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Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015)
Study Guide and Solutions Manual to Accompany Organic Chemistry Sixth Edition Jim Parise Department of Chemistry and Biochemistry University of Notre Dame Marc Loudon Department of Medicinal Chemistry and Molecular Pharmacology Purdue University ROBERTS & COMPANY PUBLISHERS Greenwood Village, Colorado Roberts and Company Publishers, Inc 4950 South Yosemite Street, F2 #197 Greenwood Village, Colorado 80111 USA Internet: Telephone: www.roberts-publishers.com (303) 221-3325 Facsimile: (303) 221-3326 ORDER INFORMATION Telephone: (800) 351-1161 or (516) 422-4050 Facsimile: (516) 422-4097 Internet: www.roberts-publishers.com Cover art: Quade Paul © 2016 by Roberts and Company Publishers Reproduction or translation of any part of this work beyond that permitted by Section 107 or 108 of the 1976 United States Copyright Act without permission of the copyright owner is unlawful Requests for permission or further information should be addressed to the Permissions Department at Roberts and Company Publishers ISBN: 978-1-936221-86-8 10987654321 Contents PREFACE ABOUT THE AUTHORS xv CHEMICAL BONDING AND CHEMICAL STRUCTURE OF study Guide Links 1 Formal Charge 12 13 Vector Addition Review Structure-Drawing Conventions 2 ©) Further Explorations Ww Dipole Moments 1.2 Electron Density Distribution in Orbitals [iP solutionsto Problems Solutions to In-Text Problems Solutionsto Additional Problems 6 12 ALKANES 25 (2 study Guide Links 21 Nomenclature of Simple Branched Compounds ® 25 25 Further Explorations 21 26 Atomic Radii and van der Waals Repulsion 26 UWU Reaction Review 26 Gy Solutionsto Problems 27 Solutions to In-Text Problems Solutions to Additional Problems ACIDS AND BASES 27 33 THE CURVED-ARROW NOTATION U7 study Guide Links 43 3.1 The Curved-Arrow Notation 43 3.2 Rules for Use of the Curved-Arrow Notation 44 33 Identification of Acids and Bases 45 ©) Further Explorations 3.1 Inductive Effects 43 46 46 iv CONTENTS Ge solutions to Problems 47 Solutions to In-Text Problems 47 Solutions to Additional Problems 56 INTRODUCTION TO ALKENES STRUCTURE AND REACTIVITY (f Study Guide Links 41 ferent Ways to Draw the Same Structure n 42 43 Drawing Structures from Names Solving Structure Problems 72 73 41 Relationship 74 42 Sources of Heats of Formation 75 44 Stepwise View of Rearrangements ©) Further Explorations 43 45 between Free Energy and Enthalpy Molecular Orbital Description of Hyperconjugation Activation Energy val 76 78 UUU reaction Review 79 82 [BY solutions to Problems Solutions to In-Text Problems 82 Solutions to Additional Problems 94 ADDITION REACTIONS OF ALKENES 107 (Wf Study Guide Links 5.1 5.2 53 Transition Elements and the Electron-Counting Rules How to Study Organic Reactions Solving Structure Problems 107 107 108 112 ©) Further Explorations Mechanism of Organoborane Oxidation 5.2 Mechanism of Ozonide Conversion into 53 114 Carbonyl Compounds us Bond Dissociation Energies and Heats of Reaction 116 UUU Reaction Review "7 G9 Solutions to Problems 122 Solutions to In-Text Problems 122 Solutions to Additional Problems 131 PRINCIPLES OF STEREOCHEMISTRY L® 71 147 study Guide Links 147 61 Finding Asymmetric Carbons in Rings 147 62 Stereocenters and Asymmetric Atoms 148 63 Using Perspective Structures 148 ©) Further Explorations 150 6.1 Terminology of Racemates 150 6.2 Isolation of Conformational Enantiomers 150 G9 Solutions to Problems 152 Solutions to In-Text Problems 152 Solutions to Additional Problems 157 CONTENTS CYCLIC COMPOUNDS STEREOCHEMISTRY OF REACTIONS (Uf Study Guide Links 7.1 Relating Cyclohexane Conformations © 167 Reactions of Chiral Molecules Analysis of Reaction Stereochemistry 168 169 74 75 Stereoselective and Stereospe When Stereoselectivity Matters 169 169 Further Explorations 71 Other Ways Alkenelike Behavior of Cyclopropanes 172 73 Optical Activity 173 74 Stereochemistry of Organoborane Oxidation 173 of Designating Relative Configuration 71 Gf solutionsto Problems 175 Solutions to In-Text Problems Solutions to Additional Problems 175 186 NONCOVALENT INTERMOLECULAR