Workbook for Organic Chemistry Supplemental Problems and Solutions

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This page intentionally left blank This page intentionally left blank WORKBOOK FOR ORGANIC CHEMISTRY SUPPLEMENTAL PROBLEMS AND SOLUTIONS

Jerry A. Jenkins Otterbein College

Printed in the United States of America

ISBN-13: 978-1-4292-4758-0 ISBN-10: 1-4292-4758-4

First printing

W.H. Freeman and Company 41 Madison Avenue New York, NY 10010 Houndmills, Basingstoke RG21 6XS England www.whfreeman.com/chemistry TABLE OF CONTENTS

PREFACE v About the author vi | Acknowledgments vi | Selected concepts/reactions locator vii TIPS viii | Common abbreviations ix

CHAPTER 1 THE BASICS 1 1.1 Hybridization, formulas, physical properties 1 | 1.2 Acids and bases 4 | 1.3 Resonance 7

CHAPTER 2 ALKANES 11 2.1 General 11 | 2.2 Nomenclature 12 | 2.3 Conformational analysis, acyclic 13

CHAPTER 3 CYCLOALKANES 15 3.1 General 15 | 3.2 Nomenclature 16 | 3.3 Conformational analysis, cyclic 18

CHAPTER 4 REACTION BASICS 21

CHAPTER 5 ALKENES AND CARBOCATIONS 27 5.1 General 27 | 5.2 Reactions 30 | 5.3 Syntheses 36 | 5.4 Mechanisms 39

CHAPTER 6 ALKYNES 49 6.1 Reactions 49 | 6.2 Syntheses 50 | 6.3 Mechanisms 53

CHAPTER 7 STEREOCHEMISTRY 55 7.1 General 55 | 7.2 Reactions and stereochemistry 61

CHAPTER 8 ALKYL HALIDES AND RADICALS 65 8.1 Reactions 65 | 8.2 Syntheses 66 | 8.3 Mechanisms 67

CHAPTER 9 SN1, SN2, E1, AND E2 REACTIONS 69 9.1 General 69 | 9.2 Reactions 71 | 9.3 Syntheses 76 | 9.4 Mechanisms 78

CHAPTER 10 NMR 87

CHAPTER 11 CONJUGATED SYSTEMS 93 11.1 Reactions 93 | 11.2 Syntheses 96 | 11.3 Mechanisms 98

CHAPTER 12 AROMATICS 103 12.1 General 103 | 12. Reactions 105 | 12.3 Syntheses 109 | 12.4 Mechanisms 111

CHAPTER 13 ALCOHOLS 117 13.1 Reactions 117 | 13.2 Syntheses 120 | 13.3 Mechanisms 124

CHAPTER 14 ETHERS 129 14.1 Reactions 129 | 14.2 Syntheses 133 | 14.3 Mechanisms 134

CHAPTER 15 ALDEHYDES AND KETONES 139 15.1 Reactions 139 | 15.2 Syntheses 149 | 15.3 Mechanisms 154

CHAPTER 16 CARBOXYLIC ACIDS 167 16.1 Reactions 167 | 16.2 Syntheses 169 | 16.3 Mechanisms 172

CHAPTER 17 CARBOXYLIC ACID DERIVATIVES 177 17.1 Reactions 177 | 17.2 Syntheses 186 | 17.3 Mechanisms 193 iv • Table of Contents Workbook for Organic Chemistry

CHAPTER 18 CARBONYL Į-SUBSTITUTION REACTIONS AND ENOLATES 201 18.1 Reactions 201 | 18.2 Syntheses 204 | 18.3 Mechanisms 207

CHAPTER 19 CARBONYL CONDENSATION REACTIONS 209 19.1 Reactions 209 | 19.2 Syntheses 217 | 19.3 Mechanisms 219

CHAPTER 20 AMINES 229 20.1 Reactions 229 | 20.2 Syntheses 233 | 20.3 Mechanisms 236

SOLUTIONS TO PROBLEMS 241

CHAPTER 1 THE BASICS 243

CHAPTER 2 ALKANES 251

CHAPTER 3 CYCLOALKANES 255

CHAPTER 4 REACTION BASICS 261

CHAPTER 5 ALKENES AND CARBOCATIONS 263

CHAPTER 6 ALKYNES 281

CHAPTER 7 STEREOCHEMISTRY 287

CHAPTER 8 ALKYL HALIDES AND RADICALS 295

CHAPTER 9 SN1, SN2, E1, AND E2 REACTIONS 299

CHAPTER 10 NMR 315

CHAPTER 11 CONJUGATED SYSTEMS 319

CHAPTER 12 AROMATICS 327

CHAPTER 13 ALCOHOLS 341

CHAPTER 14 ETHERS 351

CHAPTER 15 ALDEHYDES AND KETONES 357

CHAPTER 16 CARBOXYLIC ACIDS 379

CHAPTER 17 CARBOXYLIC ACID DERIVATIVES 387

CHAPTER 18 CARBONYL Į-SUBSTITUTION REACTIONS AND ENOLATES 405

CHAPTER 19 CARBONYL CONDENSATION REACTIONS 413

CHAPTER 20 AMINES 427 PREFACE WORKBOOK FOR ORGANIC CHEMISTRY SUPPLEMENTAL PROBLEMS AND SOLUTIONS

Organic Chemistry is mastered by reading (textbook), by listening (lecture), by writing (outlining, notetaking), and by experimenting (laboratory). But perhaps most importantly, it is learned by doing, i.e., solving problems. It is not uncommon for students who have performed below expectations on exams to explain that they honestly thought they understood the text and lectures. The difficulty, however, lies in applying, generalizing, and extending the specific reactions and mechanisms they have “memorized” to the solution of a very broad array of related problems. In so doing, students will begin to “internalize” Organic, to develop an intuitive feel for, and appreciation of, the underlying logic of the subject. Acquiring that level of skill requires but goes far beyond rote memorization. It is the ultimate process by which one learns to manipulate the myriad of reactions and, in time, gains a predictive power that will facilitate solving new problems.

Mastering Organic is challenging. It demands memorization (an organolithium reagent will undergo addition to a ketone), but then requires application of those facts to solve real problems (methyllithium and androstenedione dimethyl ketal will yield the anabolic steroid methyltestosterone). It features a highly logical structural hierarchy (like mathematics) and builds upon a cumulative learning process (like a foreign language). The requisite investment in time and effort, however, can lead to the development of a sense of self-confidence in Organic, an intellectually satisfying experience indeed.

Many excellent first-year textbooks are available to explain the theory of Organic; all provide extensive exercises. Better performing students, however, consistently ask for additional exercises. It is the purpose of this manual, then, to provide Supplemental Problems and Solutions that reinforce and extend those textbook exercises.

Workbook organization and coverage. Arrangement is according to classical functional group organization, with each group typically divided into Reactions, Syntheses, and Mechanisms. To emphasize the vertical integration of Organic, problems in later chapters heavily draw upon and integrate reactions learned in earlier chapters.

It is desirable, but impossible, to write a workbook that is completely text-independent. Most textbooks will follow a similar developmental sequence, progressing from alkane/alkene/alkyne to aromatic to aldehyde/ketone to carboxylic acid to enol/enolate to amine chemistry. But within the earlier domains placement of stereochemistry, spectroscopy, SN/E, and other functional groups (e.g., alkyl halides, alcohols, ethers) varies considerably. The sequence is important because it establishes the concepts and reactions that can be utilized in subsequent problems. It is the intent of this workbook to follow a consensus sequence that complements a broad array of Organic textbooks. Consequently, instructors utilizing a specific textbook may on occasion need to offer their students guidance on workbook chapter and problem selection.

Most Organic textbooks contain later chapters on biochemical topics (proteins, lipids, carbohydrates, nucleic acids, etc.). This workbook does not include separate chapters on such subjects. However, consistent with the current trend to incorporate biochemical relevance into Organic textbooks, numerous problems with a bioorganic, metabolic, or medicinal flavor are presented throughout all chapters.

To produce an error-free manual is certainly a noble, but unrealistic, goal. For those errors that remain, I am solely responsible. I encourage the reader to please inform me of any inaccuracies so that they may be corrected in future versions. Jerry A. Jenkins Otterbein College Westerville, OH 43081 [email protected]

Grindstones sharpen knives; problem-solving sharpens minds! vi • Preface Workbook for Organic Chemistry

ABOUT THE AUTHOR

Jerry A. Jenkins received his BA degree summa cum laude from Anderson University and PhD in Organic from the University of Pittsburgh (T Cohen). After an NSF Postdoctoral Fellowship at Yale University (JA Berson), he joined the faculty of Otterbein College where he has taught Organic, Advanced Organic, and Biochemistry, and chaired the Department of Chemistry & Biochemistry. Prof. Jenkins has spent sabbaticals at Oxford University (JM Brown), The Ohio State University (LA Paquette), and Battelle Memorial Institute, represented liberal arts colleges on the Advisory Board of Chemical Abstracts Service, and served as Councilor to the American Chemical Society. He has published in the areas of oxidative decarboxylations, orbital symmetry controlled reactions, immobilized micelles, chiral resolving reagents, nonlinear optical effects, and chemical education. Prof. Jenkins has devoted a career to challenging students to appreciate the logic, structure, and aesthetics of Organic chemistry through a problem-solving approach.

ACKNOWLEDGMENTS

I wish to express gratitude to my students, whose continued requests for additional problems inspired the need for this book; to Mark Santee, Director of Marketing, WebAssign, for encouraging and facilitating its publication; to Dave Quinn, Media and Supplements Editor, W. H. Freeman, for invaluable assistance in bringing this project to completion; to the production team at W.H. Freeman, specifically Jodi Isman, Project Editor, for all their assistance with the printing process; to Diana Blume, Art Director, and Eleanor Jaekel for their assistance in the cover design; and to my wife Carol, for her endless patience and support. Supplemental Problems and Solutions • vii

SELECTED CONCEPTS/REACTIONS LOCATOR

The location of problems relating to the majority of concepts and reactions in most Organic textbooks will be generally predictable: pinacol rearrangements will be found under ALCOHOLS, benzynes under AROMATICS, acetals under ALDEHYDES AND KETONES, etc. Placement of others, however, may vary from one text to another: diazonium ions may be under AROMATICS or AMINES, thiols may be under ALCOHOLS or ETHERS, the Claisen rearrangement may be under ETHERS or AROMATICS, etc. The following indicates where problems on several of these often variably placed concepts or reactions are initially encountered in Workbook for Organic Chemistry.

Selected concept/reaction Chapter

Active methylene chemistry (e.g., malonic/acetoacetic 18 ester syntheses) Brønsted-Lowry/Lewis equations 1 Carbocation rearrangements 5 cis-, trans- (geometric) isomers 3 Claisen, Cope, oxy-Cope rearrangements 14 Conformational analysis 2, 3 Curved arrow notation vi, 1 Degrees of unsaturation (units of hydrogen deficiency) 5 Diazonium ions 20 Diels-Alder reaction 11 Enamines, synthesis of 15 Enamines, reactions of 19 Epoxides, synthesis of 5 Epoxides, reactions of 14 Free radical additions 5 Free radical substitutions 8 Hydrogens, distinguishing different 2 Isocyanates, ketenes 17 Kinetic isotope effects 9 Kinetics, thermodynamics 4 Neighboring group participation 9 Nitriles 16 Organometallics (Grignard, organolithium, Gilman), 8 synthesis of Phenols 12 Polymers 5 Reaction coordinate diagrams 4 Reaction types/mechanisms 4 Resonance 1 Thiols, (di)sulfides 14 UV/VIS spectroscopy 11 viii • Preface Workbook for Organic Chemistry

TIPS (TO IMPROVE PROBLEM SOLVING)

Mechanism arrows. All reactions (except nuclear) involve the flow of electrons. Arrows are used to account for that movement. They originate at a site of higher electron density (e.g., lone pairs, S bond) and point to an area of lower electron density (e.g., positively or partially positively charged atoms).

H H O O O H O H right: wrong:

Equilibrium vs. resonance arrows. Equilibrium arrows interrelate real species (as above). Resonance arrows interrelate imaginary valence bond structures. Do not interchange them.

H H H O O O O H right:wrong: (resonance arrow) (equilibrium arrows)

Hydrogen nomenclature. The word “hydrogen” is commonly misused. Be more specific.

(H O :H O + H2

A proton (H ) is removed by hydride (H: ) to form hydrogen (H2).

H H X + HX H

A hydrogen atom (H ) is removed by a free radical species.

State of association/dissociation. Correct identification of the appropriate charge state on a species in a particular environment is important. Generally speaking, alkoxides (hydroxide), carboxylates, carbanions, enolates, amines, etc., exist under alkaline conditions. Protons, carboxylic acids, carbocations, enols, etc., exist under acidic conditions. For example, hydroxide does not exist in an acidic solvent OH H3O OH wrong H2O -H right OH2 and a proton is not directly available in base.

O H OO OH O OR +H H ROH H wrong H H) OR +ROH, -RO right Supplemental Problems and Solutions • ix

COMMON ABBREVIATIONS

The following abbreviations and symbols are used throughout this workbook:

Ac acetyl (CH3CO-) AcOH acetic acid * chiral center or isotopic label B: base Bn benzyl (PhCH2-) Bu butyl (C4H9-) CA conjugate acid CB conjugate base ' heat energy D-A or (4+2) Diels-Alder DB double bond(s) DCC dicyclohexylcarbodiimide DIBAH diisobutylaluminum hydride DMF dimethylformamide DMSO dimethyl sulfoxide EAS electrophilic aromatic substitution ee enantiomeric excess equiv equivalent(s) Et ethyl (CH3CH2-) F-C Friedel-Crafts [H] reduction ~H+ proton shift HMPA hexamethylphosphoramide HSCoA coenzyme A hQ light energy H-V-Z Hell-Volhard-Zelinsky reaction inv inversion of configuration L leaving group LDA lithium diisopropylamide mCPBA m-chloroperbenzoic acid Me methyl (CH3-) NAS nucleophilic acyl (or aryl) substitution NBS N-bromosuccinimide NGP neighboring group participation NR no reaction Nu: nucleophile [O] oxidation PCC pyridinium chlorochromate Ph phenyl (C6H5-) Pr propyl (C3H7-) py pyridine Ra-Ni Raney nickel ret retention of configuration rds rate determining step taut tautomerization THF tetrahydrofuran TMS tetramethylsilane or trimethylsilyl Ts tosyl (p-toluenesulfonyl) TsOH tosyl acid (p-toluenesulfonic acid) TS transition state W-K Wolff-Kishner reduction X halogen (XS) excess This page intentionally left blank PROBLEMS This page intentionally left blank CHAPTER 1 THE BASICS

1.1 Hybridization, formulas, physical properties

1. SeldaneTM is a major drug for seasonal allergies; RelenzaTM is a common antiviral. c a HO OH OH O O HO OH N OH b N H d NH O NH H N 2 SeldaneTM 2 RelenzaTM TM TM a. Complete the molecular formula for each. Seldane : C___H___NO2 Relenza : C___H___N4O7 b. Draw all the lone electron pairs in both structures. c. Which orbitals overlap to form the covalent bonds indicated by arrows a, b, and c?

a ______b ______c ______d. What is the hybridization state of both oxygens in SeldaneTM and of nitrogen d in RelenzaTM?

2. Place formal charge over any atom that possesses it in the following structures: a. :C C: b. HCO: c. :O NO: d. the conjugate base of NH2CH3

N e. O f. OO H H

BenadrylTM (antihistamine) zingerone (a constituent of the spice ginger)

3. a. One type of carbene, [:CH2], a very reactive species, has the two unshared electrons in the same orbital and is called “singlet” carbene. Identify the orbital and predict the HCH bond angle.

b. Another type of carbene is called “triplet” carbene and has a linear HCH bond angle. Identify the orbitals housing the two lone electrons.

HO OH

4. a. Which has the higher bp? NH or N b. lower mp? or HO OH

catechol hydroquinone

1.1 Hybridization, formulas, physical properties 2 • Chapter 1 The Basics

5. Must the indicated carbon atoms in each of the following structures lie in the same plane?

H H H a. b. c. d. H H

H3CCH3 H3C H e. (CH3)3C f. C C C g. h. CCCC all four carbons HH H CH3

6. Which species in each pair has the higher molecular dipole moment (P)? a. CHCl3 or CFCl3 b. CH3NH2 or CH3NO2 c. CO2 or SO2

7. Penicillin V and the antiulcerative cimetidine (TagametTM – the first billion dollar ethical drug) have the structures below:

a O H b N C N S N d c N HN S N N O H H CO2H N

penicillin V cimetidine a. Complete the molecular formulas for each.

penicillin V: C_____H_____N_____O_____S cimetidine: C_____H_____N_____S b. Identify the type of orbital (s, p, sp, sp2, sp3) that houses the lone electron pairs on the atoms indicated by arrows a, b, and c in the above structures.

a ______b ______c ______c. The bond between the carbonyl carbon and nitrogen (indicated by arrow d) is somewhat stronger than a single but weaker than a double bond. Given that fact, what type of orbital houses the lone pair of electrons on that nitrogen? (Suggestion: do this problem after studying resonance.)

d. How many lone pairs of electrons are in each structure?

penicillin V: ______cimetidine: ______

1.1 Hybridization, formulas, physical properties Problems • 3

8. Sumatriptan is often prescribed for the treatment of migraines. Prostacyclin is a platelet aggregation inhibitor. HO2C

H O O N MeHN S

O NMe2 HO OH

sumatriptan prostacyclin a. Complete the molecular formulas for each.

sumatriptan: C____H____N____O____S prostacyclin: C____H____O____ b. Sumatriptan contains _____ sp2 and _____ sp3 carbons; prostacyclin contains _____ sp2 and _____ sp3 carbons. c. Sumatriptan and prostacyclin possess _____ and _____ lone pairs of electrons, respectively.

9. RozeremTM is prescribed for the treatment of insomnia, ChantixTM for smoking cessation, and RitalinTM for ADHD.

O N N O HH N O H NH N O

RoseremTM ChantixTM Ritalin TM a. What is the molecular formula for each?

RozeremTM ______ChantixTM ______RitalinTM ______b. How many lone pairs of electrons are there in each?

RozeremTM ______ChantixTM ______RitalinTM ______

10. Theobromine (Greek theobroma – “food of the gods”) is a constituent of cocoa. How many lone pairs of electrons are in its structure? How many lone pairs of electrons are in the plasticizer melamine?

O NH2 CH3 HN N N N

O N N H2N N NH2

CH3 theobromine melamine

1.1 Hybridization, formulas, physical properties 4 • Chapter 1 The Basics

11. Which functional groups are present in each of the following medicines?

OH O O CCH O HO2C F a. O O b. c. N NN H NH NH2 HO

TamifluTM (antiviral) CiproTM (antibiotic) YasminTM component (OCP)

1.2 Acids and bases

1. What is the strongest base that can exist in ammonia?

Sodium hydride (NaH) is, in fact, a stronger base than the above answer. Write a reaction to describe what happens when NaH is added to NH3. Use arrows to show the flow of electrons.

2. Which is the stronger base: (CH3)2NH or CH3-O-CH3?

3. Using curved arrow notation, write Lewis acid/base equations for each of the following. Remember to place formal charge on the appropriate atoms. a. O +AlCl3

b. Ph3P: + BF3

c. ON +BH3

4. Place formal charge on all appropriate atoms. Label the reactants on the left of the arrow as Lewis acids (LA) or Lewis bases (LB) and draw curved arrows to show the movement of electron pairs in each reaction. a. H3CO + CH3CH2 Cl: CH3 OCH2CH3 + Cl

b. H2CCH2 + BF3 CH2 CH2 BF3

1.2 Acids and bases Problems • 5

c. H3COH + :CH2 CH3 H3CO + H3CCH3

d. :Cl Cl: + AlCl3 Cl + AlCl4

S e. CH NCS: :NH 3 + 3 CH3 NC NH3

5. Lynestrenol, a component of certain oral contraceptives, has the structure

Ha O Hb C C

a. Calculate the molecular formula: C___H___O. b. The pKas of hydrogens a and b are about 16 and 25, respectively, and the pKa of ammonia is about 35. Write a Brønsted-Lowry equation for the reaction of the conjugate base of lynestrenol with ammonia.

c. Is the Keq for the above reaction about equal to, greater than, or less than 1?

6. The structure of ibuprofen (A) and acetaminophen (B) are drawn below.

HO NH CO2H O

A B a. Write a reaction for the conjugate base of A with B.

1.2 Acids and bases 6 • Chapter 1 The Basics b. Identify the weak and strong acids and bases. c. Is Keq about equal to, less than, or greater than 1?

7. Which compound has the lowest pKa?

a. EtOH b. HOAc c. H2O d. PhOH e. H2 f. NH3

8. Which species has the ability to quantitatively (completely) remove the proton Ha (pKa 22) from R CCH ? a

a. hydroxid e b. CB of NH3 c. CA of hydride d. CB of EtOH

9. Stress levels in horses may be monitored by measuring urine estradiol. Comment on the Keq for the reaction of the conjugate base of nitromethane (pKa 10.3) with estradiol.

CH NO HO 3 2 estradiol nitromethane

10. Pyridinium chloride is drawn below. a. Place the appropriate formal charge on the atoms that bear it.

Cl N H

biu. The pKas for pyridin m chloride and sodium bicarbonate (NaHCO3) are 5.2 and 10.2, respectively. Write a Brønsted-Lowry equation for the reaction of pyridinium chloride with the conjugate base of bicarbonate. Use curved arrow notation to show the flow of electrons.

c. Is Keq greater than, less than, or about one?

1.2 Acids and bases Problems • 7

1.3 Resonance

1. Identify the type of orbital housing the electrons specified by the arrows.

CH2 O H3C CO N H

2. Which species has the lower pKa, HCN or HOCN?

3. How many nuclei can reasonably bear the charge in each of these ions?

a. HO CH NH2

O c. O

d. H2COCH3

4. The compound below can be protonated at any of the three nitrogen atoms to give a guanidinium ion derivative (creatine phosphate and the amino acid arginine possess this moiety). One of these nitrogens is much more basic than the others, however. Draw the conjugate acids resulting from such protonation, then identify the conjugate acid which is most stable. Why?

H3CNHCNH2 NH

1.3 Resonance 8 • Chapter 1 The Basics

5. Draw a resonance structure that is more stable than the one given. Use curved arrows to derive.

H N a. OOO b. ozone

OH H c. CCN: d. H

6. How many nuclei can reasonably bear the charge in each of the following ions?

O a. b. N H

CH2 c. d.

7. Recalling that resonance is a stabilizing force, explain why the pKa of Ha in A is (only!) about 10.

HHa

OO A

8. Either o xygen in acetic acid (HOAc) could, in theory, be protonated to produce two different conjugate acid forms. Draw each and explain which is more favored.

1.3 Resonance Problems • 9

9. How many nuclei can reasonably bear the charge or odd electron in each of the following?

a. b. c. N N H

O d. e. f. O NCH2

. . CH3O g. h. H i. Cl

10. B’s molecular dipole moment (P) is larger than A’s. Explain.

O O

A B

11. Bioluminescence in fireflies is a result of the conversion of chemical energy (in ATP) to light energy. Specifically, ATP, O2 , and the enzyme luciferase cause luciferin (~ 9 mg can be collected from about 15,000 fireflies!) to be oxidatively decarboxylated to an electronically excited oxyluciferin. Relaxation of the latter to its ground state is accompanied by the emission of light (fluorescence). Subsequent regeneration reactions then recycle oxyluciferin back to luciferin. Draw the two resonance structures of the CB of oxyluciferin in which either oxygen bears the negative charge.

CO2H O N N N N ATP, O2 + hv S HO S luciferase HO S S -CO luciferin 2 oxyluciferin

1.3 Resonance This page intentionally left blank CHAPTER 2 ALKANES

2.1 General

1. Which compound has the highest mp?

1. n-octane 2. 2,5-dimethylhexane 3. 2,3,4-trimethylpentane

4. bicyclo[2.2.2]octane 5. all have the same number of carbons and would melt at the same T

2. Which compound has the highest bp?

1. n-pentane 2. neopentane (dimethylpropane) 3. isopentane

3. Dodecahedrane, one of the three Platonic solids (tetrahedron, hexahedron, and dodecahedron), is a regular polyhedron consisting of twelve cyclopentane rings (think soccer ball). Eicosane is a straight-chain 0 0 compound. Although both are C20 hydrocarbon alkanes, one melts at 420 and the other at 37 . Explain.

4. How many constitutional (structural) isomers exist for

a. C6H14? b. C7H16?

