(Solution Manual) Organic Chemistry

Solomons,Fryhle,Antilla, 11

Chemistry

Solution Manual

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Author(s): Solomons, Fryhle, Antilla

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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 eit her the prior written permission of the Publisher, or authorization through pay - ment 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 State s 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 xi INTRODUCTION ―Solving the Puzzle‖ or ―Structure Is Everything (Almost)‖ xiii CHAPTER 1 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1 Solutions to Problems 1 Quiz 15 CHAPTER 2 FAMILIES OF CARBON COMPOUNDS: FUNCTIONAL GROUPS, INTERMOLECULAR FORCES, AND INFRARED (IR) SPECTROSCOPY 18 Solutions to Problems 18 Quiz 30 CHAPTER 3 ACIDS AND BASES: AN INTRODUCTION TO ORGANIC REACTIONS AND THEIR MECHANISMS 34 Solutions to Problems 34 Quiz 44 CHAPTER 4 NOMENCLATURE AND CONFORMATIONS OF ALKANES AND CYCLOALKANES 46 Solutions to Problems 46 Quiz 62 CHAPTER 5 STEREOCHEMISTRY: CHIRAL MOLECULES 65 Solutions to Problems 65 Quiz 82 vi CONTENTS vii CHAPTER 6 IONIC REACTIONS –NUCLEOPHILIC SUBSTITUTION AND ELIMINATION REACTIONS OF ALKYL HALIDES 85 Solutions to Problems 85 Quiz 103 CHAPTER 7 ALKENES AND ALKYNES I: PROPERTIES AND SYNTHESIS. ELIMINATION REACTIONS OF ALKYL HALIDES 106 Solutions to Problems 106 Quiz 128 CHAPTER 8 ALKENES AND ALKYNES II: ADDITION REACTIONS 130 Solutions to Problems 130 Quiz 153 CHAPTER 9 NUCLEAR MAGNETIC RESONANCE AND MASS SPECTROMETRY: TOOLS FOR STRUCTURE DETERMINATION 157 Solutions to Problems 157 Quiz 180 CHAPTER 10 RADICAL REACTIONS 182 Solutions to Problems 182 Quiz 200 CHAPTER 11 ALCOHOLS AND ETHERS 203 Solutions to Problems 203 Quiz 225 viii CONTENTS CHAPTER 12 ALCOHOLS FROM CARBONYL COMPOUNDS: OXIDATION - REDUCTION AND ORGANOMETALLIC COMPOUNDS 227 Solutions to Problems 227 Quiz 257 ANSWERS TO FIRST REVIEW PROBLEM SET 259 (First Review Probl em Set is available only in WileyPlus, www.wileyplus.com) CHAPTER 13 CONJUGATED UNSATURATED SYSTEMS 278 Solutions to Problems 278 Quiz 299 SUMMARY OF REACTIONS BY TYPE, CHAPTERS 1–13 301 METHODS FOR FUNCTIONAL GROUP PREPARATION, CHAPTERS 1–13 305 CHAPTER 14 AROMATIC COMPOUNDS 308 Solutions to Problems 308 Quiz 323 CHAPTER 15 REACTIONS OF AROMATIC COMPOUNDS 325 Solutions to Problems 325 Quiz 355 CHAPTER 16 ALDEHYDES AND KETONES. NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 357 Solutions to Problems 357 Quiz 386 CONTENTS ix CHAPTER 17 CARBOXYLIC ACIDS AND THEIR DERIVATIVES: NUCLEOPHILIC ADDITION -ELIMINATION AT THE ACYL CARBON 389 Solutions to Problems 389 Quiz 418 CHAPTER 18 REACTIONS AT THE α CARBON OF CARBONYL COMPOUNDS: ENOLS AND ENOLATES 421 Solutions to Problems 421 Quiz 446 CHAPTER 19 CONDENSATION AND CONJUGATE ADDITION REACTIONS OF CARBONYL COMPOUNDS: MORE CHEMISTRY OF ENOLATES 448 Solutions to Problems 448 Quiz 482 CHAPTER 20 AMINES 488 Solutions to Problems 488 Quiz 526 CHAPTER 21 PHENOLS AND ARYL HALIDES: NUCLEOPHILIC AROMATIC SUBSTITUTION 530 Solutions to Problems 530 Quiz 547 ANSWERS TO SECOND REVIEW PROBLEM SET 550 (Second Review Problem Set is available only in WileyPlus, www.wileyplus.com) CHAPTER 22 CARBOHYDRATES 566 Solutions to Problems 567 Quiz 592 x CONTENTS CHAPTER 23 LIPIDS 596 Solutions to Problems 596 Quiz 607 CHAPTER 24 AMINO ACIDS AND PROTEINS 610 Solutions to Problems 610 Quiz 625 CHAPTER 25 NUCLEIC ACIDS AND PROTEIN SYNTHESIS 626 Solutions to Problems 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 13C 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 Special Topic H Electrocyclic and Cycloaddition Reactions 654 APPENDIX A EMPIRICAL AND MOLECULAR FORMULAS 659 Problems 661 Additional Problems 662 Solutions to Problems of Appendix A 663 APPENDIX B ANSWERS TO QUIZZES 667 APPENDIX C MOLECULAR MODEL SET EXERCISES 682 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 chal - lenge. 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 ap- proached in a logical and systematic way, you will find that with the right study habits, mastering or - ganic chemistry can be a deeply satisfying experi - ence. 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, there - fore, 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 al - ready have some understanding of the assigned material. Your time in class will clarify and ex - pand 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 prob - lems. 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 care - fully study the Solved Problems. If you under - stand 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 prob - lems 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 referrin g to flash cards. There is a good reason for this. Organic struc - tures, mechanisms, and reactions are complex. If you simply examine them, you may think you un- derstand them thoroughly, but that will be a mis- perception. 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 photo - graphic memories) through writing. Only by writ - ing 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 three - dimensional 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 chem - istry. We virtually guarantee that your grade in the course will be dire ctly 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 con - cepts 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 com - pletely 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 sim - ply 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 in- structor. 