O− (Dieckmann condensation) EtO (1) HO−, Η2Ο, heat O (2) Η3Ο+ OEt O OEt D O O O O HO (3) heat −2 CΟ2 HO O Ο HCl (acid-catalyzed aldol condensation) 17. OH OH O Ο Ο Ο OEt, base (Michael addition) A O O O EtO O 561 base (intramolecular aldol condensation) O EtO C B O H3O+, heat + CO2 + (hydrolysis and decarboxylation of β-keto ester) D OH 562 ANSWERS TO SECOND REVIEW PROBLEM SET O O 18. O O H, HO− H (aldol addition) H (Cannizzaro reaction) H A OH O O OH Cl Cl O OH B H, HO− H NH3 Cl O C Cl O O NH2 O O NH2 O Meprobamate O O OH O 19. SOCl2 OH O O O A O O Cl O 2NH N O O O B C C6H5 + C6H5 20. C6H5 H2 NO2 (Diels-Alder reaction) O NO2 C6H5 H (−H2O) NH2 B A N C Pt pressure C6H5 H2 Ni pressure HN Fencamfamine ANSWERS TO SECOND REVIEW PROBLEM SET (1) OsO4 21. OH (2) NaHSO3 OH 563 CrO3 O OH A Infrared band in 3200–3550 cm−1 region OH O Infrared band in 1650–1730 cm−1 region B Notice that the second step involves the oxidation of a secondary alcohol in the presence of a tertiary alcohol. This selectivity is possible because tertiary alcohols do not undergo oxidation readily (Sections 12.4). 22. Working backward, we notice that methyl trans-4-isopropylcyclohexanecarboxylate has both large groups equatorial and is, therefore, more stable than the corresponding cis isomer. This stability of the trans isomer means that, if we were to synthesize the cis isomer or a mixture of both the cis and trans isomers, we could obtain the desired trans isomer by a base-catalyzed isomerization (epimerization): OCH3 H OCH3 O −OH O H (more stable trans isomer) (cis isomer or mixture of cis and trans isomers) Working backwards, we could synthesize a mixture of the desired isomers from phenol in the following way: OH OCH3 O O CH3OH, HA MgBr (1) CO2 Mg (2) H3O+ Et2O OH Br H2 PBr3 cat. pressure OH OH , HF (Friedel-Crafts alkylation) 564 ANSWERS TO SECOND REVIEW PROBLEM SET 23. The molecular formula indicates that the compound is saturated and is an alcohol or an ether. The IR data establish the compound as an alcohol. The 13 C spectrum of Y contains only four signals indicating that some of the carbons in Y are equivalent. The information from DEPT spectra helps us conclude that Y is 2-ethyl-1-butanol. (b) ( ( 2 (a) (a) δ 11.1 (b) δ 23.0 (c) δ 43.6 (d) δ 64.6 (d) OH (c) Notice that the most downfield signal is a CH2 group. This indicates that this carbon atom bears the OH group and that Y is a primary alcohol. The most upfield signals indicate the presence of the ethyl groups. 24. That Z decolorizes bromine indicates that Z is an alkene. We are told that Z is the more stable isomer of a pair of stereoisomers. This fact suggests that Z is a trans alkene. That the 13 C spectrum contains only three signals, even though Z contains eight carbon atoms, indicates that Z is highly symmetric. The information from DEPT spectra indicates that the upfield signals of the alkyl groups arise from equivalent isopropyl groups. We conclude, therefore, that Z is trans-2,5-dimethyl-3-hexene. (a) (a) d 22.8 (b) d 31.0 (c) d 134.5 (a) (b) (c) H 25. (a, b) OH A O O OH O OH (1) HA O (2) A−, −HA (proton transfers) B OH ANSWERS TO SECOND REVIEW PROBLEM SET 565 26. Compounds and reagents A—M, regarding the synthesis of dianeackerone: (a) OCH3 N N OCH3 N H N O B A C O O OH D E O O I O O F O G (1) CH3MgI (2) H3O + H H H CrO 2 4 O Br O O J LAH K PCC M H2, Pd L (b) Br (c) (3S,7R)-Dianeackerone O 22 CARBOHYDRATES SUMMARY OF SOME REACTIONS OF MONOSACCHARIDES CO2H Br2 (CHOH)n H2O Aldonic acid CH2OH CO2H HNO3 Aldaric acid (CHOH)n CO2H CH C6H5NHNH2 O H C C NNHC6H5 NNHC6H5 Osazone (CHOH)n−1 (CHOH)n CH2OH CH2OH CHO CN Open-chain form of aldose HCN (Kiliani-Fischer synthesis) NaBH4 (CHOH)n+1 Cyanohydrin several steps (CHOH)n+1 CH2OH CH2OH CH2OH Aldose with one more carbon Alditol (CHOH)n CH2OH CHO (1) Br2/H2O (2) H2O2/Fe3+ (CHOH)n−1 (Ruff degradation) CH2OH HO HO CH2OH O CHOH OH Cyclic form of D-glucose 566 CH3OH HA HO HO Aldose with one less carbon atom CH2OH O OH CHOCH3 Methyl glucoside (CH3)2SO4 HO− CARBOHYDRATES CH3O CH3O O O H3O+ CH2OCH3 O H2 O CHOCH3 567 CH3O CH3O H CH3O H H CHOH O CH3 H C CH2OCH3 O CH3 OCH3 H OCH3 OH CH2OCH3 SOLUTIONS TO PROBLEMS 22.1 (a) Two, CHO (b) Two, CH2OH C * CHOH O * CHOH * CHOH * CHOH CH2OH CH2OH (c) There would be four stereoisomers (two sets of enantiomers) for each general structure: 22 = 4. CHO CHO H C OH HO C H HO C H H C OH HO C H H C OH 22.2 CHO CHO H C OH HO C H CH2OH D CH2OH L CH2OH D CH2OH L CH2OH CH2OH CH2OH CH2OH C O C O C O C O H C OH HO C H HO C H H C OH H C OH HO C H H C OH HO C H CH2OH D CH2OH CH2OH CH2OH L D L 568 CARBOHYDRATES 22.3 (a) HO OH OH O OH OH H OH (b) O H H HO H OH OH OH CH2OH OH OH 22.4 (a) O HO HO and HO OH OH D-(+)-Glucose 2-Hydroxybenzyl alcohol OH OH (b) HO HO H OH O O H O + OH H H O HO HO OH O + OH Salicin OH HO OH OH H − O+ HO HO O H O HO HO + OH OH OH2 OH A −, −HA O HO HO OH OH 22.5 Dissolve -glucose in ethanol and then bubble in gaseous HCl. OH OH O HO HO H OH OH Cl O + HO O OH O HO HO −H2O H H HO O OH OH OH HO HO + OH OH HO HO H O HO HO + OH O O OH + HO HO O OH + O + OH2 CARBOHYDRATES 569 22.6 α--Glucopyranose will give a positive test with Benedict’s or Tollens’ solution because it is a cyclic hemiacetal. Methyl α--glucopyranoside, because it is a cyclic acetal, will not. O 22.7 (a) Yes OH C (b) HO HO H H (c) Yes H H OH OH O C (d) OH H H OH OH C O OH C O OH D-Mannaric acid O (e) No O H C (f ) HNO3 H OH HO H OH C HO H H OH C O OH D-Tartaric acid CH2OH D-Threose (g) The aldaric acid obtained from -erythrose is meso-tartaric acid; the aldaric acid obtained from -threose is -tartaric acid. O H 22.8 HO H H O OH O H O or OH HO H and O OH O OH H O OH H O OH H OH H OH OH H H O H OH O OH H OH or O OH OH H H H 22.9 One way of predicting the products from a periodate oxidation is to place an on each carbon atom at the point where C C bond cleavage has occurred: C OH C OH C IO4− + OH OH OH C OH O OH group 570 CARBOHYDRATES Then if we recall (Section 16.7A) that gem-diols are usually unstable and lose water to produce carbonyl compounds, we get the following results: C O H O H C O + H2O C O + H2O OH OH C Let us apply this procedure to several examples here while we remember that for every C C bond that is broken 1 mol of HIO4 is consumed. OH H + (a) HIO4 + OH H O O OH −2 H2O OH O 2 H H H H H O H H OH + + OH O + (b) HO 2 HIO4 OH H H O −3H2O H + OH OH + OH O O H OH H O H H H H OCH3 (c) HO OCH3 OH + HIO4 OH H OH + CH3O OH CH3O H OH H H −2 H2O O + CH3O O CH3O H H CARBOHYDRATES H 571 H OH H O O H + OH + + (d) HO OH 2 HIO4 O −2 H2O OH H OH OH H + OH + HO O O OH O O O HO + OH O + (e) OH H OH 2 HIO4 OH −2 H2O OH 2 + O + HO OH H O (f ) H H −2 H2O HIO4 HO OH H H H H HO H HO OH HO OH (g) O O + OH + HIO4 OH OH + OH OH O −2 H2O H + O 572 CARBOHYDRATES O H + HO (h) CHO H H OH OH CH2OH H + 3 HIO4 −3 H2O OH + HO H + O OH H OH 3H OH + HO HO H D-Erythrose O OH H H OH O HO OH H H OH H 22.10 Oxidation of an aldohexose and a ketohexose would each require 5 mol of HIO4 but would give different results. CHO CHOH HCO2H + HCO2H + HCO2H (5 HCO2H + HCHO) + HCO2H CHOH + HCO2H CHOH CHOH + 5 HIO4 + HCHO CH2OH Aldohexose CH2OH C O CHOH + 5 HIO4 CHOH CHOH CH2OH Ketohexose HCHO + CO2 + HCO2H (3 HCO2H + 2 HCHO + CO2) + HCO2H + HCO2H + HCHO CARBOHYDRATES 573 22.11 (a) Yes, -glucitol would be optically active; only those alditols whose molecules possess a plane of symmetry would be optically inactive. (b) H O H H H H OH OH OH OH OH OH OH OH OH H H H H NaBH4 OH Plane of symmetry OH Optically inactive H H HO HO H O OH H H OH OH H HO HO H NaBH4 OH H H OH OH Plane of symmetry OH Optically inactive 22.12 (a) HO H H OH CH O C H OH OH OH HO H H C6H5NHNH2 NNHC6H5 NNHC6H5 H OH OH OH (b) This experiment shows that -glucose and -fructose have the same configurations at C3, C4, and C5. 22.13 (a) H HO HO O H H OH L-(+)-Erythrose H H HO L-Glyceraldehyde OH H OH L-(+)-Threose H (b) O HO O H OH 574 CARBOHYDRATES 22.14 (a) H O H H OH OH OH D-(−)-Erythrose HCN CN H H H CN OH OH OH OH HO H H Epimeric cyanohydrins (separated) (1) Ba(OH)2 (2) H3O+ O OH H H H OH OH OH OH (1) Ba(OH)2 (2) H3O+ O HO H H OH O O H OH Epimeric γ-aldonolactones H H OH OH OH OH HO O H OH H Na-Hg, H2O pH 3-5 Na-Hg, H2O pH 3-5 H H H H OH OH OH −H2O O O OH HO H H Epimeric aldonic acids −H2O HO H OH OH OH O HO H H H H OH OH OH CARBOHYDRATES (b) O O H H H H OH OH OH OH H H H HNO3 O H HO H H H OH OH OH OH OH OH OH OH O Optically inactive D-(−)-Ribose O 575 OH HO H H H OH OH OH O Optically active HNO3 D-(−)-Arabinose 22.15 A Kiliani-Fischer synthesis starting with -(−)-threose would yield I and II. H H O H HO H O HO HO H OH H OH OH H H OH OH I II D-(+)-Xylose D-(−)-Lyxose I must be -(+)-xylose because, when oxidized by nitric acid, it yields an optically inactive aldaric acid: I HNO3 HO O H HO H OH H OH HO O Optically inactive II must be -(−)-lyxose because, when oxidized by nitric acid, it yields an optically active aldaric acid: O HO II HNO3 HO HO H HO H H OH O Optically active 576 CARBOHYDRATES O H 22.16 HO HO HO O H H H H H HO HO OH H H OH L-(+)-Ribose H HO H HO OH L-(+)-Arabinose O O H H H HO H OH H OH OH H OH OH L-(−)-Xylose L-(+)-Lyxose 22.17 Since -(+)-galactose yields an optically inactive aldaric acid, it must have either structure III or structure IV. H O H H H H OH OH OH OH OH O HO H H H H HNO3 OH OH OH OH HO O Optically inactive III O H H HO HO H OH H H OH HO HNO3 OH O H HO HO H OH H H OH HO O Optically inactive IV A Ruff degradation beginning with III would yield -(−)-ribose O H III Br2 H2O2 H2O Fe2(SO4)3 H H H OH OH OH OH D-(−)-Ribose A Ruff degradation beginning with IV would yield -(−)-lyxose: thus, -(+)-galactose must have structure IV. H IV Br2 H2O2 H 2O Fe2(SO4)3 HO HO H O H H OH OH D-(−)-Lyxose 22.18 -(+)-glucose, as shown here. O O OH H OH H OH OH H H O The other γ-lactone of D-glucaric acid O Na-Hg H HO H H OH OH H OH OH OH O H HO H H H OH O H OH OH Na-Hg pH 3-5 H HO H H O OH H OH OH OH D-(+)-Glucose CARBOHYDRATES 22.19 HO O OH OH O O O HO cat. H2SO4 (−2 H2O) OH OH OH O O O O H3O O OH HO O + KMnO4 HO− O D-Galactose O O 577 O + OH HO 2 O OH O D-Galacturonic acid Problems Carbohydrate Structure and Reactions O 22.20 (a) H (b) OH O CHOH CHOH CHOH CHOH CHOH OH CHOH OH O H (c) O (CHOH)n HO or H C (CHOH)n HO OH (e) O OH (f ) O OH CHOR O (CHOH)n CH OH H OH OH (d) C (CHOH)n (CHOH)n OH O OH 578 CARBOHYDRATES OH O OH (CHOH)n (g) CH O CH (h) CHOH CH or CHOH OH O O CHOH CHOH CHOH CHOH CHOH CH OH OH CH (i) OH CHOH O CHOH or CHOH O CH CHOH CHOH CH CHOH CHOH OH (j) Any sugar that has a free aldehyde or ketone group or one that exists as a cyclic hemiacetal. The following are examples: OH OH OH OH O H CH C C O (CHOH)n (CHOH)n CHOH O or (CHOH)n CHOH CH OH OH OH OH (CHOH)n O CH OH OH CH O (k) CHOH (l) CHOR CHOH CHOH CH CHOH O CHOH CHOR CHOH (m) Any two diastereomers that differ in configuration at only one chirality center. (See Section 22.8.) -Erythrose and -threose are examples. O H H H OH OH OH D-Erythrose O HO H H H OH OH D-Threose CARBOHYDRATES 579 (n) Cyclic sugars that differ in configuration only at the hemiacetal or acetal carbon. The following are examples: OH OH O O HO OH HO (o) HO and HO OH CH OH (p) Maltose is an example: NNHC6H5 OH NNHC6H5 C OH O OH HO (CHOH)n HO O OH O OH HO OH OH (q) Amylose is an example: OH O OH OH O OH O HO OH O OH OH O OH OH OH n (r) Any sugar in which all potential carbonyl groups are present as acetals (i.e., as glycosides). Sucrose (Section 22.12A) is an example of a nonreducing disaccharide; the methyl -glucopyranosides (Section 22.4) are examples of nonreducing monosaccharides. OH OH 22.21 (a) (b) O HO OH OH OH OH OH OCH3 (c) O CH3O OCH3 O HO OCH3 OCH3 OCH3 580 CARBOHYDRATES H 22.22 OH O H and H H HO OH OH OH O H H OH H H H H OH OH OH OH H O H + H H HIO4 H OH OCH3 O H H O H HO H H O OCH3 OH H O H H H + H O H 2 HIO4 H OCH3 + H H O O H OH OCH3 O OH OH The above would apply in the same way to the β anomers. A methyl ribofuranoside would consume only 1 mol of HIO4 ; a methyl ribopyranoside would consume 2 mol of HIO4 and would also produce 1 mol of formic acid. 22.23 One anomer of -mannose is dextrorotatory ([α]25 D = +29.3); the other is levorotatory ([α]25 = −17.0). D 22.24 The microorganism selectively oxidizes the −CHOH group of -glucitol that corresponds to C5 of -glucose. O 1 H HO H H OH H 2 3 4 5 6 OH H OH OH OH H HO H H H2 Ni D-Glucose OH H OH OH OH O2 Acetobacter suboxydans H HO H OH OH OH H OH O H OH H OH O HO H HO OH D-Glucitol L-Sorbose 22.25 -Gulose and -idose would yield the same phenylosazone as -sorbose. CH NNHC6H5 OH C HO H HO NNHC6H5 H OH H OH Same phenylosazone O C6H5NHNH2 HO H HO H OH H OH L-Sorbose O HO HO H HO H H H OH H OH L-Gulose O H HO H HO H OH H OH H OH L-Idose CARBOHYDRATES 22.26 H H H OH CH O C OH OH OH OH C6H5NHNH2 H H H NNHC6H5 O NNHC6H5 OH OH OH OH D-Psicose OH OH OH OH OH D-Allose OH O H H OH OH HO HO H H H H H H C6H5NHNH2 581 CH C C6H5NHNH2 HO HO H NNHC6H5 O H HO HO H NNHC6H5 C6H5NHNH2 H H OH OH H OH H H OH OH D-Galactose D-Tagatose 22.27 A is -altrose, B is -talose, C is -galactose. O H HO H H H OH OH H OH OH H2 Ni D-Altrose O OH OH HO H H H OH OH HO HO HO H H H H OH OH H HO OH H2 Ni HO HO HO H H H OH OH D-Talose Same alditol B C6H5NHNH2 C6H5NHNH2 C H H H H A CH H NNHC6H5 NNHC6H5 OH OH OH OH CH C Different phenylosazones HO HO H NNHC6H5 NNHC6H5 H H OH OH 582 CARBOHYDRATES O H HO HO H H CH OH H C6H5NHNH2 H C OH OH HO HO NNHC6H5 O NNHC6H5 C6H5NHNH2 H H H H H HO HO HO OH OH H H OH OH Same phenylosazone D-Galactose H D-Talose C B H2, Ni H2, Ni OH H HO HO OH H H H OH OH OH HO HO HO Different alditols H H H H OH OH (Note: If we had designated -talose as A, and -altrose as B, then C is -allose). 