INTERACTIONS Lf study Guide Links 81 © Common Nomenclature and the n-Prefix 207 207 Trouton’s Rule 207 {Gg Solutionsto Problems 209 Solutions to In-Text Problems Solutions to Additional Problems THE CHEMISTRY OF ALKYL 209 215 HALIDES 225 (Wf study Guide Links 91 9.2 9.3 ® Deducing Mechanisms from Rate Laws 225 225 Ring Carbons as Alkyl Substituents Diagnosing Reactivity Patterns in Substitution 226 and Elimination Reactions 226 Further Explorations 9.1 92 93 207 207 Further Explorations 8.1 Reaction Rates Absolute Rate Theory Mechanism of Formation of Grignard Reagents 228 228 228 229 UU Reaction Review 230 {9 Solutionsto Problems Solutions to In-Text Problems Solutions to Additional Problems 235 235 246 THE CHEMISTRYOF ALCOHOLS AND THIOLS 265 OP study Guide Links 10.1 ® More on Half-Reactions 265 265 Further Explorations 10.1 10.2 Solvation of Tertiary Alkoxides 167 167 72 73 7.2 v Mechanism of Sulfonate Ester Formation 266 266 267 vi_ _ CONTENTS 10.3 Symmetry Relationships among Constitutionally Equivalent Groups 267 UU Reaction Review Gg solutions to Problems Solutions to In-Text Problems Solutions to Additional Problems 11 268 275 275 286 THE CHEMISTRY OF ETHERS, EPOXIDES, GLYCOLS, AND SULFIDES 303 (7 Study Guide Links 11.1 Learning New Reactions from Earlier Reactions 112 ‘Common Intermediates from Different Starting Materials 303 303 303 ©) Further Explorations W1 Mechanism of OsO, Addition 304 304 UU Reaction Review G9 Solutions to Problems Solutions to In-Text Problems Solutions to Additional Problems 12 305 315 315 329 INTRODUCTION TO SPECTROSCOPY INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 351 ©) Further Explorations 351 12.1 12.2 The Vibrating Bond in Quantum Theory FT-IR Spectroscopy 351 353 12.3 The Mass Spectrometer 354 Gg solutions to Problems Solutions to In-Text Problems Solutions to Additional Problems 13 355 355 360 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY (2 Study Guide Links 13.1 13.2 Approachesto Problem Solving More NMR Problem-Solving Hints 367 367 367 367 ©) Further Explorations 369 13.1 Quantitative Estimation of Chemical Shifts 369 13.2 Fourier-Transform NMR 370 Gy solutions to Problems 375 Solutions to In-Text Problems 375 Solutions to Additional Problems 386 CONTENTS 14 THE CHEMISTRY OF ALKYNES 401 (2 study Guide Links 401 14.1 Functional Group Preparations 401 14.2 Ammonia, Solvated Electrons, and Amide lon 401 UWU reaction Review By Solutionsto Problems 15 402 408 Solutions to In-Text Problems 408 Solutionsto Additional Problems 414 DIENES, RESONANCE, AND AROMATICITY 425 CO study Guide Links 15.1 ATerminology Review 425 425 ©) Further Explorations 426 15.1 More on UV Spectroscopy 426 15.2 The Molecular Orbitals of Benzene 426 UUU Reaction Review Solutions to Problems Solutions to In-Text Problems Solutions to Additional Problems 16 428 432 432 445 THE CHEMISTRY OF BENZENE AND ITS DERIVATIVES (7 study Guide Links NMR of Para-Substituted Benzene Derivatives 469 16.2 Different Sources of the Same Reactive Intermediate 470 16.3 Reaction Conditions and Reaction Rate 470 an UUU Reaction Review G9 solutionsto Problems 17 469 469 16.1 Solutions to In-Text Problems Solutionsto Additional Problems 476 476 488 ALLYLIC AND BENZYLIC REACTIVITY 507 (% Study Guide Links 507 17.1 Synthetic Equivalence 507 171 Addition versus Substitution with Bromine 508 Further Explorations UUW Reaction Review 508 509 13? solutionsto Problems Solutions to In-Text Problems Solutionsto Additional Problems _vii 513 513 523 viii CONTENTS 18 THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS TRANSITION-METAL CATALYSIS 547 U2 study Guide Links 18.1 18.2 Contrast of Aromatic Substitution Reactions The Cumene Hydroperoxide Rearrangement 547 548 ©) Further Explorations Vl 18.1 The Fries Rearrangement 547 548 548 Reaction Review 550 Gy solutions to Problems 559 Solutions to In-Text Problems Solutions to Additional Problems 19 559 577 THE CHEMISTRY OF ALDEHYDES AND KETONES CARBONYL-ADDITION REACTIONS 609 (Wf? Study Guide Links 609 19.1 Lewis Acid Catalysis 609 19.2 19.3 19.4 Reactions That Form Carbon-Carbon Bonds Alcohol Syntheses Hemiacetal Protonation 609 610 