5. How many different kinds (constitutional) of hydrogens are in

a. 2,3-dimethylpentane? b. 2,4-dimethylpentane?

c. 3-ethylpentane? d. 2,2,4-trimethylpentane?

e. 2,5,5-trimethylheptane? f. 4-ethyl-3,3,5-trimethylheptane?

2.1 General 12 • Chapter 2 Alkanes

2.2 Nomenclature

Give the IUPAC name for each of the following. Be certain to specify stereochemistry when relevant.

I Et 1. CH CHNO2 2. s-Bu t-Bu Br

3. 4. Et n-Pr

5. 6.

7. 8.

i-Pr 9. 10. isohexyl iodide

n-pentyl

i-Bu

11. 12. t-Bu n-Pr neopentyl

Give the correct IUPAC names for problems 13 – 16.

13. 2-isopropyl-4-methylheptane 14. 3-(1-methylbutyl)octane

2.2 Nomenclature Problems • 13

15. 3-s-butyl-7-t-butylnonane 16. tetraethylmethane

17. Draw structural formulas, using bond line notation, for the following:

a. neopentyl alcohol (R-OH) b. isobutyl n-pentyl ether (R-O-R’) c. allyl bromide (R-X)

2.3 Conformational analysis, acyclic

1. The rotational energy barrier about the C-C bond in EtBr is 3.7 kcal/mole. What is the energy cost of eclipsing a C-H and C-Br bond?

2. Draw Newman projections of the a. most stable conformer , looking down the C2-C3 bond, of 2-cyclopentyl-6-methylheptane

b. gauche conformer of 1-phenylbutane, looking down the C1-C2 bond (use two-letter abbreviations for R groups).

3. Give the common name for (a) and the IUPAC name for (b).

OH s-Bu Me H H Et a. b. HH H t-Bu Me (R-OH = alkyl alcohol)

2.3 Conformational analysis, acyclic 14 • Chapter 2 Alkanes

4. Draw the conformer of isopentane that corresponds to the highest minimum in a plot of the potential energy vs. rotation about the C2-C3 bond (use a Newman projection).

rot'n about C2 - C3 bond

5. The molecular dipole moment (P) for FCH2CH2OH is much larger than that for FCH2CH2F. Use conformational analysis to explain.

2.3 Conformational analysis, acyclic CHAPTER 3 CYCLOALKANES

3.1 General

1. Which compound has the highest molecular dipole moment (u)?

Cl a. b. anti conformer of 2,3-dichlorobutane c. d. cis-1,3-dichlorocyclobutane C2Cl2 Cl

2. How many constitutional (structural) isomers exist for

a. dichlorocyclopentane?

b. C6H12 that have a cyclopropyl ring in their structure?

3. How many cis/trans stereoisomers exixt for

a. dichlorocyclopentane?

b. diphenylcyclohexane?

c. 2-chloro-4-ethyl-1-methylcyclohexane?

4. How many different kinds [constitutional and geometric (cis/trans)] of hydrogens are there in

a. 1-ethyl-1-methylcyclopropane?

b. allylcyclobutane?

c. methylcyclobutane?

3.1 General 16 • Chapter 3 Cycloalkanes

d. chlorocyclopentane?

e. vinylcyclopentane?

5. Which bicyclic compound is least strained? . . . a. b. . c. d. .

6. Three structural isomers are possible for methylbicyclo[2.2.1]heptane. One of them has two stereoisomeric forms. Draw structures for all four isomers.

7. In view of the previous problem, how many structural and geometric isomers exist for methylbicyclo[2.2.2]octane?

3.2 Nomenclature

Give the IUPAC name for each of the following. Be certain to specify stereochemistry when relevant.

1. isoamyl 2.

3. (three names!) 4.

3.2 Nomenclature Problems • 17

5. 6. Br

F 7. 8.

t-butyl I 9. 10. neopentyl

F 11. Ph 12.

13. 14.

roof-methylhausane (!)

15. 16.

3.2 Nomenclature 18 • Chapter 3 Cycloalkanes

3.3 Conformational analysis, cyclic

1. Draw the most stable conformer of

Me Me

= = OH OH

i-Pr i-Pr

menthol neomenthol

2. In each of the following predict whether Keq is about equal to, greater than, or less than one: a. trans-1,3-diphenylcyclohexane "flipped" conformer

b. i-Pr n-Pr "flipped" conformer

(if i-Pr is equatorial)

H c. H Me "flipped" conformer Et

3. Which has the most negative heat of combustion ('Hcomb) in each of (a), (b), or (c)? a. t-Bu t-Bu t-Bu

3.3 Conformational analysis, cyclic Problems • 19

Me Me Me c. Et s-Bu Et s-Bu Et s-Bu

4. a. Which has the least negative heat of combustion ('Hcomb)?

Et Et Et Et Me Me Me Me

Et Et Et Et

b. Which two structures in (a) are the same compound?

5. Many alkyl halides undergo loss of HX in the presence of base. For example, chlorocyclohexane gives cyclohexene when treated with sodium hydroxide. The reaction mechanism generally requires both the leaving proton and halide to occupy axial positions, a process known as a trans-diaxial elimination. Therefore, which do you think would react faster, cis-1-chloro-2-t-butylcyclohexane or trans-1-chloro-2-t- butylcyclohexane?

6. Trans-4-fluorocyclohexanol exists largely in a chair conformation, whereas the cis-isomer favors a twist-boat conformation. Explain.

3.3 Conformational analysis, cyclic 20 • Chapter 3 Cycloalkanes

7. Glucose, like cyclohexane, exists in a chair conformation. Two configurations of glucose are possible; they are drawn below:

OOH OOH HO HO and HO OH HO OH OH OH

1 2 a. Complete the chair conformations below to show the most stable conformer of 1 and 2.

O O

1 2 b. Which configuration would you predict would be less stable, i.e., burn with a more negative heat of combustion?

8. One of the chair conformations of cis-1,3-dimethylcyclohexane is 5.4 kcal/mol more stable than the other. If the steric strain of 1,3-diaxial interactions between hydrogen and methyl is 0.9 kcal/mol, what is the strain cost of a 1,3-diaxial interaction between the two methyl groups?

9. a. How many cis/trans stereoisomers exist for 1,2,3,4,5,6-hexamethylcyclohexane?

b. For three of those stereoisomers, Keq = 1 for conformational chair-chair flipping. Draw them.

c. Of those three, which is the least stable?

d. Which stereoisomer would be least likely to undergo conformational flipping?

3.3 Conformational analysis, cyclic CHAPTER 4 REACTION BASICS

1. Which type of reaction – addition, elimination, rearrangement, substitution, reduction [H], or oxidation [O] – best describes each of the following?

OLi a. MeLi + O Me

OH O b. H H OH O

c. RCO2R + NH2R RCONHR + HOR

O d. + H2NMe NMe +H2O

H e. + Me NOH O 2 N

O O f.

OH OH OH

g. RHC CHR RC CR

h. +H2O OH

i. PhCO2H PhCHO

k. BnOH PhCHO

4. Reaction Basics 22 • Chapter 4 Reaction Basics

l. Ac2O + H2O 2AcOH

m. CHCl3 + KO-t-Bu :CCl2 + t-BuOH + KCl

n. + HC CH

o. Br 2 2 Br

Ph Ph p. Ph Ph

q. OH O

r. vinyl chloride C2H2

O CO2H

s. O + H2O CO2H O

t. isohexyl alcohol isohexane

2. Imagine a 2-step (A to B and B to C) endothermic reaction for which 'Go values for each step are, respectively, +3 and +7 kcal/mole. The 'G± value for the rate determining step is 11 kcal/mole. (a) Draw a potential energy diagram for this reaction. (b) What is the 'G± value for the conversion of C to B?

'Go

4. Reaction Basics Problems • 23

3. A simplified mechanism for the exothermic substitution reaction below involves two steps: O OH O slow fast + HOR' R Cl + HCl R Cl R OR' OR' a. Draw an overall energy diagram and label the transition state(s), intermediate, 'G± for the rate determining step, and 'Go.

'Go

o b. The overall Keq for the conversion of RCOCl to RCO2R’ could be calculated from 'G according to the equation: Keq = ______c. If 'G± is known, the rate of the reaction could be calculated according to the equation:

rate = ______

4. Bromoform (A) in the presence of base (:B-) can form a very reactive intermediate, dibromocarbene (B), which can rapidly add to olefins to produce gem-dibromocyclopropane derivatives. The following summarizes the two-step mechanism: :B (1) Br3CH Br3C: + H-B A -Br (2) Br3C: Br2C: B Br Br

(3) Br2C: + a. Assuming that 'Go for the overall reaction is +2.5 kcal/mol and that step (2) is rate-determining, draw a reaction energy diagram that depicts all three steps.

'Go

. b. Calculate Keq for this reaction (R = 2 cal/mol K, T = 300).

4. Reaction Basics 24 • Chapter 4 Reaction Basics

5. Consider the following reaction mechanism for A in equilibrium with B:

H H OO OO OO

+ HOH2 + HOH2

A + H2O B 75% a. The overall reaction is an example of a(n) ______(type) reaction that occurs by a(n)

______mechanism. b. Draw curved arrows to show the electron flow that has occurred in each step. c. Calculate Keq, assuming only A and B are present (note: B is formed in 75% yield).

Keq = ______

o If Keq is known, then 'G = ______. d. Which species is (are) nucleophilic in this reaction? e. Draw a qualitative energy diagram for the reaction (assume the first step is slower than the second). Label the transition state(s) and intermediate.

'Go

6. Consider the following reaction:

I + MeOH OMe +HI

The rate law for the reaction may be expressed as: rate = k[A]. Given that methyl alcohol is not in the rate law, propose a reaction for the rate determining step.

4. Reaction Basics Problems • 25

7. Below are reactions we shall examine in more detail later. Classify the mechanisms as polar/ionic, free radical, or pericyclic (concerted). a.

hausene

D D2 / Pt c. D

+H +Cl d. H Cl

e. =

Br Br Br +Br Br f. -Br Br

Cl2 g. + HCl hv Cl

H Et S H h. -Cl S -H S Et Et Cl

4. Reaction Basics 26 • Chapter 4 Reaction Basics

Ph Ph Ph i. H BH2 HBH2 HBH2

:B H) O O j. Cl -BH, -Cl

4. Reaction Basics CHAPTER 5 ALKENES AND CARBOCATIONS

5.1 General

1. Nomenclature. Give the complete IUPAC name for the following:

H H a. b. H H Cl

c. 4-vinyldecane (an incorrect name!) d.

2. Identify each of the olefins below as (E)- or (Z)-:

CO2H NC vinyl a. b. CH2OH H2NH2C t-Bu

OO Ph NH O c. d. NH2 O O O O

O NH2 Ph SH e. f. H CH2F O

3. a. How many alkenes, C7H12, could you treat with H2 / Pt to prepare methylcyclohexane?

b. Which would have the least negative heat of hydrogenation?

4. How many geometric isomers exist for 2,4-heptadiene?

5.1 General 28 • Chapter 5 Alkenes and Carbocations

5. Which carbocation is the most stable? O OMe

6. Degrees of unsaturation (units of hydrogen deficiency). TM a. The antidepressant fluoxetine (Prozac ), C17H18F3NO, when treated with H2 / Ni gives a structure with molecular formula C17H30F3NO. It contains no triple bonds. How many rings are in fluoxetine?

TM b. Cipro is an antibacterial that is used to treat anthrax. Its molecular formula is C17H18FN3O3. The drug has four rings and no triple bonds. How many double bonds does it contain?

c. RU 486 is an abortion medication. Its molecular formula is C28H35NO2. Its structure contains five double bonds and one triple bond. How many rings are in RU 486?

d. The COX-2 inhibitor rofecoxib (VioxxTM), an anti-inflammatory agent, has been taken off the market because of potential increased cardiovascular risk. Its molecular formula is C17H14O4S. There are three rings and no triple bonds in rofecoxib. How many double bonds are there? (Note: for each sulfur atom, subtract four hydrogen atoms to arrive at the equivalent hydrocarbon formula.)

e. The antibiotic floxacillin, C19H17ClFN3O5S, contains eight double bonds. How many rings are present? (In this case, treat sulfur as you would oxygen.)

TM f. The antidepressant Paxil has the molecular formula C19H20FNO3. Upon exhaustive hydrogenation TM (H2/Pt) a compound C19H32FNO3 is formed. How many double bonds and how many rings are in Paxil ?

5.1 General Problems • 29

7. How many stereoisomers exist for 2,4-hexadiene? for 2-chloro-2,4-hexadiene?

8. Draw structural formulas for each of the following:

a. (Z)-3-methyl-2-phenyl-2-hexene b. propylene dichloride

c. styrene bromohydrin d. trans-cyclohexene glycol e. isobutylene epoxide

9. Draw an energy vs. progress of reaction diagram for the exothermic reaction of vinylcyclobutane with HCl to yield 1-chloro-1-methylcyclopentane. Be certain the number of intermediates is clearly indicated.

'Go

10. Draw the most

a. important contributing resonance structure of the conjugate acid of 6-methyl-1,3,5-heptatriene

b. stable intermediate in the following reaction:

MeOH, H

11. The following 1,2-hydride shift does not occur. Why?

H H H ~ H: H

adamantyl carbocation

5.1 General 30 • Chapter 5 Alkenes and Carbocations

12. Which reaction demonstrates NEITHER regiospecificity nor stereospecificity?

HF Cl2 a. trans-2-pentene b. 1-pentene (XS) NaBr

Cl2 D2 / Pt c. cyclobutene d. 1-ethylcyclopropene H2O

13. Why, and how, does E-pinene readily isomerize to D-pinene in the presence of an acid catalyst?

E-pinene D-pinene

5.2 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

HCl 1. Ph

NO 2 HI 2.

H3O 3.

1. Cl2 / ' 4. cyclopentane 2. KOMe, MeOH 3. Br2, CHCl3

5.2 Reactions Problems • 31

NMe3 5. HI

HF 6. F

DCl 7. vinylcyclohexane

HBr 8.

9. OH H O

(complete)

HBr 10. CCl3

Et DBr 11.

1. H2 / Pd 12. cyclopentene 2. Br2 / hv

5.2 Reactions 32 • Chapter 5 Alkenes and Carbocations

H, EtOH 13.

14. HF MeO D

Cl HI 15.

Cl / H O 16. 2 2

1. B2D6 17. 2. H2O2, OH

Cl2 18. propylene (XS) NaI

Et 1. Hg(OAc)2, PhOH 19. 2. NaBH4

(XS) CH2I2 20. H2C CCH2 Zn(Cu) allene

5.2 Reactions Problems • 33

1. KMnO4, OH 21. AcO 2. HIO4

1. O3 22. 2. H3O, Zn

HBr, di-t-butyl peroxide 23. Ph

diazomethane 24. (E)-3-hexene hQ

1. base 2. OsO4 25. cyclopentyl bromide

3. NaHSO3

IN3 26. 3-methyl-1-butene

1. BD , THF 27. 3 2. H O , OH HO 2 2

cholesterol

O H, MeOH 28.

5.2 Reactions 34 • Chapter 5 Alkenes and Carbocations

29. H

(complete)

HIO 30. styrene glycol 4

OH 1. H2SO4 31. 2. KMnO4, OH

H HCl 32. CCO H

OR O 1. BH3, THF O 33. O 2. H O , OH 2 2 O

O artemisinin (antimalarial)

1. Br2, H2O 34. estrone 2. base HO

H 35.

O (complete)

5.2 Reactions Problems • 35

1. OR 36. chlorocyclopentane 2. mCPBA 3. EtOH, H

Br , s-BuOH 37. 2

HIO4 38.

pregnenolone

39. Draw the structure of the largest carbon-containing product in the following reaction:

OH KMnO4, H

vitamin A

1. OsO4 40. O 2. NaHSO3

O 3. HIO4

O incensole acetate (found in frankincense)

HCl 41. two 1,2-shifts

(complete)

5.2 Reactions 36 • Chapter 5 Alkenes and Carbocations

1,1-elimination cyclohexene 42. PhCH Cl + n-BuLi 2 -HCl (a very strong base)

(CH3)2CI2 (1 equiv) 43. Cl O O Zn(Cu) Cl O permethrin (insect repellent)

5.3 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

2. cyclohexyl alcohol cyclohexyl chloride

3. t-BuCl t-BuF

Cl 4.

5.3 Syntheses Problems • 37

5. D

6. cycloheptane HO OH

Cl 7. Br

H 8. allylbenzene O

9. ethylene bromocyclopropane

10. cyclopentyl alcohol

OH 11.

5.3 Syntheses 38 • Chapter 5 Alkenes and Carbocations

Br O 12. O

D 13. cyclohexane Br

Br 14. Br

15. isobutane isobutyl alcohol

CO2H 16.

CO2H Cl

17. t-butyl chloride isobutylene chlorohydrin

18. OO

5.3 Syntheses Problems • 39

OEt 19.

20. Br

21. ethylene HO OH OO

22. t-butyl bromide (only source of carbon) di-t-butyl ether

23. cyclobutyl alcohol hausane (bicyclo[2.1.0]pentane)

5.4 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates. NO WORDS!

H 1.

5.4 Mechanisms 40 • Chapter 5 Alkenes and Carbocations

H 2.

H OMe 3. MeOH isoprene

H 4.

I I2 5.

O CO2H O

6. Isobutylene in the presence of excess propylene and a trace of acid yields C7H14. Deduce this product.

5.4 Mechanisms Problems • 41

H 7.

1. Hg(OAc)2 8. O OH 2. NaBH4

9. H

(a C11 olefin)

H N I N 10. 2 -HI I

N N 11. + H2C NN diazomethane

5.4 Mechanisms 42 • Chapter 5 Alkenes and Carbocations

12. The reaction of 3-bromocyclohexene with HBr yields only trans-cyclohexene dibromide, i.e., no cis- product is formed. In contrast, 3-methylcyclohexene reacts with HBr to yield a mixture of cis- and trans- stereoisomers, as well as a tertiary alkyl halide. Explain with appropriate structures and arrows.

13. The natural products caryophyllene and isocaryophyllene (odor somewhere between cloves and turpentine) are stereoisomers that differ in the configuration of a double bond. They have the molecular formula C15H24. Catalytic hydrogenation of either yields the same compound, C15H28. Ozonolysis, followed by zinc and aqueous acid, yields A and an other aldehyde. Suggest structures for the caryophyllenes.

O O O H A

14. Treatment of an unknown alkene with Hg(OAc)2 in H2O/THF, followed by a NaBH4 workup, produces an alcohol isomeric to one obtained by hydroboration-oxidation of the same alkene. Reduction of the alkene affords the compound C5H12, while ozonolysis yields an aldehyde, CH3CHO, as one of the products. Deduce the structure of the alkene.

15. Partial catalytic hydrogenation of C5H8 (A) yields a mixture of B, C, and D. Ozonolysis, followed by a + reductive work-up (Zn, H3O ), of B gives no new products. When treated in the same way, C gives formaldehyde and 2-butanone and D gives formaldehyde and isobutyraldehyde. Provide structures for compounds A through D. What is the common name of A?

O O O

HH H

formaldehyde 2-butanone isobutyraldehyde

5.4 Mechanisms Problems • 43

16. E-Myrcene, C10H16, found in bayleaves and hops, is an intermediate in the manufacture of perfumes. When treated with H2/Pt, 2,6-dimethyloctane is formed (E-myrcene has no triple bonds). Treatment of E-myrcene with ozone, followed by an acidic zinc work-up, yields A (C5H6O3), acetone (Me2CO), and two equivalents of formaldehyde. What are the structures of E-myrcene and A?

17. Reaction of A, C10H16, with H2/Pd yields B. When treated with KMnO4, a brown precipitate forms. When A is treated with ozone followed by zinc in acid, compound C and another product are produced. What are the structures of A and the other ozonolysis product?

O O

H O

B C

18. Draw a. the structure of the monomer that would give the following polymer by an addition mechanism:

CO2Me CO2Me

CO2Me CO2Me b. a segment (three or four repeating units) of poly(styrene).

19. t-Butyl vinyl ether is polymerized commercially by a cationic process for use in adhesives. Show the mechanism for linking three monomeric units.

' 20. 2 CH2N2 ethylene + 2 N2

5.4 Mechanisms 44 • Chapter 5 Alkenes and Carbocations

CH2N2, hQ 21.

H 22.

H 23.

isocomene (from goldenrod)

24. Hydride shifts and alkyl migrations occur in many enzyme-catalyzed reactions in all living species – including you as you are working these problems! Below is one such biochemical reaction (see 14.3, 6 for perhaps the very best example). Account for the formation of all intermediates leading to the product. (Hint: positive sulfur, like positive oxygen, is a good leaving group, i.e., it easily leaves a carbon to which it is attached, taking with it both bonding electrons.)

R R' S SAM (S-adenosylmethionine, a common methylating agent in all of us)

CH3

D DD D CO2H CO2H

C8H17 C8H17 oleic acid (a fatty acid)

5.4 Mechanisms Problems • 45

1. Hg(OAc)2, H2O 25. O 2. NaBH4

H 26.

27. Elaidic acid (C18H34O2), a fatty acid, is present in processed foods such as margarine and may contribute to elevated levels of cholesterol. Reaction of elaidic acid with Simmons-Smith reagent produces compound I, whereas reaction with acidic permanganate yields II and III. What is the structure of elaidic acid? Indicate stereochemistry.

R OO O R' HO OH OH I II III

28. Compound A (C10H18O) reacts with H2SO4 to give B (C10H16) and an isomer C. Ozonolysis of B yields a diketone; ozonolysis of C yields D. (a) Draw structures for A, B, and C. (b) Describe a simple chemical color test that would differentiate A from B or C.

CHO ______

A B C D

5.4 Mechanisms 46 • Chapter 5 Alkenes and Carbocations

OH H 29.

limonene (volatile in lemons and oranges)

Hint: some alcohols can be protonated to form oxonium ions which may then “leave” as water to give a carbocation.

30. Aziridines (B) are nitrogen analogs of epoxides and can be made from azides (A) by the following reaction: R ' RN3 + Me N Me + N2 A B

Recalling the mechanism of generating carbene from diazomethane, and the fact that nitrogen is an excellent “leaving group,” (a) draw the resonance structure of A that best illustrates how it can decompose to extrude N2 and (b) supply electron flow arrows to show the structure of the reactive intermediate derived from A that reacts with cis-2-butene to give B.

N2 + B A

31. H

(an olefin - complete)

1. Cl2, H2O 32. styrene (vinylbenzene) 2. base 3. dry HCl

5.4 Mechanisms Problems • 47

In 33. CCl3 + BrCCl3 (a free radical initiator) Br

34. Compound A, C16H30O, is a sex attractant (pheremone) for the male silkworm moth. Given the data from the following three experiments, deduce the structure of A, clearly showing its stereochemistry. a. Catalytic hydrogenation of A yields C16H34O. b. Ozonolysis of A, followed by treatment with zinc and acid, yields compounds B, C, and D.

OO O H HO H H H O B C D c. Incomplete reaction of A with diazomethane (CH2N2) gives a mixture of E and F (the C11 and C9 substituents contain one oxygen atom). Note: this experiment establishes the stereochemistry of A. C3 C11 C9

C5 E F A = ______

35. 1,4-Cyclohexadiene undergoes isomerization to 1,3-cyclohexadiene in the presence of acid. Two mechanisms are possible: protonation followed by deprotonation (path a) vs. protonation followed by a 1,2- hydride shift and subsequent deprotonation (path b):

HH H H -H H path a H

path b 1,2-H: H -H shift H

If 3,3,6,6-tetradeuterio-1,4-cyclohexadiene is treated with acid, 1,2,5,5-tetradeuterio-1,3-cyclohexadiene is formed. Which path is favored? (Note: C-H bonds are slightly weaker than C-D bonds.)

5.4 Mechanisms 48 • Chapter 5 Alkenes and Carbocations

36. Carbonyl groups greatly affect the acidity of nearby (D-) protons. For example, the pKa of cyclohexane is about 60, but the pKa of Ha in cyclohexanone is about 20. This dramatic increase in acidity is largely a consequence of resonance stabilization of the conjugate base of the latter (for an example of the additive effect on pKas of 1,3-dicarbonyls, see problem 1.3, 7), and allows an easy exchange of D- hydrogen for deuterium atoms by the following mechanism:

O :B O O DOD O Ha D +B: +D2O -BH - OD cyclohexanone

Under the same conditions, however, species A does not undergo hydrogen-deuterium exchange. Explain. Hint: consider the geometric constraints of olefinic moieties.