22.28 O H H HO H OH OH H OH OH H HO H NaBH4 or H2/Pt D-Xylose 22.29 O H HO H H OH H OH OH D-Xylitol O H OH H OH OH OH H HO H H Br2 H2O O OH OH H OH OH OH pyridine (epimerization) D-Glucose O − H2O HO HO H H O H H OH OH O Na-Hg pH 3-5 HO HO H H H H H OH OH OH D-Mannose HO HO H H OH H H OH OH OH CARBOHYDRATES 583 22.30 The conformation of -idopyranose with four equatorial OH groups and an axial CH2 OH group is more stable than the one with four axial OH groups and an equatorial CH2 OH group. OH HO OH OH O HO OH OH OH O HO Less stable More stable OH 4 Axial OH groups 1 Equatorial CH2OH group 4 Equatorial OH groups 1 Axial CH2OH group Structure Elucidation 22.31 (a) The anhydro sugar is formed when the axial cyclic acetal. OH HO OH OH O OH HA (−H2O) O OH OH CH2 OH group reacts with C1 to form a HO β-D-Altropyranose OH OH O O OH OH HO Anhydro sugar Because the anhydro sugar is an acetal (i.e., an internal glycoside), it is a nonreducing sugar. Methylation followed by acid hydrolysis converts the anhydro sugar to 2,3,4-tri-Omethyl--altrose: O O O (CH3)2SO4 O HO OH OH Anhydro β-Daltropyranose O HO− CH3O OCH3 OCH3 CH3O H H3O+ H H H H OCH3 OCH3 OH OH 2,3,4-Tri-Omethyl-D-altrose 584 CARBOHYDRATES (b) Formation of an anhydro sugar requires that the monosaccharide adopt a chair conformation with the CH2 OH group axial. With β--altropyranose this requires that two OH groups be axial as well. With β--glucopyranose, however, it requires that all four OH groups become axial and thus that the molecule adopt a very unstable conformation: Highly unstable conformation OH OH O OH HO HO O OH HA (−H2O) OH OH HO OH O β-D-Glucopyranose OH O HO O OH Anhydro-β-D-glucopyranose 22.32 1. The molecular formula and the results of acid hydrolysis show that lactose is a disaccharide composed of -glucose and -galactose. The fact that lactose is hydrolyzed by a β-galactosidase indicates that galactose is present as a glycoside and that the glycosidic linkage is beta to the galactose ring. 2. That lactose is a reducing sugar, forms a phenylosazone, and undergoes mutarotation indicates that one ring (presumably that of -glucose) is present as a hemiacetal and thus is capable of existing to a limited extent as an aldehyde. 3. This experiment confirms that the -glucose unit is present as a cyclic hemiacetal and that the -galactose unit is present as a cyclic glycoside. 4. That 2,3,4,6-tetra-O-methyl--galactose is obtained in this experiment indicates (by virtue of the free OH at C5) that the galactose ring of lactose is present as a pyranoside. That the methylated gluconic acid obtained from this experiment has a free OH group at C4 indicates that the C4 oxygen atom of the glucose unit is connected in a glycosidic linkage to the galactose unit. Now only the size of the glucose ring remains in question, and the answer to this is provided by experiment 5. 5. That methylation of lactose and subsequent hydrolysis gives 2,3,6-tri-O-methyl-glucose—that it gives a methylated glucose derivative with a free OH at C4 and C5 demonstrates that the glucose ring is present as a pyranose. (We know already that the oxygen at C4 is connected in a glycosidic linkage to the galactose unit; thus, a free OH at C5 indicates that the C5 oxygen atom is a part of the hemiacetal group of the glucose unit and that the ring is six membered.) CARBOHYDRATES OH 22.33 OH HO O H H OH H O CH2 H O OH H H H OH H HO OH H HO 585 O or HO OH O O HO OH HO OH OH 6-O-(α--Galactopyranosyl)--glucopyranose We arrive at this conclusion from the data given: 1. That melibiose is a reducing sugar and that it undergoes mutarotation and forms a phenylosazone indicate that one monosaccharide is present as a cyclic hemiacetal. 2. That acid hydrolysis gives -galactose and -glucose indicates that melibiose is a disaccharide composed of one -galactose unit and one -glucose unit. That melibiose is hydrolyzed by an α-galactosidase suggests that melibiose is an α--galactosyl--glucose. 3. Oxidation of melibiose to melibionic acid and subsequent hydrolysis to give -galactose and -gluconic acid confirms that the glucose unit is present as a cyclic hemiacetal and that the galactose unit is present as a glycoside. (Had the reverse been true, this experiment would have yielded -glucose and -galactonic acid.) Methylation and hydrolysis of melibionic acid produces 2,3,4,6-tetra-O-methyl-galactose and 2,3,4,5-tetra-O-methyl--gluconic acid. Formation of the first product—a galactose derivative with a free OH at C5—demonstrates that the galactose ring is six membered; formation of the second product—a gluconic acid derivative with a free OH at C6—demonstrates that the oxygen at C6 of the glucose unit is joined in a glycosidic linkage to the galactose unit. 4. That methylation and hydrolysis of melibiose gives a glucose derivative (2,3,4-tri-Omethyl--glucose) with free OH groups at C5 and C6 shows that the glucose ring is also six membered. Melibiose is, therefore, 6-O-(α--galactopyranosyl--glucopyranose. 22.34 Trehalose has the following structure: OH O H HO H OH H H H OH H H H O OH H HO O H or OH OH α -D-Glucopyranosyl-α -D-glucopyranoside OH HO HO O OH HO O OH O OH OH 586 CARBOHYDRATES We arrive at this structure in the following way: 1. Acid hydrolysis shows that trehalose is a disaccharide consisting only of -glucose units. 2. Hydrolysis by α-glucosidases and not by β-glucosidases shows that the glycosidic linkages are alpha. 3. That trehalose is a nonreducing sugar, that it does not form a phenylosazone, and that it does not react with bromine water indicate that no hemiacetal groups are present. This means that C1 of one glucose unit and C1 of the other must be joined in a glycosidic linkage. Fact 2 (just cited) indicates that this linkage is alpha to each ring. 4. That methylation of trehalose followed by hydrolysis yields only 2,3,4,6-tetra-O-methyl-glucose demonstrates that both rings are six membered. 22.35 (a) Tollens’ reagent or Benedict’s reagent will give a positive test with -glucose but will give no reaction with -glucitol. (b) -Glucaric acid will give an acidic aqueous solution that can be detected with blue litmus paper. -Glucitol will give a neutral aqueous solution. (c) -Glucose will be oxidized by bromine water and the red brown color of bromine will disappear. -Fructose will not be oxidized by bromine water since it does not contain an aldehyde group. (d) Nitric acid oxidation will produce an optically active aldaric acid from -glucose but an optically inactive aldaric acid will result from -galactose. (e) Maltose is a reducing sugar and will give a positive test with Tollens’ or Benedict’s solution. Sucrose is a nonreducing sugar and will not react. (f) Maltose will give a positive Tollens’ or Benedict’s test; maltonic acid will not. (g) 2,3,4,6-Tetra-O-methyl-β--glucopyranose will give a positive test with Tollens’ or Benedict’s solution; methyl β--glucopyranoside will not. (h) Periodic acid will react with methyl α--ribofuranoside because it has hydroxyl groups on adjacent carbons. Methyl 2-deoxy-α--ribofuranoside will not react. 22.36 That the Schardinger dextrins are nonreducing shows that they have no free aldehyde or hemiacetal groups. This lack of reaction strongly suggests the presence of a cyclic structure. That methylation and subsequent hydrolysis yields only 2,3,6-tri-O-methyl--glucose indicates that the glycosidic linkages all involve C1 of one glucose unit and C4 of the next. That α-glucosidases cause hydrolysis of the glycosidic linkages indicates that they are α-glycosidic linkages. Thus, we are led to the following general structure. CARBOHYDRATES 587 OH O O OH HO O OH HO O OH OH O OH HO OH HO O n O O OH n = 3, 4, or 5 Note: Schardinger dextrins are extremely interesting compounds. They are able to form complexes with a wide variety of compounds by incorporating these compounds in the cavity in the middle of the cyclic dextrin structure. Complex formation takes place, however, only when the cyclic dextrin and the guest molecule are the right size. Anthracene molecules, for example, will fit into the cavity of a cyclic dextrin with eight glucose units but will not fit into one with seven. For more information about these fascinating compounds, see Bergeron, R. J., “Cycloamyloses,” J. Chem. Educ. 1977, 54, 204–207. 22.37 Isomaltose has the following structure: OH O H HO H OH H H OH OH H HO or HO O H HO O H OH H OH H OH 6-O-(α-D-Glucopyranosyl)-D-glucopyranose O OH O HO HO O OH OH (1) The acid and enzymic hydrolysis experiments tell us that isomaltose has two glucose units linked by an α linkage. (2) That isomaltose is a reducing sugar indicates that one glucose unit is present as a cyclic hemiacetal. (3) Methylation of isomaltonic acid followed by hydrolysis gives us information about the size of the nonreducing pyranoside ring and about its point of attachment to the reducing ring. The formation of the first product (2,3,4,6-tetra-O-methyl--glucose)—a compound with an OH at C5—tells us that the nonreducing ring is present as a pyranoside. The formation of 2,3,4,5-tetra-O-methyl--gluconic acid—a compound with an OH at C6— shows that the nonreducing ring is linked to C6 of the reducing ring. 588 CARBOHYDRATES (4) Methylation of maltose itself tells the size of the reducing ring. That 2,3,4-tri-O-methyl-glucose is formed shows that the reducing ring is also six membered; we know this because of the free OH at C5. 22.38 Stachyose has the following structure: OH O HO H OH Hydrolysis here by an α-galactosidase yields D-galactose and raffinose H H O H H HO HO O H OH D-Galactose Hydrolysis here by an α-galactosidase yields sucrose H H H H HO D-Galactose O O H H H OH O HO Raffinose OH H H OH H OH H O HO OH H D-Fructose D-Glucose Sucrose Raffinose has the following structure: OH Hydrolysis here by an α-galactosidase O H HO yields D-galactose and sucrose H Hydrolysis here by an invertase H OH O H yields melibiose and fructose OH H HO O H H H Melibiose O D-Galactose H OH H H HO O OH HO OH H OH H D-Fructose D-Glucose Sucrose The enzymatic hydrolyses (as just indicated) give the basic structure of stachyose and raffinose. The only remaining question is the ring size of the first galactose unit of stachyose. That methylation of stachyose and subsequent hydrolysis yields 2,3,4,6-tetra-O-methyl-galactose establishes that this ring is a pyranoside. CARBOHYDRATES 589 22.39 Arbutin has the following structure: OH O OH O OH OH or O HO O HO HO OH OH p-Hydroxyphenyl-β-D-glucopyranoside OH Compounds X, Y, and Z are hydroquinone, p-methoxyphenol, and p-dimethoxybenzene, respectively. OH (a) (b) (a) Singlet δ 7.9 [2H] (b) Singlet δ 6.8 [4H] OH (a) X Hydroquinone OH (a) (b) OCH3 (c) Y p-Methoxyphenol (a) Singlet δ 4.8 [1H] (b) Multiplet δ 6.8 [4H] (c) Singlet δ 3.9 [3H] OCH3 (a) (b) OCH3 (a) Z p-Dimethoxybenzene (a) Singlet δ 3.75 [6H] (b) Singlet δ 6.8 [4H] 590 CARBOHYDRATES The reactions that take place are the following: OH OH O O OH O H3O+ or OH + HO OH β-glucosidase HO OH OH HO X Hydroquinone OH D-Glucose OH OCH3 O Arbutin (CH3)2SO4 (excess) OCH3 O H3O+ OCH3 HO− CH3O OCH3 OCH3 O OH + OCH3 HO CH3O Y OCH3 2,3,4,6-Tetra-O-methyl-D-glucose p-Methoxyphenol (CH3)2SO4 HO− OCH3 p-Methoxyphenol CH3O OCH3 Z p-Dimethoxybenzene 22.40 Aldotetrose B must be -threose because the alditol derived from it (-threitol) is optically active (the alditol from -erythrose, the other possible -aldotetrose, would be meso). Due to rotational symmetry, however, the alditol from B (-threitol) would produce only two 13 C NMR signals. Compounds A-F are thus in the family of aldoses stemming from -threose. Since reduction of aldopentose A produces an optically inactive alditol, A must be -xylose. The two diastereomeric aldohexoses C and D produced from A by a KilianiFischer synthesis must therefore be -idose and -gulose, respectively. E and F are the alditols derived from C and D, respectively. Alditol E would produce only three 13 C NMR signals due to rotational symmetry while F would produce six signals. 22.41 There are four closely spaced upfield alkyl signals in the 13 C NMR spectrum (δ 26.5, δ 25.6, δ 24.9, δ 24.2), corresponding to the four methyls of the two acetonide protecting groups. (The compound is, therefore, the 1,2,5,6-bis-acetonide of mannofuranose, below.) O O HO O O O CARBOHYDRATES 591 22.42 The final product is the acetonide of glyceraldehyde (below); two molar equivalents are formed from each molar equivalent of the 1,2,5,6-bis-acetonide of mannitol. O O H O Challenge Problems 22.43 The β-anomer can hydrogen bond intramolecularly, as in: O CH3 H O O H H H H O In contrast, the α-anomer can only hydrogen bond intermolecularly, leading to a higher boiling point. 22.44 Ο OH H H Ο Η O OMs O H H H3O+ Ο H Η H2O Ο II H MsO H OH H OH H OH OH HO − H2O O − H MsO H OH H OH H OH OH [III] H O H H H OH H OH OH H OH H H OH O H + H OH OH OH IV 22.45 (a) The proton at Cl. (b) Because of the single neighboring hydrogen (at C2). (c and d). OAc H OAc H O O AcO AcO H H H AcO AcO H β-anomer V OAc H H AcO AcO H α-anomer VI H OAc 592 CARBOHYDRATES QUIZ 22.1 Supply the appropriate structural formula or complete the partial formula for each of the following: (a) (b) O (c) H (d) H O C C C C C C C OH OH A ketotetrose (e) A D-sugar An L-sugar (f) (g) An aldose (h) H O H H HO H OH OH H OH OH D-Gulose α-D-Gulopyranose β-D-Gulopyranose The compound that gives the same osazone as D-gulose The compound that gives the same aldaric acid as D-gulose 22.2 Which of the following monosaccharides yields an optically inactive alditol on NaBH4 reduction? H O HO HO H H H H OH OH OH A H O HO H HO H H OH H OH OH B H O H HO HO H OH H H OH OH C H O HO HO HO H H H H OH OH D Answer: CARBOHYDRATES 593 22.3 Give the structural formula of the monosaccharide that you could use as starting material in the Kiliani-Fischer synthesis of the following compound: O Kiliani-Fischer synthesis H HO H H OH H OH OH + epimer 22.4 The -aldopentose, (a), is oxidized to an aldaric acid, (b), which is optically active. Compound (a) undergoes a Ruff degradation to form an aldotetrose, (c), which undergoes oxidation to an optically inactive aldaric acid, (d). What are the structural formulas of (a), (b), (c), and (d)? (a) (b) HNO3 β-Pyranose form of (a) Ruff degradation (c) (d) HNO3 22.5 Give the structural formula of the β-pyranose form of (a) in the space just given. 22.6 Complete the following skeletal formulas and statements by filling in the blanks and circling the words that make the statements true. 594 CARBOHYDRATES The Haworth and conformational formulas of the β-cyclic hemiacetal O of H HO HO H H H OH H OH O O are OH (a) and (b) (Fill in groups as appropriate) D-Mannose This cyclic hemiacetal is (c) reducing, nonreducing; on reaction with Br2 /H2 O it gives an optically (d) active, inactive (e) aldaric, aldonic acid. On reaction with dilute HNO3 it gives an optically (f) active, inactive (g) aldaric, aldonic acid. Reaction of the cyclic hemiacetal with (h) converts it into an optically (i) active, inactive alditol. 22.7 Outline chemical tests that would allow you to distinguish between: O (a) Glucose H HO H H H OH H OH OH OH O and galactose H HO HO H H OH H H OH OH CARBOHYDRATES (b) Glucose and fructose HO H H 595 OH O H OH OH OH 22.8 Hydrolysis of (+)-sucrose (ordinary table sugar) yields (a) -glucose (b) -mannose (c) -fructose (d) -galactose (e) More than one of the above. 