610 19.5 19.6 19.7 Mechanism of Carbinolamine Formation Dehydration of Carbinolamines Mechanism of the Wolff-Kishner Reaction 610 612 612 19.1 IR Absorptions of Cyclic Ketones 613 ® Further Explorations UUU Reaction Review 615 G3? solutions to Problems Solutions to In-Text Problems Solutions to Additional Problems 20 613 624 624 638 THE CHEMISTRY OF CARBOXYLIC ACIDS 659 (® study Guide Links 20.1 Reactions of Bases with Carboxylic Acids 659 20.3 Mechanism of Acid Chloride Formation: 660 204 20.5 More on Synthetic Equivalents Mechanism of Anhydride Formation 661 661 20.2 ® 659 Resonance Effect on Carboxylic Acid Acidity 660 Further Explorations 20.1 20.2 20.3 20.4 ‘Chemical Shifts of Carbonyl! Carbons, More on Surfactants Orthoesters — Mechanism of the LiAIH4 Reduction of Carboxylic Acids 663 663 663 664 665 UU Reaction Review Solutions to Problems Solutions to In-Text Problems Solutions to Additional Problems 666 672 672 678 21 THE CHEMISTRY OF CARBOXYLIC ACID DERIVATIVES (f? Study Guide Links 21.1 21.2 ® Solving Structure Problems Involving Nitrogen-Containing Compounds 699 699 21.3 Basicity of Nitriles Mechanism of Ester Hydrolysis 700 21.4 215 Another Look at the Friedel-Crafts Reaction Esters and Nucleophiles 701 701 700 Further Explorations 211 702 NMR Evidence for Internal Rotation in Amides 702 21.2 Cleavage of Tertiary Esters and Carbonless Carbon Paper 702 21.3 Reaction of Tertiary Amines with Acid Chlorides 705 III) Reaction Review Solutionsto Problems 22 706 a Solutions to In-Text Problems 715 Solutions to Additional Problems 725 THE CHEMISTRY OF ENOLATE IONS, ENOLS, AND a@, B-UNSATURATED CARBONYL COMPOUNDS 749 (Uf Study Guide Links 22.1 © lonization versus Nucleophilic Reaction at the 749 Carbonyl Carbon 749 22.2 Kinetic versus Thermodynamic Stability of Enols 750 22.3 22.4 22.5 22.6 Dehydration of B-Hydroxy Carbonyl Compounds Understanding Condensation Reactions Variants of the Aldol and Claisen Condensations Further Analysis of the Claisen Condensation 751 751 752 752 22.7 Synthetic Equivalents in Conjugate Addition 753 Further Explorations 22.1 22.2 22.3 Malonic Ester Alkylation Alkylation of Enolate lons Derived from Ketones Conjugate Addition of Organocuprate Reagents 754 754 754 755 UU Reaction Review Solutionsto Problems Solutions to In-Text Problems Solutions to Additional Problems 23 699 756 767 767 793 THE CHEMISTRY OF AMINES 823 OP study Guide Links 823 23.1 Nitration of Aniline 823 23.3 Formation and Decarboxylation of Carbamic Acids 824 23.1 Alkyl Group Polarization in lonization Reactions 825 23.2 23.3 Structuresof Amide Bases Mechanism of Diazotization 826 826 23.2 Mechanism of the Curtius Rearrangement Further Explorations UUW Reaction Review 823 825 828 SOLUTIONS TO PROBLEMS + CHAPTER 69 Because the major acidic species in solution in each case is H;O*, and 10° M of this species is present in each case, both solutions have the same pH value of ‘The most acidic species that can exist in a solvent is the conjugate acid of the solvent; in water, this is HyO*, Likewise, the most basic species that can exist in a solvent is the conjugate base of the solvent; in water this is “OH Because acids or bases stronger than the solvent react to give the conjugate acid or base of the solvent, respectively, their greater acidity or basicity is not reflected in the pH of the resulting solution This effect is sometimes termed the leveling effect (b) 3.59 of solvent ‘The amide ion is strong enough to react completely with water, and this reaction results in a 10° M solution of hydroxide ion—that is, a solution pH = 11 This is identical to the solution made from 10° M hydroxide itself, except for the ammonia by-product, which reacts only slightly with water The firs step in the reaction is shown below: H = | ‘The second step involves donation of an electron pair from the nucleophile and Lewis base “OH to the electrondeficient electrophile and Lewis acid BH; Z H H—B H 3.60 To solve the various parts of this problem, recognize that astatine, At, is a halogen that is below iodine on the periodic table, and apply trends in the periodic table (a) (b) (c) Because bond dissociation