:B, D2O H O D A O

5.4 Mechanisms CHAPTER 6 ALKYNES

6.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. NaH 1. 3-penten-1-yne 2. D2O

H, HgSO4, PhOH 2. 1-octyne

1. B2H6 3. phenylacetylene 2. H2O2, OH

1. OMe, HOMe 4. n-BuCl 2. Cl2

3. (XS) NaNH2 . 4. BH3 THF 5. H2O2, OH

5. RC C: Cl

1. Li / NH3 6. 2. HBr, di-t-butyl peroxide

1. H2 / Pd(Pb) 2. BH 7. isopropylacetylene 3 3. H2O2, OH

1. NaH 2. CH (CH ) Cl 8. 1-decyne 3 2 12 3. Lindlar catalyst muscalure (pheremone for house fly)

1. (XS) NaNH2 9. 1,1-dichlorobutane 2. H3O, HgSO4

6.1 Reactions 50 • Chapter 6 Alkynes

Cl , H O 10. CCH 2 2

1. (XS) HI 11. PhC CH 2. Zn(Cu), cyclopentene

O Cl Cl

PCl5 1. (XS) NaNH2 12. 2. D2O OMe OMe

1. LiNH2 (2 equiv) 13. HCHCC 2OH

2. n-C5H11Br (1 equiv) 3. H

1. NaNH2 (1 equiv) 2. n-Pr-I 14. acetylene

3. NaNH2 4. t-Bu-Cl

6.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

Br 1. OH

Br O

2. acetylene n-pentyl bromide

6.2 Syntheses Problems • 51

3. vinyl chloride methyl vinyl ether

4. acetylene (E)-3-octene

5. t-butylacetylene 2-chloro-2,3-dimethylbutane

7. propyne

8. propyne n-propyl bromide

O 9. CCH

6.2 Syntheses 52 • Chapter 6 Alkynes

10. styrene (E)-1-phenyl-1-butene

11. diphenylacetylene cis-1,2-diphenylcyclopropane

PhCHO

Et Cl 12. 3-hexyne Cl Et

n-Bu Et 13. Cl

ClCl

14. acetylene (odor of cheddar cheese)

15. acetylene O

disparlure (pheremone for female gypsy moth)

6.2 Syntheses Problems • 53

16. 1-pentyne 2 OH

6.3 Mechanisms

Br C HO C 1. O [I ] O O O O OH OH O O Br I

O CH3 OH C H O 2. C 3

1. NaH 2. ethylene epoxide 3. CCH 3. H OH

6.3 Mechanisms 54 • Chapter 6 Alkynes

Me2N Me2N O OH CH3 1. H3CC C: 4. H H

H 2. H H O O

mifepristone (RU-486)

5. In the presence of very strong base an internal triple bond in any position of a straight chain alkyne will shift to the terminus of the chain, a process known as the acetylene zipper reaction:

strong base RCH3 R CH2 CC:

6.3 Mechanisms CHAPTER 7 STEREOCHEMISTRY

7.1 General 1. Which of the following molecules are chiral?

Me H Me S a. b. c. H Et CO2H

O Cl Ph Me O d. e. f. CO H Me Ph 2 Cl

g. h. HCCCCCHCCHCHCHCH2CO2H (an antibiotic)

Me H Me Ph i. j. k. H

OH adamantane (an antiviral agent)

Br HH Br HO O l. m. the C2-epimer of n. Cl N NMe2 Ph Ph loperamide (ImodiumTM - antidiarrheal)

2. How many chiral carbons are there in each of the following molecules?

H Bn N S N a. O b. N N O CO H 2 O O penicillin G strychnine

7.1 General 56 • Chapter 7 Stereochemistry

O O N CO2Me O c. d. O OPh O OCH3 O cocaine aflatoxin B1

3. Identify each chiral center as (R)- or (S)-.

H HO H Ph OH CH2CH2NH2 NH2 Me Br a. HO b. c. N Br H H Me (-)-norepinephrine

Ph O O CO2H d. H2N H e. f. H Ph HS N H Me NH 2 H captopril (antihypertensive) O O OMe

g. Me OH misoprostol (CytotecTM - promotes cervical ripening)

4. Identify each of the following pairs of structures as identical, enantiomers, or diastereomers.

Me H CH2OH CHO a. b. Cl H Et Cl H CHO HO CH2OH Et Me HO H

Et H H F c. Me F d. H Me HEt Me Me

7.1 General Problems • 57

CHO H e. HOH HO CHO f. AcO CH CH 3 3 OAc

vinyl H Cl g. h. Cl H O O Cl H Cl H vinyl D-pinene (from pine resin)

Et O O O Et i. j. Et Me Me O Et

5. How many “kinds” of hydrogens (enantiomeric and diastereomeric hydrogens are different!) are there in a. isohexane? b. (R)-2-chloropentane? c. (S)-4-chloro-1-pentene?

6. Nomenclature. Give the complete IUPAC name for the following: Cl Me H Bn allyl Br a. b. H OH c. s-Bu vinyl CH2Cl n-Pr

vinyl Et H i-Pr Me H Me d. I e. s-Bu H f. Et H H OMe allyl Me

7.1 General 58 • Chapter 7 Stereochemistry

7. How many a. pairs of enantiomers exist for bromochlorocyclopentane?

b. geometric diastereomers exist for 1,3-dichloro-2,4-dimethylcyclobutane?

c. pairs of enantiomers are possible for chlorofluorocyclobutane?

OH OH d. meso stereoisomers and how many enantiomeric pairs exist for ? Cl

e. meso stereoisomers exist for 2,3,4,5-tetrachlorohexane?

8. a. D-Xylose is a common sugar found in maple trees. Because it is much less likely to cause tooth decay than sucrose, D-xylose is often used in the manufacture of candy and gum. D-Xylose is the C4-epimer of the enantiomer of A. Draw its structure.

CHO HO H H H OH MeHN Me H OH H OH

CH2OH Ph A (-)-ephedrine b. Ephedrine, a very potent dilator of the air passages in the lungs, has been used to treat asthma. The naturally occurring stereoisomer, isolable from the plant Ephedra sinica, is levorotatory ([D] = -400) and has the configuration above.

(i) Assign (R)- or (S)- configuration to each chiral center.

(ii) If a solution of (+) and (-) ephedrine has a specific rotation of +100, what percentage of the mixture is dextrorotatory enantiomer?

7.1 General Problems • 59

9. Optically pure quinine has a specific rotation of -1700. What percent of levorotatory form is present in an optically impure sample whose [D] is +680? How many chiral carbons are there in quinine?

N N

OH quinine

10. (S)-Naproxen is an active non-steroidal anti-inflammatory drug (NSAID), but the (R)-enantiomer is a harmful liver toxin. Assign the configuration for the (S)-enantiomer.

CO2H MeO

naproxin

11. For each of the molecules below, indicate whether it is capable of enantiomerism only (E), diastereomerism only (D), or both enantiomerism and diastereomerism (ED).

Ph H Me a. b. c. d. e.

O CO2H Ph O Cl O f. Me g. h. S

12. Thalidomide was used as a sedative and anti-nausea drug for pregnant women in Europe (1959-62). Unfortunately, it was sold as a racemate and each enantiomer has a different biochemical activity. One enantiomer, the (S)-form, is a teratogen that was responsible for thousands of serious birth defects. Which of the following is (R)-thalidomide? H O O O O H N N O O N vs. N H H O O

7.1 General 60 • Chapter 7 Stereochemistry

13. Another example of different enantiomers having remarkably different biochemical activities is penicillamine. The (S)-form has anti-arthritic properties, whereas the (R)-form is toxic. Which form is the following configuration?

HS Me H CO2H NH 2 penicillamine

14. Taxol is an anticancer agent active against ovarian and breast tumors. (a) How many chiral carbons are in taxol? (b) If the specific rotation of optically pure taxol is -120o, and a synthetic preparation of taxol containing only its two enantiomers shows a specific rotation of +24o, what is the percentage of dextrorotatory enantiomer in the mixture?

AcO O OH

HO O Ph N H O HO O Ph O AcO O O Ph taxol

15. Compound A below has _____ chiral carbons, _____ meso stereoisomers, and _____ pair(s) of enantiomers.

CO2H

Cl O

O HO OH A B PGE2 (a prostaglandin)

The number of stereoisomers possible for B is _____ (do not change cis/trans configurations of the olefins).

7.1 General Problems • 61

16. The antibiotic cephalosporin C has a specific rotation of +103o in water.

O N S H H2N N O O HO2C CO2H O

cephalosporin C a. What is the maximum number of stereoisomers for the above structure? b. If a synthetic sample of cephalosporin C has an optical rotation of +82o, what percent of the enantiomers is levorotatory?

7.2 Reactions and stereochemistry

1. Draw the stereochemical formula for the major organic product(s) in the following reactions by completing the Fisher projections.

Me H OH 1. OsO4

Ph 2. NaHSO3 Me Ph

Me Br2 / H2O

2. Have the following reactions proceeded with syn- or anti- stereochemistry?

A AB Me H a. cis-2-butene HB Me Me D Me b. CC C D D Me D Me C

7.2 Reactions and stereochemistry 62 • Chapter 7 Stereochemistry c. Fumarase catalyzes the following reaction in mitochondria:

CO2H HO2C H D2O DO H H HD CO2H fumarase CO2H

malic acid

3. For each of the following reactions, (a) how many fractions could be collected by fractional distillation or recrystallization, and (b) for each fraction describe whether it is one enantiomer (E), a racemate (R), or a meso compound (M).

Ph Ph Cl HCl a.

HBr

Br b. (Z)-3-hexene 2

KMnO , H c. 4

HI d. (S)-3-phenyl-1-butene

H e. HF Cl Me

D / Ni f. 2

7.2 Reactions and stereochemistry Problems • 63

Me H, MeOH g. H H

1. BH3 h. 2. H2O2, OH

Et OH

H3O i. H

H2 / Pt

1. Hg(OAc)2, H2O

2. NaBH4

1. OsO4 j.

2. NaHSO3

1. mCPBA

2. H3O

O MeOH, H k.

4. Outline syntheses for the following conversions that ensure the indicated stereochemical outcomes.

Br a. Ph Ph racemic Ph Ph Br

7.2 Reactions and stereochemistry 64 • Chapter 7 Stereochemistry b. 2-butyne meso-2,3-dibromobutane

OH c. racemic

d. trans-2-butene meso-glycol only

OH 5. Consider the structure NHMe , and answer the following: Ph a. How many stereoisomers are possible? b. Two of the structures are the decongestants ephedrine and pseudoephedrine:

OH H OH H N N Ph Me Ph Me Me Me

ephedrine pseudoephedrine

Which stereochemical term best describes their structural relationship? c. The HCl salt of ephedrine has a specific rotation of -34o. What would you predict for the specific rotation of the HCl salt of pseudoephedrine? d. Both ephedrine and pseudoephedrine can be dehydrated to an olefin, which upon hydrogenation produces methamphetamine (“speed,” “meth”).

i. How many stereoisomers exist for the olefin?

ii. How many stereoisomers are possible for “meth”?

7.2 Reactions and stereochemistry CHAPTER 8 ALKYL HALIDES AND RADICALS

8.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. How many different dichlorides could be isolated by ordinary physical methods (e.g., fractional distillation) from the following reaction? Would each, as collected, be optically active or inactive?

H Cl2, hv

2. Calculate the maximum % of (R)-2-bromopentane that could be formed from the reaction of bromine with n-pentane.

1. Br2, ' 2. Mg 3. 2,3-dimethylbutane

3. D2O

1. conc HCl 2. Li 3. CuI 4. OH 4. allyl iodide

1. Cl2, hQ 2. Li 5. cyclobutane 3. CuI 4. vinyl iodide 5. HI

1. Br2, ' 6. propane 2. Mg 3. phenylacetylene

1. Li 2. CuI 3. n-PrBr 4. NBS, peroxides 7. bromobenzene

5. KOH 6. Br2 / H2O

8.1 Reactions 66 • Chapter 8 Alkyl Halides and Radicals

8.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. Br

O 2. O

Br 3. isopentane

4. iodobenzene

Me 5. cyclopentane Bn

CHO 6. chlorobenzene

3 ways! 7. cyclohexene deuteriocyclohexane

8.2 Syntheses Problems • 67

8.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

O2, ROOR 1. OOH

Cl Cl 1. CH N , hv 2. Cl , hv H CCCH 2 2 2 Cl 2. 2 2 ++ Cl allene spiropentane Cl (propose a mechanism for step 2)

3. Bergman reaction: D D '

D D

4. Alkyl nitrite esters (RO-NO) readily undergo photolytic homolysis. The Barton reaction utilizes this fact to functionalize the remote G-position of steroids. Use conformational analysis to explain. R R NO G CH2 hv H H D AcO AcO H H O OH NO

8.3 Mechanisms 68 • Chapter 8 Alkyl Halides and Radicals

5. a. The vinylcyclopropane – cyclopentene rearrangement proceeds by a free radical mechanism. Explain. Hint: the cyclopropyl C-C bond is easily homolyzed.

b. Predict the product:

6. Aspirin, as well as other non-steroidal anti-inflammatory drugs (NSAIDS), blocks the synthesis of certain inflammation-mediating prostaglandins by inhibiting the enzyme cyclooxygenase (COX – see 5.1, 6d). COX converts arachidonic acid to the prostaglandin PGG2, which subsequently undergoes reduction to give PGH2. Other prostaglandins derive from the latter. Outline a mechanism for the synthesis of PGG2. Hint: begin by a free radical removal of one of the doubly allylic hydrogen atoms.

CO H 2 O COX enzyme CO2H O H H 2 O2 R O PGG2 OH arachidonic acid

[H] O other prostaglandins CO2H O

OH PGH2

8.3 Mechanisms CHAPTER 9 SN1, SN2, E1, AND E2 REACTIONS

9.1 General For problems 1 – 9, circle the

1. reaction that will go faster: ethanol a. AcO + allyl chloride

HMPA b. AcO + allyl chloride

2. structure with the poorest leaving group: a. R-SH b. R-NH2 c. R-OAc d. R-OH

3. stronger nucleophile:

a. N b. Et3N

4. alkyl halide most reactive by an SN2 pathway:

Br a. Br b. c. Br

5. solvent that will maximize the rate of the reaction of Et3N with n-BuBr: a. DMSO b. MeOH c. PhH d. chloroform

6. halide that will react more rapidly by an E2 pathway: a. Me b. Me

Me Br Me Br

7. approximate value of kH / kD when PhCHBrCH3, vs. PhCDBrCD3, is allowed to react with potassium t- butoxide: a. 1 b. <1 c. >1

9.1 General 70 • Chapter 9 SN1, SN2, E1, and E2 Reactions

8. reaction that will yield the more stereochemically pure product(s):

Et Br methanolysis (S ) a. (or diastereomer) N

MeO , MeOH (S ) b. (R)-2-bromopentane (or enantiomer) N

9. change in rate of reaction if the concentration of Ph2CHBr is tripled and the concentration of ethanol is doubled: a. rate is unaffected b. rate triples c. rate doubles d. rate increases 5-fold e. rate increases 6-fold ______

10. Which would be the reaction of choice (higher yielding) for each of these syntheses?

OMe O-t-Bu a. Br Br

OH OH b. Br Br

c. + OR vs. SR (which?) to maximize SN Br

11. Which reaction would be expected to show a primary hydrogen kinetic isotope effect?

KO-t-Bu a. H(D) Cl t-BuOH

KOH b. H(D) Cl H(D) MeOH

KOMe c. Cl MeOH (D)H H(D)

9.1 General Problems • 71

12. The following reaction might be envisioned as occurring by an intramolecular SN2 process. However, kinetic evidence indicates a bimolecular mechanism. Explain.

SO 2 SO2 O O CH3 C: C CH H 3 TsO TsO H

9.2 Reactions Identify (if not already stated) each reaction as largely SN1, SN2, E1, or E2 – then draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

MeOH 1. n-octyl bromide + KOCH3

2. 3-iodo-3-methylpentane + sodium ethoxide / EtOH

3. potassium t-butoxide + sec-BuCl

4. 2-bromo-3-methylbutane + lithium diisopropylamide

KCN / DMF 5. n-hexyl iodide

methanolysis (RT) 6. Cl

refluxing EtOH 7. Br

O acetolysis (S ) 8. N Cl

9.2 Reactions 72 • Chapter 9 SN1, SN2, E1, and E2 Reactions

9. n-propyl bromide + Me2NH

10. isopropyl bromide + sodium t-butoxide

sodium acetate / DMF 11. 3-iodopentane

Cl NaSH (1 equiv) 12. Cl

ethanolysis (S ) 13. N

Br OAc / Ag 14. Ph

Et 15. Me H OEt (E) Cl H Me

OMe / MeOH 16.

17. E2 D D Cl

9.2 Reactions Problems • 73

CH3 18. DH E2 Br H Ph

19. acetone OH Br

triphenylphosphine 20. 4-iodo-1-pentane

t-butyl alcohol (SN)

methanolysis 21. SCl

NMe 2 ' 22. I

23. conjugate base of H2Se +

OH NHMe PhCH2Cl (1 equiv) 24. Ph

ephedrine

HO Me3O BF4 25. NMe3 choline

9.2 Reactions 74 • Chapter 9 SN1, SN2, E1, and E2 Reactions

OH 1. TsCl 26. 2. OH (S ) H N

PhNH 27. S 2

HO NHMe Br 28. + (1 equiv) OH F OH

epinephrine

29. refluxing MeOH Cl

IF (XS) NaSePh 30.

O EtBr (1 equiv) 31. H N NEt 2 O 2

NovocaineTM

1. NaNH2 32. Ph CCH 2. cyclohexyl bromide

MeOH (E1) 33.

9.2 Reactions Problems • 75

KSCN 34. Ph OTs

Br acetolysis 35. RT

1. MeI 36. S 2. refluxing EtOH

KO-t-Bu / t-BuOH 37. O Br AromasinTM (an aromatase inhibitor used in breast cancer therapy)

38. NN(XS) MeI

paraquat (an herbicide)

39. Br H Me EtO a. O O EtOH H

Me H Br EtO b. O O EtOH H

9.2 Reactions 76 • Chapter 9 SN1, SN2, E1, and E2 Reactions

9.3 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

Ph Ph

1. Ph Ph OH Br Ph

OCH3 Ph

t-Bu Br

2. t-Bu

3. Br HD H

Me Ph Ph 4. Me H IH Me Me H

Et I 5.

OPh 6. Br +

9.3 Syntheses Problems • 77

7. Cl OPh

Br 8. Br

9. EtO

10.

O Br

CO2H

Ph Ph 11. Ph Ph

OTs 12. Ph OTs Ph

13.

9.3 Syntheses 78 • Chapter 9 SN1, SN2, E1, and E2 Reactions

14. OH Cl

S 15. ethylene S

via an alkyne H 16. O

17. I Br

D D 18. Br D H

9.4 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

Br S 1. NaSH / HCO3 Br

9.4 Mechanisms Problems • 79

acetolysis 2.

Cl OAc

Me Me N Me N C N 3. Br C N

Cl NEt2 OH 4. Ph Ph NEt2 OH

O H dil OH O 5. H O O Cl OH (Note: retention of configuration!)

vinyl I S 6. DMF SH

7. n-butyl bromide + O N H O pyridine N-oxide

9.4 Mechanisms 80 • Chapter 9 SN1, SN2, E1, and E2 Reactions

8. When treated with hydroxide, trans-A yields B. However, when cis-A is treated with hydroxide, no B is observed. Explain.

O HO Cl

A B

______

Problems 4 and 5 above illustrate the concept of “neighboring group participation” (NGP), wherein an internal nucleophilic atom (e.g., N and O, respectively, in those examples) facilitates the ejection of the leaving group by an intramolecular SN2 attack to form an unstable intermediate. This type of mechanism is often evidenced by (1) rearrangement (problem 4), (2) stereochemistry (problem 5), or (3) kinetic data (problem 9 below). Problems 9 – 16 are additional examples. Account for the observations mechanistically.

9. Unlike most primary alkyl halides the molecules below, types of sulfur and nitrogen mustard gases, do NOT undergo second order hydrolysis, but rather first order: -d[RX]/dt = k[RX]. Yet their rates of hydrolysis are enormously faster than those of most primary alkyl halides.

Cl Cl HO OH S S H2O

Cl Cl HO OH N N

10. Compound II undergoes acetolysis at 75o about 103 times more rapidly than I and yields a racemate. Explain. What stereochemical outcome would you predict for the product from I?

OTs OTs HOAc

OAc OAc a racemate I II

9.4 Mechanisms Problems • 81

O OAc O HOAc 11. OTs O + OAc

60% 40%

12. Paquette (OSU) observed that II undergoes solvolysis, e.g., acetolysis about 104 times more rapidly than I.

X X

I II

Cl 13. OH undergoes ethanolysis 5,700 times more rapidly than Cl OH .

14. Sometimes a carbon-carbon double bond can act as a neighboring group nucleophile. For example, II undergoes acetolysis ~ 1011 times faster than I and does so with retention of configuration. Explain.

OTs OTs

I II

15. In view of the previous problem, account for the following:

HOAc OTs AcO NaOAc

9.4 Mechanisms 82 • Chapter 9 SN1, SN2, E1, and E2 Reactions

16. DNA is stable in dilute aqueous hydroxide solution, but RNA rapidly hydrolyzes. A mechanistic clue is provided in the observation that hydrolysis of the latter yields not only 3’-phosphates but also 2’- phosphates. Explain.

O O O O

RO OP CH2 NR"2 RO OP CH2 NR"2 RO OP CH2 NR"2 RO OP CH2 NR"2 O O O O O O dil OH, H O O O 2 + 3' 2' O O O O O OH O OH OH O R'O P R'O P O P P O O O O O O

DNA RNA a 3'-phosphate a 2'-phosphate

______

' 17. racemic camphene + HCl Cl

camphene hydrochloride

O O OP adenine adenine O O O 18. OOP O + PPi O OH OH O P O OH O POH O O ATP cAMP

9.4 Mechanisms Problems • 83

Some terpene chemistry…

19. The biosynthesis of terpenes (natural products constructed from the essence of n units of isoprene) begins with a “head-to-tail” coupling of two derivatives of isoprene, dimethylallyl pyrophosphate (DMA- PP) and isopentenyl pyrophosphate (I-PP) to form geranyl pyrophosphate (G-PP):

O-PP O-PP base H2O a. + O-PP OH

DMA-PP I-PP G-PP geraniol O O (a monoterpene) PP = P OPOH (-OPP is a good leaving group) O O

H b. geraniol OH

terpineol

c. A similar coupling of G-PP with I-PP yields the C15 -sesquiterpene farnesyl pyrophosphate (F-PP) to produce a C20-diterpene:

O-PP I-PP O-PP O-PP F-PP [O] (a C20-diterpene) H3O x2 x2 OH triterpenes (C30) A (e.g., squalene => cholesterol) tetraterpenes (C40) H (e.g., lycopene, E-carotene)

vitamin A (retinol)

Outline a mechanism for the coupling and for the conversion of the diterpene A to vitamin A.

9.4 Mechanisms 84 • Chapter 9 SN1, SN2, E1, and E2 Reactions d. F-PP can isomerizes to nerolidol pyrophosphate (N-PP). F-PP and N-PP undergo a “head-to-head” reductive coupling by an E1 reaction to form the C30-triterpene squalene. Outline the mechanisms for each of these events. Hint: reductive coupling is initiated by hydride attack on N-PP as shown below.

O-PP :H

O-PP F-PP N-PP

reductive coupling

squalene

20. The most common methylating agent in biochemistry is SAM (S-adenosylmethionine), formed by an SN reaction between the amino acid methionine and ATP. An example of a metabolic methylation is the conversion of norepinephrine (the prefix “nor” means one-less-carbon-than) to epinephrine. Formulate a mechanism for producing SAM and draw the structure of epinephrine.

O S O OP adenine S adenine O O O + OOP H2N H2N O OHOH CO H CO2H OHOH 2 O P OH O methionine ATP SAM

OH HO NH2

HO norepinephrine

epinephrine

9.4 Mechanisms Problems • 85

TsOH EtOH 21. Ph2CNN Ph2CHOEt -N2

O 22. PhCH Cl + :P(OMe) 2 3 PhCH2 P(OMe)2 + MeCl

KOH, H2O 23. HCCl3 + KI HCCl2I (Note: reaction does NOT occur in the absence of KOH!)

9.4 Mechanisms This page intentionally left blank CHAPTER 10 NMR

Deduce the structures in problems 1 - 17 from the 1H NMR and IR information.