22.9 Select the reagent needed to perform the following transformation: OH OH O O OH OCH3 ? HO HO OH OH OH OH O (a) CH3OH, KOH (b) (d) CH3OH, HCl (e) 2O CH3OCH3, HCl (c) (CH3)2SO4, HO− 23 LIPIDS SOLUTIONS TO PROBLEMS 23.1 (a) There are two sets of enantiomers, giving a total of four stereoisomers: H Br H Br OH OH 5 7 Br 5 H O 7 Br O H + Br 5 + Br H H OH 7 5 H Br O erythro 7 Br OH H O threo Br− (a) (b) 5 OH 7 H (b) + Br2 O H H (a) H OH 6 Br O H + H OH H 6 O Br Br O H H H 6 OH H OH 4 (±)-threo-9,10-Dibromohexadecanoic acids 596 O Br 4 Br Br Br 4 O Br H (b) Br 4 OH 6 4 6 LIPIDS 597 Formation of a bromonium ion at the other face of palmitoleic acid gives a result such that the threo enantiomers are the only products formed (obtained as a racemic modification). The designations erythro and threo come from the names of the sugars called erythrose and threose (Section 22.9A). 23.2 (a,b) H H H β-Selinene (a sesquiterpene) Zingiberene (a sesquiterpene) Caryophyllene (a sesquiterpene) Squalene (a triterpene) H O + O (1) O3 23.3 (a) H (2) Me2S H + O O Myrcene O H + H H O (b) H O H H (1) O3 O + (2) Me2S O O Limonene (c) α-Farnesene (see Section 23.3) O O (1) O3 + (2) Me2S H O O O O O + + H H + H O H H 598 LIPIDS O (1) O3 (d) Geraniol (see Section 23.3) O + (2) Me2S H O O + O (1) O3 (e) Squalene (see Section 23.3) O + 2 (2) Me2S OH H H H O O + 4 H O 23.4 (a) O (b) + CO2 (+ further oxidation products) Cl H (c) (d) H H OH OH Cl (+ rearranged products) 23.5 Br2 or KMnO4 at room temperature. Either reagent would give a positive result with geraniol and a negative result with menthol. (a) (a) H (e) H 23.6 (e) H H (a) (e) H H (a) H(a) H (e) H (a) CH3 (e) H H H (a) H H (a) (a) H (e) H (e) H (a) 5α Series (e) H H (a) H (a) CH3 R LIPIDS 599 (a) (e) H (a) H CH3 H(a) (a) H (a) (e) H (a) H H(e) H H H H(a) (a) H (e) H(e) H3C (a) H3C H(a) H (a) O H H H 3α-Hydroxy-5α-androstan-17-one (androsterone) ∗ OH C CH H H H HO H3C (b) HO 23.8 CH3 R 5β Series H(e) 23.7 H H(a) H(a) H(e) (e) H O 17α-Ethynyl-17β-hydroxy-5(10)-estren-3-one (norethynodrel) CH3 ∗ H3C H ∗ ∗ H3C ∗ H ∗ ∗ ∗ H H H Absolute configuration of cholesterol (5-cholesten-3β-ol) 23.9 Estrone and estradiol are phenols and thus are soluble in aqueous sodium hydroxide. Extraction with aqueous sodium hydroxide separates the estrogens from the androgens. 23.10 (a) H H3C H H3C Br2 H H HO HO Br Cholesterol Br 5α,6β-Dibromocholestan-3β-ol 600 LIPIDS (b) H3C H3C H H H H2O HO HO OH O OH Cholestan-3β,5α,6β-triol 5α,6α-Epoxycholestan-3β-ol (prepared by epoxidation of cholesterol; cf. Section 23.4G) (c) H H3C H3C H2CrO4 acetone H H H HO O H H 5α-Cholestan-3β-ol (prepared by hydrogenation of cholesterol; cf. Section 23.4G) H3C (d) H H3O+ H 5α-Cholestan-3-one H3C BH3:THF (cf. Sect. 23.4G) H H H HO HO H Cholesterol B O OD H3C H H HO H D 6α-Deuterio-5α-cholestan-3β-ol (e) H H3C H H3C HBr H H HO HO O 5α,6α-Epoxycholestan-3β-ol (cf. (b) above) OH Br 6β-Bromocholestan-3β,5α-diol LIPIDS 23.11 (a) HO O O OH + OH 601 + H3PO4 + OH R OH R' + + (b) HO + OH HO O O OH N(CH3)3 X − + H3PO4 + OH R OH R' + (c) OH HO O O + OH + n + H H3PO4 OH R' + + N(CH3)3 X − HO O 23.12 (a) 16 O OH HA OH + O OH SOCl2 16 H2O O OH Cl 16 16 O O O (b) + O 16 O 16 NH2 HO OH Cl + O 16 HCl (from a) O (c) 16 O Cl NH3 (excess) 16 O (d) 16 NH2 + NH4Cl O Cl CH3 (CH3)2NH (excess) 16 N CH3 O (e) NH2 (from c) 16 (1) LiAlH4, Et2O NH2 16 (2) H2O O (f ) 16 NH2 Br2 HO− 15 O (g) O LiAlH O 16 NH2 Cl (from a) Et2O, −78 °C 3 16 H + + (CH3)2NH2 Cl− + HCl 602 LIPIDS HO O (h) 16 + (1) LiAlH4, Et2O O (from a) OH 16 (2) H2O O Cl 16 O 16 O 16 O (i) 16 (1) LiAlH4, Et2O OH OH 16 (2) H2O O 16 H2 O O (j) + OH 16 Ni pressure OH OH (1) CH3MgI, Et2O (2) H O, H O+ H (from g) 16 2 16 3 O H2CrO4 acetone 16 (k) 16 PBr3 OH Br 16 (from h) (l) 16 (1) NaCN Br (from k) 23.13 (a) OH (2) H3O+ , heat 16 O H 11 Br2 H3O+ Br H (2) H3O+ 11 O OH 11 OH (1) HO−, heat OH (2) H3O+ 11 O Br OH 11 (1) NaCN (2) H3O+ O (from a) OH 11 O (from a) CN OH 11 O Br (d) OH 11 O O (from a) (c) Br O Br (b) (1)Ag2O (1) NH3 (excess) (2) H3O+ + NH2 NH3 O− OH or 11 11 O O LIPIDS Br OH 4 OH Br2 6 23.14 (a) 7 5 O (E) or (Z ) 603 O Br O OH 6 (b) 4 OH OH 6 (c) 4 4 7 5 O OH HCl 7 5 O (E) or (Z ) O− OH OH 6 (1) OsO4, pyridine (2) NaHSO3/H2O O (E) or (Z ) (d) OH 14 Ni O (E) or (Z ) H2 O Cl + Cl OH 7 5 O 23.15 Elaidic acid is trans-9-octadecenoic acid (trans-octadec-9-enoic acid): O 7 7 OH It is formed by the isomerization of oleic acid. 23.16 A reverse Diels-Alder reaction takes place. heat + 23.17 (a) OH 9 7 O O and 9 7 OH (b) Infrared spectroscopy (c) A peak in the 675–730 cm−1 region would indicate that the double bond is cis; a peak in the 960–975 cm−1 region would indicate that it is trans. 604 LIPIDS CH3 CH2 23.18 α-Phellandrene β-Phellandrene Note: On permanganate oxidation, the CH2 group of β-phellandrene is converted to CO2 and thus is not detected in the reaction. Roadmap Syntheses H 23.19 + NaNH2 :− Na+ liq. NH3 −33 °C A I Cl Cl NaCN B CN C KOH H2O O OK D O H3O+ OH E H2 Pd-BaSO4 (Lindlar’s catalyst) OH O Vaccenic acid LIPIDS 23.20 Br F 605 :− Na+ + H H F F (1) NaNH2, liq. NH3 Cl F Cl (2) I NaCN G CN F (1) KOH (2) H3O+ H O OH F I H2 OH F Ni2B (P-2) O 23.21 O H H3C H C6H5 H H3C OOH H O H 5α-Cholest-2-ene H A CH3 Br Br H H3C HBr H HO H H HO H B H H Here we find that epoxidation takes place at the less hindered α face (cf. Section 23.4G). Ring opening by HBr takes place in an anti fashion to give a product with diaxial substituents. O 23.22 (a) H (b) BuLi, Et2 O (d) NC (c) Br (e) Michael addition using a basic catalyst. NO2 606 LIPIDS 23.23 First: an elimination takes place, O O + R3N − NH + R3N 2 + NH3 + Then a conjugate addition occurs, followed by an aldol addition: O − − NH 2 +HA O O then dehydration − NH 2 aldol O O O HA, heat −H2O OH O A cyperone Challenge Problems OH O OH O O 23.24 12 H O 12 A C O O O O O OH 12 Cl E O O + 12 O 12 D O O Pahutoxin OH 12 B N(CH3)3 Cl − LIPIDS 23.25 (a, b) The reaction is an intramolecular transesterification. H O + H OH O OH 14 14 O O H H OClO3 (1) A −, −HA (2) HA OH O 14 + O O O OH OH O −HOClO3 F O + 14 O H H − OClO 3 QUIZ 23.1 Write an appropriate formula in each box. (a) A naturally occurring fatty acid (c) A solid fat (e) A synthetic detergent (b) A soap (d) An oil (f ) 5α-Estran-17-one − OClO 3 607 608 LIPIDS 23.2 Give a reagent that would distinguish between each of the following: (a) Pregnane and 20-pregnanone (b) Stearic acid and oleic acid (c) 17α-Ethyny1-1,3,5(10)-estratriene-3,17β-diol (ethynylestradiol) and 1,3,5(10)estratriene-3,17β-diol (estradiol) 23.3 What product would be obtained by catalytic hydrogenation of 4-androstene? LIPIDS 609 23.4 Supply the missing compounds: Br 4 NaNH2 :− Na+ H (a) (b) I Cl 5 (d) (c) CN 4 (1) KOH, H2O, heat 6 (f ) (e) (2) H3O+ palmitoleic acid 23.5 The following compound is a: (a) Monoterpene (b) Sesquiterpene (d) Triterpene (e) Tetraterpene (c) Diterpene 23.6 Mark off the isoprene units in the previous compound. 23.7 Which is a systematic name for the steroid shown here? (a) 5α-Androstan-3α-ol (b) 5β-Androstan-3β-ol (c) 5α-Pregnan-3α-ol H (d) 5β-Pregnan-3β-ol (e) 5α-Estran-3α-ol HO H CH3 H 24 AMINO ACIDS AND PROTEINS SOLUTIONS TO PROBLEMS O 24.1 (a) O O HO (b) OH O −O O− + NH 3 O O (c) NH2 predominates at the isoelectric point rather than O− HO + NH 3 O O −O because of the acid-strengthening inductive effect of the α-aminium group. OH + NH 3 (d) Since glutamic acid is a dicarboxylic acid, acid must be added (i.e., the pH must be made lower) to suppress the ionization of the second carboxyl group and thus achieve the isoelectric point. Glutamine, with only one carboxyl group, is similar to glycine or phenylalanine and has its isoelectric point at a higher pH. 24.2 The conjugate acid is highly stabilized by resonance. + NH HA R N H NH2 NH2 NH2 R R N H NH2 N NH2 + H NH2 R + N H 610 NH2 AMINO ACIDS AND PROTEINS O O N OEt EtO NK + 24.3 (a) O EtO O O O Br O EtO O O O OEt ONa NaOH heat N Br 611 OEt O O O O− O H N O O HCl heat O− O O− + CO2 + NH3 O DL-Leucine O− OH + OH O O O OEt N (b) OEt O O O OEt ONa CH3I NaOH heat N O O O O OEt O O− H N O O O O O− O− HCl heat O− + NH3 DL-Alanine + CO2 OH + OH O 612 AMINO ACIDS AND PROTEINS O O O O OEt OEt ONa N (c) C 6H 5 OEt O N Br C6H5 OEt O O O O O− H N NaOH heat O O O− C6H5 O HCl heat + CO2 C6H5 O− O O− + NH3 DL-Phenylalanine O OH + OH O H 24.4 (a) CN NH3 HCN O H3O+ NH2 O Phenylacetaldehyde O− + NH 3 DL-Phenylalanine O (b) O H CH3SH + base O CN CH3S NH2 H CH3S H3O+ CH3S NH3 HCN O− + NH3 DL-Methionine 24.5 Because of the presence of an electron-withdrawing 2,4-dinitrophenyl group, the labeled amino acid is relatively nonbasic and is, therefore, insoluble in dilute aqueous acid. The other amino acids (those that are not labeled) dissolve in dilute aqueous acid. AMINO ACIDS AND PROTEINS NO2 H O N+ 24.6 (a) H3 O N N F O2N O− HCO3− O H VAG NO2 H H O N N N H O2N NO2 H O + Labeled valine (separate and identify) O O− + O2N N+ H3 + O− NH3 Alanine Glycine NO2 H O N heat O OH NO2 H H3O+ O− O N (b) O O N OH O− + + O2N O2N Labeled valine ε-Labeled lysine O + N+ H3 O− Glycine H O 24.7 N C S Phenyl isothiocyanate + H2N O N N O− O H SCH3 H HN MIR N NH2 HO− NH3 613 614 AMINO ACIDS AND PROTEINS H H N N H O N N S O H3O+ O− O H H SCH3 N HN NH2 S N N H + H N H2N O O O− O SCH3 Phenylthiohydantoin derived from methionine H N HN NH2 IR S (1) N C O − S, HO N N O+ (2) H3 H + H2N O− O H Phenylthiohydantoin derived from isoleucine N HN + R NH3 24.8 (a) Two structures are possible with the sequence ECG. Glutamic acid may be linked to cysteine through its α-carboxyl group. + NH H O 3 HO N O O SH or through its γ -carboxyl group, + H NH3 O −O N O O O− N H N O O− O H SH (b) This result shows that the second structure is correct, that in glutathione the γ -carboxyl group is linked to cysteine. AMINO ACIDS AND PROTEINS 615 24.9 We look for points of overlap to determine the amino acid sequence in each case. (a) ST TO PS PSTO (b) AC CR RV LA LACRV 24.10 Sodium in liquid ammonia brings about reductive cleavage of the disulfide linkage of oxytocin to two thiol groups; then air oxidizes the two thiol groups back to a disulfide linkage: R R R liq. NH3 S O2 SH Na S S S SH R R R 24.11 Removal of the Fmoc group initially involves an elimination reaction promoted by piperidine to form the carbamic acid derivative of the amino acid and 9-methylidenefluorene, which reacts further with piperidine by nucleophilic addition to form the byproduct. Spontaneous decarboxylation of the carbamic acid generates CO2 and the free amino acid. R O N H R O H O OH N H + + − N O H OH N O O H H R O + HO OH N N O H N H + R CO2 + OH H2N O 616 AMINO ACIDS AND PROTEINS O + O 24.12 H3N O− + HO− O O Di-tert-butyl carbonate Glycine H O O ( N (1) (2) O OH O Boc-G Cl H O O N O O + H3N OEt Valine O Mixed anhydride H O O Boc-GV H O −CO2, − N H O O N O H Mixed anhydride )3N Cl OEt OEt H 3N Alanine O N N O N O H − CF3 O H O + CO2 + H3 OH O OH 25 °C Boc-GVA N+ O− O H O O + O O O (1) (2) O OH O N H OH ( O N OEt O− O N N O H GVA )3N O O− AMINO ACIDS AND PROTEINS 24.13 (a) 2 C6H5 Cl O + H2N NH2 Benzyl chloroformate O O OH O N N H H O O O O ( O (1) (2) O C6H5 Cl OEt O C6H5 HO− 25 °C Lysine O C6H5 O− O O N N H H )3N OEt O− O + O H3N O C6H5 −CO2, − O −O NH O O C6H5 O O N N H H C6H5 O HBr O OH, cold O −O O 2 C6H5 Br + 2 CO2 + NH + H3N NH2 KI OH 617 618 AMINO ACIDS AND PROTEINS (b) 3 C6H5 O O− O O Cl H N H2N + HO− NH2 25 °C NH C6H5 C6H5 O H N O O OH O O O N N H NH C6H5 O H N O O C6H5 ( )3N O Cl O O O + N H NH O− O O N OEt OEt O C6H5 O (1) (2) O H3N C6H5 −CO2, − OH O OH C6H5 C6H5 O H N O O O O NH N N NH O Br OH, cold O H N + 3 C6H5 C6H5 H 3 CO2 + H3N + HBr O O O H N O− NH2 NH RA 24.14 The weakness of the benzyl-oxygen bond allows these groups to be removed by catalytic hydrogenolysis. 24.15 Trifluoroacetic acid protonates the carbonyl group of the ester linkage joining the resin to the peptide. Heterolysis of the ester linkage then yields the relatively stable benzyl-type carbocation at the point of attachment. Acid hydrolysis of the amide linkages (peptide bonds) requires more stringent conditions because the fragments produced are not similarly stabilized. O H Polymer CF3 O peptide O O O Polymer peptide O O OH + Polymer O + + O peptide OH H2O Polymer O OH AMINO ACIDS AND PROTEINS Η Ο 24.16 + ΟΗ N ΗΟ 1. Add Ala-Fmoc. Fmoc Diisopropylcarbodiimide Η Ο N O 2. Purify by washing. Fmoc 3. Remove protecting group. Piperidine in DMF Ο 4. Purify by washing. NH2 O Η Ο N HO Fmoc 5. Add Phe-Fmoc. C6H5 and Diisopropylcarbodiimide C6H5 Η Ο Fmoc N O 6. Purify by washing. Ν Ο H 7. Remove protecting group. Piperidine in DMF C6H5 Η Ο N O 8. Purify by washing. ΝH2 Ο H Ο N O HO O N H Fmoc and Diisopropylcarbodiimide 9. Add protected Lys. (Side-chain amino group protected by Boc.) 619 620 AMINO ACIDS AND PROTEINS C6H5 Ο Η Ο Η N N Ν O Ο Fmoc Η 10. Purify by washing. H N O Ο 11. Remove Fmoc protecting group. Piperidine in DMF C6H5 Ο Η Ο Ν Ο H H N CF3 + 13. Detach tripeptide and remove Lys side-chain Boc group. ΟΗ Ο OH O Ο Ο −O 12. Purify by washing. NH2 N O C6H5 Ο Η NH2 N Ν Ο H + KFA H3N 14. Isolate product. AMINO ACIDS AND PROTEINS 621 Problems Structure and Reactivity 24.17 (a) Isoleucine, threonine, hydroxyproline, and cystine. + CO2− H H (b) H3N CH3 + CO2− + H CH3 H3N H and CH2 CH2 CH3 CH3 CO2− H3N H H OH CO2− + and H3N HO H H CH3 CH3 − CO2 + H2N H H CO2− + H2N and HO OH H H (With cystine, both chirality centers are α-carbon atoms; thus, according to the problem, both must have the -configuration, and no isomers of this type can be written.) (c) Diastereomers O Br O O− 24.18 (a) + NH 3 HO Br H (c) C6H5 N O O O− (b) OEt + NH 3 622 AMINO ACIDS AND PROTEINS 24.19 (a) CO2− + CO2CH3 CH2OH CO2CH3 H Cl− CH2Cl A B (C4H10ClNO3) (C4H9Cl2NO2) + HCl CH3OH H H3N CH2OH (−)-Serine (1) H3O+, H2O, heat CO2− + Na-Hg H H3N (2) HO− PCl5 Cl− H H3N CO2− + H3N dil H3O+ CH2Cl + H3N H CH3 L-(+)-Alanine C (C3H6ClNO2) CO2CH3 HO− (b) B H2N CO2CH3 NaSH H H2N H CH2Cl CH2SH D E (C4H8ClNO2) (C4H9NO2S) − (1) H3O+, H2O, heat CO2 H CH2SH + H3N (2) HO− L-(+)-Cysteine + CO2− (c) H3N CO2− NaOBr, HO− Hofmann rearrangement H CH2CNH2 H2N O O F C [from part (a)] ONa CN O OEt O OH (95% yield) N H CH2Cl O + H (C3H7N2O2) OEt OEt CO2− H3N CH2NH2 O L-(−)-Asparagine 24.20 (a) H + NH3 CN N O OEt H G concd HCl reflux 6 h (66% yield) O O O + O− HO + NH 3 DL-Glutamic acid + 2 OH OH + + NH 4 + CO2 AMINO ACIDS AND PROTEINS (b) O O O OEt OEt O NH2 H2, Ni CN N H O 623 OEt 68 °C, 1000 psi (90% yield) N O OEt H G O − OH O EtO concd HCl reflux 4 h (97% yield) N HN O O H H + −O Cl− NH3 + NH 3 DL-Ornithine hydrochloride O + + OH + CO2 OH 24.21 At pH 2–3 the γ -carboxyl groups of polyglutamic acid are uncharged. (They are present as –CO2 H groups.) At pH 5 the γ -carboxyl groups ionize and become negatively charged. (They become γ -CO2 − groups.) The repulsive forces between these negatively charged groups cause an unwinding of the α helix and the formation of the random coil. Peptide Sequencing 24.22 We look for points of overlap: FS PGF PP SPF RP FR RPPGFSPFR Bradykinin 24.23 1. This experiment shows that valine is the N-terminal amino acid and that valine is attached to leucine. (Lysine labeled at the ǫ-amino group is to be expected if lysine is not the N-terminal amino acid and if it is linked in the polypeptide through its α-amino group.) 