energy decreases down a column of the periodic table, the H—At bond should be weaker than the H—I bond The bond dissociation energy of H—At is smaller Because electron affinities decrease down a column of the periodic table, the electron affinity of At should be smaller than that of I Because dissociation energies dominate the dissociation constant within a column of the periodic table, H—At should be a stronger acid than H—I Introduction to Alkenes Structure and Reactivity STUDY GUIDE LINKS 41 Different Ways to Draw the Same Structure The discussion in the text requires you to realize that there are several different ways to draw the same alkene stereoisomer For example, two equivalent ways to represent cis-2-butene are the following: Hs H \ , C=C / oN CH; H H \ H c I ° Qe ‘CH HC two ways of drawing cis-2- butene If this is not clear, think of the structure as an object Rotate the structure 180° about a horizontal axis through the C=C bond, as follows: H H \ HC CHs When we perform such a rotation, the atom labels, for example, CH, are maintained “right side up”; that is, they are not inverted Similarly, two equivalent ways of drawing trans-2-butene are as follows: HAC, H H CH, \ / / c=¢ c=C H / CH; H;C / \ H different ways of drawing trans-2-butene Their equivalence can be demonstrated by a 180° rotation about a vertical axis 71 72 CHAPTER + INTRODUCTION TO ALKENES STRUCTURE AND HC REACTIVITY H The two structures in text Eq 4.4 (Study Problem 4.1, text p 133) are examples of identical structures drawn differently Some students think that structures such as these are stereoisomers, but in fact they represent the same molecule As these structures demonstrate, the ultimate test of identity of two structures is their congruence—the superimposability of each atom of one structure and an identical atom of the other Rud 42 Drawing Structures from Names When asked to draw a complicated structure from a name, as in Problem 4.6(b), you should always strive to be systematic The natural tendency is to try to write the finished structure immediately Instead, you should take it one step at a time: Write the carbon skeleton of the principal chain; not be concerned with stereochemistry Add the substituents to the principal chain Add the hydrogens Decide on relative group priorities at the double bonds to which an E or Z configuration must be assigned Write them on your structure Redraw your structure with proper stereochemistry It helps to maintain proper 120° bond angles at the double bonds Let's illustrate with an example Suppose you are asked to draw the structure of (2£,4Z)-3isobutyl-2,4-hexadiene Follow the above steps in order The principal chain contains six carbons with double bonds at carbons and 4: C=C There is an isobutyl group at carbon Add the hydrogens within this group, since you won't have to manipulate this group further c—c=c—c=c—c | CH,CH(CH:), isobutyl group Rewrite, adding the missing hydrogens HyC —CH=C—CH=CH—CH, CH,CH(CH,), Indicate the relative priorities of the groups at each double bond For example, focusing on the C2-C3 double bond (carbons in boldface below), the relative priorities of the attached groups are as follows: STUDY GUIDE - CHAPTER 73 higher priority at carbon-3 V lower priority at carbon-2 higherseveral priority LINKS J Pric—Gr—€—ch—ch—oh CH,CH(CH;), —_ : (You should complete the priorities for the C4-C5 double bond.) Redraw the structure with proper stereochemistry HC CH,CH(CH,), \ / c=c CH, H C=C / \ H H (2E,4Z)-3-isobutyl-2,4-hexadiene You should work Problem 4.6(b) using this step-by-step procedure In many cases, it is easier to work a complicated problem if it can be broken into smaller “chunks.” When you learn to bypass the anxiety created by a complex problem and adopt this, approach, you have made significant progress toward becoming a good problem solver Rud 43 Solving Structure Problems A number of problems ask you to deduce structures that are initially unknown by piecing together chemical data Problems 4.53 and 4.54 (text p 178) are of this type This study guide link illustrates a systematic approach to this type of problem by beginning the