1. C6H12: G0.9 (t, 3H), 1.6 (s, 3H), 1.7 (s, 3H), 2.0 (p, 2H), 5.1 (t, 1H); no long-range coupling evident.

2. C6H12Cl2O2: G1.3 (t, 6H), 3.6 (q, 4H), 4.4 (d, 1H), 5.4 (d, 1H).

-1 1 3. C8H18O2: IR (3405 cm ). H NMR į 1.3 (s, 12H), 1.5 (s, 4H), 1.9 (s, 2H).

-1 1 4. C10H14O: IR (3200 cm ). H NMR į 1.2 (s, 6H), 1.6 (s, 1H), 2.7 (s, 2H), 7.2 (s, 5H).

5. C5H10O4: į 3.2 ( (s, 6H), 3.8 (s, 3H), 4.8 (s, 1H).

6. C8H9BrO: į 1.4 (t, 3H), 3.9 (q, 2H), 6.7 (d, 2H), 7.4 (d, 2H).

7. C3H5ClF2: į 1.75 (t, 3H), 3.63 (t, 2H).

8. C9H10: į 2.04 (m, 2H), 2.91 (t, 4H), 7.17 (s, 4H).

10. NMR 88 • Chapter 10 NMR

9. C8H9Br: į 2.0 (d, 3H), 5.3 (q, 1H), 7.6 (m, 5H).

10. C4H6Cl2: į 2.18 (s, 3H), 4.16 (d, 2H), 5.71 (t, 1H).

11. C9H11Br: į 2.15 (m, 2H), 2.75 (t, 2H), 3.38 (t, 2H), 7.22 (s, 5H).

12. C9H10O3: į 2.3 (t, 2H), 4.1 (t, 2H), 7.3 (m, 5H), 11.0 (br s, 1H).

13. C6H11Br: į 1.0 (s, 9H), 5.5 (d, 1H, J = 17 Hz), 6.6 (d, 1H, J = 17 Hz).

14. C8H14: į 1.7 (s, 6H), 1.8 (s, 6H), 6.0 (s, 2H).

-1 1 15. C6H11FO2: IR (3412 cm ). H NMR į 1.2 (s, 6H), 2.2 (s, 3H), 3.8 (d, 1H), 4.1 (s, 1H).

-1 1 16. C7H14O2: IR (1610 cm ). H NMR į 1.0 (s, 9H), 2.1 (m, 2H), 3.8 (br s, 1H), 4.0 (t, 1H), 8.6 (t, 1H).

10. NMR Problems • 89

-1 1 17. C11H12O2: IR (1705 cm ). H NMR į 2.2 (s, 3H), 2.5 (s, 3H), 5.8 (m, 1H), 7.1 (d, 2H), 7.9 (d, 2H), 9.8 (s, 1H).

______

18. What is the maximum multiplicity for either of the methylene protons in the proton NMR for F H Cl ? H CH3 F

19. The structure below represents two diastereomeric compounds, A and B. Compound A gives a singlet proton NMR for the methylene group, but B gives a multiplet for the same group. What are the structures of A and B?

Br Me, Br Me

20. Trans-3-bromo-1-phenyl-1-propene shows a spectrum in which the vinylic proton at C2 is coupled with the C1 proton (J = 16 Hz) and the C3 protons (J = 8 Hz). What is the expected multiplicity for that proton? Use a spin tree diagram to explain.

21. a. What is the multiplicity of the chemical shift at highest field in the proton NMR of (R)-1,2-dichloro- 2-fluoropropane?

b. Use a spin tree diagram to explain why the lowest field chemical shift appears as a triplet.

10. NMR 90 • Chapter 10 NMR

22. What is the maximum multiplicity for Ha in the amino acid phenylalanine?

Ha CO2H Ph

NH2

phenylalanine

23. A compound has only two singlets in its 1H NMR spectrum: į 1.4 and 2.0 with relative intensities of 3:1. Its 13C NMR spectrum has chemical shifts at į 22, 28, 80, and 170. A strong absorption in its IR occurs at 1740 cm-1. Draw a possible structure for the compound.

24. The following questions relate to deuterated cholesterol, drawn below:

DO a. Predict the theoretical multiplicity of the lowest field proton.

13 b. What is the maximum number of C chemical shifts that would be expected for the C8H17 alkyl side chain?

0 25. Treatment of 2,3-dibromo-2,3-dimethylbutane with SbF5 (a very strong Lewis acid) in SO2 at -60 - 1 yields SbF6 and a substance whose H NMR shows only a singlet at į 2.9. Draw the structure of that substance.

26. What is the multiplicity of the methylene group in the following compound?

OO PP (i-Pr-O)2 (O-i-Pr)2

10. NMR Problems • 91

27. Below is the structure and partial 1H NMR for an organoplatinum compound. Platinum has three isotopes: 195Pt (I = ½, 34% natural abundance), 194Pt (I = 0), and 196Pt (I = 0) – the latter two account for the remaining 66% natural abundance. (Note: aromatic proton resonances are not shown.)

PPh3 Cl Pt H

PPh3

G -13.6 -16.1 -19.6 ppm a. Explain the relative amplitude and multiplicity of the signal at G -16.6. Clearly explain JH, ? by using a spin tree diagram.

b. Explain the amplitude and multiplicity of the two signals at G -13.6 and -19.6. Again, clearly explain JH,? by using a spin tree diagram.

c. What do the very negative chemical shift values of the signals suggest about the magnetic environment of the resonating proton?

10. NMR 92 • Chapter 10 NMR

28. Pettit (UT) observed that the protonation of cyclooctatetraene (COT) yields a carbocation (homotropylium ion) that possesses homoaromatic stabilization. (Homoaromatics refers to S systems that are interrupted by a saturated center but in which the geometry still permits significant overlap of the p orbitals across a gap.)

H a Hb

H H2SO4 H =

COT homotropylium ion

The 1H NMR of the homotropylium ion shows a remarkable chemical shift difference of 5.5 ppm for geminal protons Ha (G 0.5) and Hb (G 5.0). Each appears as a pseudoquartet. Explain both the location of the chemical shifts and multiplicities of these protons.

1 10 11 29. The H NMR spectrum of NaBH4 is shown below. Boron has two isotopes: B (I = 3) and B (I = 3/2) whose natural abundances are 20% and 80%, respectively. Interpret the spectrum.

H Na HBH H

G 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 ppm

10. NMR CHAPTER 11 CONJUGATED SYSTEMS

11.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

HBr (1 equiv) 1. (1,4-addition)

DCl (1 equiv) 2. (1,4-addition)

D-farnesene (in waxy coatings of apple skins)

3. isoprene + MeO2CCO2Me

4. retro D-A ' O

HBr (1 equiv) 5.

(product of thermodynamic control)

6. ' 2-butyne +

DBr (1 equiv) 7. 3-methyl-1,3,5-hexatriene ROOR (1,4-addition)

8. N O + N Ph N O Cookson's dienophile

11.1 Reactions 94 • Chapter 11 Conjugated Systems

9. + (Z)-1,2-diphenylethene N H

1. NBS, ROOR 2. KOMe (E2) 10. cyclohexene 3. phenylacetylene

1. ' (retro D-A) 11. 4-vinylcyclohexene 2. trans-2-butene

C 12. CH intra D-A

1. vinyl chloride 13. O 2. KO-t-Bu

1. CO2Me 14. 1,3-cyclohexadiene

2. O3 3. Zn, H

15. ' (retro 4+2)

11.1 Reactions Problems • 95

1. ' (retro D-A) 16. 2. cis-1,2-diphenylethylene

CO2H 17. + ' HO2C

fumaric acid

18. + estrone MeO O

. 19. . HO2CCO2H OAc ' '

+ OAc

CO2Me 20. CO2Me

O 21. N ' SO2 +

O2S

11.1 Reactions 96 • Chapter 11 Conjugated Systems

22. base 4 + 2 C19H24O2

MeO

NMe3 I

MeO Br R CHO 23. Si Me2

OMe R' N 4 + 2 24. +

Me3SiO RR

Danishefsky's diene

11.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. cyclohexane via a conjugated diene

2. cyclohexane bicyclo[2.2.2]octane

Me 3. cyclohexene

11.2 Syntheses Problems • 97

4. vinylcyclohexane

5. A Diels-Alder dimerization of A gives the indicated product. Draw the structure of A.

O H (4 + 2) A O

6. Draw the structures of the starting materials that may be used to synthesize the following product:

Me N

(4 + 2) ? O

O O

7. The Alder-ene reaction, like the Diels-Alder, is a concerted (pericyclic) reaction:

R C R ' C Z Z = C, O Z RH R H

How could the following compound be prepared by an ene reaction?

' CO2Et + OH

11.2 Syntheses 98 • Chapter 11 Conjugated Systems

11.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

1. Phenolphthalein in solutions below pH 8.5 is colorless, but in solutions above pH 8.5 is a deep red- purple color. Explain.

O H

HO phenolphthalein

2. '

3. Similar to the Diels-Alder the following electrocyclic reaction is generally concerted (pericyclic) and readily reversible. '

Explain the observed conversions:

O O a. ' + O O

O O

O O b. '

11.3 Mechanisms Problems • 99

4. The structure of pyridine is shown below:

..N pyridine a. Describe the longest wavelength Omax electronic transition in terms of VV SS or n. b. Comment on the probability of that transition. What term in the Beer-Lambert equation reflects this probability? c. Draw the conjugate acid of pyridine. How would that transition in (a) be affected?

5. Compound A, upon standing in acid, yields a new isomeric compound B whose 1H NMR is G 1.7 (s, 3H), 1.8 (s, 3H), 2.3 (br s, 1H), 4.1 (d, J = 8 Hz, 2H), 5.5 (t, J = 8 Hz, 1H). Draw the structure of compound B and give its mechanism of formation.

OH A

6. One approach to synthesizing the sesquiterpene occidentalol, found in New England white cedar trees, begins with a forward Diels-Alder reaction, followed by a retro-Diels-Alder, to form A. Explain.

Me O Me + ' O O O CO2Me H HOH CO2Me

A occidentalol

11.3 Mechanisms 100 • Chapter 11 Conjugated Systems

7. An early stage reaction in Paquette’s (OSU) total synthesis of dodecahedrane employed the following “domino” Diels-Alder:

R R

O O

' 8.

O O tautomerize 9. ' (ene reaction)

(see 11.2, 7)

Ts NC CN N N Ts N hQ N -N2 CN CN

11.3 Mechanisms Problems • 101

11. The degradation of heme proceeds by way of the bile pigments biliverdin and bilirubin, green and red, respectively. Elevated levels of the latter produce jaundice. Bilirubin, a principal antioxidant in blood plasma, is formed by reducing biliverdin. Label the structures below as biliverdin or bilirubin and identify the site of reduction in the former. Explain the difference in color of the two pigments.

OH OH N OH N OH NH H N N H N N N CO H CO2H 2

CO H CO2H 2

12. Depending upon the number of S electrons in a pericyclic process, reversible cycloaddition reactions may be classified as thermally “allowed” or “forbidden” (a theoretical prediction of the probability that such a reaction will occur). The Diels-Alder reaction is the most common example of a thermally allowed (4+2) cycloaddition. Examples of thermally forbidden reactions include (2+2) and (4+4) cycloadditions (they do occur, however, under photochemical conditions). Formation of the dibenzenes below could be envisioned by a cycloaddition mechanism. Identify each as (2+2), (4+2), or (4+4). Which would be expected to undergo a thermal retro-cycloaddition to benzene most rapidly?

a. b. c.

11.3 Mechanisms This page intentionally left blank CHAPTER 12 AROMATICS

12.1 General

1. Circle the compounds that would be expected to have aromatic character.

O N B N a. b. c. d. N N O

H N O N e. O f. g. h. N

H O H B H N N HN i. j. k. B B H N H O N H H

l. carbocation in the reaction of SbF5 Cl

2 Li m. product in the reaction of 2 Li +

2 MeLi n. product in the reaction of 2 MeH +

Br o. product in the reaction of Zn -ZnBr2 Br calicene

2. Which would have the largest molecular dipole moment (P)?

a. b. c.

12.1 General 104 • Chapter 12 Aromatics

3. Which nitrogen atom is least basic in purine and most basic in ZofranTM?

O N N N N CH3 N N H N CH3 purine ZofranTM (antiemetic)

4. One of the following ketones is unstable and undergoes a Diels-Alder reaction rapidly. Which?

a. O b. O c. O

5. Which of the following compounds would most easily form its conjugate base?

a. b. c. d.

6. Which would undergo an SN1 reaction most readily?

O O O

a. b. c. Cl Cl Cl

7. Circle the more(most) basic electron pair in each of the following: O H N a. N b. c. N N O

8. Use a Frost mnemonic to explain why 7-chloro-1,3,5-cycloheptatriene gives a singlet 1H NMR spectrum when dissolved in a solvent containing a Lewis acid.

anti- bonding

0 bonding

12.1 General Problems • 105

9. Which ketone has the largest molecular dipole moment (ȝ )?

O O a. b. c. O d. O

10. A in the presence of HBF4 forms a salt. Explain.

O HBF4 Ph Ph A

11. Explain the regioselectivity of the following addition:

HCl Cl

12. The 1H NMR spectrum for the following [14]annulene compound shows two major chemical shifts. Simulate their approximate location and predict the integration of each.

a [14]annulene

12.2 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

NHCOPh 1. fuming H2SO4

HONO / H SO 2. o-methylphenol 2 2 4

12.2 Reactions 106 • Chapter 12 Aromatics

.. PH2 H2SO4, SO3 3.

4. Cl2 / Fe N H

1. PhCH2CH2Br, AlCl3 2. NBS, R O 5. benzene 2 2 3. KOMe, MeOH

Br / CCl 6. 2 4

Br2, Fe

NBS, R2O2

Se Me ICl, Fe 7.

Br2, FeBr3 8. Ph NO

Cl2, FeCl3 9. SH

Cl , BF 10. 2 3 N

12.2 Reactions Problems • 107

H2SO4 11. 2 Cl Cl + formaldehyde

Cl (C13H6Cl6 - hexachlorophene, a disinfectant)

O 1. ' (-CO2, -N2) 12. NN 2. 1,3-cyclohexadiene

13. picric acid

F CN NH 14. 3

NMe2 1. MeLi 15. 2. H Br

Cl 1. HNO , H SO 16. 3 2 4 2. NaOMe, MeOH CF3

H 17. toluene + O -H2O

(a bicyclic C13 compound)

12.2 Reactions 108 • Chapter 12 Aromatics

18. H

BHT (C15H22O - a food preservative)

Br2 / Fe 19. OO

D-pyrone

1. fuming HNO3 (x2!) 20. F3CBr 2. i-Pr2NH

Trifluralin BTM (a pre-emergent herbicide)

1. MeI, AlCl3 2. NBS, ROOR 21. anisole 3. KOH

(flavor in licorice) partial 1H NMR: G7.2 (d, 2H), 7.9 (d, 2H)

22. Although iodination of aromatic rings does not occur as readily as bromination, it can be observed when activating substituents are present, e.g., in the biosynthesis of the hormone thyroxine:

I I

I2 HO HO O CO2H CO2H catalyst NH2 I I NH2 tyrosine thyroxine

12.2 Reactions Problems • 109

1. [O] 2. Cl , BF 3. [H] 23. 2 3 N N N O (complete- pyridine N-oxide contrast to 12.2, 10)

12.3 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. benzene Cl

D D 2. benzene

3. benzene DD

4. benzene 1,2-diphenylethane

5. Cl

12.3 Syntheses 110 • Chapter 12 Aromatics

6. PhH

8. benzene o-nitrobenzoic acid

9. benzene

10. NHMe N

11. benzene

CO2H ibuprofen

12. acetone, phenol HO OH HO

12.3 Syntheses Problems • 111

13. 2,4-D and 2,4,5-T are the active agents in the defoliant Agent OrangeTM. How could they be prepared from the indicated starting materials? Cl Cl Cl OCH2CO2 Cl Cl Cl Cl , Cl , Cl Cl Cl Cl OCH2CO2 2,4-D 2,4,5-T

NH OH 14. O

NO 2 TylenolTM

O H2N CO H 2 H2N 15. p-hydroxybenzoic acid O O O NEt2

proparacaine (a local anesthetic)

12.4 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

H 1. styrene

2. Epoxides, because of ring strain, are much more reactive than most ethers. Account for the following: O , H anisole O OH

12.4 Mechanisms 112 • Chapter 12 Aromatics

AlCl3 3. (XS) N,N-dimethylaniline + COCl2 phosgene Me2N NMe2 Michler's ketone

4. The Kolbe reaction is used industrially to convert phenol to salicylic acid, an immediate precursor to aspirin.

OH OH OAc 1. OH CO2H 2. Dry Ice CO2H 3. H

salicylic acid aspirin

Br2 / AlCl3 5. + isobutylene

1.R CX , AlCl3 6. 2. HBr

12.4 Mechanisms Problems • 113

OO OH AlCl3 7. R R

Cl Cl Cl O Cl NaOH 8. Cl Cl Cl O Cl

dioxin

BF 9. 3 Br MeO MeO

10. Formyl chloride, A, does NOT exist; therefore, one cannot do a Friedel-Crafts type acylation to produce benzaldehyde. However, the latter can by synthesized by the reaction of benzene with carbon monoxide and HCl (a process known as the Gatterman-Koch reaction). Outline a mechanism.

O CHO CO / HCl HCl A

H2O2 11. toluene OH (a convenient way to substitute an hydroxyl group CF3SO3H onto an aromatic ring)

12.4 Mechanisms 114 • Chapter 12 Aromatics

12. Dyes such as indigo blue (see 19.1, 34) do not bond well to cotton and tend to wash off after repeated laundering; they are known as surface dyes. On the other hand, reactive dyes bind covalently to cotton, resulting in greater color retention (‘fastness’). The following process illustrates the latter. An amino- containing dye is initially bound to cyanuryl chloride to give a product that subsequently is allowed to react with the hydroxyl groups of cotton. Show a mechanism for this process that illustrates how cyanuryl chloride serves to crosslink the dye with cotton. What type of reaction describes each step?

N N 1. dye-NH2

Cl N Cl 2. cotton-OH cyanuryl chloride

13. Malaria, which claims over one million lives per year, mostly children and largely in Africa, could be eradicated with the judicious use of DDT. Banned in the US in 1972, in large part because of Rachael Carson’s 1962 book The Silent Spring, exhaustive scientific review has since shown DDT, in moderation, not only to be safe for humans and the environment, but also the single most effective anti-malarial agent ever formulated. Although the World Health Organization and the US have now reversed their anti-DDT stance, emotional opposition to the pesticide remains so fierce that its use continues to be resisted – at the cost of millions of unnecessary deaths.

DDT is easily prepared as follows:

O H SO H + 2 chlorobenzene 2 4 Cl3CH Cl Cl CCl3 DDT

1. BF3 14. + Cl 2. H

12.4 Mechanisms Problems • 115

H 15. CHO

OH OH

16. When poly(styrene) is treated with chloromethyl methyl ether and SnCl4 (a strong Lewis acid), Merrifield resin (named after Nobel laureate Bruce Merrifield who pioneered in vitro peptide syntheses) is formed.

ClCH2 OCH3

SnCl4

CH2Cl poly(stryene) Merrifield resin

1. Br 2. :B 3. H O 2 C H Br 2 17. 7 8 2 C7H7Br O

ditropyl ether

Me H OTs HOAc 18. Ph H S (acetolysis) Me N (a racemate) Hint: recall the concept of neighboring group participation (9.4, 9-16) in some nucleophilic substitution reactions; even aromatic rings are sometimes capable of acting as a “neighboring group.”

12.4 Mechanisms 116 • Chapter 12 Aromatics

Cl 19. CHCl3 N OR H N

H 20.

OMe OMe

H PO / H SO 21. 3 4 2 4

PO3H2

12.4 Mechanisms CHAPTER 13 ALCOHOLS

13.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. NaBH4 2. NH4Cl (a weak acid) 1. benzyl methyl ketone 3. PCl3 4. KO-t-Bu

1. H2 / Pd

2. H2SO4

1. i-PrMgBr 2. 3-ethyl-3-pentanol

Br 2.

OH 3. 1. H2SO4

2. H3O

HCl

1. TsCl 4. 5-hydroxy-2-heptanone 2. NaOAc

1. NaH 5. 1-hexen-3-ol 2. O MeO S OMe O

O 1. PhMgCl 6. Ph O 2. H

1. LiAlH4 2. H

13.1 Reactions 118 • Chapter 13 Alcohols

1. Li 2. diisopropyl ketone 7. iodomethane 3. H

1. HBr 2. LDA 8. 2-butanol . 3. BH3 THF 4. H2O2, HO

O O 1. NaBH4 OMe 9. 2. H

Ph O 1. LiAlH4 O 2. H

H2 / Pt

Me POCl 10. H OH 3 Me D pyridine H

1. NaOH 11. O NMe 2. CH3I (1 equiv)

HO morphine codeine

OH OH Jones reagent 12. HO HO

13.1 Reactions Problems • 119

1. Br2, H2O 2. Me SiCl 13. 3 3. Li 4. acetone 5. H3O

O 1. (XS) MeLi 14. BnO OBn 2. H

1. LiAlH4 15. p-hydroxybenzoic acid 2. (XS) HBr

O O 1. NaBH4 16. OAc 2. H

O cortisone acetate 1. LiAlH4 2. H

OH H 17. HO (pinacol rx)

H 18. Ph OH OH

O O

aromatase 1. NaBH4 19. 2. H O HO

andostenedione estrone estradiol

13.1 Reactions 120 • Chapter 13 Alcohols

O OH 1. SOCl2, Et2O 20. 2. Mg 3. H

patchouli alcohol (used as a fragrance)

O Ph OEt 1. n-PrMgCl 21. N 2. H Me

DemerolTM (narcotic analgesic)

OCH3 O 1. toluene, ' 22. + 2. H TMSO

Danishefsky's diene

13.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

O D

1. O

O CHO 2. cyclohexanol

13.2 Syntheses Problems • 121

OH 3. 1-butene Ph

OH 4. H

H OH

racemic s-BuCl

*OH

H (* = 18O)

5. n-butane HO OH

6. cyclohexane O

7. vinyl chloride 1,3,5-hexatriene

8. n-hexyl alcohol CN

13.2 Syntheses 122 • Chapter 13 Alcohols

CO2H 9.

10. p-chlorophenol p-hydroxybenzaldehyde (via a Grignard)

OH O

11.

HO HO estradiol

OH 12.

13. n-butane OH

Cl OH 14. OH

OH 15.

13.2 Syntheses Problems • 123

O OH O 16. Cl

OH OH

17.

HO CH C 18. HO HO

OH OH C CH 19.

Me HO O estradiol O OH C CH

HO O

EnovidTM constituents (OCPs)

20.

13.2 Syntheses 124 • Chapter 13 Alcohols

21. Berson (Yale) discovered that the bicyclic carbocation A undergoes a clever rearrangement in which the cyclopropyl ring circumambulates around the cyclopentenium ring. Beginning with B, synthesize a deuterium-labeled species that would support this observation.

etc.

O A B

13.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

1. dil H SO OH 2 4 OH

(a cyclic ether)

1. NaBH4 2.

2. H2SO4

OH retinal

H 3. cycloheptene glycol

C7H12O IR: 1729 cm-1 1H NMR: G (d, 1H), plus other chemicals shifts

13.3 Mechanisms Problems • 125

Ph H Ph 4. OH OH O

H 5. + OH

H H 6. glycerol O

H 7. OH OH (an aldehyde)

2. MgCl 8. 1. H3PO4

CO2H 3. H3O O 4. HBr 5. Me2NH NMe2 amitriptyline (an antidepressant)

13.3 Mechanisms 126 • Chapter 13 Alcohols

9. Aflatoxin B1 is one of the most potent carcinogens known. In the presence of water and acid, compound A is formed.

O O O O O H O H3O O O H O OCH O OCH 3 O 3 H

aflatoxin B1 A

CH 3 CH3 CH3 10. H Br HBr H Br Br H + racemate HO H Br H H Br CH 3 CH3 CH3

Note: retention at C2,3 inversion at BOTH C2,3! This observation by Winstein (UCLA) provided stereochemical support for the concept of neighboring group participation (see 9.4, 9-16).