2. This experiment shows that alanine is the C-terminal amino acid and that it is linked to glutamic acid. At this point, then, we have the following information about the structure of the heptapeptide. V L (A, K, F) E A Sequence unknown 624 AMINO ACIDS AND PROTEINS 3. (a) This experiment shows that the dipeptide, A, is: L K (b) The carboxypeptidase reaction shows that the C-terminal amino acid of the tripeptide, B, is glutamic acid; the DNP labeling experiment shows that the N-terminal amino acid is phenylalanine. Thus, the tripeptide B is : F A E Putting these pieces together in the only way possible, we arrive at the following amino acid sequence for the heptapeptide. VL LK FAE EA VLKFAEA Challenge Problem 24.24 The observation that the 1 H NMR spectrum taken at room temperature shows two different signals for the methyl groups suggests that they are in different environments. This would be true if rotation about the carbon-nitrogen bond was not taking place. CH3 δ 2.95 δ 8.05 H C O N CH3 δ 2.80 We assign the δ 2.80 signal to the methyl group that is on the same side as the electronegative oxygen atom. The fact that the methyl signals appear as doublets (and that the formyl proton signal is a multiplet) indicates that long-range coupling is taking place between the methyl protons and the formyl proton. That the two doublets are not simply the result of spin-spin coupling is indicated by the observation that the distance that separates one doublet from the other changes when the applied magnetic field strength is lowered. [Remember! The magnitude of a chemical shift is proportional to the strength of the applied magnetic field, while the magnitude of a coupling constant is not.] That raising the temperature (to 111 ◦ C) causes the doublets to coalesce into a single signal indicates that at higher temperatures the molecules have enough energy to surmount the energy barrier of the carbon-nitrogen bond. Above 111 ◦ C, rotation is taking place so rapidly that the spectrometer is unable to discriminate between the two methyl groups. AMINO ACIDS AND PROTEINS 625 QUIZ 24.1 Write the structural formula of the principal ionic species present in aqueous solutions at pH 2, 7, and 12 of isoleucine (2-amino-3-methylpentanoic acid). At pH = 2 At pH = 7 (a) At pH = 12 (b) (c) 24.2 A hexapeptide gave the following products: F O2N Hexapeptide NO2 − H3O+ N O2N HN HCO3 NO2 O OH O OH = Proline (P) Hexapeptide Hexapeptide 3N HCl, 100 °C 1N HCl, 80 °C 2 G, 1 L, 1 F, 1 P, 1 Y FGY + GFG + PLG + LGF The structure of the hexapeptide (using abbreviations such as G, L etc.) is 25 NUCLEIC ACIDS AND PROTEIN SYNTHESIS SOLUTIONS TO PROBLEMS 25.1 Adenine: NH2 N NH N N N H N N N N NH H N N H H N H N and so on Guanine: OH O H N N N N N H N N N NH2 OH N NH2 H N N N H H and so on NH Cytosine: NH2 NH NH H H N N N N N O OH N O and so on H H Thymine (R = CH3) or Uracil (R = H): O R N N H 626 OH H O R OH R N N H O N N and so on OH NUCLEIC ACIDS AND PROTEIN SYNTHESIS 627 25.2 (a) The nucleosides have an N-glycosidic linkage that (like an O-glycosidic linkage) is rapidly hydrolyzed by aqueous acid but is one that is stable in aqueous base. (b) HO N O H HO +H3O+ N O + + H2O −H3O+ OH Nucleoside HO OH + O H N + H Heterocyclic base OH −H2O +H2O HO HO + OH2 O −H3O + H2O + O O NH2 O O O O C6H5 O addition EtO O EtO O O O O NH EtO HN Michael NH + C6H5 O OH OH Deoxyribose O H OH +H3O+ OH 25.3 C6H5 HO OH OEt O amide formation − OH H EtO H H N N − O N OH N O 25.4 (a) The isopropylidene group is part of a cyclic acetal and is thus susceptible to hydrolysis by mild acid. 628 NUCLEIC ACIDS AND PROTEIN SYNTHESIS (b) It can be installed by treating the nucleoside with acetone and a trace of acid and by simultaneously removing the water that is produced. HO HO base O cat. HA −H2O + O OH OH 25.5 (a) Cordycepin is HO O SEt H H H H H H OH (3'-Deoxyadenosine) Cl Ad SOCl2 S O H OH H O Ad O H (b) HO base O O O O SEt H Ad −SO2, −Cl− H H H O S O Cl C1 S O O O H HO HO Ad Ad O O O Ad −SO2, −HCl H2 Et H H H EtS + Raney Ni S H H H H H H OH OH H H H H 2'-Deoxyadenosine H H O H −10 34Å 25.6 (a) 3 × 10 base pairs × × 10 m ∼ =1m 10 base pairs Å g (b) 6 × 10−12 ovum × 6.5 × 109 ova = 4 × 10−2 g 9 O−H N 25.7 (a) N H N O CH3 H−N N N N−H O H H Lactim form Thymine of guanine (b) Thymine would pair with adenine and thus adenine would be introduced into the complementary strand where guanine should occur. NUCLEIC ACIDS AND PROTEIN SYNTHESIS 629 25.8 (a) A diazonium salt and a heterocyclic analog of a phenol. + NH2 N N HONO N N OH N2 Η2O N N R N −N2 N N N R N N R O H N N N N Hypoxanthine nucleotide R H N H N H−N O N (b) N−H N N N O H Cytosine Hypoxanthine (c) Original double strand A T First replication H C A T Second replication No errors in daughter strands H Errors C G C 630 NUCLEIC ACIDS AND PROTEIN SYNTHESIS 25.9 H H O N N H N N N R N N O R Uracil (in mRNA) 25.10 (a) Adenine (in DNA) ACC CCC AAA AUG UCG (b) T P K M S Amino acids (c) UGG GGC UUU UAC AGC Anticodons (a) R AGA I AUA C UGC Y UGG V GUA Amino acids mRNA (b) TCT TAT ACG ACC CAT DNA (c) UCU UAU ACG ACC CAU Anticodons 25.11 mRNA Problems Nucleic Acid Structure 25.12 O N NH N N NH2 O HO NH2 O O P OH N O− N O O OH OH O NUCLEIC ACIDS AND PROTEIN SYNTHESIS 631 O 25.13 NH O N O HO NH2 O O N O− P N O N N O OH Mechanisms 25.14 Ionization of the benzoyl group at the anomeric carbon (C1) is assisted by the carbonyl oxygen of the C2 benzoyl group, resulting in a stabilized cation blocked from attack on the α-face. HO HO O O O OBz + OBz O O Ph OBz O Ph 632 NUCLEIC ACIDS AND PROTEIN SYNTHESIS 25.15 O O O O CH3O NH2 N NH2 CH3O OCH3 CH3 O OH NH N CH3 O Mitomycin A NH OH Leuco-aziridinomitosene Protonation and ring opening O OH + O O NH2 O OH CH3O NH2 CH3O + N CH3 N CH3 OH OH NH2 NH2 Resonance-stabilized cation intermediate Alkylation by N 2 of a deoxyguanosine in DNA A− CH3O CH3O N NH2 HN O OH 2'-Deoxyribose N CH3 OH NH2 Tautomerization O DNA NH2 O O A monoadduct with DNA CH3O Second alkylation by N 2 of another deoxyguanosine in DNA DNA N 2'-Deoxyribose O NH O N DNA NH CH3O N HN OH CH3 N CH3 O NH N N NH2 OH A cross-linked adduct with DNA N DNA 2'-Deoxyribose DNA NH2 H 1-Dihydromitosene A N A N OH DNA NH2 H CH3 N N NH O O O O O Deprotonation H NH2 NUCLEIC ACIDS AND PROTEIN SYNTHESIS Rib 25.16 633 Rib N N N N N N G2102 (2061) NH2 N O H H G2482 (2447) O N H N N H N O O H O H − O O N Rib N O H O H N O O O H O A2485 (2450) O O O P O − A2485 (2450) − O P N NH N O N− O N H N N Rib N N N NH H H N A2486 (2451) N NH2 N O OH O − O O P O− O O P O OH O O P O− O P O− O A SPECIAL TOPIC 13 C NMR Spectroscopy A.1 Analysis of the molecular formula C5 H10 O indicates an IHD = 1. Structural possibilities are: a C=C, a C=O, or a ring. The δ 211.0 signal must be due to a carbonyl group. The remaining C4 H10 is represented by only two signals in the alkyl group region, suggesting symmetry in the molecule and two unique carbons. Two CH3 CH2 groups must be present, with the CH3 giving rise to the signal at δ ∼10 and the CH2 giving the signal at δ ∼37. O Hence, the structure is A.2 Qualifying structures are: O O and H Each will give four 13 C NMR signals, one of which is due to the carbonyl group (δ 211.0). OH A.3 (a) Br (b) 4 signals Cl 6 signals (c) 6 signals A.4 (a) 5 signals (b) 7 signals (c) 8 signals 634 (d) 6 signals (e) 4 signals SPECIAL TOPIC A A.5 (a) Cl 4 signals Br 6 signals (b) Cl (c) Br Cl 4 signals 635 B SPECIAL TOPIC Chain-Growth Polymers B.1 (a) Ph Ph Ph Ph Ph Atactic polystyrene (Ph = C6H5) Ph Ph Ph Ph Ph Ph Syndiotactic polystyrene (Ph = C6H5) Ph Ph Ph Ph Ph Ph Isotactic polystyrene (Ph = C6H5) Ph (b) The solution of isotactic polystyrene. 636 C SPECIAL TOPIC Step-Growth Polymers SOLUTIONS TO PROBLEMS OH O2 cat. C.1 (a) O O HNO3 cat. + OH HO O + N2O O O OH 2 NH3 (b) HO + O− NH4 −O O + NH4 O O heat −2 H2O NH2 350 °C H2N 4 H2 NC CN catalyst catalyst O NH2 H2N Cl2 (c) Cl O 2 HCl CN Cl H2 NC CN 4 H2 NC (d) 2 NaCN Cl catalyst 2 NaCN 4 H2 catalyst NH2 H2N Cl H2N Ni NC CN NH2 637 638 SPECIAL TOPIC C C.2 (a) HO −O B− + OH OH + HB O OCH3 −O + RO OH O O− O OH O OCH3 RO OH O RO O O + CH3O− [R = CH3 or HO ] CH3O− + HB B− CH3OH + + OH O (b) OCH3 OCH3 HA RO A− RO O + HO − HO OH OH O OH OH OH + O H O RO +OH O OH RO CH3 CH3 O O +OH O − CH3OH OH A− RO + CH3OH HA O O [R = CH3 or HO O ] RO O OH SPECIAL TOPIC C 639 O C.3 (a) OCH3 CH3O HO + OH O (b) By high-pressure catalytic hydrogenation O C.4 etc. O O O etc. O O O O O O OH O O O O etc. O O O O O O O O O O etc. O O O O O O O O etc. O + Cl C.5 HO pyridine Cl −HCl OH O O O n Lexan SPECIAL TOPIC C C.6 (a) The resin is probably formed in the following way. Base converts the bisphenol A to a double phenoxide ion that attacks a carbon atom of the epoxide ring of each epichlorohydrin: O O 640 Cl + + − Ο Ο Cl − Cl Cl O O Ο Ο − O O then O O − O O − −2 Cl− O− O O O OH n Cl O O O (b) The excess of epichlorohydrin limits the molecular weight and ensures that the resin has epoxy ends. SPECIAL TOPIC C 641 (c) Adding the hardener brings about cross linking by reacting at the terminal epoxide groups of the resin: H N H2N OH [polymer] + NH2 O H H N N N O OH OH etc. [polymer] OH [polymer] HO H H N N OH N OH etc. [polymer] H H O O O O O O CH3 (b) To ensure that the polyester chain has O C.7 (a) N O N H n CH2 OH end groups. C.8 Because the para position is occupied by a methyl group, cross-linking does not occur and the resulting polymer remains thermoplastic (See Section C.4.) 642 SPECIAL TOPIC C C.9 OH + O + OH H OH A− HA H H + OH OH H H + OH OH OH H H (as before) OH HO OH OH H H (as before) OH HO OH HO HA OH2 −H2O + OH OH OH OH + HO OH HO H A− OH OH OH Bakelite + OH OH H etc. D SPECIAL TOPIC Thiols, Sulfur Ylides, and Disulfides SOLUTIONS TO PROBLEMS O + CH2 D.1 (a) (b) O + CH2 S + D.2 (a) S (c) + CH3SCH3 O S(CH3)2 + CH3SCH3 NH2 Br − SH (b) NH2 S S− Na+ (d) S (e) NH2 D.3 O S(CH3)2 Br + S NH2 (1) OH SH (2) H O−, H2O H2O2 S D.4 OH Br2 S Br Br SH OH NaSH HS OH 643 SPECIAL TOPIC D O O D.5 (a) Cl O 644 (b) SOCl2 SH (c) 2 C6H5 and excess KOH (d) H3 O+ O (e) OH S S H H (This step is the Friedel-Crafts alkylation of an alkene.) E SPECIAL TOPIC Thiol Esters and Lipid Biosynthesis SOLUTION TO PROBLEM E.1 HA H O −H2O H2 O + Farnesol + + + H A− Bisabolene 645 F SPECIAL TOPIC Alkaloids SOLUTIONS TO PROBLEMS F.1 (a) The first step is similar to a crossed Claisen condensation (see Section 19.2B): O O ONa OEt + O N N O N CH3 N CH3 (b) This step involves hydrolysis of an amide (lactam) and can be carried out with either acid or base. Here we use acid. H O O N N H3O+ CH3 H2O N HO H heat O CH3 N (c) This step is the decarboxylation of a β-keto acid; it requires only the application of heat and takes place during the acid hydrolysis of step (b). (d) This is the reduction of a ketone to a secondary alcohol. A variety of reducing agents can be used, including sodium borohydride, for example. OH H N NaBH4 CH3 N (e) Here we convert the secondary alcohol to an alkyl bromide with hydrogen bromide; this reagent also gives a hydrobromide salt of the aliphatic amine. Br H N HBr H heat N 646 + Br − CH3 SPECIAL TOPIC F 647 (f) Treating the salt with base produces the secondary amine; it then acts as a nucleophile and attacks the carbon atom bearing the bromine. This reaction leads to the formation of a five-membered ring and (±) nicotine. H Br N base −HBr N base −HBr CH3 N N CH3 F.2 (a) The chirality center adjacent to the ester carbonyl group is racemized by base (probably through the formation of an enolate anion that can be protonated to form both enantiomers; cf. Section 18.3A). CH3 N (b) O H O C6H5 H F.3 (a) OH CH3 N O OH OH HO Tropine H (b) Tropine is a meso compound; it has a plane of symmetry that passes through the CHOH group, the five-membered ring. H2C H2C NCH3 group, and between the two CH CH2 NCH3 CH CH2 – groups of the CH2 CHOH plane of symmetry 648 SPECIAL TOPIC F (c) CH3 N H F.4 OH CH3 CH3 N H3C + CH3 N N I− CH3I − H2O (1) Ag2O/H2O (2) heat HO Tropine H C8H13N CH3 H3C C9H16NI + − N(CH3)3 I N CH3I (1) Ag2O/H2O (2) heat C9H15N C10H18NI F.5 One possible sequence of steps is the following: O O H + CH3NH2 H −H2O, +H + H +N +H2O, −H + CH3 H O −HA O O OH OH enolization O O O OH O OH H Mannich reaction (See Section 19.8) SPECIAL TOPIC F 649 O H H CH3 OH HO O + −H2O, +H + N O +H2O, −H CH3 + OH HO O O enolization O O O + N −H + CH3 HO OH O N Mannich reaction (See Section 19.8) OH −2 CO2 O OH N CH3 O O O H CH3 N O N CH3 O Tropinone F.6 CH3O CH3O NH2 CH3O + O HO− Cl CH3O O N H P4O10 CH3O heat (−H2O) CH3O CH3O C20H25NO5 CH3O CH3O CH3O N CH3O CH3O CH3O Dihydropapaverine N Pd heat (−H2) CH3O CH3O CH3O Papaverine 650 SPECIAL TOPIC F F.7 A Diels-Alder reaction was carried out using 1,3-butadiene as the diene component. CH3O CH3O O CH3O NC heat CH3O O O H O NC + F.8 Acetic anhydride acetylates both OH groups. O O O H O H N O CH3 Heroin s F.9 (a) A Mannich-type reaction (see Section 19.8). (b) −H2O, +H+ O + HN(CH3)2 H + CH2 N(CH3)2 H N H H N(CH3)2 N(CH3)2 + N H −H+ N H Gramine G SPECIAL TOPIC Carbon-Carbon Bond-Forming and Other Reactions of Transition Metal Organometallic Compounds CO2H H3CO G.1 (a) (b) CO2CH3 O G.2 (a) + X X = Cl, Br or I X (b) + NC G.3 O B(OH)2 G.4 H + Br CF3 G.5 (a) O (b) t-Bu 651 652 SPECIAL TOPIC G O O (d) (c) CO2Et O2N t-Bu I + G.6 (a) Bu3Sn t-Bu + (b) Bu3Sn I OH 6 H3CO G.7 (a) (b) CO2CH3 Cl + G.8 (a) H Cl O Br (b) EtO + H Si(CH3)3 Br O O G.9 (a) N C6H5 OTDBMS H (b) + + N O O (c) O O (d) O OH O N + 2 C6H5 + SPECIAL TOPIC G G.10 Li Br (a) 2 (b) 2 Li CuLi + CuLi + CuI 2 653 CuLi Br I G.11 A syn addition of D2 to the trans alkene would produce the following racemic form. H EtO2C CO2Et H + D2 Rh(Ph3P)3Cl D D H EtO2C H CO2Et + H CO2Et H D D EtO2C H SPECIAL TOPIC Electrocyclic and Cycloaddition Reactions SOLUTIONS TO PROBLEMS H.1 According to the Woodward-Hoffmann rule for