solution to Problem 4.53 If the formulas of any of the unknowns are given, deduce all the information you can from the formulas Begin with the unsaturation numbers (Sec 4.3, text p 144) In Problem 4.53, compound X has an unsaturation number U = 2, and compound Y has U = This means that X has one ring and one double bond, or two double bonds (It has to have at least one double bond, because it undergoes addition.) Compound Y results from addition of HBr, because the formula of Y is equal to that of X plus the elements of HBr Write all the information in the problem in equation form This process gives you the entire problem at a glance x oa Y ‘CHy 1,1-dimethyleyclopentane CoH) ;Br U=1 U=1 (a single compound) If there is a structure given explicitly anywhere in the problem, even at the end of the problem, deduce what you can from this structure 1,1-Dimethylcyclopentane is given explicitly in the problem as the catalytic hydrogenation product of X Barring rearrangements, the structure provides the real key to solving the problem: it provides the carbon skeleton of compound X It follows that one of the degrees of FURTHER EXPLORATIONS + CHAPTER 75 the reaction The quantity TAS° is the amount of heat that would be produced when the reaction is harnessed with 100% efficiency to work We can interpret the entropy change in more useful and descriptive terms than simply a discrepancy between total energy and work Entropy measures molecular randomness, or freedom of motion For example, consider the following reaction: — > H,C=CHCH,CH, I-butene cyclobutane The experimentally determined AH” for the reaction at 25 °C (or 298 K) is -26.53 kJ mol” (-6.34 kcal mol), and the AG? for the reaction at 25 °C is -40.13 kJ mol (-9.59 kcal mol”) Solving Eq FE4.1, TAS* at T = 298 K is 13.60 kJ mol" (3.25 keal mol"); that is, AS? is positive A positive AS° means that there is more randomness or freedom of motion in the products than the reactants What is it about I-butene that is more “random” than cyclobutane? 1-Butene has two carbon carbon single bonds Internal rotation can occur readily about these bonds, just as in butane However, in cyclobutane, internal rotations cannot occur because the various carbon atoms are constrained into a small ring (If this is not clear to you, make a model of cyclobutane and convince yourself that internal rotation about its carbon-carbon bonds cannot occur.) Thus, there is greater freedom of motion in 1-butene than in cyclobutane Because of the internal rotations in I-butene, the hydrogens on the carbons connected by single bonds can move through a greater volume of space than the same hydrogens in cyclobutane Thus, their positions are more random Hence, the randomness—or entropy—in 1-butene is greater than that in cyclobutane Because of entropy—the “randomness factor”—the energy that controls the position of a chemical equilibrium (AG’) is different from the energy stored in chemical bonds (AH’) Eq SG4.1 shows that an increase of randomness in a reaction—a positive AS’ —gives the product an additional advantage in a chemical equilibrium over and above that which results from the formation of more stable bonds In terms of the heat-work interpretation, we can get more work from this reaction than is available from bond changes, because heat is absorbed from the surroundings that goes toward the increased randomness of the product 42 Sources of Heats of Formation Heats of formation are not obtained by direct measurement Rather, they are calculated from more readily available data by applying Hess’s law Two types of data that are commonly used are heats of combustion and heats of hydrogenation To illustrate, suppose that we want to calculate the heat of formation of rrans-2-butene from the following known heats of combustion: hydrogen, ~241.8 kJ mol"!; carbon, -393.5 kJ mol"; and srans-2-butene, -2530.0 kJ mol" The combustion reaction (Sec 2.7) is the reaction of each species with oxygen to give water (in the combustion of hydrogen), carbon dioxide (in the combustion of carbon), and both water and carbon dioxide (in the combustion of hydrocarbons) Hess's law allows us to express