Similarly, Br Br HBr 11. or only trans-product is formed! OH OH

12. A step in the biosynthesis of the amino acid valine:

O OH OH H 1. [H] CO2H CO H 2 2. (-H O) CO2H 2 NH2 O valine

13.3 Mechanisms Problems • 127

13. H2SO4 OH

14. The conversion of ethylene glycol to acetaldehyde under acidic conditions could occur by one of two pathways: (1) dehydration to an enol followed by tautomerization, or (2) a pinacol-like rearrangement. In view of the following experiment, which pathway is suggested? O H O H2C CD2 OH OH DH2C D NOT H3C D

15. Cyclohexene glycol in the presence of acid forms cyclohexanone. Similar to problem 14, two pathways are possible: dehydration/tautomerization vs. a pinacol-like rearrangement:

H OH H - H O + H taut 2 ~ H: H - H2O O OH OH OH - H O H cyclohexene glycol

Synthesis of deuterium-labeled glycol A, when treated with acid, yields B:

OH D D H D D O OH A B a. Which pathway is consistent with this observation?

b. Suggest a preparation of A from cyclohexene.

13.3 Mechanisms 128 • Chapter 13 Alcohols

O H OH 16. ( = 13C) OH O

Cl 1. Li 2. acetone 17. C4H8 + 3. H

(1H NMR shows only a singlet at G 8.2)

OH Cl D :PPh3 D 18. D D CCl4

19. OH H

D-pinene (a constituent in oil of turpentine - interestingly, the dextrorotatory form is found in North American oils and the levorotatory form in European oils)

13.3 Mechanisms CHAPTER 14 ETHERS

14.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. HBr (1 equiv) 1. 2. TsCl O 3. KOAc, 18-crown-6

(XS) HI 2. benzyl phenyl ether

3. phenyl mercaptan + Me3S I

OH 4. KOH

O 1. HF 5. 2. PCC

NaCN 6. 2-isopropyloxirane MeOH

1. styrene epoxide 7. PhLi 2. H2SO4

OMe 1. HI (1 equiv) 2. CrO3, H 8.

3. NaBD4 4. H

14.1 Reactions 130 • Chapter 14 Ethers

1. PhCO3H 9. 2. PhOH, H

H, MeOH 10.

11. The fungicide flutriazole can be synthesized by the following scheme:

F MgI 2. 1. ClCH2COCl F

AlCl3 3. H

F N N 4. base 5. N

F OH 6. H N N

N flutriazole

1. mCPBA 12. Ph 2. MeNH2

ephedrine (bronchodilator)

13. The Claisen rearrangement of allyl phenyl ethers:

1. NaOH OH 2. Br

3. ' (Claisen)

14.1 Reactions Problems • 131

14. The Claisen rearrangement can be generalized to include allyl vinyl ethers:

' O O H

Draw the expected Claisen rearrangement product for each of the following:

O CO H a. 2 '

OBn b. A stage in the biosynthesis of aromatic amino acids (draw the structure of prephenic acid and give a mechanism for its conversion to phenylpyruvic acid):

HO CO2H ' H CO2H O Claisen O

HO2C chorismic acid prephenic acid phenylpyruvic acid

15. Mechanistically similar to the Claisen rearrangement is the Cope rearrangement:

This specific example became known as the “degenerate Cope,” a moniker that did not particularly please its discoverer, Prof. A. Cope! Of course, the degeneracy can be removed:

' Cope

14.1 Reactions 132 • Chapter 14 Ethers

16. Going back to problem 14.1, 13, if the ortho positions are blocked the initial Claisen rearrangement product may be followed by a Cope rearrangement. Fill in the brackets.

Claisen Cope

' '

~ H

17. A slight variation of problem 14.1, 15 is the oxy-Cope rearrangement:

HO HO O ' tautomerize H

Predict the oxy-Cope product for the reaction below:

SH 18. 2 [O]

H2NCO2H cysteine cystine [crystallization in kidneys can lead to one type of calculi (stone)]

OH SH [O] 19. HS OH dithiothreitol C4H8O2S2

14.1 Reactions Problems • 133

14.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. 3-methylpentane 3-methoxy-3-methylpentane

2. cyclohexene trans-cyclohexene glycol

OTs

S 4. propylene 2 diallyldisulfide (found in garlic)

O 5. cyclohexane via an epoxide H

6. cyclohexene oxide cyclohexane

14.2 Syntheses 134 • Chapter 14 Ethers

O O

7. N HS N

CO2H CO2H

captopril (antihypertensive)

S S 8.

CO2H CO2H asparagusic acid (isolable from asparagus)

14.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

BF3, Et2O 1. styrene epoxide CHO

1. LDA 2. methyloxirane allyl alcohol 2. H

3. H O OH

14.3 Mechanisms Problems • 135

OH OOO OH, MeOH O O OH 4.

5. Complex ladder polyether natural products, so named for their rung-like structure, are the active toxins found in harmful algal blooms known as red tides, which cause devastating ecological damage. Brevetoxin B is an example. HO H

O H O O H H H O O H H H H O O O H O O HH O O O H HHH brevetoxin B

Twenty years ago Nakanishi (Columbia) proposed such products arise biosynthetically from an elaborate cascade of epoxide ring-opening reactions that zip up the polyether structure. The following reaction, discovered by Jamison (MIT) in 2007, supports this hypothesis.

H HO H H H H O O O O H2O O O H HO O O H H H H

6. The biosynthesis of steroids involves an absolutely gorgeous (!) polycyclization reaction of squalene epoxide, followed by two sequential 1,2-hydride shifts and two 1,2-methide shifts to form lanosterol (lanosterol is then converted to cholesterol, the precursor to most other steroid hormones):

HO O

squalene epoxide lanosterol

14.3 Mechanisms 136 • Chapter 14 Ethers

7. Biochemical hydroxylation of aromatic compounds proceeds via arene oxides, which subsequently undergo ring opening to form phenols: H OH O O H OH cytochrome P450 + H (a) - H

benzene oxide A

(b) 1,2-H: shift tautomerization

H O O H H - H H H

Phenol could be formed from intermediate A simply by an E1-like loss of a proton (path a) or, alternatively, by a pinacol-like rearrangement followed by tautomerization (path b). Support for path b was provided by chemists at the NIH who observed the following conversion:

D O OH H D H3O

Explain. Account for the role of the methyl substituent. (This rearrangement of an arene oxide has now become known as the NIH shift!)

CO3H

R OH R OH O 1. 2. BF ,EtO R 8. R' Cl R' 3 2 O R' OH

Step 2 illustrates a semi-pinacol type rearrangement. Propose a mechanism for that step.

14.3 Mechanisms Problems • 137

PhO N S DBN N S 9. H PhO H O N O N O O O

O CH2Cl

Note: DBN (1,5-diazabicyclo[4.3.0]non-5-ene) is a sterically hindered nitrogen base that favors elimination over substitution:

H N N N HA N A

DBN

14.3 Mechanisms This page intentionally left blank CHAPTER 15 ALDEHYDES AND KETONES

15.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. CrO3, H 1. OH 2. hydrazine, H

1. Ph3P 2. MeLi 2. PhOCH2Br 3. methyl ethyl ketone a vinyl ether (see 15.1, 12, 13 and 15.3, 3, 33 for examples of their reactivity)

O H3O 3. O Ph

4. + opsin-NH2

11-cis-retinal (a protein) rhodopsin

O OH H3O 5. Ph

OH 1. PCC 6. 2. H3O 3. HOEt, H

O 1. KO-t-Bu / t-BuOH Cl 7. 2. HCl

15.1 Reactions 140 • Chapter 15 Aldehydes and Ketones

1. NaBD4 8. methyl n-propyl ketone 2. H

3. H2SO4 (E1)

1. KMnO4 2. semicarbazide 9.

3. H2 / Pt (XS) HO

vitamin D

O 1. ethylene glycol, H CH 2. DIBAH, -78o 10. OMe 3. Ph3P=CMe2 O 4. H3O citronellal

1. Ph3P O 11. p-nitrobenzaldehyde

2. H3O

a fluorescent "spy dust" ingredient

1. Ph3P-CHOCH3 12. O

2. H3O

an aldehyde

13. Using the above reaction (12) as a model, how could you prepare pentanal from butanal?

15.1 Reactions Problems • 141

OMe 1. CH2I2 / Zn (Cu) 2. H3O 14. O 3. Ph3P=CHC=CH2 OMe

1. H3O 15. cyclopropanecarbaldehyde hydrazone 2. EtMgI 3. H

1. H3O 16. acetophenone diphenyl ketal 2. H2NOH

O CHO H3O 17.

a heterocycle

1. Ph3P 2. n-BuLi 18. O O Br 3. butanal 4. H3O

O H O 19. 3 OO

2+ 1. H3O, Hg 20. t-butylacetylene 2. hydrazine, OH

15.1 Reactions 142 • Chapter 15 Aldehydes and Ketones

1. Br2, H2O

2. O , H 21. 3. Li 4. ethylene oxide 5. H3O

1. HONO2 / H2SO4 22. 2. H NOH, H CHO 2 O nitrofuroxime (used in treating urinary tract infections)

H3O 23. Et O

O multistriatin (European elm bark beetle pheremone)

OH / ROH 24. 2-oxopropanal (intra-Cannizzaro)

CHO

1. N D , OD, D O 25. 2 4 2 OMe 2. HI OH

vanillin

CO2H CN CO2H O O Ph O OH 26. OH H H3O OH + HCN + ?? OH OH OH OH This reaction, with the release of the very toxic HCN, provides a defense mechanism for millipedes.

15.1 Reactions Problems • 143

O H3O 27. O

frontalin (insect pheremone)

MeOH, H 28. 3-oxobutanal

-1 C6H12O3 IR: 1715 cm 1H NMR: G 2.2 (s, 3H), 2.8 (d, 2H), 3.4 (s, 6H), 4.9 (t, 1H)

OOOMe H3O 29.

O 30. H3O O

safrole (odor of sassafras)

NH2 1. H 31. H CO2H 2. H2 / Pt O proline (an amino acid)

AcO O OH

32. 1. LiAlH4 MeO 2. H3O MeO cortisone acetate dimethyl ketal

15.1 Reactions 144 • Chapter 15 Aldehydes and Ketones

O O OH 33. H H

OAc 1. LiAlH4 34. OAc 2. H3O

HO O HO O 35. H3O O

O F flunisolide (anti-inflammatory in allergy medication)

(XS) hydroxylamine 36. 1,2-cycloheptanedione

heptoxime (used in quantitative determination of Ni)

H3O 37. OOOO

paraformaldehyde

38. Chain degradation of a hexose:

OH OH O 1. NH2OH, H H 2. Ac2O (dehydrates OH OH OH an oxime)

15.1 Reactions Problems • 145

O Me N N N mild acid 39. H O

O O

tadalafil (CialisTM)

HH O NN 40. Me2N S mild acid MeNH2 +

NO2

ranitidine (ZantacTM - antiulcerative)

N N H3O 41. N H S

olanzapine (ZyprexaTM - antipsychotic)

1. PCC 2. Me2CuLi 42. OH 3. H3O

citral

H N

H3O 43. O O

F paroxetine (PaxilTM - antidepressant)

15.1 Reactions 146 • Chapter 15 Aldehydes and Ketones

44. Ammonia is produced in the mitochondria primarily by the oxidation of glutamate to produce an imine, which is subsequently hydrolyzed:

NH2 [O] H O O CCO 3 2 2 + NH4

glutamate

D-ketoglutarate

O O 45. H3O 4 steps

HO O diosgenin (from Mexican yams) progesterone

1. PCC 46. 2. MeLi EtO 3. H3O EtO

testosterone diethyl ketal 17-methyltestosterone (an anabolic steroid)

47. A step in Woodward's (Harvard) synthesis of strychnine:

N O 1. HC CNa / THF

N 2. H2 / Lindlar catalyst H O O dehydrostrychninone

15.1 Reactions Problems • 147

48. Aldehyde protons are non-acidic. However, if the aldehyde is converted to a 1,3-dithiane (the sulfur analog of an acetal), the proton can then be quantitatively removed by NaNH2 or organolithiuim bases. The resultant anion (Corey-Seebach reagent) readily undergoes SN2 or carbonyl addition reactions. Subsequent hydrolysis of the product unmasks the starting carbonyl.

O S S R n-BuLi R 1. R'X O + R H SH SH H S S 2. H3O R R' Li a dithiane Corey-Seebach reagent

Predict the products of the following reactions:

1. HS(CH2)3SH, H 2. MeLi a. benzaldehyde 3. EtI 4. H3O

1. n-BuLi S 2. n-decyl bromide b. S 3. H3O

1. HS SH , H c. acetaldehyde 2. NaNH2 3. cyclohexanone 4. H3O

49. The amino acid serine can undergo a retro-aldol-like reaction (see CARBONYL CONDENSATION REACTIONS) to form glycine and formaldehyde; in cells this reaction is catalyzed by a derivative of pyridoxine (vitamin B6):

OH O retro-aldol H2NCO2H + H2N CO2H HH

serine glycine

(cont. on next page)

15.1 Reactions 148 • Chapter 15 Aldehydes and Ketones

Catabolic reactions that produce formaldehyde, as above, generally occur in the presence of another vitamin derivative, tetrahydrofolic acid (FH4). The later detoxifies formaldehyde by reacting with it to produce A. On the other hand, many anabolic reactions require formaldehyde as a building block (e.g., biosyntheses of the nucleoside bases). In those instances A undergoes hydrolysis to yield FH4 and formaldehyde in situ. Draw the structure of FH4.

H H2N NN

N H3O FH4 + CH2O H N FH4 + CH2O O N catabolism anabolism

A O

HN CO2H

CO2H

[Note: Unlike us, bacteria can synthesize FH4 de novo from precursors such as p-aminobenzoic acid (PABA). Sulfa drugs are effective competitive inhibitors to enzymes that utilize PABA, thus destroying the ability of the bacteria to synthesize FH4.]

N N N H 50. NH2 O -H O Cl 2

XanaxTM (anxiolytic)

15.1 Reactions Problems • 149

15.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

HO CO2H 1. cis-2-butene

NH2

OCH 3 OH 2. H D

3. cyclopentanone CO2H

O Cl 4. D

5. benzaldehyde Ph CO2H

15.2 Syntheses 150 • Chapter 15 Aldehydes and Ketones

O O O 6. Ph C CH2Cl C Ph O

OH O

OMe

OO OH O 7. OMe OMe

8. via a Wittig

O 9. MVK (methyl vinyl ketone) OMe

15.2 Syntheses Problems • 151

via a Wittig 10. OH O O

O H 11. + Br O HO

CHO

12. Hydrazones can be deprotonated by strong bases to give carbanions that act as nucleophiles, e.g.,

NR NR N 2 N 2 n-BuLi (H -H

OOH

How could this observation be used to form Ph from acetone and benzaldehyde?

CHO CHO

13.

O O

14. H

15.2 Syntheses 152 • Chapter 15 Aldehydes and Ketones

OH OH C CH 15.

O O

major component in OCPs

O Ph Et

16. Ph , benzaldehyde Ph

R R TM (R = -OCH2CH2NMe2) tamoxifen (Nolvadex - antiestrogen)

O 17. 1-butene

18. CO2Me CO2Me

H O 19. H OH O chrysomelidial (secreted by larvae of some beetles in self-defense)

15.2 Syntheses Problems • 153

20. The hotness of chili peppers can be quantified by determining their Scoville heat units (SHUs). An SHU is the amount of dilution needed before the chili is undetectable. The hottest, according to the Guinness Book of World Records, is the bhut jolokia from India, firing up at around 1,041,427 SHU, i.e., a drop of extract needs about a million drops of water! (Jalapeño and Tabasco range a mild 5,000 – 25,000 and 100,000 – 200,000, respectively, on the SHU scale.) The active ingredient is capsaicin. Formulate a synthesis of the carboxylic acid moiety from 6-bromo-1-hexanol.

HO HO Br O OMe N O H

capsaicin

21. Similar to benzyl carbon-oxygen single bonds, carbon-sulfur single bonds readily undergo hydrogenolysis. This observation provides a more gentle reduction of aldehyde or ketone carbonyls than the highly alkaline Wolff-Kishner or acidic Clemmensen reductions. Complete the following illustration of this approach:

O H / Ra-Ni Ph Ph 2 PhCH2Ph (hydrogenolysis)

a dithiane (see 15.1, 48) (or thioketal)

O 22. O via an enamine H N O

15.2 Syntheses 154 • Chapter 15 Aldehydes and Ketones

15.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

18 H, H2 O 1. O 18O

H O 2. 3 H N 2 CHO N

3. Vinyl ethers, unlike ordinary ethers, hydrolyze rapidly in water with just a trace of added acid. Draw the products and mechanism for

TsOH / H2O ?? O

O H 4. + hydrazine N N H H

Cl OH OMe / HOMe 5. O OMe

H MeO

15.3 Mechanisms Problems • 155

OO R H2NOH, H 6. RR R N O

HO O O O O H 7. OH

8. Fugu, a fish, is a Japanese delicacy. Unfortunately it produces a very toxic substance, tetrodotoxin (an adult fugu contains enough to kill 30 people), in organs that must be removed by a licensed chef. To become a fugu chef requires training for years with a master and culminates in a battery of state- administered exams, including eating a fugu prepared by oneself ! Though the risk of fugu poisoning is practically nil, if prepared by a master, a handful of diners succumb to fugu each year; perhaps that is why Japan’s Imperial Family is forbidden from tasting one of their country’s choicest dishes. Deduce the structure and outline the mechanism of the carboxylic acid produced when tetrodotoxin is treated with aqueous acid. O

HO O H O H2N N OH H3O N HO H OH OH tetrodotoxin

9. E. J. Corey (Harvard) found that sulfur ylids, similar to the Wittig reagent, can be prepared as follows:

O 1. SN2 O S + CH3I Me2 S CH2 2. n-BuLi

When treated with cycloheptanone a 70% yield of A is obtained. Explain, showing clearly how the intermediate betaine’s behavior to form an epoxide differs from that of a typical Wittig intermediate.

15.3 Mechanisms 156 • Chapter 15 Aldehydes and Ketones

O OH HCl 10.

Cl O OH

O H3O 11. acetaldehyde O

OH H O 12. 3 O O

diazomethane 13. cyclohexanone cycloheptanone

Ph N H NHPh 14. + acetaldehyde H N NHPh Ph

15.3 Mechanisms Problems • 157

15. The Vedejs olefin inversion reaction readily converts cis-to-trans or trans-to-cis stereoisomers:

1. mCPBA (Hint: think Wittig-like) 2. Ph3P

:P(OMe)3 16. propylene epoxide propylene

HO HO OH O NHR 17. H O + NH2R OH OH OH OH

H 18. 2 phenol + acetone C15H16O2 bisphenol A (a starting material in the synthesis of LexanTM)

O Et n-PrNH2 19. H N

15.3 Mechanisms 158 • Chapter 15 Aldehydes and Ketones

20. Another protecting group for alcohols (in addition to TMS or vinyl ether derivatives) is MOM (methoxymethyl). MOM is stable to base, but can be cleaved upon treatment with mild acid. The following sequence illustrates its use:

Cl Cl 1. Li 1. NaH 2. CH2O OH 2. ClCH2OCH3 O HO OH 3. H3O MOM a. Draw the structure of the MOM derivative and explain its mechanism of formation.

b. Outline the mechanism of the last step. What other two organic products are formed from the MOM group?

Br O H3O 21.

O OH

SCH3 22. MeSH 0o O OH

O H O, OH 23. CN 2 NC H OH

15.3 Mechanisms Problems • 159

OOEt O 1. MeLi 24. 2. H3O

E-vetivone

Et H H O O N 25. EtNH2

26. Outline the mechanism for steps 2 and 3. TMS O

2. Ph OMe Ph 1. (Danishefsky's diene) N PhCHO + PhNH2 N O Ph Ph 3. H

27. Aromatic aldehydes cannot be prepared by direct Friedel-Crafts acylation (formyl chloride is unstable). One alternative is the Gatterman-Koch reaction (12.4, 10). Other options include the following two reactions: a. the Reimer-Tiemann reaction

OH OH 1. CHCl3, OH CHO 2. H

15.3 Mechanisms 160 • Chapter 15 Aldehydes and Ketones b. the Vilsmeier reaction (outline a mechanism for both steps)

OH Me Me O PCl POCl 2 2 phenol CHO N H 3 N Cl 1. Me Me C O H 2. H3O salicaldehyde Vilsmeir reagent

28. Another approach to enhancing the acidity of an aldehyde proton (see 15.1, 48 – Corey-Seebach reaction) is illustrated by the benzoin condensation reaction:

O OH CN 2 PhCHO Ph Ph benzoin

29. NH formaldehyde, H N

OH OH

HO OH HO O H O 30. OH OH OH OH OH

E-D-ribose D-D-ribose (Carbohydrate chemists call this process "mutarotation" and refer to the two epimeric diastereomers as "anomers.")

15.3 Mechanisms Problems • 161

31. The final step in the urea cycle:

NH2 H O NH2 NNH H3O HO C NH 2 H2NNH2 + 2 HO2C NH2 arginine urea ornithine

32. Fluorescamine reacts with amines to give a highly fluorescent derivative. As little as a nanogram of an amino acid, for example, can be detected by this method. (Warning: do not attempt this one alone!)

Ph R Ph O N

O RNH2 O OH O H CO H O 2

fluorescamine highly fluorescent derivative

33. Unlike other types of phospholipids, plasmalogens undergo hydrolysis to produce not only fatty acids but also fatty aldehydes. Explain the formation of the latter.

(CH ) CH O O 2 n 3 R OC H3O O POR' O O a plasmalogen (platelet activating factor)

34. Although ketones are generally not reactive with most oxidizing agents, they are readily oxidized to esters when treated with peracids (Baeyer-Villager reaction).

O O + RCO H Ph Ph 3 Ph OPh

15.3 Mechanisms 162 • Chapter 15 Aldehydes and Ketones

35. Many historians of chemistry credit the discovery of molecular rearrangements to the benzilic acid rearrangement:

OO KOH, EtOH OH Ph CO2 Ph Ph Ph benzil CB of benzilic acid

Discovered by Liebig in 1838, it is a rare example of a rearrangement under alkaline conditions (most require acidic environments). Because of (1) disagreements over atomic weights at the time (the “conventional” weights for carbon and oxygen were thought to be 6 and 8!), and (2) the (erroneous as we now know) dogma propagated by Kekulé that carbon skeletal rearrangements could not occur in the course of chemical reactions, many wrong structures for benzilic acid were proposed -- until Baeyer finally got it right nearly forty(!) years later in 1877. a. Propose a mechanism for the benzilic acid rearrangement.

b. Baeyer observed a benzilic acid-type rearrangement when phenanthrenequinone is treated with base. Draw the expected product.

O 1. NaOH

O 2. H

phenanthrenequinone

c. Another more modern benzilic acid-type rearrangement:

O O 1. MgX O OH Ph Ph 2. H Ph Ph

15.3 Mechanisms Problems • 163

Problems 36 – 40 illustrate the dienone – phenol-type rearrangements.

O OH

36. H

RR R R

OH HO 37. base acid

OTs

30% HClO 38. 4 O HO

However (!),

H SO 39. 2 4 O OH

15.3 Mechanisms 164 • Chapter 15 Aldehydes and Ketones

And, lastly, a steroid dienone – phenol rearrangement:

HO HO O O O OH O OH OH H 40.

prednisone (anti-inflammatory)

Problems 41 and 42 illustrate the Favorskii-type rearrangement.

O CO2R CO2R Cl OR 41. + (=13C tag) HOR

50% 50%

Br O OR CO2R 42. Br HOR

OMe

H3O 43. + NMe2 OH NMe O O 2

15.3 Mechanisms Problems • 165

R N 44. Br2 / H2O NR CHO

45. Sheehan’s (MIT) classic total synthesis of penicillin V involved a condensation step between the following reactants. Formulate a mechanism. (Note: the product is simply a nitrogen – sulfur analog of an acetal.)