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 conside r 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 re- call 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 struc - ture 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 appreci - ation for the structure of the molecule s 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 re- action. 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 wate r, 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 , 2py, 2pz, 3s, 3px , 3py, 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 2 p orbitals. The ground state configuration for the eight electrons of oxygen is, therefore 1s2 2s2 2px 2 2py1 2pz1 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, py, 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 an s orbital z 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. O O x x H 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 th e four groups of bonding electrons in methane (CH 4) 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 105   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 2 px , 2py, and 2 pz 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. H H   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 CH 4 Now let’s consider the structure of methane (CH 4). 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 t he 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 H H x x    H C x H x  H H C H H C H H H All H -C-H angles are 109.5   The structure at the far right above uses the dash -wedge notation to indicate three dimen - sions. 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 viewe r. 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 CH 3OH Now let’s consider a molecule that incorporates structural aspects of both water and methane. Methanol (CH 3OH), 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 v alence 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 nonbo nding 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 O H H H H C H H O 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 ox ygen. The same would be true for a methanol molecule. The lower case Greek δ (delta) means ―partial.‖   O H H C H H O H H    O 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 co - valent bond. The trend among elements is that of increasing electronegativity toward the upper right corner of the periodic table. (Fluorine is the most electronegative ele - ment.) 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. Car- bon is more electronegative than hydrogen, although only slightly. Oxygen is significan tly 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. H H    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 associ ated 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 CH 3 region of the molecule. H  H C H H O    Now let’s consider methane. Methane is a nonpolar molecule. This is evident first be - cause there are no nonbonding electron pairs, and secondly because there is relatively little electronegativity difference between the hydrogens and the central carbon. F urthermore, what little electronegativity difference there is between the hydrogens and the central car - bon 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 Cl C Cl − 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. Net dipole is zero. 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 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 m olecules. 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 fo rces (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 labo- ratory. 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). H H C H H INTRODUCTION xxi O  O H H  O H H    H H  O hydrogen bonds H H The specific attraction between a partially positive hydrogen atom attached to a het - eroatom (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 polarity - induced 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 intermole cular attraction between methanol molecules and a lower boiling point since less energy is required to overcome the lesser intermolecular attractive forces. H H C H H O H  H C H H  O H  H C H H O The CH 3 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 for ces, 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 C H 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 interac - tions. Since methane is a gas under ordinary conditions, for the purposes of this discussion let’s consider a close relative of methane –hexane. Hexane (C 6H14) 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 CH 3 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 sal ts 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 st ructure 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 differ - ent physical properties to these three compounds. Water, a small molecule with strong intermolecular forces, is a moderately high boiling liquid. Methane, a small molec ule 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 in - termolecular 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