electrocyclic reactions of 4n π -electron systems (Section H.2A), the photochemical cyclization of cis, trans-2,4-hexadiene should proceed with disrotatory motion. Thus, it should yield trans-3,4-dimethylcyclobutene: hv H CH3 CH3 H cis, trans-2,4-Hexadiene CH3 CH3 H (+) H + enantiomer trans-3,4-Dimethylcyclobutene CH3 (+) CH3 (+) − CH3 CH3 H3C H.2 (a) H disrotatory 200° C − (+) H3C ( −) H (−) H (−) CH 3 ( −) ( +) H CH3 (+) ψ2 of a hexadiene (Section H.2A) (b) This is a thermal electrocyclic reaction of a 4n π -electron system; it proceeds with conrotatory motion. hv H.3 H3C CH3 H H H conrotatory disrotatory CH3 CH3 trans,trans-2,4-Hexadiene heat H cis-3,4-Dimethylcyclobutene H CH3 CH3 H cis,trans-2,4-Hexadiene 654 SPECIAL TOPIC H 655 Here we find that two consecutive electrocyclic reactions (the first photochemical, the second thermal), provide a stereospecific synthesis of cis,trans-2,4-hexadiene from trans,trans-2,4-hexadiene. H.4 (a) This is a photochemical electrocyclic reaction of an eight π -electron system—a 4n π -electron system where n = 2. It should, therefore, proceed with disrotatory motion. hv (disrotatory) CH3 H H CH3 CH3 CH3 H H cis-7,8-Dimethyl-1,3,5-cyclooctatriene (b) This is a thermal electrocyclic reaction of the eight π -electron system. It should proceed with conrotatory motion. heat (conrotatory) CH3 H H cis-7,8-Dimethyl-1,3,5-cyclooctatriene CH3 H CH CH3 3 H H.5 (a) This is conrotatory motion, and since this is a 4n π -electron system (where n = 1) it should occur under the influence of heat. H3CO2C heat CO2CH3 H CO2CH3 H (conrotatory) CO2CH3 H H (b) This is conrotatory motion, and since this is also a 4n π -electron system (where n = 2) it should occur under the influence of heat. heat (conrotatory) H CH3 H3C H + enantiomer (c) This is disrotatory motion. This, too, is a 4n π -electron system (where n = 1); thus it should occur under the influence of light. CH3 H H CH3 hv (disrotatory) H H H H 656 SPECIAL TOPIC H H.6 (a) This is a (4n + 2) π-electron system (where n = 1); a thermal reaction should take place with disrotatory motion: heat CH3 H H CH3 CH3 H3C (disrotatory) H H (b) This is also a (4n + 2) π -electron system; a photochemical reaction should take place with conrotatory motion. CH3 H3C H hv (conrotatory) H CH3 H CH3 H H.7 Here we need a conrotatory ring opening of trans-5,6-dimethyl-1,3-cyclohexadiene (to produce trans,cis,trans-2,4,6-octatriene); then we need a disrotatory cyclization to produce cis-5,6-dimethyl-1,3-cyclohexadiene. CH3 H hv (conrotatory) H CH3 trans-5,6-Dimethyl-1,3cyclohexadiene CH3 H H CH3 heat (disrotatory) trans,cis,trans-2,4,6Octatriene CH3 H3C H H cis-5,6-Dimethyl-1,3cyclohexadiene Since both reactions involve (4n + 2) π -electron systems, we apply light to accomplish the first step and heat to accomplish the second. It would also be possible to use heat to produce trans,cis,cis-2,4,6-octatriene and then use light to produce the desired product. SPECIAL TOPIC H 657 H.8 The first electrocyclic reaction is a thermal, conrotatory ring opening of a 4n π -electron system. The second electrocyclic reaction is a thermal, disrotatory ring closure of a (4n + 2) π -electron system. trans heat H cis H (conrotatory) H cis H cis All three double bonds are involved in the second reaction. cis heat (disrotatory) This double bond is not involved in the first reaction. A H B H H.9 (a) There are two possible products that can result from a concerted cycloaddition. They are formed when cis-2-butene molecules come together in the following ways: CH3 CH3 hv H H3C H3C H H CH3 H hv H H3C CH3 CH3 H H3C H H CH3 H H H H H CH3 CH3 CH3 CH3 H H H CH 3 CH3 658 SPECIAL TOPIC H (b) There are two possible products that can be obtained from trans-2-butene as well. CH3 H H hv H H H CH3 hv H H3C CH3 CH3 H H CH3 H H.10 This is an intramolecular [2 + 2] cycloaddition. CN CN CH3 H3C H.11 (a) CN CN H H CN CN H3C (b) H CN CN H CH3 + Enantiomer H H CH3 CH3 H H H CH3 CH3 CH3 CH3 CH3 CH3 H H3C H CH3 A APPENDIX Empirical and Molecular Formulas In the early and middle 19th century methods for determining formulas for organic compounds were devised by J. J. Berzelius, J. B. A. Dumas, Justus Liebig and Stanislao Cannizzaro. Although the experimental procedures for these analyses have been refined, the basic methods for determining the elemental composition of an organic compound today are not substantially different from those used in the nineteenth century. A carefully weighed quantity of the compound to be analyzed is oxidized completely to carbon dioxide and water. The weights of carbon dioxide and water are carefully measured and used to find the percentages of carbon and hydrogen in the compound. The percentage of nitrogen is usually determined by measuring the volume of nitrogen (N2 ) produced in a separate procedure. Special techniques for determining the percentage composition of other elements typically found in organic compounds have also been developed, but the direct determination of the percentage of oxygen is difficult. However, if the percentage composition of all the other elements is known, then the percentage of oxygen can be determined by difference. The following examples will illustrate how these calculations can be carried out. EXAMPLE A A new organic compound is found to have the following elemental analysis. Carbon Hydrogen Nitrogen Total: 67.95% 5.69 26.20 99.84% Since the total of these percentages is very close to 100% (within experimental error), we can assume that no other element is present. For the purpose of our calculation it is convenient to assume that we have a 100-g sample. If we did, it would contain the following: 67.95 g of carbon 5.69 g of hydrogen 26.20 g of nitrogen In other words, we use percentages by weight to give us the ratios by weight of the elements in the substance. To write a formula for the substance, however, we need ratios by moles. 659 660 APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS We now divide each of these weight-ratio numbers by the atomic weight of the particular element and obtain the number of moles of each element, respectively, in 100 g of the compound. This operation gives us the ratios by moles of the elements in the substance: C 67.95 g = 5.66 mol 12.01 g mol−1 H 5.69 g = 5.64 mol 1.008 g mol−1 N 26.20 g = 1.87 mol 14.01 g mol−1 One possible formula for the compound, therefore, is C5.66 H5.64 N1.87 . By convention, however, we use whole numbers in formulas. Therefore, we convert these fractional numbers of moles to whole numbers by dividing each by 1.87, the smallest number. C 5.66 = 3.03 which is ∼ 3 1.87 H 5.64 = 3.02 which is ∼ 3 1.87 N 1.87 = 1.00 1.87 Thus, within experimental error, the ratios by moles are 3 C to 3 H to 1 N, and C3 H3 N is the empirical formula. By empirical formula, we mean the formula in which the subscripts are the smallest integers that give the ratio of atoms in the compound. In contrast, a molecular formula discloses the complete composition of one molecule. The molecular formula of this particular compound could be C3 H3 N or some whole number multiple of C3 H3 N; that is, C6 H6 N2 , C9 H9 N3 , C12 H12 N4 , and so on. If, in a separate determination, we find that the molecular weight of the compound is 108 ± 3, we can be certain that the molecular formula of the compound is C6 H6 N2 . FORMULA C3 H3 N C6 H6 N2 C9 H9 N3 C12 H12 N4 MOLECULAR WEIGHT 53.06 106.13 (which is within the range 108 ±3) 159.19 212.26 The most accurate method for determining molecular weights is by high-resolution mass spectrometry (Section 9.17A). A variety of other methods based on freezing point depression, boiling point elevation, osmotic pressure, and vapor density can also be used to determine molecular weights. APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS 661 EXAMPLE B Histidine, an amino acid isolated from protein, has the following elemental analysis: Carbon Hydrogen Nitrogen Total: Difference 46.38% 5.90 27.01 79.29 20.71 (assumed to be oxygen) 100.00% Since no elements, other than carbon, hydrogen, and nitrogen, are found to be present in histidine, the difference is assumed to be oxygen. Again, we assume a 100-g sample and divide the weight of each element by its gram-atomic weight. This gives us the ratio of moles (A). (A) (B) (C) C H N O 46.38 12.01 5.90 1.008 27.01 14.01 20.71 16.00 3.86 1.29 5.85 = 5.85 1.29 1.93 = 1.93 1.29 1.29 = 1.29 1.29 = 3.86 = 2.99 × 2 = 5.98 ∼ 6 carbon atoms = 4.53 × 2 = 9.06 ∼ 9 hydrogen atoms = 1.50 × 2 = 3.00 = 3 nitrogen atoms = 1.00 × 2 = 2.00 = 2 oxygen atoms Dividing each of the moles (A) by the smallest of them does not give a set of numbers (B) that is close to a set of whole numbers. Multiplying each of the numbers in column (B) by 2 does, however, as seen in column (C). The empirical formula of histidine is, therefore, C6 H9 N3 O2 . In a separate determination, the molecular weight of histidine was found to be 158 ± 5. The empirical formula weight of C6 H9 N3 O2 (155.15) is within this range; thus, the molecular formula for histidine is the same as the empirical formula. PROBLEMS A.1 What is the empirical formula of each of the following compounds? (a) Hydrazine, N2 H4 (d) Nicotine, C10 H14 N2 (b) Benzene, C6 H6 (e) Cyclodecane, C10 H20 (c) Dioxane, C4 H8 O2 (f) Acetylene, C2 H2 A.2 The empirical formulas and molecular weights of several compounds are given next. In each case, calculate the molecular formula for the compound. 662 APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS EMPIRICAL FORMULA (a) CH2 O (b) CHN (c) CCl2 MOLECULAR WEIGHT 179 ± 5 80 ± 5 410 ± 10 A.3 The widely used antibiotic, penicillin G, gave the following elemental analysis: C, 57.45%; H, 5.40%; N, 8.45%; S, 9.61%. The molecular weight of penicillin G is 330 ± 10. Assume that no other elements except oxygen are present and calculate the empirical and molecular formulas for penicillin G. ADDITIONAL PROBLEMS A.4 Calculate the percentage composition of each of the following compounds. (a) C6 H12 O6 (b) CH3 CH2 NO2 (c) CH3 CH2 CBr3 A.5 An organometallic compound called ferrocene contains 30.02% iron. What is the minimum molecular weight of ferrocene? A.6 A gaseous compound gave the following analysis: C, 40.04%; H, 6.69%. At standard temperature and pressure, 1.00 g of the gas occupied a volume of 746 mL. What is the molecular formula of the compound? A.7 A gaseous hydrocarbon has a density of 1.251 g L−1 at standard temperature and pressure. When subjected to complete combustion, a 1.000-L sample of the hydrocarbon gave 3.926 g of carbon dioxide and 1.608 g of water. What is the molecular formula for the hydrocarbon? A.8 Nicotinamide, a vitamin that prevents the occurrence of pellagra, gave the following analysis: C, 59.10%; H, 4.92%; N, 22.91%. The molecular weight of nicotinamide was shown in a separate determination to be 120 ± 5. What is the molecular formula for nicotinamide? A.9 The antibiotic chloramphenicol gave the following analysis: C, 40.88%; H, 3.74%; Cl, 21.95%; N, 8.67%. The molecular weight was found to be 300 ± 30. What is the molecular formula for chloramphenicol? APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS 663 SOLUTIONS TO PROBLEMS OF APPENDIX A A.1 (a) NH2 (b) CH A.2 EMPIRICAL FORMULA (c) C2 H4 O EMPIRICAL FORMULA WEIGHT (d) C5 H7 N (e) CH2 (f) CH MOLECULAR WEIGHT EMP. FORM. WT. MOLECULAR FORMULA 30 179 ∼ =6 30 C6 H12 O6 (b) CHN 27 80 ∼ =3 27 C3 H3 N3 (c) 83 410 ∼ =5 83 C5 Cl10 (a) CH2 O CCl2 A.3 If we assume that we have a 100-g sample, the amounts of the elements are WEIGHT Moles (A) C 57.45 57.45 = 4.78 12.01 4.78 = 15.9 ∼ = 16 0.300 H 5.40 5.40 = 5.36 1.008 5.36 = 17.9 ∼ = 18 0.300 N 8.45 8.45 = 0.603 14.01 0.603 = 2.01 ∼ =2 0.300 S 9.61 9.61 = 0.300 32.06 0.300 = 1.00 = 1 0.300 O∗ 19.09 19.09 = 1.19 16.00 1.19 = 3.97 ∼ =4 0.300 100.00 (B) (∗ by difference from 100) The empirical formula is thus C16 H18 N2 SO4 . The empirical formula weight (334.4) is within the range given for the molecular weight (330 ± 10). Thus, the molecular formula for penicillin G is the same as the empirical formula. A.4 (a) To calculate the percentage composition from the molecular formula, first determine the weight of each element in 1 mol of the compound. For C6 H12 O6 , C6 = 6 × 12.01 = 72.06 72.06 = 0.400 = 40.0% 180.2 H12 = 12 × 1.008 = 12.10 12.10 = 0.0671 = 6.7% 180.2 O6 = 6 × 16.00 = 96.00 MW 180.16 96.00 = 0.533 = 53.3% 180.2 (MW = molecular weight) 664 APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS Then determine the percentage of each element using the formula Percentage of A = (b) Weight of A × 100 Molecular Weight 24.02 75.07 5.04 75.07 14.01 75.07 32.00 75.07 C2 = 2 × 12.01 = 24.02 H5 = 5 × 1.008 = 5.04 N = 1 × 14.01 = 14.01 O2 = 2 × 16.00 = 32.00 = 0.320 = 32.0% = 0.067 = 6.7% = 0.187 = 18.7% = 0.426 = 42.6% Total = 75.07 (c) 36.03 = 0.128 = 12.8% 280.77 5.04 = 0.018 = 1.8% 280.77 239.70 = 0.854 = 85.4% 280.77 C3 = 3 × 12.01 = 36.03 H5 = 5 × 1.008 = 5.04 Br3 = 3 × 79.90 = 239.70 Total = 280.77 A.5 If the compound contains iron, each molecule must contain at least one atom of iron, and l mol of the compound must contain at least 55.85 g of iron. Therefore, g of Fe 1.000 g × mol 0.3002 g of Fe g = 186.0 mol MW of ferrocene = 55.85 A.6 First, we must determine the empirical formula. Assuming that the difference between the percentages given and 100% is due to oxygen, we calculate: C 40.04 H 6.69 O 53.27 100.00 The empirical formula is thus CH2 O. 40.04 = 3.33 12.01 6.69 = 6.64 1.008 53.27 = 3.33 16.00 3.33 3.33 6.64 3.33 3.33 3.33 =1 ∼ =2 =1 APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS 665 To determine the molecular formula, we must first determine the molecular weight. At standard temperature and pressure, the volume of 1 mol of an ideal gas is 22.4 L. Assuming ideal behavior, MW 1.00 g = where MW = molecular weight 0.746 L 22.4 L (1.00)(22.4) MW = = 30.0 g 0.746 The empirical formula weight (30.0) equals the molecular weight; thus, the molecular formula is the same as the empirical formula. A.7 As in Problem A.6, the molecular weight is found by the equation MW 1.251 g = 1.00 L 22.4 L MW = (1.251)(22.4) MW = 28.02 To determine the empirical formula, we must determine the amount of carbon in 3.926 g of carbon dioxide, and the amount of hydrogen in 1.608 g of water. C     12.01g C 3.926 g CO2 = 1.071 g carbon 44.01 g CO2 H    2.016 g H 0.180 g hydrogen = 1.608 g H2O 18.016 g H2O 1.251 g sample The weight of C and H in a 1.251-g sample is 1.251 g. Therefore, there are no