the formation of trans-2-butene from its elements as the sum of three combustion reactions Identical species on opposite sides of the equations cancel, and the resulting equations and their enthalpies are added algebraically: Equations AH? (ki mol") 4H, + 20, ——> 44,0 4(-241.8) = -967.2 +40, —» 4£0, = -1574.0 4(-393.5) + 44,0 —» 60, + 40 4X0, Sum: 4C + 4H, ——> trans-2-butene trans-2-butene +2529.6 ? =-11.6ks mot" 76 CHAPTER + INTRODUCTION TO ALKENES STRUCTURE AND REACTIVITY Because four moles of both H and C are required, their respective enthalpies of combustion must be multiplied by Note also that the combustion of trans-2-butene must be written in reverse so that the formation equation comes out with trans-2-butene on the right Consequently, the sign of the enthalpy of this combustion is also reversed As this example illustrates, heats of formation derived from combustion are in many cases small differences between large numbers and are therefore subject to considerable uncertainty Another reaction that has been used for obtaining heats of formation is catalytic hydrogenation In this reaction, hydrogen is added to an alkene double bond in the presence of a catalyst (Sec 4.9A) For trans-2-butene, this reaction is as follows: Hs H ac’ H + Hy ‘OH, CH.CHCH,CHs butane trans-2-butene The overall AH” of this reaction can be measured with excellent precision, and is typically much smaller than a heat of combustion, (For example, the AH® of hydrogenation of trans-2-butene is =115.5 kJ mot.) The heats of formation of simple alkanes are known with good precision For example, the heat of formation of butane is -127.1 kJ mol The heat of formation of H3 is zero by definition Hence, it is possible to calculate the heat of formation of the alkene (try it!) See Problems 4.15 (text p 149) and 4.17 (text p 151) for calculations of heats of formation from combustion and hydrogenation data, respectively Fortunately, calculations like these are not necessary every time a heat of formation is needed, because such calculations have already been done to provide the data available in standard tables of heats of formation, 43 Molecular Orbital Description of Hyperconjugation A molecular orbital treatment shows why hyperconjugation is a stabilizing effect If you understand the molecular orbital treatment of zr bonds (Sec 4.1B, text pp 128-130), you should have no difficulty with the treatment that follows We'll use as our starting orbitals the empty 2p orbital of the carbocation and the sp*-Is obond orbital of the adjacent C—H bond Both orbitals contain a node When the axes of the two orbitals are coplanar, they can overlap, as shown by the overlap lines in the following diagram This overlap is hyperconjugation hyperconjugative overlap \ C-H abond q , 2p orbital hyperconjugative overlap / To construct molecular orbitals from these two orbitals, imagine both bonding overlap and antibonding overlap, and construct an orbital interaction diagram as shown in Fig FE4.3.1 FURTHER EXPLORATIONS + CHAPTER 77 antibonding MO node = Z 2p orbital bonding MO Figure FE4.1_ An orbital interaction diagram showing a molecular-orbital interpretation of hyperconjugation (The nodes are shown in the molecular orbitals only.) Two electrons must be distributed between the two molecular orbitals With opposite spin, both of these electrons can be accommodated in the bonding molecular orbital, as shown in the figure Because this orbital has a lower energy than the original o-bond orbital, molecular orbital formation (that is, hyperconjugation) lowers the energy of the system It is not strictly correct to use a bond orbital as a starting orbital for this treatment, because a bond orbital is not a proper molecular orbital of the carbocation, In fact, a full-fledged molecularorbital treatment of the fert-butyl cation would start with all the carbons and hydrogens in the proper geometry and allow all their valence orbitals to interact Adopting such an approach gives a more rigorous and complete molecular orbital picture, but, as noted in Sec 1.9A on text pp 35-36, discrete bonds disappear and the descriptive value of the treatment is