O O HS C O H H N + CO2R H NCOH N O 2 2 S O penicillamine HN RO2C CO2H

46. Woodward (Harvard) envisioned the biosynthesis of strychnine as beginning with a condensation of derivatives of the amino acids tryptophan (trp) and phenylalanine (phe). Sketch a likely sequence of events.

trp NH N 2 NH H O H N strychnine H phe HO OH HO OH

15.3 Mechanisms 166 • Chapter 15 Aldehydes and Ketones

47. Thioketones, in the presence of aqueous acid, form hydrates via an intermediate ketone:

S HO OH Br Br H3O Br Br

Br Br Br Br

48. Many aldehydes autooxidize in air. For example, a white powder (benzoic acid) may often be seen around the cap of a bottle of previously opened benzaldehyde (liquid). Such autooxidation is thought to proceed by the addition of O2 to a molecule of benzaldehyde via a free radical process to form perbenzoic acid. The perbenzoic acid then reacts with a second molecule of benzaldehyde to form two molecules of benzoic acid. Outline a mechanism for the second step. Hint: recall the Baeyer-Villager oxidation of ketones to esters (15.3, 34). O

O O H O O OH H 2 O OH 2

49. An impressive biomimetic conversion in Johnson’s (Stanford) total synthesis of progesterone (see 19.3, 16 for the final stage):

O O O O 1. O H O H O 2. aq K2CO3 H H H H CF3CO2H OH

progesterone

15.3 Mechanisms CHAPTER 16 CARBOXYLIC ACIDS

16.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

KMnO , H 1. N 4

N nicotine niacin 1. NaOH 2. phenylacetic acid 2. Me3O BF4

1. NaBH4 3. 3-oxobutanoic acid 2. H

Ph OPh 1. H3O 4. OO 2. CrO3, H

1. NaOH (1 equiv) 5. J-bromobutyric acid 2. ' 3. LiAlH4 4. H3O

1. KOH OH 2. acrylic acid (propenoic acid) 6. 3. BH3 4. H3O

1. NaCN 2. PhMgCl 7. benzyl chloride

3. H3O

O 1. SOCl2 8. NH2 2. DIBAH, -78o 3. H3O

16.1 Reactions 168 • Chapter 16 Carboxylic Acids

9. N 1. (XS) PhLi Ph CO2H OH 2. H Ph OH

fexofenadine (AllegraTM - antihistaminic)

OAc 1. LiAlH4 10. 2. H CO2H aspirin

1. KCN, H 2. H3O 11. O 3. BH3 4. H3O testosterone

CO2H 1. EtLi 12. Ph NMe2 2. H Ph methadone

1. H O 13. 3 CO2H CO2 + 2. PCC

14. A reaction in the biosynthesis of the amino acid leucine:

CO2H 1. [O] CO H 2 2. (-CO ) HO 2 CO2H H2N

leucine

16.1 Reactions Problems • 169

15. The alkaloid cocaine, isolable from coca leaves, can be converted to tropinone, a precursor to the antispasmodic atropine (see 20.3, 12). Deduce the structure of tropinone.

O N N OMe 1. OH CO2H 3. Jones reagent 2. H 4. ' O

O Ph OH

cocaine ecgonine tropinone

CO2H 1. NaOH (2 equiv) 16. 2. MeI (1 equiv) 3. H

17. Chemical structures for medicinals that contain acid-base components are routinely drawn incorrectly in prescription information supplied by drug companies. For example, sumatriptan succinate, an active ingredient of TreximetTM (prescribed for migraines) is drawn as shown below. Draw its correct structure.

CH2CH2N(CH3)2 CH3NHSO2CH2 COOH . CH2 = N CH2 H COOH

16.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. benzoic acid PhCH2CO2H

2. propylene pentanedioic acid

16.2 Syntheses 170 • Chapter 16 Carboxylic Acids

3. OH acetone + CO2

4. o-chloroacetophenone

CO2H

CO H 5. 2

styrene ibuprofen (MotrinTM - antipyretic)

CO2Na 6. 1-butanol

sodium valproate (used in the treatment of epilepsy)

CO2H O

7. OH CO2H O O HO

via a nitrile CHO 8. benzyl bromide

16.2 Syntheses Problems • 171

NC NN 9. O NC O

10. pentanedioic acid

11. RCO2H RCH2 R'

OOH 12. 3-oxobutanoic acid Ph

13. ethanol butanedioic acid (succinic acid)

CO2H 14.

CO2H

16.2 Syntheses 172 • Chapter 16 Carboxylic Acids

15. acetylene hexanoic acid

OH OH OH O 16.

16.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

H 1. CO2H O

OH OH 2. CO2H '

-CO2 O O

tetrahydrocannabinolic acid THC

O Cl Cl OH 3. Cl CO2

16.3 Mechanisms Problems • 173

4. Isobutylene and carbon monoxide, in the presence of acid, give dimethylpropionic acid. Explain.

5. The carboxyl group may be protected by allowing it to react with 2-amino-2-methylpropanol to form an oxazoline derivative. Outline the mechanism. (Acid hydrolysis of the oxazoline regenerates the carboxylic acid.) O N ROH + HO R O H2N (an oxazoline)

O OMe MeOH, H 6. O CO2H O

7. The aldehyde flavorings formed in the roasting of cocoa beans is caused by the Strecker degradation of amino acids:

NH2 O OO H3O, ' + R CO2H RH+ CO2 + NN

8. Strecker also developed a synthesis of amino acids:

O R' O H3O N R H + R'NH2 + CN H OH R

16.3 Mechanisms 174 • Chapter 16 Carboxylic Acids

9. The biosynthesis of the amino acid phenylalanine involves an acid-catalyzed decarboxylation of prephenic acid:

CO2H HO H CO2H CO H -CO 2 CO2H O 2 O H2N prephenic acid phenylalanine

10. Ninhydrin reacts with amino acids to give a blue dye which can be colorimetrically assayed. Sketch the intermediates. O O O O OH -H2O O + NH CHRCO H OH 2 2 N O O O O + CO + RCHO ninhydrin 2 a blue dye

11. The vitamin niacin is used to form nicotinamide adenosine dinucleotide, which readily shuttles between its oxidized (NAD+) and reduced (NADH) forms. The latter serves as a cellular equivalent to NaBH4. The essential portions of the structures are shown below. Outline a mechanism for the cellular conversion of pyruvate to lactate. (Note: like NaBH4, NADH cannot reduce carboxylic acid carbonyls.)

O HH O NH [H] 2 NH2 O OH NADH N CO [O] N 2 CO2 R R

NAD+ NADH pyruvate lactate

12. The cellular biosynthesis of glucose (gluconeogenesis) begins with the conversion of oxaloacetate (OAA) to phosphoenolpyruvate (PEP!) via a decarboxylation-phosphorylation pathway. Provide arrows. OO CO O OO 2 guanosine OCO2 + OPOPOPO CO2 + O + GDP OH OO P OHO O OAA GTP PEP

16.3 Mechanisms Problems • 175

13. Unlike ȕ-ketocarboxylic acids, Į-ketocarboxylic acids do NOT undergo mild thermal decarboxylation. However, the enzyme pyruvate decarboxylase (PDC) gently converts pyruvate to acetaldehyde at 37o. The key is provided by an essential cofactor, a derivative of vitamin B1 (thiamine). The activity of thiamine resides in the thiazolium ring, shown below. A mechanistic clue was offered by Breslow’s (Columbia) discovery that Ha rapidly undergoes exchange with deuterium when thiamine is dissolved in D2O, suggesting that Ha is relatively acidic. Propose a mechanism for thiamine-assisted decarboxylation of D- ketocarboxylic acids. (Hint: begin with the conjugate base of thiamine, then consider how the thiazolium nitrogen can serve as an ‘electron sink’ to accept the electrons from decarboxylation.)

R' O PDC O N S R + CO2 CO2 thiamine H Ha pyruvate thiazolium ring

14. Another biochemical approach to decarboxylation: Vitamin B6 (pyridoxine) is a precursor to the coenzyme PLP (pyridoxal phosphate), a catalyst for many reactions, such as decarboxylations, that involve amino acids. Outline a mechanism. (Hint: form an imine from PLP and the amino acid, then consider the role of the pyridinium nitrogen as an ‘electron sink.’)

HO CO2H NH2 R OH PLP NH2 NH N -CO2 NH N N H histidine histamine PLP

15. Determination of the molecular mass of acetic acid in a nonpolar solvent, e.g., hexane, yields a value of 120. Explain.

16.3 Mechanisms This page intentionally left blank CHAPTER 17 CARBOXYLIC ACID DERIVATIVES

17.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

O Et NH (1 equiv) / pyridine 1. OEt 2 Cl O

2. butyric anhydride + methylamine

EtOH (XS), H 3. cyclopropyl cyclohexanecarboxylate

4. propane -1,3-diol phosgene

1. PCl3 5. oxalic acid 2. LiAlH(O-t-Bu)3 3. H

H O 6. N,N-diisopropylpropionamide 3

1. isopropyl magnesium bromide 7. phenyl hexanoate 2. H3O

O MeOH, H 8. O

1. LiAlH4 2. H 9.

3. Ac2O CO2H

17.1 Reactions 178 • Chapter 17 Carboxylic Acid Derivatives

saponification 10. H-caprolactone

O NH S Bn H2O, OH 11. N O CO2H penicillin G

O Et OR 12. + urea Et OR O

VeronalTM (a barbiturate, sedative)

O N 1. OH O 2. H 13. 3. Jones reagent 4. ' Ph O

cocaine O

O OH SCoA 14. . NH2CH2CO2H (glycine) HO . OH

cholyl coenzyme A (a rare example of cis-fused A-B rings in steroids) glycocholate (a major bile salt)

N 1. H3O 15. 2. CH2N2 N

O O strychnine

17.1 Reactions Problems • 179

OH 1. H3O (a lactone) 16. OEt 2. PhMgCl 3. H O

O O 1. LiAlH4 17. 2. H O O Spanish fly

H 1. phosgene 18. N N 2. LiAlH4 H 3. H

protein O OH 19. OPF Me

SarinTM (a cholinesterase inhibitor)

H CO2H

N 1. SOCl2 20. 2. diethylamine

N LSD H lysergic acid

O O

21. S Cl + H2N N O H

OrinaseTM (for diabetes) O H O, OH 22. NH 2 S O O saccharin

17.1 Reactions 180 • Chapter 17 Carboxylic Acid Derivatives

acetic anhydride 23. p-hydroxyaniline (1 equiv) acetaminophen (TylenolTM - antipyretic)

Al2O3, ' aniline 24. acetic acid H2CCO (-H O) 2 ketene

O O H 25. + [ethylene glycol]n MeO OMe

dimethyl phthalate n DacronTM

26. LexanTM, a high-molecular weight “polycarbonate,” is manufactured by mixing bisphenol A (see 15.3, TM 18) with phosgene (COCl2) in the presence of pyridine. Draw a partial structure for Lexan .

HO OH

bisphenol A

27. Me N C O + methyl isocyanate

(active ingredient in the insecticide SevinTM)

O 1. Li 2. CuI 28. O 3. benzoyl chloride 4. H O Cl 3

O NHMe 1. propionic anhydride 29. 2. LiAlH4 F3C 3. H

fluoxetine (ProzacTM - antidepressant)

17.1 Reactions Problems • 181

HO O

O O

O NH 30. 3

simvastatin (ZocorTM - antilipemic)

O C CH 31. 1. saponification

2. H3O N OH

Ortho TriCyclinTM (OCPs)

O MeO 1. LiAlH4 32. N H HO 2. H3O

capsaicin (active agent in cayenne pepper)

O H N 2 S

O 1. H2O, OH N 33. N 2. SOCl2 CF3 3. urea

celecoxib(CelebrexTM - anti-inflammatory)

F S O HO O O H O 34. 3 F O F fluticasone propionate (FlonaseTM - anti-inflammatory)

17.1 Reactions 182 • Chapter 17 Carboxylic Acid Derivatives

1. PhLi 35. CO2Na 2. H3O MeO

naproxen sodium (AleveTM - anti-inflammatory)

O HN N N S H3O 36. N N O N OEt

sildenafil (ViagraTM - treatment of ED)

1. base, ' 37. Br CO2H

2. NH2

NH OH 38. n

H-caprolactam Nylon 6 (a polyamide)

NH 1. ethyl chlorocarbonate 39. N 2. Et2NH diethylcarbamazine (anthelmintic)

O OEt

N H3O 40. CO2 Cl

loratadine (ClaritinTM - antihistaminic)

17.1 Reactions Problems • 183

NH 1. ethyl benzoate 41. 2. LiAlH4 3. H

Cl Cl sertraline (ZoloftTM - antidepressant)

OMe

CO2H 1. SOCl2 42. H2N OMe S 2. N O CO2H methicillin [an estimated 90,000 people in the US fall ill each year from MRSA (methicillin resistant Staphylococcus aureus)]

CF3 O N H

43. H3O CF3

O N H dutasteride (AvodartTM - treatment of BPH)

CO2H O H O N 3 44. H2N H OMe O Ph aspartame

O Cl N N exhaustive hydrolysis 45. N N O O N

zopiclone (LunestaTM - sedative, hypnotic)

17.1 Reactions 184 • Chapter 17 Carboxylic Acid Derivatives

O 1. NaBH4 46. CO2H 2 H2O + C6H8O4 2. H pyruvic acid

47. Chain elongation of a tetrose sugar:

1. HCN, CN OH O 2. H3O 3. SOCl2 H

OH OH 4. LiAlH(t-BuO)3 5. H

(see chain degradation of a sugar, 15.1, 38)

48. Consider the reaction of amino acid A with amino acid B. Four possible products are possible: A-A, B-B, A-B, and B-A, if simply A and B are heated together. A more rational synthesis of, for example, A-B is to first treat A with t-butyl chlorocarbonate (C), which has the effect of eliminating (blocking) the nucleophilicity of the nitrogen in A. The blocked species is termed a t-BOC amino acid (t-butoxy- carbonyl). R O R R' O N + H2N H OH O Cl H2NCO2H H NCOH 2 2 OR' A B A-B C a. Draw the product of the reaction of A with C.

b. The t-BOC-A is then condensed with B to yield a derivative of A-B. A-B is formed by treating that derivative with mild acid. Show the mechanism of removing the blocking group to form A-B. (Hint: CO2 and isobutylene are by-products.)

17.1 Reactions 186 • Chapter 17 Carboxylic Acid Derivatives

O H 53. O glucose + glucose OH

tuliposide tulipaline (a J-lactone) (found in tulip bulbs) (produced when bulbs are damaged - a fungicide)

N O 1. SOCl2 NN 2. ammonia 54.

3. SOCl2 FCO2H O LevaquinTM (antibacterial)

OH O N

N H3O N 55. H S N ' CO2 + MeNH3 + H3N + ?? S S OO (gives a positive Tollens' test) meloxicam (MetacamTM - anti-inflammatory)

17.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. sec-butyl acetate

O O

2. R NH2 R H

RCH3

17.2 Syntheses Problems • 187

O O

NH N 3.

O 5. 1,3-cyclopentadiene, acetylenedicarboxylic acid O

O OH 6. CO2H H O

O 7. benzamide Ph O

8. butanal 2-pentanone

17.2 Syntheses 188 • Chapter 17 Carboxylic Acid Derivatives

9. Following is an outline for the synthesis of diazepam (ValiumTM). Supply the appropriate reagents for each step. O O NHCH 3 NCH 3 NCH3 Cl O Cl Cl

CH3 O H N NCH Cl 3 O Cl O Cl

CH3 H C O 3 O N N NH O 2 Cl Cl N

ValiumTM

10. 3-oxohexanedioic acid O

11. methyl benzoate methyl phenylacetate

17.2 Syntheses Problems • 189

12. Following is an outline for the synthesis of fluoxetine (ProzacTM). Supply reagents for each step.

O O

NaO N ON O NH2

O O F3C F3C

O N O N OEt H CH3 H O F3C F3C

ProzacTM (antidepressant)

Ph 13. Ph N N EtO O DemerolTM (analgesic)

14. Following is an alternative synthesis of ProzacTM (see 17.2, 12). The reagent for step 5 is indicated; supply reagents for all the other steps. Outline a mechanism for step 5.

O ONMe2 HO NMe2

O O NMe2 Cl NMe2 5. Cl OEt F C mechanism? 3

Me Me O N OEt O N H O F3C F3C

ProzacTM mechanism:

17.2 Syntheses 190 • Chapter 17 Carboxylic Acid Derivatives

Et HO C 15. 2

ibuprofen

O O O

16. OH CO2H

O OH

17. O

Cl CONEt 18. 2

DEET (N,N-diethyl-m-toluamide - insect repellent)

19. Melatonin mediates circadian rhythm, the 24-hour sleep-wake cycle. Because its biosynthesis is inhibited by light, it is produced in the brain when the eye is not receiving light. Outline a synthesis from the neurotransmitter serotonin. H O NH2 N H HO MeO O O N

N N H H H

serotonin (5-HT) melatonin RoseremTM

Insomnia affects one in every eight people. RoseremTM, a selective melatonin receptor agonist, is an example of several drugs approved to treat short- and long-term insomnia.

17.2 Syntheses Problems • 191

20. The two monomers (B and C) for the synthesis of Nylon 66 can be prepared from a sugar derivative A. Supply the necessary reagents. Br O O CHO catalyst O (- CO) Br A

O Cl Cl CN O B Nylon 66 CN

H2N NH2 C

21. Name the following polymer and devise a synthesis for it. Remember, OH is not an appropriate starting monomer. Why?

OH OH OH OH

22. Some members of the morphine family of opium alkaloids…

RO CH3O CH3O

O O O OH N N N

R'O O O

R, R' = H (morphine) hydrocodone (a component oxycodone (HCl salt = OxyContinTM, of VicodinTM) a component of PercosetTM) R = Me, R' = H (codeine)

R, R' = Ac (heroin)

How can the following conversions be accomplished?

a. morphine codeine

17.2 Syntheses 192 • Chapter 17 Carboxylic Acid Derivatives b. morphine heroin

c. codeine hydrocodone

d. In aqueous solution codeinone exists in dynamic equilibrium with its EJ-unsaturated isomer, neopinone, hydration of which yields oxycodone. Write a mechanism for the equilibration.

CH3O CH3O

H H3O O O oxycodone N N

O O codeinone neopinone

OH H N O2C N CO H 23. 2 N N OH

NMe2 CO2H O AmbienTM (sedative)

17.2 Syntheses Problems • 193

17.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

H2*O, H 1. J-butyrolactone (show location of the labeled oxygen)

1. (XS) RMgX 2. O t-BuOH + ?? O 2. H Cl

3. Lactic acid (D-hydroxypropanoic acid) forms a cyclic compound, C6H8O4. Formulate a structure for this compound. Why does lactic acid not form a simple D-lactone?

*O TsOH 4. O an alkene + ?? (show location of the labeled oxygen) Et

PhMgCl (1 equiv) O 5. ethyl 5-oxohexanoate

17.3 Mechanisms 194 • Chapter 17 Carboxylic Acid Derivatives

6. Phenylisothiocyanate (A, PITC, Edman reagent) can be used to sequence proteins, i.e., to determine the order of amino acids (primary structure). For example, treatment of dipeptide B with A in the presence of acid yields C (a phenylthiohydantoin, or PTH, derivative of the amino acid). Characterization of C identifies the first (from the N-terminal end) amino acid, in this case alanine.

valine residue

S O Ph H N H2N OH NH + Ph N C S + N H H2NCO2H PITC O O A B C valine alanine residue PTH-alanine

O 1. Me3O BF4 O 7. NH 2 2. H3O OMe

8. The Swern oxidation:

a. "activation" of DMSO step: b. oxidation step: HOH O O O + O - CO2 RR S + RR Me Me Cl Cl - CO NR3 DMSO oxalyl chloride

a chlorosulfonium salt

17.3 Mechanisms Problems • 195

9. Similar to the Swern is the Corey-Kim oxidation:

O HOH + O S + Cl N RR RR NR3 O N-chlorosuccinimide

a chlorosulfonium salt

10. The biosynthesis of pyrimidine bases, e.g., uracil, begins with the formation of dihydoorotic acid. Formulate a mechanism. O O OO CO H H H 2 N N P H N H NOO + 2 2 CO H OH 2 O N CO2H O N H H carbamoyl phosphate aspartic acid dihydroorotic acid uracil

11. The antimalarial mefloquine can be synthesized from substituted 4-quinolones by the following sequence of reactions. Outline a mechanism for step 1 and draw the structures in brackets.

O Br 2. Li 1. POBr3 3. CO2

N CF3 4. H H N CF3 0 CF 5. (- 60 ) 3 CF 3 NLi

6. H3O

HO N H 7. [H]

N CF3 CF3

mefloquine

17.3 Mechanisms 196 • Chapter 17 Carboxylic Acid Derivatives

12. Acid halides react with diazomethane to give diazomethyl ketones, which, like diazomethane, decompose to give carbenes. O CH N O O 2 2 .. hQ RCl H + N2 RCNN RC.. H a diazomethyl ketone a. Formulate a mechanism.

b. This reaction was used in the synthesis of twistane. Draw the structures in brackets.

1. SOCl2 hQ

2. CH2N2 - N2 CO2H 0 [D]D +48

1. H2 / Pd 2. Wolff-Kishner =

twistane 0 [D]D +434

13. Cyanogen bromide (CNBr) specifically cleaves certain peptide (amide) bonds to yield a lactone:

H O R' R H N 1. NCBr N O N H N O R H + 2 2. H3O O R' S O

17.3 Mechanisms Problems • 197

14. Draw the structure in brackets and give a mechanism for the conversion of A to strychnone.

OH N H (Baeyer-Villager oxid)

N H2O2, H

O O pseudostrychnine

O N H O N O H3O

N N

O O O O

strychnone A

15. A step in Woodward’s (Harvard) classic total synthesis of strychnine:

Ac N Ac N

Ac2O, pyridine N CO H Me 2 N O OAc O

17.3 Mechanisms 198 • Chapter 17 Carboxylic Acid Derivatives

16. The final step in Sheehan’s (MIT) total synthesis of penicillin V involved the formation of a strained E- lactam. To accomplish this he employed a new reagent, dicyclohexylcarbodiimide (DCC), first reported from his lab two years earlier to smoothly form amides from an aqueous mixture of a carboxylic acid and an amine at room temperature. [That important advance in the state of the art for forming amide bonds was subsequently utilized by Merrifield (Rockefeller) in his solid-phase approach to synthesizing proteins by linking amino acids together through amide (peptide) bonds.] Propose a mechanism for the lactamization reaction.

OPh OPh N S NCN H N S DCC H OCON 2 ON H CO2 O CO2 salt of penicillin V

OH ONa O 1. PCl3 (2 equiv), O P OH O P OH 17. HN2 H3PO3 (1 equiv) pH 4.3 OH OH OH 2. H O -aminobutyric acid 2 OOHP J H2N H2N OOHP (GABA) OH OH (a bisphosphonic acid) FosamaxTM (alendronate sodium, bone resorption inhibitor)

Hint: phosphorous acid exists in two tautomeric forms; use the nucleophilic form to attack the product of the reaction of GABA with PCl3. This one is rather challenging.

17.3 Mechanisms Problems • 199

18. Tertiary alcohols are weakly nucleophilic because of steric hindrance near the hydroxyl group and, therefore, do not readily undergo Fischer esterification. One approach to form acetate esters of such alcohols is to allow them to react with isopropenyl acetate in the presence of an acid catalyst. Hint: the actual acetylation step involves an SN1-like reaction of the alcohol with an acylium ion. OMe OMe O

O , H

O OH O O O

19. In contrast to phenyl acetate, the conjugate base of aspirin (acetylsalicylic acid) readily undergoes hydrolysis in water, suggesting kinetic enhancement by the latter’s carboxylate moiety. Consider two possible pathways and outline a mechanism for each. a. The carboxylate anion acts as a nucleophile, attacking the acetate ester to form a mixed anhydride, which is subsequently hydrolyzed by water:

b. The carboxylate anion acts as a base, removing a proton from water to form hydroxide, which subsequently attacks the ester:

c. Experimental evidence indicates that when the reaction is conducted in the presence of 18O-labeled water, no label is found in the salicylic acid product. Which pathway is supported by this experiment?

H O 1. H NH2 2. NaBH4 N 20. + CO2Me NO N 3. H H H

17.3 Mechanisms This page intentionally left blank CHAPTER 18 CARBONYL Į-SUBSTITUTION REACTION AND ENOLATES

18.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. LDA 1. O 2. n-PrBr

1. OMe, MeOH 2. benzoyl bromide 2. methyl 3-oxopentanoate 3. H3O, '

1. EtOCO2Et, LDA 2. OEt 3. cyclohexanone 3. EtI 4. H3O, '

1. base 2. benzoic anhydride 4. diethyl malonate

3. H3O, '

Cl CN 1. OEt 5. + NC-CH2-CO2Et 2. H3O, '

NO2

6. +NaCH(CO2R)2

OTs

1. H3O 2. CrO3, H 7. (E)-3-pentene-2-one 3. NaH (2 equiv) 4. benzyl chloride (1 equiv) 5. H

18.1 Reactions 202 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

1. OEt 2. isobutylene epoxide 8. diethyl malonate

3. H3O, '

1. (XS) NaOEt 2. Br(CH2)4Br 9. ethyl acetoacetate

3. H3O, ' (C7H12O) O 1. LDA 2. PhSeBr 10. 3. H2O2 4. MeOH, H

O 1. Br2, H O 2. KO-t-Bu / t-BuOH 11.