other elements present. To determine the empirical formula, we proceed as in Problem A.6 except that the sample size is 1.251 g instead of 100 g. C 1.071 = 0.0892 12.01 H 0.180 = 0.179 1.008 0.0892 =1 0.0892 0.179 =2 0.0892 The empirical formula is thus CH2 . The empirical formula weight (14) is one-half the molecular weight. Thus, the molecular formula is C2 H4 . A.8 Use the procedure of Problem A.3. C 59.10 59.10 = 4.92 12.01 4.92 = 6.02 ∼ = 6 0.817 H 4.92 4.92 = 4.88 1.008 4.88 = 5.97 ∼ = 6 0.817 666 APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS N O 22.91 22.91 = 1.64 14.01 13.07 13.07 = 0.817 16.00 100.00 1.64 =2 0.817 0.817 =1 0.817 The empirical formula is thus C6 H6 N2 O. The empirical formula weight is 122.13, which is equal to the molecular weight within experimental error. The molecular formula is thus the same as the empirical formula. A.9 C 40.88 40.88 = 3.40 12.01 3.40 = 5.5 0.619 H 3.74 3.74 = 3.71 1.008 3.71 =6 0.619 6 × 2 = 12 Cl 21.95 21.95 = 0.619 35.45 0.619 =1 0.619 1×2=2 N 8.67 8.67 = 0.619 14.01 0.619 =1 0.619 1×2=2 O 24.76 100.00 24.76 = 1.55 16.00 1.55 = 2.5 0.619 5.5 × 2 = 11 2.5 × 2 = 5 The empirical formula is thus C11 H12 Cl2 N2 O5 . The empirical formula weight (323) is equal to the molecular weight; therefore, the molecular formula is the same as the empirical formula. B APPENDIX Answers to Quizzes EXERCISE 1 O 1.1 (d) (d) 1.2 (e) 1.3 (d) 1.4 1.5 (c) 1.6 CH3 1.7 CH3CHCH2CH3 and CH3 CH3 1.8 C C CH3 O− CH3 CH3 H H H C C H 1.9 CH3 Cl (a) sp 2 (b) sp 3 1.10 (a) + 1 1.11 (b) 0 O (c) 0 1.12 H3C Cl C N H H3C H OH 1.13 (e) 0 D (c) − 1 H C (d) trigonal planar F OH OH OH unstable OCH3 O O O H O OH unstable 667 668 APPENDIX B ANSWERS TO QUIZZES EXERCISE 2 2.1 (e) 2.2 (a) 2.3 (b) O CH3 N 2.4 (a) (c) Cl (b) OH CH3 Cl Cl O CH3 (d) H (e) O O (f) Cl CH3 CH3 N (g) CH3 2.5 or O OH 2.6 (a) O (d) CH3 (b) N H (c) O OH CH3 (e) N H 2.7 (a) Isopropyl phenyl ether (b) Ethylmethylphenylamine (c) Isopropylamine EXERCISE 3 3.1 (a) 3.2 (c) 3.6 (b) 3.7 3.8 (CH3 )2 NH 3.3 (b) 3.4 (e) 3.5 (b) OH APPENDIX B 3.9 (a) CH3 CH2 Li ANSWERS TO QUIZZES (b) D2 O CH3 CH3 (b) CH3CHCH2OH 3.10 (a) CH3 Li (c) CH3CHCH2OLi EXERCISE 4 4.1 (c) 4.2 (c) 4.6 (a) 4.7 (a) Cl 4.8 4.3 (b) (a) 4.4 Cl (a) 4.5 (b) Br (b) or H Br H (c) H H Br 4.9 H2 , Pt, pressure or H2 , Ni, pressure 4.10 EXERCISE 5 5.1 (a) 5.2 (b) H CH3 5.3 (b) and 5.4 (e) H CH3 5.6 5.7 H CH3 5.8 (E and Z ) 5.9 (d) 5.10 (e) 5.5 (b) Br 669 670 APPENDIX B ANSWERS TO QUIZZES EXERCISE 6 6.1 (b) 6.2 (b) 6.3 (a) Br 6.4 Br 6.5 A Br Br H = B Br = CN C H = D OCH3 = 6.6 (b) 6.7 A = B Br 2 H2 Pt pressure = H H C = H EXERCISE 7 7.1 (c) 7.2 (d) 7.3 (a) 7.4 (a) Li, C2 H5 NH2 , −78 ◦ C, then NH4 Cl (b) H2 /Ni2 B(P−2) or (c) H2 /Ni, pressure or H2 /Pd/ CaCO3 (Lindlar’s catalyst) H2 /Pt, pressure (using at least 2 molar equivalents of H2 ) (d) C2 H5 ONa/C2 H5 OH, heat (e) (CH3 )3 COK/(CH3 )3 COH, heat > 7.5 7.6 (a) > Br (b) − Na + (e) > − Na + Br (d) Br (c) H APPENDIX B ANSWERS TO QUIZZES EXERCISE 8 8.1 (e) 8.2 (c) 8.7 (c) 8.8 (b) 8.3 (e) 8.4 (a) 8.5 (d) 8.6 (c) EXERCISE 9 9.1 (a) (b) Br Br Br Br O (d) 9.2 (c) 9.3 (a) (c) (e) 9.4 (b) 9.5 (c) 9.6 (c,d,e) NO2 9.7 EXERCISE 10 10.1 (d) 10.2 (b) 10.3 (c) 11.2 (a) 11.3 (e) 10.4 (b) . 10.5 10.6 Six EXERCISE 11 11.1 (d) ONa C= O B= O D= O OH O 11.4 A = OSO2CH3 671 672 APPENDIX B ANSWERS TO QUIZZES EXERCISE 12 12.1 (b) 12.3 A 12.2 (a) = MgBr or Li B = NaH C = CH3 I 12.4 A = OH B = PCC or Swern oxidation C O = H MgBr D 12.5 A = O = O or OR (if excess CH3MgBr is used) EXERCISE 13 13.1 (d) 13.2 (c) 13.3 (c) 13.4 (c) 14.2 (a) 14.3 (b) 14.4 (b) EXERCISE 14 14.1 (e) 14.5 Cl 14.6 Azulene EXERCISE 15 15.1 (a) 15.2 (a) 15.3 (b) 13.5 (b) APPENDIX B ANSWERS TO QUIZZES 673 CH3 NO2 15.4 (a) A = SO3 /H2 SO4 B = SO3H OH O NO2 C = H2 O, H2 SO4 , heat (b) A = SOCl2 or PCl5 D = B = + AlCl3 C = Zn(Hg), HCl, reflux D = Br2 /FeBr3 or Wolff-Kishner reduction EXERCISE 16 16.1 (d) 16.2 (b) 16.3 (b) Br 16.4 (a) = A B = NaCN (1) DIBAL-H, hexane, −78 ◦ C C (b) A = PCC or Swern oxidation B (2) H2 O = OH , H3O+ HO C = Mg Ο (c) A = (C6 H5 )3 P (d) A = 16.5 −H2O = + P(C6H5)3 Br− (1) CH3MgBr (2) H2CrO4 or Swern oxidation CH3 (excess) B Cl2 heat, hv, or peroxide B = HCN Cl H Cl HO− H2O C = C = (1) LiAlH4 , Et2 O (2) H2 O OH H OH O H The gem-diol formed in the alkaline hydrolysis step readily loses water to form the aldehyde. 674 APPENDIX B ANSWERS TO QUIZZES 16.6 The general formula for an oxime is OH N Both carbon and nitrogen are sp2 hybridized; the electron pair on nitrogen occupies one sp2 orbital. Aldoximes and ketoximes can exist in either of these two stereoisomeric forms: OH (R′)H (R′)H or N R N R OH This type of stereoisomerism is also observed in the case of other compounds that possess the C N group, for example, phenylhydrazones and imines. EXERCISE 17 17.1 (b) 17.2 (d) 17.3 (d) 17.4 A = 3-Chlorobutanoic acid B = Methyl 4-nitrobenzoate or methyl p-nitrobenzoate C = N-Methylaniline 17.5 (a) A = (1) KMnO4 , HO− , heat B = SOCl2 or PCl5 (2) H3 O+ O C O N(CH3)2 = D O− Na+ = O E O O = F N H = O 18 (b) A CH3 O = O C = O− Na+ B = CH3 18 OH APPENDIX B ANSWERS TO QUIZZES O O N NH2 = (c) A B = C = 17.6 (b) EXERCISE 18 18.1 (a) 18.2 A = B = Br N H C = + N 18.3 (c) Br− 18.4 (e) 18.5 (b) 19.2 (e) 19.3 (b) EXERCISE 19 19.1 (c) O 19.4 (a) A = O EtO OEt O C = O = = EtO K+ O OEt OH D = HO O − EtO O O E B 675 OEt O OH H 676 APPENDIX B ANSWERS TO QUIZZES OEt (b) A O O O O H = B OEt = O OH O O OH = O (c) A OEt OH O C OEt OH O = B = I O − Li+ (d) A = B Br = N H C = + Br− N O O 19.5 (a) (b) (c) CH3 MgI, Et2 O OH O (e) Zn(Hg)/HCl or Wolff-Kishner reduction (d) + simple addition 19.6 (e) APPENDIX B O ANSWERS TO QUIZZES O 19.7 (a) (b) OH 677 (c) HCN OH O 19.8 (a) O (b) OH H (d) LiAlH4 , Et2 O (c) H (E and Z ) (e) H2 , Ni, pressure OCH3 (g) (f) CH3 OH (excess), HA H (h) OCH3 O O (1) (i) OEt, LDA, −78 °C (2) H2O EXERCISE 20 20.1 (d) 20.2 (e) 20.3 (a) (2) (b) (4) (c) (3) CH3 20.4 (a) A = HNO3 /H2 SO4 B = C = NaNO2 , HCl, 0−5 ◦ C O2N N(CH3)2 D = CuCN E = LiAlH4 , Et2 O F = O − B (b) A = NaN3 N = + N NH2 N C = Br NH2 D E = H3 PO2 = Br 20.5 (a) (2) Br (b) (2) (c) (1) (d) (1) (e) (2) (f) (2) 678 APPENDIX B ANSWERS TO QUIZZES EXERCISE 21 2 1. 1 (a) (d) 2 1. 2 2 1. 3 (b) (e) 2 1. 4 OH 2 1. 5 A = B = KNH2 , liq. NH3 , −33 ◦ C Br + 21.6 Br OH 21.7 (a) (1) (b) (1) CH3Br (c) (2) EXERCISE 22 OH 22.1 (a) H (b) O H (c) O OH CHOH OH CHOH O CHOH OH on either side CHOH CHOH HO OH on either side H OH OH H H OH (d) O H (e) HO (CHOH)n OH n = 1,2,3... O HO H HO H H H OH H OH OH OH O O OH OH OH OH OH (g) (f) HO OH (h) O HO H HO HO H H OH H H OH OH APPENDIX B ANSWERS TO QUIZZES 679 22.2 C 22.3 O H HO H H OH OH O 22.4 (a) HO H H H O (b) HO H H H OH OH OH OH (c) O H OH (d) H H H OH OH O O OH OH H H OH O OH OH OH OH OH O 22.5 HO OH OH OH OH O 22.6 (a) OH (b) OH HO (c) reducing HO HO (d) active OH O OH HO (e) aldonic (f) active (g) aldaric (h) NaBH4 (i) active 22.7 (a) Galactose NaBH4 Galactose (b) dil HNO3 optically inactive alditol; glucose → optically active alditol optically inactive aldaric acid; glucose → optically active aldaric acid HIO4 oxidation different products: O Fructose 2 mol Glucose 1 mol H O H O 22.8 (e) 22.9 (d) H + + CO2 3 O H + 5 H OH H OH 680 APPENDIX B ANSWERS TO QUIZZES EXERCISE 23 O O 23.1 (a) (b) OH or C16 or C18 12 O (c) O 12 O (d) O 12 O 16 O O 7 5 7 5 7 7 O O O ONa O O 14 H3C SO3Na (e) H (f) 13 O H H H H 23.2 (a) I2 /OH− (iodoform test) (b) Br2 (c) Ethynylestradiol shows IR absorption at ∼3300 cm−1 for the terminal alkyne hydrogen. 23.3 5α-Androstane H 23.4 (a) 4 4 (c) 4 − Na+ (b) 5 Cl (d) KCN OH (f ) H2/Pd O (e) 4 23.5 (b) Sesquiterpene 6 APPENDIX B 23.6 23.7 (e) EXERCISE 24 O O 24.1 (a) OH + NH 3 O− NH2 24.2 PLGFGY O− + NH 3 O (c) (b) ANSWERS TO QUIZZES 681 C APPENDIX Molecular Model Set Exercises The exercises in this appendix are designed to help you gain an understanding of the threedimensional nature of molecules. You are encouraged to perform these exercises with a model set as described. These exercises should be performed as part of the study of the chapters shown below. Chapter in Text 4 5 7 22 24 13 14 Accompanying Exercises 1, 3, 4, 5, 6, 8, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21 2, 7, 9, 13, 24, 25, 26, 27 9, 19, 22, 28 29 30 31 23, 27 The following molecular model set exercises were originally developed by Ronald Starkey. Refer to the instruction booklet that accompanies your model set for details of molecular model assembly. Exercise 1 (Chapter 4) Assemble a molecular model of methane, CH4 . Note that the hydrogen atoms describe the apexes of a regular tetrahedron with the carbon atom at the center of the tetrahedron. Demonstrate by attempted superposition that two models of methane are identical. Replace any one hydrogen atom on each of the two methane models with a halogen to form two molecules of CH3 X. Are the two structures identical? Does it make a difference which of the four hydrogen atoms on a methane molecule you replace? How many different configurations of CH3 X are possible? Repeat the same considerations for two disubstituted methanes with two identical substituents (CH2 X2 ), and then with two different substituents (CH2 XY). Two colors of atomcenters could be used for the two different substituents. Exercise 2 (Chapter 5) Construct a model of a trisubstituted methane molecule (CHXYZ). Four different colored atom-centers are attached to a central tetrahedral carbon atom-center. Note that the carbon now has four different substituents. Compare this model with a second model of CHXYZ. Are the two structures identical (superposable)? 682 APPENDIX C MOLECULAR MODEL SET EXERCISES 683 Interchange any two substituents on one of the carbon atoms. Are the two CHXYZ molecules identical now? Does the fact that interchange of any two substituents on the carbon interconverts the stereoisomers indicate that there are only two possible configurations of a tetrahedral carbon atom? Compare the two models that were not identical. What is the relationship between them? Do they have a mirror-image relationship? That is, are they related as an object and its mirror image? Exercise 3 (Chapter 4) Make a model of ethane, CH3 CH3 . Does each of the carbon atoms retain a tetrahedral configuration? Can the carbon atoms be rotated with respect to each other without breaking the carbon-carbon bond? Rotate about the carbon-carbon bond until the carbon-hydrogen bonds of one carbon atom are aligned with those of the other carbon atom. This is the eclipsed conformation. When the C H bond of one carbon atom bisects the H C H angle of the other carbon atom the conformation is called staggered. Remember, conformations are arrangements of atoms in a molecule that can be interconverted by bond rotations. In which of the two conformations of ethane you made are the hydrogen atoms of one carbon closer to those of the other carbon? Exercise 4 (Chapter 4) Prepare a second model of ethane. Replace one hydrogen, any one, on each ethane model with a substituent such as a halogen, to form two models of CH3 CH2 X. Are the structures identical? If not, can they be made identical by rotation about the C C bond? With one of the models, demonstrate that there are three equivalent staggered conformations (see Exercise 3) of CH3 CH2 X. How many equivalent eclipsed conformations are possible? Exercise 5 (Chapter 4) Assemble a model of a 1,2-disubstituted ethane molecule, CH2 XCH2 X. Note how the orientation of and the distance between the X groups changes with rotation about the carboncarbon bond. The arrangement in which the X substituents are at maximum separation is the anti-staggered conformation. The other staggered conformations are called gauche. How many gauche conformations are possible? Are they energetically equivalent? Are they identical? Exercise 6 (Chapter 4) Construct two models of butane, . Note that the structures can be viewed as dimethyl-substituted ethanes. Show that rotations about the C2, C3 bond of butane produce eclipsed, anti-staggered, and gauche-staggered conformations. Measure the distance between C1 and C4 in the conformations just mentioned. [The scale of the Darling Framework Molecular Model Set, for example, is: 2.0 inches in a model corresponds to approximately 1.0 Å (0.1 nm) on a molecular scale.] In which eclipsed conformation are the C1 and C4 atoms closest to each other? How many eclipsed conformations are possible? 684 APPENDIX C MOLECULAR MODEL SET EXERCISES Exercise 7 (Chapter 5) Using two models of butane, verify that the two hydrogen atoms on C2 are not stereochemically equivalent. Replacement of one hydrogen leads to a product that is not identical to that obtained by replacement of the other C2 hydrogen atom. Both replacement products have the same condensed formula, CH3 CHXCH2 CH3 . What is the relationship of the two products? Exercise 8 (Chapter 4) Make a model of hexane, . Extend the six-carbon chain as far as it will go. This puts C1 and C6 at maximum separation. Notice that this straight-chain structure maintains the tetrahedral bond angles at each carbon atom and therefore the carbon chain adopts a zigzag arrangement. Does this extended chain adopt staggered or eclipsed conformations of the hydrogen atoms? How could you describe the relationship of C1 and C4? Exercise 9 (Chapters 5 and 7) Prepare models of the four isomeric butenes, C4 H8 . Note that the restricted rotation about the double bond is responsible for the cis-trans stereoisomerism. Verify this by observing that breaking the π bond of cis-2-butene allows rotation and thus conversion to trans-2-butene. Is any of the four isomeric butenes chiral (nonsuperposable with its mirror image)? Indicate pairs of butene isomers that are structural (constitutional) isomers. Indicate pairs that are diastereomers. How does the distance between the C1 and C4 atoms in trans-2-butene compare with that of the anti conformation of butane? Compare the C1 to C4 distance in cis-2-butene with that in the conformation of butane in which the methyls are eclipsed. 