diminished However, in such a full-fledged molecular-orbital approach, the overlap described here can be discerned as a component of the complete molecular orbitals, 44 Stepwise View of Rearrangements As noted in Sec 4.7, the methide or hydride shift is a one step rearrangement of a carbocation However, it may be easier for you to understand this process by thinking of it as if it occurs in a stepwise fashion (it is important to note here that this process is most certainly one step, however, we are imagining it as if it were two steps) If we reimagine the one-step reaction shown in Eq 4.27a on p 160 as a two step reaction, it 78 CHAPTER + INTRODUCTION TO ALKENES STRUCTURE AND REACTIVITY would proceed as follows: (rors Cx ‘CH, CHs CH, ~~ Ho, — ad CHs The CH; group “leaves”, or detaches from, the main chain with the electron pair that was holding it to the rest of the ion What's left behind is a dication (a “di-cation” which is exceptionally unstable, which is why this process is most certainly imaginary) The methide, or :CH;~, now has a choice: rebond with the carbon from which it came, or bond with the adjacent carbon Since the original carbon looks like a tertiary carbocation, this carbon is more stable than the adjacent one, which looks like a secondary carbocation So the methide ion donates its electron pair to the secondary carbon, yielding a more stable tertiary carbocation This stepwise method of thinking is especially useful when considering ring expansion reactions When a carbocation can rearrange to one that is more hyperconjugated, has less ring strain, or both, it usually does Ke ah or =o secondary tertiary carbocation carbocation Again, it’s best to write the mechanism in one step with one arrow showing the shift + [> — Remember that you should only consider a rearranged product as major if the rearranged carbocation intermediate is significantly more stable (through hyperconjugation, resonance, or ring strain relief) than its precursor 45 Activation Energy You may have learned in general chemistry about the Arrhenius equation, which treats rates in terms of an energy barrier called the activation energy, Ey Kate = Ae7E2®T = A(10-F2250R7) (FE4.2) In this equation, k,, is the rate constant, which is the rate measured under standard conditions of M concentration of all reactants and catalysts, (We'll discuss the rate constant in Chapter 9; but, for now, think of it as the rate.) The factor A is called the preexponential factor The purpose of this Further Exploration is to relate the activation energy to the standard free energy of activation The Arrhenius equation originated as an empirical way to describe the variation of rate with temperature The view was that the activation energy is an energy barrier, but its description in thermodynamic terms was not provided Eyring theory, or transition-state theory, actually envisioned a barrier characterized by a structure—the transition state The transition state can in principle be described in terms of its thermodynamic properties, such as free energy, enthalpy, and entropy As transition-state theory emerged as an important theoretical framework for discussing FURTHER EXPLORATIONS + CHAPTER 79 reaction rates, it became of interest to relate the variables in the Arrhenius equation to those in transition-state theory It can be shown that the activation energy of the Arthenius equation can be related to the enthalpy of activation: E,= AH + RT (FE4.3) The enthalpy of activation is the enthalpy of the transition state minus that of the starting materials Because many typical organic reactions have AH“ values that are substantially greater than RT (which equals about 2.5 kJ mol”), then, for these reactions, E,= AH® (FE4.4) It can also be shown that the preexponential factor in the Arrhenius equation is related to the entropy of activation: (FE4.5) where kg is the Boltzmann constant (= 1.38 x 10° J K"'=R/N, where N = Avogadro’s number), is the absolute temperature in kelvins, and h = Planck’s constant = 6.63 x 10™ J sec Because AG* = AH* — TAS*, the standard free energy of activation can be calculated from the AH** and TAS* derived