O 3. LiMe2Cu 4. H3O

1. Br2, H 2. (CN)2CH: 12. t-butyl methyl ketone

3. H3O, '

1. OEt, HOEt 2. butanoyl chloride 13. diethyl malonate 2 CO2 + 3. H3O, '

O 1. a. Br2, OH b. H 2. PCl3, Br2 14. 3. MeOH

O 15. + NaOEt HOEt HH (pKa 15) (C8H10)

18.1 Reactions Problems • 203

O 1. LDA 16. 2. Cl , H Br 2

1. HCl (1,4-addition)

2. OO 17.

isoprene 3. H3O, '

O 1. Cl2, H 18. 2. OH (SN2) 3. KMnO4 1. Cl2, H 2. KO-t-Bu

3. H3O 4. KMnO4

1. KOH, EtOH 19. CH2(CO2Me)2 LiH (XS) 2. '

Cl 95% 91%

1. SOCl2 2. Me2NH 3. m-chloroperbenzoic acid

OO OO 1. base tautomerize 20.

2. OH O O O

coumadin (WarfarinTM - an anticoagulant)

18.1 Reactions 204 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

18.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. butyric acid ethylpropanedioic acid

2. dimethyl malonate G-valerolactone (valeric acid is a common name for pentanoic acid)

3. dimethyl malonate butanedial

4. dimethyl malonate NH O N O H seconal (a sedative)

5. dimethyl malonate 2-benzylbutanoic acid

6. dimethyl malonate + styrene Ph OH

7. dimethyl malonate + methyl acrylate Br Br (acrylic acid is propenoic acid)

8. ethyl acetoacetate s-Bu-CONH2

18.2 Syntheses Problems • 205

OO 9. ethyl acetoacetate OH

10. ethyl acetoacetate O

OO Cl Cl 11. ethyl acetoacetate

12. ethyl acetoacetate s-butyl methyl ketone 2-methylbutanoate

13. cyclopentanone

OAc

14. cyclopentanol CO2H

OO 15. 2,4-pentanedione

18.2 Syntheses 206 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

16. methyl acetoacetate 2,4-pentanedione

via an organoselenium cmpd 17. 3-pentanone ethyl vinyl ketone OO

Cl O

18. chlorobenzene N H

WellbutrinTM (an antidepressant)

OH O N H OH 19. O , OH

propranolol (a E-adrenergic blocker, developed by Sir James Black, recipient of '88 Nobel Prize in medicine; greatest breakthrough in pharmaceuticals for heart illness since discovery of digitalis approximately 200 years ago)

S H NNNa 20. dimethyl malonate O O

sodium pentothal (used to induce pre-surgical anesthesia in combination with sedatives)

Ph Ph 21. CN N O N NC O Et meperidine (an analgesic)

18.2 Syntheses Problems • 207

18.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

O OH

1. H O

OH OH OH H

(a central reaction in glycolysis catalyzed by the enzyme TIM, triose isomerase)

O O 1. EtO, HOEt 2. ethyl acetoacetate O 2. propylene oxide

3. 1,3-Diphenyl-1,3-propanedione gives a positive iodoform test even though it is not a methyl ketone. In addition to CHI3, two equivalents of benzoate are formed. Explain.

O O H 4. H racemization

O O OMe 5. MeOH O O O

18.3 Mechanisms 208 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

O H 6. O O

7. The vitamin biotin is necessary for many metabolic carboxylation reactions. It reacts initially with CO2 to form unstable A, which then “donates” CO2 to a substrate. Outline the mechanism for carboxylation of pyruvic acid to oxaloacetic acid (OAA).

O O O HN NH CO , ATP O 2 HN N OH O CO2H CO2H HO2C S CO2H HO2C S

biotin A pyruvic acid OAA

OH H O 8. O O O PO O H OH

TMS O Me3N O 9. + Me3Si Cl

18.3 Mechanisms CHAPTER 19 CARBONYL CONDENSATION REACTIONS

19.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. OH, (-H2O) 1. benzaldehyde + acetophenone 2. NaCH(CO2R)2

3. H3O

H 2. isobutryaldehyde

O OEt, EtOH 3. + CHO

4. OMe, MeOH OMe

1. OMe 5. dimethyl heptanedioate CO2 + 2. H3O, '

OH, ROH 6. (conj. add'n + retro-aldol)

O OH 7. O cis-jasmone (a perfume)

1. BrCH CO Me, t-BuO 8. propionaldehyde 2 2

2. H3O, ' (Darzen's cond.)

19.1 Reactions 210 • Chapter 19 Carbonyl Condensation Reactions

1. base 9. MeO CHO + propionic anhydride 2. acid (Perkin cond.)

OCH3 OCH3

base 10. + ethyl vinyl ketone O

(complete)

OH 11. benzaldehyde + CH3NO2

OMe, MeOH (retro-Claisen) 12. CO2Me O

1. NaOEt / EtOH (-H2O) 13. cyclopentanone

2. NH2NHPh

O 1. OMe, MeOH O 2. Cl 14. CO2 + C5H6O MeO O 3. H3O

OH 1. CH (CO Me) , OMe 15. acetone 2 2 2 (aldol) 2. H3O, ' C6H10O C8H12O2

19.1 Reactions Problems • 211

CHO OH (-H2O) 16. + 3-pentanone CHO

OH OH O retro-aldol 17. OH OH H

1. methyl vinyl ketone 18. N

2. H3O 3. NaOH (aldol)

OH 19. acetaldehyde + (XS) formaldehyde

C(CH2OH)4 -- pentaerythritol

O PhCHO, H 20.

CHO 21. 1. base + CH2(CO2Et)2 OH 2. H

C12H10O4

O CHO KOH 22. + CHO O (a pentacyclic dione)

19.1 Reactions 212 • Chapter 19 Carbonyl Condensation Reactions

23. A reaction in the biosynthesis of the amino acid leucine: O

O 1. SCoA (aldol)

CO2H 2. hydrolysis

24. Trans-resveratrol, isolable from red wine, has been implicated as a cardioprotective and can be synthesized as follows:

MeO CHO MeO 1. OEt + CN + CO2 2. H3O OMe

HO OH 3. BBr3, RT (hydrolysis)

trans-resveratrol (82%) OH

O 1. H3O 25. 2. KO-t-Bu (retro-aldol)

26. Forward and retro-aldol-like reactions that occur in plants:

CO2 O O retro-aldol CO SCoA 2 O C H hydrolysis HO 2 malate O2C glyoxylate CO2 - CO2 succinate

19.1 Reactions Problems • 213

O O + OEt / EtOH 27. O (Robinson annulation)

28. The biosynthesis of glucose involves aldolase, an enzyme that catalyzes both forward and retro-aldol reactions. The forward process illustrates a mixed aldol wherein the enzyme initially binds with A, promoting its tautomerization and subsequent reaction with B to form a ketohexose: O O aldolase + H OH OH OH OH A B

29. The Krebs Cycle begins with an aldol-like condensation of a thioester (acetyl coenzyme A) with oxaloacetate, followed by hydrolysis:

O O H3O + CO2 SCoA O2C

oxaloacetate

citric acid

30. The following sequence illustrates how fatty acids are catabolized to acetyl coenzyme A, a process known as E-oxidation. Fill in the brackets.

O H3O [O] SCoA (conj. add'n)

HSCoA O (retro-Claisen) + SCoA

19.1 Reactions 214 • Chapter 19 Carbonyl Condensation Reactions

31. Excessive accumulation of acetyl CoA can lead to metabolic ketosis by the following pathway:

H3O

Claisen A

O [H] - CO2 2 SCoA

C B

[A, B, and C (unfortunately!) are referred to as “ketone bodies;” accumulation of acids A and C lowers blood pH (acidosis).]

______

32. A cortisone story…

Cortisone is one of 43 steroids found in adrenal cortical glands. It was first isolated by Kendall (Mayo Clinic) in 1934 (extraction of ~ 1,000 lbs of beef adrenal glands yielded only 85 – 200 mg of cortisone). One of the earliest total syntheses of cortisone was published by Sarett (Merck) in 1952. The following reactions illustrate his strategy. a. The initial sequence of reactions formed the A-B-C rings. Draw the missing structures.

O HO

O Diels-Alder 1. H2 / Ni (1 equiv) H C OH B H 2. a. LiAlH4 EtO b. H EtO

C 1. H3O (=> a ketone) A B 2. methyl vinyl ketone, base (Robinson annulation)

(complete)

19.1 Reactions Problems • 215 b. Construction of the D ring began as follows. Fill in the bracketed structure and outline the mechanism for step 3.

R 1. X , t-BuO R C 3. mild acid C O 2. a. EtO CCMgX H

b. H CO2Et

c. Subsequent selective reduction followed by tosylation produced the indicated structure, which was then treated with the sequence of reagents shown. Draw the product of step 2 and give the mechanism for step 3.

O O O C CD

H OTs

1. a. OsO4 b. NaHSO3

2. KIO4 3. OMe / MeOH

d. The above product was then subjected to the following steps. Draw structures for the critical intermediates in steps 2 -4. OAc O OO O O 1. O 2. I , OH O 2 O O CD RO OR 3. OH, ' CD O CD H OR RO 4. KOAc H H

19.1 Reactions 216 • Chapter 19 Carbonyl Condensation Reactions e. The final four steps yielded cortisone. Deduce the structure of cortisone acetate. OAc

O 1. HCN O 2. (-H O) CD 2

H 3. KMnO4, OH (- HCN)

O O OH 4. H3O

cortisone cortisone acetate

______

CO2Me 1. OMe, MeOH 33.

2. H3O, ' CO2Me

O O H

H CH3NO2 N 34. NaOMe / MeOH NO2 N H O C H N O Na 8 7 2 5 indigo blue (probably oldest known coloring agent - used to dye bluejeans)

35. A step involving an intra-Michael reaction in Corey’s (Harvard) synthesis of longifolene, a component of Indian turpentine oil:

Et N O 3

longifolene

19.1 Reactions Problems • 217

19.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

via an aldol 1. ? + ? 1,3-diphenyl-1-propanol

O O via an enamine 2. cyclohexanone

via an aldol CHO 3. ? OH

via an aldol O 4. ?

5. diethyl ketone

6. acetone

O CO2R 7. CO2R CO2R O

19.2 Syntheses 218 • Chapter 19 Carbonyl Condensation Reactions

8. cyclohexanone 4-benzyl-1,3-cyclohexadione

9. 1-pentene

CHO

10. phenylacetaldehyde PhCH(CH2OH)2

O via a Robinson annulation 11. ? + ? O

12.

via a Wittig* 13. cyclohexanone, acetone O *Why not via an aldol?

O O via a Robinson annulation 14. O O

19.2 Syntheses Problems • 219

O 15. H

16. acetone

19.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

O O H 1.

O O

2. + CH2N2 + N2

O OO NaOEt / EtOH 3. OEt O

19.3 Mechanisms 220 • Chapter 19 Carbonyl Condensation Reactions

O O 1. CH NH 3 2 OCH 4. 2 methyl acrylate 3 2. OMe N

OH OH 5. p-chlorobenzaldehyde + CHBr3 Cl CO2

1. NaOEt, EtOH, MVK CO2Et 2. H O, OH 6. 2 O 3. ' O

O O

H 1. acetone, OEt 7. 2. H

geranial (a step in the commercial synthesis of vitamin A)

OO OO OEt + PhCO2Et + ethyl acetate 8. OEt Ph OEt

19.3 Mechanisms Problems • 221

OH O O OH 9. + methyl vinyl ketone CHO OH

HCO2 +

10. The anaerobic breakdown of glucose (glycolysis) involves the following isomerization and retro-aldol:

HO O HO OH OH O OH H O aldolase OH H + OH OH OH OH OH OH OH OH O

D-D-glucose D-D-fructose

R1 R2NCO2R R2NCO2R CO2R [Br ] 11. ' r (from NBS) % 1 NH NH

12. The following illustrates the Stobbe reaction. Hint: a key intermediate is a J-lactone.

CO2R O base Ph CO2R + Ph Ph CO CO2R Ph 2

19.3 Mechanisms 222 • Chapter 19 Carbonyl Condensation Reactions

O O 1. OMe / MeOH 13. 2. H O

O O O

OR 14. O O

CO2R CO2R

O EtO O

EtO OEt, EtOH 15. O O

O O

16. The final stage of Johnson’s (Stanford) historic total synthesis of progesterone (give a mechanism for step 2):

O O

1. ozonolysis

2. aq KOH O

progesterone

19.3 Mechanisms Problems • 223

1. NH, H

17. + 2. EtI O O O 3. H2O Et Et

O O

base 18. CO2Me

CO2Me

1. NaH 19. O 2. H2O CO2Me

CO2Me

20. Humulones are found in hops. When boiled, the insoluble humulones isomerize to the soluble isohumulones, which give beer its distinctive bitterness. (Caution: difficult!)

OH O O O R H, H2O R HO O ' HO HO OH O

humulone: R = i-Bu cis- and trans-isohumulones cohumulone: R = i-Pr adhumulone: R = s-Bu

19.3 Mechanisms 224 • Chapter 19 Carbonyl Condensation Reactions

21. Woodward’s (Harvard) total synthesis of the alkaloid strychnine included the following steps:

NMe2 OMe 1. CH2O, Me2NH OMe N H H OMe N H OMe 92% 2. MeI 3. NaCN, DMF

CO2Et N CN

OMe 4. LiAlH4, THF OMe N N 5. ?? H H OMe OMe 97% a. Step 1 is an example of the (name) ______reaction. b. Outline the mechanism for steps 2 and 3.

c. Supply the missing reagent in step 5.

22. Enzyme-catalyzed mixed aldol reactions are very common in metabolism. The beginning sequence in the de novo synthesis of aromatic amino acids, for example involves the following steps. Fill in the structures and write a mechanism for step 4.

OH O O OH HO P H O 1. hydrolysis 2. O OH CO2 phosphoenolpyruvate (PEP) an enol

3. - H2O O OH HO 4. H phenylalanine, tyrosine CO2H HO2C

HO OOHO

19.3 Mechanisms Problems • 225

23. The biosynthesis of porphyrin rings (e.g., heme) begins with an annulation reaction that involves an aldol reaction and imine formation in the dimerization of G-aminolevulinic acid (ALA) to form porphobilinogen.

CO2H O H CO H NH2 2 2 HO2C N ALA H3N H porphobilinogen

24. Several steps in Sheehan’s (MIT) total synthesis of penicillin V are shown below.

CO H O CO2H 2 1. ClCH2COCl 2. Ac2O O NH NH 2 N valine O Cl

O OSH 3. SH, OMe isomerization O Me N CO2Me MeOH H N a. Propose a mechanism for step 2.

b. Propose a mechanism for step 3.

25. The biosynthesis of fatty acids begins with a Claisen-like reaction: O O OO + SR SR CO2 + SR O O

19.3 Mechanisms 226 • Chapter 19 Carbonyl Condensation Reactions

26. The CD – CE bond in E-hydroxyketones is easily cleaved via a retro-aldol reaction; the carbonyl – CD bond is unreactive. In D-hydroxyketones, however, the CD – CE bond is unreactive; but, in the presence of thiamine, the carbonyl – CD bond can be cleaved (2).

OH O O O retro-aldol (1) + E D H R'

R N S O O 1. thiamine H R (2) (CB of thiamine) O + OH 2. O OH R H

Recalling the mechanism for thiamine-assisted decarboxylation of D-ketocarboxylic acids (problem 16.3, 13), formulate a mechanism for reaction (2).

27. The biosynthesis of cholesterol begins with the formation of HMG-CoA (3-hydroxy-3-methylglutaryl coenzyme A): OH O O SCoA 3 SCoA CO2 HMG-CoA a. Formulate a mechanism.

19.3 Mechanisms Problems • 227 b. HMG-CoA is subsequently reduced to mevalonate by an enzyme, HMG-CoA reductase. Because this reaction is the major control (rate-limiting) step, considerable research has been devoted toward developing a class of medicines that inhibits the action of this enzyme, notably the statins [e.g., atorvastatin (LipitorTM)].

OH OH PPPOO a reductase ATP HMG-CoA CO 2 CO2 mevalonate O P = PO OH O O PP = POPO O OH

Mevalonate then undergoes phosphorylation and decarboxylation to form I-PP (isopentenyl pyrophosphate) and DMA-PP (dimethylallyl pyrophosphate) – recall problem 9.4, 19a. Outline the mechanisms for decarboxylation to form I-PP and isomerization of the latter to form DMA-PP.

PPPOO - CO2 OPP OPP

CO2 I-PP DMA-PP

O OH OH O

N N OH H

F LipitorTM

19.3 Mechanisms This page intentionally left blank CHAPTER 20 AMINES

20.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. base 2. ClCH(CO2Et)2 3. base 1. phthalimide 4. isopropyl chloride 5. H O 3 alanine

1. (XS) CH3I 2. Ag2O, H2O, ' 3. OsO4 2. 2-methyl-2-hexylamine 4. NaHSO3 5. (XS) COCl2 6. NH3 TM C9H18N2O4 - Miltown

1. (XS) MeI 2. Ag2O, H2O 3. N-ethylcyclohexylamine 3. '

4. 1. (XS) MeI 2. Ag2O, H2O N 3. ' H coniine (toxin in hemlock, killed Socrates)

1. (XS) CH I 5. 3 O 2. Ag2O, H2O, ' N

HO morphine

Me 1. H2O2 6. Ph2N H H Me 2. ' Et

20.1 Reactions 230 • Chapter 20 Amines

1. SOCl2 2. NaN3 7. cyclopropanecarboxylic acid

3. ', H2O

OH OH 1. (XS) MeI 8. 2. Ag2O, H2O, ' NHMe HO epinephrine

D Hofmann elimination H 9. NMe2 H Cope elimination

H 10. 3-pentanone + dimethylamine + formaldehyde (Mannich rx)

NH2

11. N HONO O N H O H 2

cytosine uracil

12. Ph 1. Li 2. CO2 3. H Cl 4. SOCl2 5. NH3 6. Br2, OH, H2O phentermine (a diet drug)

20.1 Reactions Problems • 231

OH NH Br2, OH 13. Me 2 NH3 + O H2O

1. NaNH2 2. ethylene oxide 3. PBr3 14. Ph2CHOH 4. Me2NH

diphenhydramine (BenadrylTM - antihistamine)

1. HBr, ROOR 2. potassium phthalimide 15. 3. hydrazine amphetamine (CNS stimulant)

Me2N OH 1. MeI 16. MeO 2. Ag2O, H2O, ' 3. H EffexorTM (antidepressant) (gives a positive DNP test)

Ph 1. SOCl2 2. NH OH 3 17. 3. Br2, OH, H2O O OH 4. I (2 equiv)

DetrolTM (treatment of urinary incontinence)

1. a. Br2, PBr3 b. H2O 2. KO-t-Bu 18. CO2H 3. HCN, CN 4. H2 / Pt pregabalin (LyricaTM - first treatment approved for fibromyalgia)

20.1 Reactions 232 • Chapter 20 Amines

1. Br2 2. NaNO2, HCl 19. p-toluidine (p-aminotoluene) 3. KI

1. Cl2, FeCl3 2. NaNH2 / NH3 20. benzonitrile

3. KNO2, H 4. CuCN 5. H3O

1. KMnO4, H 2. fuming nitric acid 3. Fe, HCl 21.

4. NaNO2, HCl 5. HBF4

1. Br2, Fe H 2. Cl2, Fe N 3. H2O, OH 22. O 4. I-Cl, Fe 5. HONO 6. H3PO2

1. NaH CO2Me 2. H3O 3. NaBH 23. + O 4 N N 4. HBr (SN1) 5. ' (-HBr) nicotine

HO O

1. H 24. O OH NH NaBH3CN, EtOH 2. HCl HO NubainTM (narcotic)

20.1 Reactions Problems • 233

20.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions.

1. methylcyclohexane 1-methyl-1-cyclohexylamine

2. cyclohexane NH2

3. isopentane 3-methyl-1-butene (via a Hofmann elimination)

4. 3-methyl-1-butene isopropylamine

OMe OMe OMe MeO OMe MeO OMe MeO OMe 5.

NH2 HO NHPh mescaline (from peyote cactus)

6. p-nitrotoluene p-nitrobenzylamine

7. p-nitrotoluene p-nitroaniline

20.2 Syntheses 234 • Chapter 20 Amines

O 8. NH2 O

9. HO3SNO2 HO3SNNN Et2

methyl orange

N N O CO2Me 10.

OCOPh O cocaine tropinone

11. toluene 2,6-dichlorotoluene

12. p-nitroaniline HO N H

acetaminophen (TylenolTM)

O NEt 13. 1-nitro-2,6-dimethylbenzene, ethylene oxide, diethylamine N 2 H

lidocaine

14. benzene anisole

20.2 Syntheses Problems • 235

NO 15. 2 NN NH2

butter yellow

H N 16. benzyl methyl ketone Ph

methamphetamine

H Me Me N N Br O 17. S S

O S O S OH OH O O

tiotropium bromide (SpirevaTM - bronchodilator)

NNO2 OH N 18. , benzene OH

para-red dye

19. benzene

OH OH

20.

NHAc propofol (intravenous anesthetic)

20.2 Syntheses 236 • Chapter 20 Amines

20.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates.

H OH N OPh 1. butanamide + phenol + Br2 O

O CO2H 1. Cl2, H2O, OH 2. NH 2. H NH2 O anthranilic acid

1. HONO CO2H 2. adjust to pH 8 3. NH2 3. 1,3-butadiene

O O 1. HCN, CN 2. H2 / Pt 4.

3. NaNO2, HCl

O 1. NH OH 5. cyclohexanone 2 2. H NH

H-caprolactam (This is an example of the Beckmann rearrangement, similar to the Hofmann and Curtius rearrangements.)

20.3 Mechanisms Problems • 237

6. Sir Robert Robinson (Oxford) observed that thebaine (a dimethylated derivative of the alkaloid morphine) forms phenyldihydrothebaine when treated with phenylmagnesium halide. Formulate a mechanism and draw the Hofmann elimination product.

CH3O CH3O Ph 1. PhMgX 1. (XS) MeI HO O N N 2. H 2. Ag2O, H2O, '

CH O 3 CH3O thebaine phenyldihydrothebaine

7. The Curtius rearrangement not only occurs with acyl halides but also alkyl azides. Draw the bracketed structure and deduce a mechanism for its formation.

Cl N3 O NaN3 ' H3O NH H 3 SN2 - N2 an alkyl azide

8. Hydrazoic acid (HN3) undergoes addition to ketones to form a product that readily rearranges to an amide (Schmidt reaction):

O O HO N3 HN3 ' NH

20.3 Mechanisms 238 • Chapter 20 Amines

9. The Fischer indole synthesis involves an isomerization known as a [3,3] sigmatropic rearrangement, shown by the arrows below: O H2SO4 PhNH-NH + 2 Ph N Ph N H H N Ph 2-phenylindole H tautomerization

[3,3]

N H N N Ph HNPh H H

Outline a mechanism for conversion of the intermediate in brackets to the indole product.

10. CH2O NH2 H NH

O H 11. acetophenone + CH O + NH 2 3 Ph N 3

20.3 Mechanisms Problems • 239

12. Atropine, an antidote to cholinesterase inhibitors (e.g., nerve gases), can be easily synthesized from tropinone. The first total synthesis of tropinone required 17 steps. Years later Robinson (Oxford) accomplished its synthesis in a one-step, one-pot reaction (Robinson-Schopf condensation)! Sketch the critical intermediates in this synthesis.

N N O CHO 1. [H] + H2NMe + 2. esterification CO2 CO2 CHO Ph O O tropinone atropine O OH

13. The Ritter reaction offers a way to prepare amides (or, by subsequent hydrolysis, amines) from good precursors to carbocations: O H2SO4 a. + acetonitrile N H

O O CN O t-Bu N b. H MeO H2SO4 MeO

14. A convenient method of synthesizing pure secondary amines involves (1) treating the sulfonamide of a primary amine with hydroxide, followed by (2) an alkyl halide, then (3) hydrolysis. Outline such an approach to preparing N-methylaniline.