1-Butene cis-2-Butene trans-2-Butene 2-Methylpropene Exercise 10 (Chapter 4) Make a model of cyclopropane. Take care not to break your models due to the angle strain of the carbon-carbon bonds of the cyclopropane ring. It should be apparent that the ring carbon atoms must be coplanar. What is the relationship of the hydrogen atoms on adjacent carbon atoms? Are they staggered, eclipsed, or skewed? Exercise 11 (Chapter 4) A model of cyclobutane can be assembled in a conformation that has the four carbon atoms coplanar. How many eclipsed hydrogen atoms are there in the conformation? Torsional strain (strain caused by repulsions between the aligned electron pairs of eclipsed bonds) can be relieved at the expense of increased angle strain by a slight folding of the ring. The deviation APPENDIX C MOLECULAR MODEL SET EXERCISES 685 of one ring carbon from the plane of the other three carbon atoms is about 25◦ . This folding compresses the C C C bond angle to about 88◦ . Rotate the ring carbon bonds of the planar conformation to obtain the folded conformation. Are the hydrogen atoms on adjacent carbon atoms eclipsed or skewed? Considering both structural and stereoisomeric forms, how many dimethylcyclobutane structures are possible? Do deviations of the ring from planarity have to be considered when determining the number of possible dimethylcyclobutane structures? Exercise 12 (Chapter 4) Cyclopentane is a more flexible ring system than cyclobutane or cyclopropane. A model of cyclopentane in a conformation with all the ring carbon atoms coplanar exhibits minimal deviation of the C C C bond angles from the normal tetrahedral bond angle. How many eclipsed hydrogen interactions are there in this planar conformation? If one of the ring carbon atoms is pushed slightly above (or below) the plane of the other carbon atoms, a model of the envelope conformation is obtained. Does the envelope conformation relieve some of the torsional strain? How many eclipsed hydrogen interactions are there in the envelope conformation? Cyclopentane Exercise 13 (Chapter 5) Make a model of 1,2-dimethylcyclopentane. How many stereoisomers are possible for this compound? Identify each of the possible structures as either cis or trans. Is it apparent that cis-trans isomerism is possible in this compound because of restricted rotation? Are any of the stereoisomers chiral? What are the relationships of the 1,2-dimethylcyclopentane stereoisomers? Exercise 14 (Chapter 4) Assemble the six-membered ring compound cyclohexane. Is the ring flat or puckered? Place the ring in a chair conformation and then in a boat conformation. Demonstrate that the chair and boat are indeed conformations of cyclohexane—that is, they may be interconverted by rotations about the carbon-carbon bonds of the ring. H H H 6 H4 5 2 1 H H H H 3 H H Chair form H H H 1 H H H H H 6 5 H H 3 2 H H 4 H H Boat form Note that in the chair conformation carbon atoms 2, 3, 5, and 6 are in the same plane and carbon atoms 1 and 4 are below and above the plane, respectively. In the boat conformation, 686 APPENDIX C MOLECULAR MODEL SET EXERCISES carbon atoms 1 and 4 are both above (they could also both be below) the plane described by carbon atoms 2, 3, 5, and 6. Is it apparent why the boat is sometimes associated with the flexible form? Are the hydrogen atoms in the chair conformation staggered or eclipsed? Are any hydrogen atoms eclipsed in the boat conformation? Do carbon atoms 1 and 4 have an anti or gauche relationship in the chair conformation? (Hint: Look down the C2, C3 bond). A twist conformation of cyclohexane may be obtained by slightly twisting carbon atoms 2 and 5 of the boat conformation as shown. 4 1 5 6 2 H 3 H H 1 H 4 2 3 H H 5 6 H H Boat form Twist form Note that the C2, C3 and the C5, C6 sigma bonds no longer retain their parallel orientation in the twist conformation. If the ring system is twisted too far, another boat conformation results. Compare the nonbonded (van der Waals repulsion) interactions and the torsional strain present in the boat, twist, and chair conformations of cyclohexane. Is it apparent why the relative order of thermodynamic stabilities is: chair > twist > boat? Exercise 15 (Chapter 4) Construct a model of methylcyclohexane. How many chair conformations are possible? How does the orientation of the methyl group change in each chair conformation? Identify carbon atoms in the chair conformation of methylcyclohexane that have intramolecular interactions corresponding to those found in the gauche and anti conformations of butane. Which of the chair conformations has the greatest number of gauche interactions? How many more? If we assume, as in the case for butane, that the anti interaction is 3.8 kJ mol−1 more favorable than gauche, then what is the relative stability of the two chair conformations of methylcyclohexane? Hint: Identify the relative number of gauche interactions in the two conformations. Exercise 16 (Chapter 4) Compare models of the chair conformations of monosubstituted cyclohexanes in which the substituent alkyl groups are methyl, ethyl, isopropyl, and tert-butyl. R H APPENDIX C MOLECULAR MODEL SET EXERCISES 687 Rationalize the relative stability of axial and equatorial conformations of the alkyl group given in the table for each compound. The chair conformation with the alkyl group equatorial is more stable by the amount shown. ALKYL GROUP CH3 CH2 CH3 CH(CH3 )2 C(CH3 )3 G◦ (kJ mol−1 ) EQUATORIAL ⇌ AXIAL 7.3 7.5 9.2 21 (approximate) Exercise 17 (Chapter 4) Make a model of 1,2-dimethylcyclohexane. Answer the questions posed in Exercise 13 with regard to 1,2-dimethylcyclohexane. Exercise 18 (Chapter 4) Compare models of the neutral and charged molecules shown next. Identify the structures that are isoelectronic, that is, those that have the same electronic structure. How do those structures that are isoelectronic compare in their molecular geometry? CH3 CH3 CH3 NH2 CH3 OH CH3 CH2 − CH3 NH3 + CH3 OH2 + CH3 NH− Exercise 19 (Chapter 7) Prepare a model of cyclohexene. Note that chair and boat conformations are no longer possible, as carbon atoms 1, 2, 3, and 6 lie in a plane. Are cis and trans stereoisomers possible for the double bond? Attempt to assemble a model of trans-cyclohexene. Can it be done? Are cis and trans stereoisomers possible for 2,3-dimethylcyclohexene? For 3,4dimethylcyclohexene? 6 1 5 2 4 3 Cyclohexene Assemble a model of trans-cyclooctene. Observe the twisting of the π -bond system. Would you expect the cis stereoisomer to be more stable than trans-cyclooctene? Is cis-cyclooctene chiral? Is trans-cyclooctene chiral? 688 APPENDIX C MOLECULAR MODEL SET EXERCISES Exercise 20 (Chapter 4) Construct models of cis-decalin (cis-bicyclo[4.4.0]decane) and trans-decalin. Observe how it is possible to convert one conformation of cis-decalin in which both rings are in chair conformations to another all-chair conformation. This interconversion is not possible in the case of the trans-decalin isomer. Suggest a reason for the difference in the behavior of the cis and trans isomers. Hint: What would happen to carbon atoms 7 and 10 of trans-decalin if the other ring (indicated by carbon atoms numbered 1–6) is converted to the alternative chair conformation. Is the situation the same for cis-decalin? 10 H H 2 9 3 1 6 8 7 4 5 H trans-Decalin H cis-Decalin Exercise 21 (Chapter 4) Assemble a model of norbornane (bicyclo[2.2.1]heptane). Observe the two cyclopentane ring systems in the molecule. The structure may also be viewed as having a methylene (CH2 ) bridge between carbon atoms 1 and 4 of cyclohexane. Describe the conformation of the cyclohexane ring system in norbornane. How many eclipsing interactions are present? Norbornane Using a model of twistane, identify the cyclohexane ring systems held in twist conformations. In adamantane, find the chair conformation cyclohexane systems. How many are present? Evaluate the torsional and angle strain in adamantane. Which of the three compounds in this exercise are chiral? Twistane Adamantane Exercise 22 (Chapter 7) An hypothesis known as Bredt’s Rule states that a double bond to a bridgehead of a smallring bridged bicyclic compound is not possible. The basis of this rule can be seen if you attempt to make a model of bicyclo[2.2.1]hept-1-ene, A. One approach to the assembly of this model is to try to bridge the number 1 and number 4 carbon atoms of cyclohexene with a methylene (CH2 ) unit. Compare this bridging with the ease of installing a CH2 bridge APPENDIX C MOLECULAR MODEL SET EXERCISES 689 between the 1 and 4 carbon atoms of cyclohexane to form a model of norbornane (see Exercise 21). Explain the differences in ease of assembly of these two models. A B Bridgehead double bonds can be accommodated in larger ring-bridged bicyclic compounds such as bicyclo[3.2.2]non-1-ene, B. Although this compound has been prepared in the laboratory, it is an extremely reactive alkene. Exercise 23 (Chapter 14) Not all cyclic structures with alternating double and single bonds are aromatic. Cyclooctatetraene shows none of the aromatic characteristics of benzene. From examination of molecular models of cyclooctatetraene and benzene, explain why there is π -electron delocalization in benzene but not in cyclooctatetraene. Hint: Can the carbon atoms of the eight-membered ring readily adopt a planar arrangement? Benzene Cyclooctatetraene Note that benzene can be represented in several different ways with most molecular model sets. In this exercise, the Kekulé representation with alternating double and single bonds is appropriate. Alternative representations of benzene, such as a form depicting molecular orbital lobes, are shown in your model set instruction booklet. Exercise 24 (Chapter 5) Consider the CH3 CHXCHYCH3 system. Assemble all possible stereoisomers of this structure. How many are there? Indicate the relationship among them. Are they all chiral? Repeat the analysis with the CH3 CHXCHXCH3 system. Exercise 25 (Chapter 5) The CH3 CHXCHXCH3 molecule can exist as the stereoisomers shown here. In the eclipsed conformation (meso) shown on the left (E), the molecule has a plane of symmetry that bisects the C2, C3 bond. This is a more energetic conformation than any of the three staggered conformations, but it is the only conformation of this configurational stereoisomer that has 690 APPENDIX C MOLECULAR MODEL SET EXERCISES a plane of symmetry. Can you consider a molecule achiral if only one conformation, and in this case not even the most stable conformation, has a plane of symmetry? Are any of the staggered conformations achiral (superposable on its mirror image)? Make a model of the staggered conformation shown here (S ) and make another model that is the mirror image of it. Are these two structures different conformations of the same configurational stereoisomer (e.g., are they conformers that can be interconverted by bond rotations), or are they configurational stereoisomers? Based on your answer to the last question, suggest an explanation for the fact that the molecule is not optically active. X H H X X X H H S E Exercise 26 (Chapter 5) Not all molecular chirality is a result of a tetrahedral chirality center, such as CHXYZ. Cumulated dienes (1,2-dienes, or allenes) are capable of generating molecular chirality. Identify, using models, which of the following cumulated dienes are chiral. H C H H H H C Cl C H H A B C Are the following compounds chiral? How are they structurally related to cumulated dienes? H H H H D E Is the cumulated triene F chiral? Explain the presence or absence of molecular chirality. More than one stereoisomer is possible for triene F. What are the structures, and what is the relationship between those structures? H C C H F APPENDIX C MOLECULAR MODEL SET EXERCISES 691 Exercise 27 (Chapters 5 and 14) Substituted biphenyl systems can produce molecular chirality if the rotation about the bond connecting the two rings is restricted. Which of the three biphenyl compounds indicated here are chiral and would be expected to be optically active? Build models of J , K , and L to help determine your answers. J a = f = CH3 a f b e a = b = CH3 K + L a = f = CH3 + b = e = N(CH3)3 b=e=H e = f = N(CH3)3 Exercise 28 (Chapter 7) Assemble a simple model of ethyne (acetylene). The linear geometry of the molecule should be readily apparent. Now, use appropriate pieces of your model set to depict the σ and both the π bonds of the triple bond system using sp hybrid carbon atoms and pieces that represent orbitals. Based on attempts to assemble cycloalkynes, predict the smallest cycloalkyne that is stable. Exercise 29 (Chapter 22) Construct a model of β--glucopyranose. Note that in one of the chair conformations all the hydroxyl groups and the CH2 OH group are in an equatorial orientation. Convert the structure of β--glucopyranose to α--glucopyranose, to β--mannopyranose, and to β-galactopyranose. Indicate the number of large ring substituents (OH or CH2 OH) that are axial in the more favorable chair conformation of each of these sugars. Is it reasonable that the β-anomer is more stable than the α-anomer of -glucopyranose? Make a model of β--glucopyranose. What is the relationship between the  and  configurations? Which is more stable? O OH HOHO O OH β-D-Glucopyranose OH H HO H H H OH H OH OH OH D-(+)-Glucose O HO HO H H H H H OH OH OH D-(+)-Mannose O H HO HO H H OH H H OH OH D-(+)-Galactose 692 APPENDIX C MOLECULAR MODEL SET EXERCISES Exercise 30 (Chapter 24) Assemble a model of tripeptide A shown here. Restricted rotation of the C N bond in the amide linkage results from resonance contribution of the nitrogen nonbonding electron pair. Note the planarity of the six atoms associated with the amide portions of the molecule caused by this resonance contribution. Which bonds along the peptide chain are free to rotate? The amide linkage can either be cisoid or transoid. How does the length (from the N-terminal nitrogen atom to the C-terminal carbon atom) of the tripeptide chain that is transoid compare with one that is cisoid? Which is more “linear”? Convert a model of tripeptide A in the transoid arrangement to a model of tripeptide B. Which tripeptide has a longer chain? O H2N H H R N N R H O O H OH R H 7.2 Å Tripeptide A R = CH3 (L-Alanine) Tripeptide B R = CH2OH (L-Serine) Exercise 31 (Chapter 13) Make models of the π molecular orbitals for the following compounds. Use the phase representation of each contributing atomic orbital shown in your molecular model set instruction booklet. Compare each model with π molecular orbital diagrams shown in the textbook. (a) π1 and π2 of ethene (CH2 CH2, or (b) π1 through π4 of 1,3-butadiene (CH2 ) CH CH (c) π1 , π2 , and π3 of the allyl (propenyl) radical (CH2 CH2, or CH ) CH2, or ) EXERCISE 32 Use your model set to construct several of the interesting representative natural product structures shown here. H H H H O H H O Progesterone Caryophyllene Longifolene APPENDIX C MOLECULAR MODEL SET EXERCISES H N H HO