empirically from the Arrhenius equation The entropy of activation AS* is a particularly interesting aspect of transition-state theory (See the interpretation of entropy in Further Exploration 4.1.) The AS°, like AS° in an equilibrium, can be interpreted in terms of molecular freedom, or randomness This interpretation provides a “window” on the structure of the transition state \yuil | REACTIONS ADDITION REACTIONS OF ALKENES ‘A ADDITION OF HYDROGEN HALIDES TO ALKENES Hydrogen halides add to alkenes in a regioselective manner so that the hydrogen adds to the carbon of the double bond with fewer alkyl substituents and the halide group to the carbon of the double bond with more alkyl substituents (Markovnikov’s rule) The products of these reactions are called alkyl halides HBr + D- CH; ——> Br ‘CH, an alkyl halide The addition of a hydrogen halide to an alkene is a regioselective reaction that occurs in two successive steps a In the first step, protonation of the alkene double bond occurs at the carbon with the fewer alkyl substituents so that the more stable carbocation is formed—the one with the greater number of alkyl substituents at the electron-deficient carbon b In the second step, the halide ion reacts at the electron-deficient carbon 80 CHAPTER + INTRODUCTION TO ALKENES STRUCTURE AND Ot > REACTIVITY CHy Hydrogen halide addition to an alkene is a regioselective reaction because addition involves the transition state that resembles the more stable of the two possible carbocation intermediates The formation of a carbocation from an alkene is a Bronsted acid-base reaction—that is, an electron-pair displacement reaction in which the bond acts as a Bronsted base towards the Bronsted acid H—X In hydrogen halide addition to alkenes, rearranged products are formed if a rearrangement-prone carbocation intermediate is involved—for example, a secondary carbocation that could rearrange to a tertiary carbocation B, CATALYTIC HYDROGENATION OF ALKENES Catalytic hydrogenation is an addition of hydrogen in the presence of a catalyst; it is one of the best ways to convert alkenes into alkanes a, Many of the common hydrogenation catalysts are insoluble in the reaction solution; these are examples of heterogeneous catalysts b Benzene rings are inert to conditions under which normal double bonds react readily, but, under conditions of high temperature and pressure, they can be hydrogenated Hy + ( proto a CH,—CH, C._ HYDRATION OF ALKENES The addition of water to the alkene double bond, called alkene hydration, is an acid-catalyzed reaction a, In many cases, the catalyzing acid is soluble in the reaction solution, Soluble catalysts are called homogeneous catalysts b Hydration is a regioselective reaction A proton is added to the carbon of the double bond with fewer alkyl substituents, and the hydroxy (OH) group is added to the carbon of the double bond with more alkyl substituents HO" cH, > + OH no + (on CH Alkene hydration is a multistep reaction a In the first step of the reaction (the rate-limiting step), a Bronsted acid-base reaction, the double bond is protonated to give a carbocation b In the next step of the hydration reaction, the carbocation reacts with the Lewis base water in a Lewis acid-base association reaction c Finally, a proton is lost to solvent in another Bronsted acid-base reaction to give an alcohol product and regenerate the catalyzing acid er cH + is b J= cH CHs e Oe 20H, ct > CH, ae, a OHCHy Hy oe REACTIONS + CHAPTER 81 In alkene hydration, rearranged hydration products are formed if a rearrangement-prone carbocation intermediate is involved—for example, a secondary carbocation that could rearrange to a tertiary carbocation Alkene hydration is a reversible reaction a Although hydration is used industrially, dehydration of alcohols to alkenes is the more common laboratory use of this reaction b The mechanism of alcohol dehydration is the reverse of the mechanism of alkene hydration (principle of microscopic reversibility) 82 CHAPTER + INTRODUCTION a, TO ALKENES STRUCTURE AND REACTIVITY SOLUTIONS TO PROBLEMS Solutions to In-Text Problems 4.1 The order of increasing bond length is d < a