H NCH3 O 693 H H H N H O H HO O Strychnine Morphine MOLECULAR MODEL SET EXERCISES — SOLUTIONS Solution 1 Replacement of any hydrogen atom of methane leads to the same monosubstituted product CH3 X. Therefore, there is only one configuration of a monosubstituted methane. There is only one possible configuration for a disubstituted methane of either the CH2 X2 or CH2 XY type. Solution 2 Interchange of any two substituents converts the configuration of a tetrahedral chirality center to that of its enantiomer. There are only two possible configurations. If the models are not identical, they will have a mirror-image relationship. Solution 3 The tetrahedral carbon atoms may be rotated without breaking the carbon-carbon bond. There is no change in the carbon-carbon bond orbital overlap during rotation. The eclipsed conformation places the hydrogen atoms closer together than they are in the staggered conformation. H H H H H H H H H Staggered conformation H H Eclipsed conformation Solution 4 H All monosubstituted ethanes (CH3 CH2 X) may be made into identical structures by rotations about the C C bond. The following structures are three energetically equivalent staggered conformations. X H H H H H H H H H H H H X X H H H H H H H The three equivalent eclipsed conformations are X H H H H H H H X H H H X H 694 APPENDIX C Solution 5 MOLECULAR MODEL SET EXERCISES The two gauche conformations are energetically equivalent, but not identical (superposable) since they are conformational enantiomers. They bear a mirror-image relationship and are interconvertible by rotation about the carbon-carbon bond. X H X H H Solution 7 H X H H H H H gauche Conformations There are three eclipsed conformations. The methyl groups (C1 and C4) are closest together in the methyl-methyl eclipsed conformation. The carbon-carbon internuclear distances between C1 and C4 are shown in the following table. The number of conformations of each type and the molecular distances in angstroms (Å) are shown. CONFORMATION NUMBER DISTANCES (Å) Eclipsed (CH3 , CH3 ) Gauche Eclipsed (H, CH3 ) Anti 1 2 2 1 2.5 2.8 3.3 3.7 The enantiomers formed from replacement of the C2 hydrogen atoms of butane are H X Solution 8 X H H H X anti Conformation Solution 6 X X H The extended chain assumes a staggered arrangement. The relationship of C1 and C4 is anti. H H H H H H H H H H H H H H APPENDIX C Solution 9 MOLECULAR MODEL SET EXERCISES 695 None of the isomeric butenes is chiral. They all have a plane of symmetry. All the isomeric butenes are related as constitutional (or structural) isomers except cis-2-butene and trans2-butene, which are diastereomers. co constitutional isomers ns tit rs uti l na io ut tit s on me on al iso iso me rs diastereomers constitutional isomers c constitutional isomers Molecular Model Set C1 to C4 Distances: COMPOUND DISTANCES (Å) cis-2-Butene trans-2-Butene Butane (gauche) Butane (anti) 2.0 3.7 2.8 3.7 Solution 10 The hydrogen atoms are all eclipsed in cyclopropane. Solution 11 All the hydrogen atoms are eclipsed in the planar conformation of cyclobutane. The folded ring system has skewed hydrogen interactions. There are six possible isomers of dimethylcyclobutane. Since the ring is not held in one particular folded conformation, deviations of the ring planarity need not be considered in determining the number of possible dimethyl structures. Solution 12 In the planar conformation of cyclopentane, all five pairs of methylene hydrogen atoms are eclipsed. That produces 10 eclipsed hydrogen interactions. Some torsional strain is relieved in the envelope conformation since there are only six eclipsed hydrogen interactions. Solution 13 The three configurational stereoisomers of 1,2-dimethylcyclopentane are shown here. Both trans stereoisomers are chiral, while the cis configuration is an achiral meso compound. 696 APPENDIX C MOLECULAR MODEL SET EXERCISES enantiomers CH3 H H CH3 trans H CH3 CH3 H trans dia rs ste me reo me o ere st rs dia H H CH3 CH3 cis Solution 14 The puckered ring of the chair and the boat conformations can be interconverted by rotation about the carbon-carbon bonds. The chair is more rigid than the boat conformation. All hydrogen atoms in the chair conformation have a staggered arrangement. In the boat conformation, there are eclipsed relationships between the hydrogen atoms on C2 and C3, and also between those on C5 and C6. Carbon atoms that are 1,4 to each other in the chair conformation have a gauche relationship. An evaluation of the three conformations confirms the relative stability: chair > twist > boat. The boat conformation has considerable eclipsing strain and nonbonded (van der Waals repulsion) interactions, the twist conformation has slight eclipsing strain, and the chair conformation has a minimum of eclipsing and nonbonded interactions. Solution 15 Interconversion of the two chair conformations of methylcyclohexane changes the methyl group from an axial to a less crowded equatorial orientation, or the methyl that is equatorial to the more crowded axial position. CH3 5 6 4 1 3 2 Axial methyl H CH3 H Equatorial methyl The conformation with the axial methyl group has two gauche (1,3-diaxial) interactions that are not present in the equatorial methyl conformation. These gauche interactions are axial methyl to C3 and axial methyl to C5. The methyl to C3 and methyl to C5 relationships with methyl groups in an equatorial orientation are anti. Solution 16 The G◦ value reflects the relative energies of the two chair conformations for each structure. The crowding of the alkyl group in an axial orientation becomes greater as the bulk of the group increases. The increased size of the substituent has little effect on the steric interactions of the conformation that has the alkyl group equatorial. The gauche (1,3-diaxial) interactions are responsible for the increased strain for the axial conformation. Since the ethyl and isopropyl groups can rotate to minimize the nonbonded interactions, their effective size is less than their actual size. The tert-butyl group cannot relieve the steric interactions by rotation and thus there is a considerably greater difference in potential energy between the axial and equatorial conformations. APPENDIX C MOLECULAR MODEL SET EXERCISES 697 Solution 17 All four stereoisomers of 1,2-dimethylcyclohexane are chiral. The cis-1,2-dimethylcyclohexane conformations have equal energy and are readily interconverted, as shown here. H H enantiomers CH3 CH3 CH3 trans H rs me dia o ere t ias ste reo me rs d CH3 trans diastereomers diastereomers H H H H CH3 cis conformational enantiomers H CH3 CH3 CH3 cis Solution 18 The structures that are isoelectronic have the same geometry. Isoelectronic structures are CH3 CH3 and CH3 NH3 + CH3 NH2 CH3 CH2 − and CH3 OH2 + Structure CH3 NH− would be isoelectronic to CH3 OH. Solution 19 Cis-trans stereoisomers are possible only for 3,4-dimethylcyclohexene. The ring size and geometry of the double bond prohibit a trans configuration of the double bond. Two configurational isomers (they are enantiomers) are possible for 2,3-dimethylcyclohexene. cis-Cyclooctene is more stable because it has less strain than the trans-cyclooctene structure. The relative stability of cycloalkene stereoisomers in rings larger than cyclodecene generally favors trans. The trans-cyclooctene structure is chiral. trans-Cyclooctene (one enantiomer) 698 APPENDIX C MOLECULAR MODEL SET EXERCISES Solution 20 The ring fusion in trans-decalin is equatorial, equatorial. That is, one ring is attached to the other as 1,2-diequatorial substituents would be. Interconversion of the chair conformations of one ring (carbon atoms 1 through 6) in trans-decalin would require the other ring to adopt a 1,2-diaxial orientation. Carbon atoms 7 and 10 would both become axial substituents to the other ring. The four carbon atoms of the substituent ring (carbon atoms 7 through 10) cannot bridge the diaxial distance. In cis-decalin both conformations have an axial, equatorial ring fusion. Four carbon atoms can easily bridge the axial, equatorial distance. Solution 21 The cyclohexane ring in norbornane is held in a boat conformation, and therefore has four hydrogen eclipsing interactions. All the six-membered ring systems in twistane are in twist conformations. All four of the six-membered ring systems in adamantane are chair conformations. Solution 22 Bridging the 1 and 4 carbon atoms of cyclohexane is relatively easy since in the boat conformation the flagpole hydrogen atoms (on C1 and C4) are fairly close and their C H bonds are directed toward one another. With cyclohexene, the geometry of the double bond and its inability to rotate freely make it impossible to bridge the C1, C4 distance with a single methylene group. Note, however, that a cyclohexene ring can accommodate a methylene bridge between C3 and C6. This bridged bicyclic system (bicyclo[2.2.1]hept-2-ene) does not have a bridgehead double bond. 1 6 2 5 3 H H 1 4 6 4 2 5 3 Bicyclo[2.2.1]hept-2-ene Solution 23 The 120◦ geometry of the double bond is ideal for incorporation into a planar six-membered ring, as the internal angle of a regular hexagon is 120◦ . Cyclooctatetraene cannot adopt a planar ring system without considerable angle strain. The eight-membered ring adopts a “tub” conformation that minimizes angle strain and does not allow significant p-orbital overlap other than that of the four double bonds in the system. Thus, cyclooctatetraene has four isolated double bonds and is not a delocalized π -electron system. Cyclooctatetraene (tub conformation) APPENDIX C MOLECULAR MODEL SET EXERCISES 699 Solution 24 In the CH3 CHXCHYCH3 system, there are four stereoisomers, all of which are chiral. enantiomers H X Y H Y H HX ers m reo dia ste X H YH reo me e st dia diastereomers D diastereomers A rs Y H X H enantiomers B C In the CH3 CHXCHXCH3 system, there are three stereoisomers, two of which are chiral. The third stereoisomer (E) (shown on page 698) is an achiral meso structure. Solution 25 If at least one conformation of a molecule in which free rotation is possible has a plane of symmetry, the molecule is achiral. For a molecule with the configurations specified, there are two achiral conformations: the eclipsed conformation E shown in the exercise and staggered conformation F. X X H H E 180° rotation X H H X H H X F T H X X X H S A model of F is identical with its mirror image. It is achiral, although it does not have a plane of symmetry, due to the presence of a center of symmetry that is located between C2 and C3. A center of symmetry, like a plane of symmetry, is a reflection symmetry element. A center of symmetry involves reflection through a point; a plane of symmetry requires reflection about a plane. A model of the mirror image of S (structure T) is not identical to S, but is a conformational enantiomer of S. They can be made identical by rotation about the C2, C3 bond. Since S and T are conformational enantiomers, the two will be present in equal amounts in a solution of this configurational stereoisomer. Both conformation S and conformation T are chiral and therefore should rotate the plane of plane polarized light. 700 APPENDIX C MOLECULAR MODEL SET EXERCISES Since they are enantiomeric, the rotations of light will be equal in magnitude but opposite in direction. The net result is a racemic form of conformational enantiomers, and thus optically inactive. A similar argument can be made for any other chiral conformation and this configuration of CH3 CHXCHXCH3 . Chemical interchange of two groups at either chirality center in meso compound E leads to a pair of enantiomers (G and H). er m eo dia ste reo ter s H XX H as E (meso) di me rs X HX H enantiomers X HH X H G Solution 26 Structures B and C are chiral. Structure A has a plane of symmetry and is therefore achiral. Compounds D and E are both chiral. The relative orientation of the terminal groups in D and E is perpendicular, as is the case in the cumulated dienes. Cumulated triene F is achiral. It has a plane of symmetry passing through all six carbon atoms. Structure F has a trans configuration. The cis diastereomer is the only other possible stereoisomer. Solution 27 Structure J can be isolated as a chiral stereoisomer because of the large steric barrier to rotation about the bond connecting the rings. Biphenyl K has a plane of symmetry and is therefore achiral. The symmetry plane of K is shown here. Any chiral conformation of L can easily be converted to its enantiomer by rotation. It is only when a = b and f = e and rotation is restricted by bulky groups that chiral (optically active) stereoisomers can be isolated. + CH3 N(CH3)3 CH3 N(CH 3)3 + A plane of symmetry Solution 28 A representation of the molecular orbitals in ethyne is given in Section 1.14 of the text. The smallest stable cycloalkyne is the nine-membered ring cyclononyne. Solution 29 As shown here, the alternative chair conformation of β--glucopyranose has all large substituents in an axial orientation. The structures α--glucopyranose, β--mannopyranose, and β--galactopyranose all have one large axial substitutent in the most favorable conformation. β--Glucopyranose is the enantiomer (mirror image) of β--glucopyranose. Enantiomers are of equal thermodynamic stability. APPENDIX C MOLECULAR MODEL SET EXERCISES ΟΗ ΟΗ ΗΟ ΟΗ Ο ΟΗ ΗΟ 701 ΟΗ ΟΗ ΟΗ Ο ΟΗ β-D-Glucopyranose ΟΗ ΗΟ Ο ΗΟ ΗΟ Ο ΟΗ ΟΗ ΟΗ ΗΟ ΟΗ ΟΗ α-D-Glucopyranose ΗΟ ΗΟ ΟΗ β-D-Galactopyranose ΟΗ ΟΗ Ο Ο ΗΟ ΟΗ ΗΟ β-D-Mannopyranose ΟΗ ΟΗ β-L-Glucopyranose Solution 30 The peptide chain bonds not free to rotate are those indicated by the bold lines in the structure shown here. The transoid arrangement produces a more linear tripeptide chain. The length of the tripeptide chain does not change if you change the substituent R groups. O H2N H R H O N N R H H O OH R H Solution 31 The models of the π molecular orbitals for ethene are shown in the Orbital Symmetry section of the Darling Framework Molecular Model Set instruction booklet. A representation of these orbitals can be found in the text in Section 1.13. The π molecular orbitals for 1,3-butadiene are shown in the text in Section 13.6C. A model of the π molecular orbitals of 1,3-butadiene is also shown in the Orbital Symmetry section of the Darling Framework Molecular Model instruction booklet. The phases of the contributing atomic orbitals to the molecular orbitals of the allyl radical can be found in the text in Section 13.2. The π molecular orbital of the allyl radical has a node at C2.