20.3 Mechanisms 240 • Chapter 20 Amines

15. The Corey-Link reaction (step 2) may be used to prepare D-amino acids:

O OH N3 NH2 1. LiCCl3 2. base 4. H2 / Pd RH R CCl RCOMe H 3 2 RCO2H 3. NaN3, MeOH H 5. hydrolysis H a. Outline a mechanism for steps 2 and 3.

b. Account for the product in a mechanistically similar reaction:

HO OO 1. :CCl3 NH O 2. 2 N H

16. Parkinson’s disease is associated with low levels of dopamine, a neurotransmitter. The enzyme monoamine oxidase (MAO) deaminates dopamine, thereby decreasing its concentration. One approach to treating Parkinson’s utilizes (-)-DeprenylTM, a “suicide inhibitor” to MAO. The mechanism first involves oxidation of the drug by a flavin cofactor of MAO, followed by a conjugate addition reaction between the reduced flavin and oxidized drug to irreversibly “kill” any future normal activity by the MAO enzyme. Outline the mechanism for formation of the adduct.

H R R H R H N N O N N O N N O [H] NH NH NH N N N O flavin cofactor reduced flavin H O H C O C H H C H N N [O] H CC N H CC Ph Ph Ph H H

DeprenylTM oxidized drug

20.3 Mechanisms SOLUTIONS TO PROBLEMS CHAPTER 1 THE BASICS

1.1 Hybridization, formulas, physical properties

TM TM 1. a. Seldane : C32H42NO2 Relenza : C12H20N4O7 c HO OH OH O a O HO OH b. N b OH N H NH d 2 O NH H2N c. a: sp3 – sp3; b: sp3 – sp2; c: sp2 – sp2 d. SeldaneTM oxygens: sp3; nitrogen d: sp2

2. a. :C C: b.HCO: c.ONO d. the conjugate base of :NH2CH3

-H H Cl H3CN N e. O f. O O H H

H 3. a. :C sp2 => bond angle ~ 1200 b. H H = CH H 2 a linear HCH bond angle implies sp hybridization; therefore, each lone electron lies in an un- hybridized p orbital with spins aligned (Hund's rule) H G+ 4. a. higher bp: NH N b. lower mp: catechol G- O this isomer is capable of intermolecular G+ catechol, unlike hydroquinone, can under- H H-bonding, thereby increasing intermolecular go intramolecular H-bonding, which decreases attractive forces and raising its bp O G- intermolecular attractions and results in lower- relative to the other amine H ing its mp relative to hydroquinone

5. a. no b. no c. yes d. yes e. yes f. no g. yes h. yes

6. a. CHCl3 b. CH3NO2 c. SO2 P F H P P CH3 S C vs. C N vs. NO OCO vs. OO Cl Cl Cl Cl H H O P linear, P = 0 bent, > 0 Cl Cl H3C P permanent charge separation (> H)

7. a. penicillin V: C16H18N2O4S cimetidine: C10H16N6S

b. arrow a: sp2 b: sp3 c: sp

1.1 Hybridization, formulas, physical properties 244 • Chapter 1 The Basics

O O

c. N N H H Ph Ph this resonance structure suggests some double bond character; electrons must be in a p orbital in order to resonate

d. lone pairs: penicillin V: 12; cimetidine: 8. O N: N S C H HN N N S O N N N O H H HO

8. a. sumatriptan: C14H21N3O2S prostacyclin: C20H32O5 b. Sumatriptan contains 8 sp2 and 6 sp3 carbons; prostacyclin contains 5 sp2 and 15 sp3 carbons.

HO c. lone pairs: sumatriptan: 7; prostacyclin: 10. H O O N O S MeHN O NMe 2 HO OH

TM TM TM 9. a. Rozerem : C16H21NO2 Chantix : C13H13N3 Ritalin : C14H20NO2

b. lone pairs: RozeremTM: 5; ChantixTM: 3; RitalinTM: 4. O N N HH H O NH NO N O

10. lone pairs: theobromine:8; melamine:6. O NH2 CH3 HN N N N

O N N H2N N NH2

CH3

11. a. alkene, amide, amine, ester, ether b. alkene, amine, arene, carboxylic acid, halide, ketone

c. alcohol, alkyne, arene.

1.1 Hybridization, formulas, physical properties Solutions • 245

1.2 Acids and bases

1. strongest base in ammonia: H2N amide anion - the CB of ammonia

H2N(H :H H2N + H2

2. stronger base: (CH3)2NH nitrogen is more electron-releasing than oxygen

AlCl3 3. a. O AlCl3 O

b. Ph3P: BF3 Ph3PBF3

c. NO BH3 NO BH3

4. a. CH3 O CH3CH2 Cl CH3 OCH2CH3 + Cl

LB LA

b. H CCH CH2 CH2 BF3 2 2 BF3 LB LA

c. H3CO (H :CH2 CH3 H3CO + H3CCH3

LA LB (also Bronsted-Lowry acid-base, respectively)

d. Cl Cl AlCl3 Cl + AlCl4

LB LA

S e. CH3 NCS :NH3 CH3 NC NH3 LA LB

pKa ~ 35 pKa ~ 16 O O H 5. a. C H O b. H) NH2 20 28 + NH2 c. Keq << 1 C C (' pK s > 4) C C a H H WB WA SA SB

1.2 Acids and bases 246 • Chapter 1 The Basics

O O H pKa ~ 5 O O NHAc O H) 6. a. + O NHAc

pKa ~ 10 i-Bu i-Bu

b. WB WA SA SB

c. Keq << 1 ('pKas > 4)

7. lowest pKa: b. a. ~ 16 b. ~ 5 c. ~ 16 d. ~ 10 e. ~ 38 f. ~ 35

Keq >> 1 8. quantitative rx: b. R CC (H NH2 R CC + :NH3 pKa 22 SB WB pKa ~ 35

a. hydroxide (Keq << 1) c. H2 (NR) d. EtO (Keq << 1)

OH pKa ~ 16 OH

9. + CH3NO2

O O pKa ~ 10 Keq ~ 1 (' pKas ~ 0) H :CH2NO2 pKa ~ 10

10. a. b. O + HCO3 N Cl N N pKa ~ 10 OO H H pKa ~ 5 SA WA

c. Keq >> 1 (' pKas > 4)

1.3 Resonance

CH2 O 1. CH3 CO p* N sp2 H p* sp2 * not sp3 per VSEPR; electrons may resonate if housed in a p orbital

1.3 Resonance Solutions • 247

-H 2. lower pKa: H-O-CN H) O C N: OCN: OCN: negative charge delocalization => more stable CB

H) C N: -H :C N: negative charge localized on C

3. a. 3. HO C NH2 HO C NH2 HO C NH2 H H H b. 1. localized charge

saturated, no p orbital

O O O O O c. 5.

O O O O O

OCH OCH 3 OCH3 OCH3 3 OCH3 d. 5.

CH CH 2 CH2 CH2 2 CH2

most basic H NH NH NH NH H 4. Me +H Me vs. Me vs. Me N NH2 N NH2 N NH2 N NH2 H H H H H no important resonance contributions

H H H 4 resonance structures NH NH NH => a more stable CA Me Me Me N NH2 N NH2 N NH2 H H H

H H N N 5. a. OOO OOO b. oxygen 'octetted,' closer charge separation carbon 'octetted,' additional S bond

1.3 Resonance 248 • Chapter 1 The Basics

OH OH H H c. CCN: CCN: d. H H negative charge is carbon 'octetted,' borne by more additional S bond electronegative atom

O O O 6. a. 2. b. 3. N N H H

CH2 c. 1.d. 3.

O O O

H(Ha -Ha 7. OO OO OO OO

A CB of A is stabilized by charge delocalization over three nuclei and, therefore, more easily formed => pKa of Ha is lowered

O H charge localized O O +H H 8. vs. H H OH O O charge delocalized, => > stability OH OH therefore, more favored CA species

9. a. 5.

NMe2 NMe2 NMe2 NMe2 NMe2

b. 5.

H H H H H

c. 4. not N N N N N nitrogen 'sextetted'

1.3 Resonance Solutions • 249

O O O d. 2. e. 3. O O

f. 4.

NCH2 NCH2 NCH2 NCH2

g. 4.

OCH3 OCH3 OCH3 OCH3 h. 4.

HCl HCl HCl HCl

i. 2.

O O O O O O 10. vs. A B these additional resonance structures increase both H and d (P = H x d); therefore, PB > PA

N N O N N O 11. O S S O S S CB of oxyluciferin

1.3 Resonance This page intentionally left blank CHAPTER 2 ALKANES

2.1 General

1. highest mp: 4. bicyclo[2.2.2]octane (most spherical)

2. highest bp: 1. n-pentane (least branched)

eicosane, mp 370 dodecahedrane, mp 4200 spherical molecules (dodecahedrane) pack more closely in the solid state than linear (eicosane) ones, therefore requiring more energy to separate (melt) them

4. constitutional isomers for

a. C6H14: 5.

b. C7H16: 9.

5. different kinds (constitutional) of hydrogens in

H Ha a H H e c Hf a. 2,3-dimethylpentane: 6. b. 2,4-dimethylpentane: 3. Hb Hb Hc Hd

Hb Hc c. 3-ethylpentane: 3. d. 2,2,4-trimethylpentane: 4. Ha Hd Ha Hc Hb

He Ha Hd Hd Hf Hb Hi Hg e. 2,5,5-trimethylheptane: 7. f. 4-ethyl-3,3,5-trimethylheptane: 10. Hb Hf H H H H c g a H j He Hc h

2.1 General 252 • Chapter 2 Alkanes

2.2 Nomenclature I

NO2 1. 2. 1 5 7 1 9 Br 3-nitro-4-ethyl-2,2,5-trimethylheptane 7-bromo-2-iodo-3-ethyl-5,6-dimethylnonane

7 4 3. 4. 3 1 8 1

4-ethyl-3,3,5-trimethylheptane 5-ethyl-3,5-dimethyloctane

8 5 4 1 5. 7 6. 1 F

4-fluoro-2-methyl-2-phenylheptane 5-ethyl-3,4-dimethyloctane

5 4 7. 9 8. 8 3 (1) (3) 1 1 5-(1,2-dimethylpropyl)nonane 2,3,7-trimethyl-4-n-propyloctane (choose path with more branching)

I 9. 10. 9

2,3-dimethyl-4-n-propylnonane 1-iodo-4-methylpentane

1 7 4 11. 10 12. 4

(1) (3) 1 Cl 5-(2-chloro-2-methylpropyl)-4-methyldecane 4-t-butyl-2,2,6-trimethyl-4-n-propylheptane not 4-t-butyl-4-isobutyl-2,2-dimethylheptane (less branching)

2.2 Nomenclature Solutions • 253

1 10 13. 14. 8 1 2,3,5-trimethyloctane 5-ethyl-4-methyldecane

7 3 15. 16. 10

3,7-diethyl-2,2,8-trimethyldecane diethylpentane (3,3- not necessary!)

OH Br 17. a. b. O c.

2.3 Conformational analysis, acyclic

Br 3.7 kcal/mol 3.7 kcal/mol H -2.0 (2 x 1.0 kcal/mol) 1. H PE 1.7 kcal/mol H HH 1.0 kcal/mol rot'n

H H 2 = 2. a. Me largest R-groups are anti 3 H

1 H H Et b. 2 = gauche: dihedral angle ~ 600 H Ph H

H OH Me H s-Bu 7 Me H OH H Et Ph 3. a. = Me b. = 1 HH H t-Bu Me H H Ph H isobutyl alcohol 4-t-butyl-3-methyl-5-phenylheptane

2.3 Conformational analysis, acyclic 254 • Chapter 2 Alkanes

H PE H H dihedral angle 4. only 600 between Me Me both methyls Me

rot'n about C2-3 bond

H 5. intramolecular hydrogen-bond F F O HH stabilizes a nearly eclipsed conformer vs. for FCH2CH2OH H HH H HH F P >> 0 P ~ 0

2.3 Conformational analysis, acyclic CHAPTER 3 CYCLOALKANES

3.1 General Cl P! Cl H H a. b. c. Cl CCCl 1. highest molecular dipole moment: d. Cl Cl HH Cl Cl P = 0

Cl Cl 2. constitutional isomers for Cl Cl a. dichlorocyclopentane: 3. Cl Cl 1,1- 1,2- 1,3-

b. C6H12 that contain a cyclopropyl ring: 6.

Cl 3. cis/trans stereoisomers for Cl a. dichlorocyclopentane: 2 pairs. Cl + trans- Cl + cis-

1,2- 1,3- Ph Ph Ph b. diphenylcyclohexane: 3 pairs. + cis- + trans-+ cis- Ph Ph Ph 1,2- 1,3- 1,4- Cl Cl Cl Cl

c. 2-chloro-4-ethyl-1-methylcyclohexane: 4.

H 4. different kinds (constitutional and geometric) of hydrogens in g Hf Hh H H a c Ha Hc a. 1-ethyl-1-methylcyclopropane: 5. b. allylcyclobutane: 9. Hi Hb He H He d Hb Hd Ha and Hb are cis- and trans- to methyl, and so are 'different'

c. methylcyclobutane: 6. d. chlorocyclopentane: 5. e. vinylcyclopentane: 8. H e H H f Ha Hc Ha Hc Cl e Ha Hc Hg Hf Hb He Hb Hd Hd Hb Hh Hd

3.1 General 256 • Chapter 3 Cycloalkanes

H H H 5. least strained: a. b. H H H H H cis- H trans- diaxial strain no strain H c. d.

ring strain prevents angle strain alkene from being planar - violates Bredt's rule

HMe Me H

6. H Me Me exo- endo-

stereoisomers

HMe

Me H + == 7. only 2. H Me Me identical structures (for now!)

3.2 Nomenclature

1 4 2 1. 1 2. 4

1-cylclopropyl-3-methylbutane 4-s-butyl-1-ethyl-2-n-propylcyclohexane

1 (2) 4 3. 4. (1) 7

t-pentylcyclopentane 4-(2-cyclohexylethyl)-3-methylheptane

(1,1-dimethylpropyl)cyclopentane

2-cyclopentyl-2-methylbutane

3.2 Nomenclature Solutions • 257

5. 6. Br

trans-1-bromo-2-s-butylcyclopentane

cis-1-isopentyl-5-n-propylcyclodecane

Cl F (4) 7. (1) 8. 1 6 (2) 3

(2-chloro-1-methylbutyl)cycloheptane 1-fluoro-6-t-butyl-3-vinylcyclooctane

1 I (2) (1) 6 9. 10. 3

9 cis-1-allyl-2-isobutylcyclohexane 5-iodo-3-(1-cyclobutylethyl)-6-ethyl- 2,2,8,8-tetramethylnonane

6 1 F 2 11. Ph 12. 7 5 trans-1-fluoro-3-phenylcyclohexane 2,6,6-trimethylbicyclo[3.1.1]heptane

6 7 5 1 4 13. 10 14. 2 2 1 9 5-methylbicyclo[2.1.0]pentane 7-allylbicyclo[4.3.1]decane

9 2 1 15. (1) 16. 8 7

(4) 2,9,9-trimethylbicyclo[5.2.0]nonane trans-1-(2,3-dimethylbutyl)-2-n-propylcycloheptane

3.2 Nomenclature 258 • Chapter 3 Cycloalkanes

3.3 Conformational analysis, cyclic

H H H OH 1. most stable conformer: Me Me O H H H H menthol neomenthol Ph

2. a. Ph degenerate structures (same energy) Ph therefore, Keq = 1 Ph

H H H H b. H H

H H > 1,3-diaxial interactions => less stable conformer therefore, Keq < 1 H H c. Me H H Et Et Me larger ethyl group in more stable equatorial position therefore, Keq > 1

3. most negative 'Hcomb (=> least stable):

a. t-Bu t-Bu t-Bu

least stable (most stable, therefore, least negative 'Hcomb)

Me Me Me Me b. Me

Me > dimethyl repulsion therefore, least stable

least stable (most stable)

3.3 Conformational analysis, cyclic Solutions • 259

4. least negative 'Hcomb (=> most stable):

a. =

all alkyl groups are equatorial therefore, most stable b. same compound!

H H t-Bu t-Bu t-Bu :B 5. most stable conformer for: Cl vs. H (-HCl) H trans- cis- Cl t-butyl group prevents 'flipping,' so chlorine cannot assume the axial position in the trans-isomer therefore, the cis-isomer would react more rapidly

OH OH F 6. F OH vs. F

trans- cis- twist-boat stabilized by intramolecular hydrogen- bonding; not possible in chair conformer (or in trans-isomer)

OH HO HO O OH O 7. a. HO OH HO OH HO HO 1 2 b. configuration 2 is less stable (one substituent must be axial) and would burn with a more negative 'Hcomb

H Me Me 5.4 kcal/mol -0.9 8. (2 Me-H 1,3-diaxial strain interactions) Me Me -0.9 3.6 kcal/mol (Me-Me 1,3-diaxial strain interaction) 5.4 kcal/mol less stable than

9. a. number of cis/trans stereoisomers: 8.

1234 567 8

3.3 Conformational analysis, cyclic 260 • Chapter 3 Cycloalkanes

b. for conformational chair-chair flipping, Keq = 1 for configurations 1, 3, and 5: 3e/3a 3a/3e

1 2 3 4

3e/3a 4e/2a 3e/3a 5e/1a

5 6 78

3e/3a 4e/2a 4e/2a 6e/0a

three 1,3-diaxial steric interactions exist between two methyl groups c. least stable: 1. 1 (only one such interaction exists in configurations 3 and 5)

d. least likely to flip: 8. 8

all methyls are equatorial all axial!

3.3 Conformational analysis, cyclic CHAPTER 4 REACTION BASICS

1. a. addition b. oxidation [O] c. substitution d. substitution

e. elimination f. reduction [H] g. oxidation [O] h. addition

i. reduction [H] j. rearrangement k. oxidation [O] l. substitution

m. elimination n. addition o. reduction [H] p. reduction [H]

q. rearrangement r. elimination s. substitution t. reduction [H]

b. 'G = 11 - 7 = +4 kcal/mol +11 kcal/mol C +7 kcal/mol 2. a. 'Go B +3 kcal/mol A B +3 kcal/mol A B C +7 rx A C +10 kcal/mol

'G for rds intermediate o o -'G / RT 3. a. 'G b. Keq = e

'Go = -'G / RT c. rate = k[conc. term(s)] = koe

'G for rds B

o o 4. a. 'G b. 'G = -RT ln Keq

Br3C: -'G / RT -2,500 / (2)(300) Keq = e = e -2 A Keq = 1.6 x 10 o rx 'G = +2.5 kcal/mol

4. Reaction Basics 262 • Chapter 4 Reaction Basics

5. a. type of reaction: rearrangement; mechanism: polar / ionic.

H H OO OO OO H) OH2 b. + HOH2 A H) B H2O

o c. Keq = [B] / [A] = 75% / 25% = 3.0; 'G = -RT ln Keq. d. nucleophiles: A, H2O, and B. e. TS

intermediate

'Go 'G0 =

rds fast 6. I OMe -I +MeOH, -H

7. a. pericyclic b. free radical c. pericyclic d. polar / ionic

e. pericyclic f. polar / ionic g. free radical h. polar / ionic

i. pericyclic j. polar / ionic

4. Reaction Basics CHAPTER 5 ALKENES AND CARBOCATIONS

5.1 General

H H 1 1 1. a. 3 b. H = H H 2 Cl Ph H 8 H Ph 5 H

(E)-3-chloromethyl-4-s-butyl-2- trans-5-phenyl-2-pentene methyl-3-octene

1 3 1 9

c. 3 9 d. 7 6

3-n-propyl-1-nonene 6-methylbicyclo[5.2.0]-3,8-nonadiene

2. a. (Z)- b. (E)- c. (Z)- d. (Z)- e. (Z)- f. (Z)-

3. a. number of alkenes: 4.

H2 Pt

b. least negative 'Hhydrogenation: most stable (trisubstituted)

4. number of geometric isomers: 4.

trans, trans- cis,cis- trans,cis- cis, trans-

OMe OMe 5. most stable carbocation:

H O N 6. a. no. deg. unsat: C17H36 - C17H20(18+3-1) = H16 => H16/2 = 8 deg. hydrogenation: C17H30F3NO - C17H18F3NO = H12 => H12/2 = 6 DB F C no. rings = 8 - 6 DB = 2. 3

fluoxetine

HN NN b. no. deg. unsat: C17H36 - C17H16(18+1-3) = H20 => H20/2 = 10 deg. no. DB = 10 - 4 rings = 6. OH F O O CiproTM

5.1 General 264 • Chapter 5 Alkenes and Carbocations

N OH c. no. deg. unsat: C28H58 - C28H34(35-1) = H24 => H24/2 = 12 deg. CC no. rings = 12 - 5 DB -1 TB(= 2 DB!) = 5.

O RU 486

d. no. deg. unsat: C17H36 - C17H10(14-0-4) = H26 => H26/2 = 13 deg. O no. DB = 13 - 3 rings = 10. O S O rofecoxib

F O S e. no. deg. unsat: C19H40 - C19H16(17+2-3) = H24 => H24/2 = 12 deg. N H no. rings = 12 - 8 DB = 4. Cl N N O O O floxacillin HO

H N

f. no. deg. unsat: C19H40 - C19H20(20+1-1) = H20 => H20/2 = 10 deg. O O hydrogenation: C19H32FNO3 - C19H20FNO3 = H12 => H12/2 = 6 DB no. rings = 10 - 6 DB = 4. O

F PaxilTM

7. number of stereoisomers for 2,4-hexadiene: 3 ; for 2-chloro-2,4-hexadiene: 4. Cl Cl

Cl Cl

Cl 8. a. b. => Cl

(Z)-3-methyl-2-phenyl-2-hexene propylene dichloride (note: no double bond) OH Br OH O c. , NOT d. e. Br OH OH styrene bromohydrin trans-cyclohexene glycol isobutylene epoxide

5.1 General Solutions • 265

+H 1,2-R: 1 9. 2 3 shift H 1 2 'G0 +Cl 1,2-H: 'G0 = Cl shift 3 rx

10. a. +H

H CA most important contributing H resonance structure H HH +H 1,2-H: b. shift most stable intermediate: benzylic carbocation

H H ~H: H 11. H

rigidity of the carbon skeleton prevents carbocation from being planar

F HF 12. neither regiospecificity nor stereospecificity: a. +

Br F Cl2 b. Cl regiospecific (XS) NaBr Cl Cl2 / H2O D / Pt c. d. 2 anti-add'n syn-add'n OH DD stereospecific stereospecific

5.1 General 266 • Chapter 5 Alkenes and Carbocations

H +H -H 13.

H E-pinene D-pinene more highly substituted double bond => more stable olefin therefore, Keq >> 1

5.2 Reactions

HCl 1,2-H: shift +Cl 1. Ph Ph Ph Ph Cl +H

O NO N NO2 2 HI 2. O I

+H 1,2-R: shift +H O 3. 2 OH -H

1. Cl2, ' 2. KOMe 3. Br2 4. Cl Br MeOH CHCl3

NMe3 NMe NMe NMe HI 3 1,2-H: 3 3 5. shift

Et Et Et Et I

HF F 6. F F F F

Cl D D D DCl 1,2-H: 7. shift

5.2 Reactions Solutions • 267

HBr 1,2-R: again! 8. shift Br

+H -H 9. O O O O (H H H

HBr 10. CCl3 CCl3 CCl3

Et Et Et Et Br DBr Br Et Br 11. + = D D D D H H H H anti-add'n syn-add'n

1. H / Pd 2. Br2 / hv 12. 2 Br

OEt +H +EtOH 13. -H

H F H OMe H HF 14. + MeO D MeO D OMe D F D syn- anti-

Cl HI Cl Cl Cl 15. I

Cl Cl / H O 16. 2 2 OH

1. B2D6 2. H O , OH 17. 2 2 DBD2 DOH

5.2 Reactions 268 • Chapter 5 Alkenes and Carbocations

I Cl Cl 2 Cl 18. G+ G+ I Et Et Et 1. Hg(OAc)2, PhOH 2. NaBH4 19. OPh OPh HgOAc

(XS) CH2I2 20. H2C CCH2 Zn(Cu) spiropentane

CHO

1. KMnO4, OH 2. HIO4 21. AcO AcO OH AcO OH CHO

1. O3 22. O 2. H3O, Zn HO

HBr +HBr 23. Ph Ph Br -Br Ph Br OO H note: opposite regioselectivity than HBr without peroxide - and no rearrangement

Et H2CN2 24. +N2 hv Et trans-

OH 1. base 2. OsO4 25. Br OH -HBr 3. NaHSO3

I IN3 I 26. N N3 N N

5.2 Reactions Solutions • 269

1. BD3.THF 27. 2. H O , OH HO 2 2 HO D OH

O O O OOMe +H +MeOH 28. -H H H

H 29. +H -H

OH O O HIO4 H 30. + HH OH

OH OH

1. H2SO4 2. KMnO4, OH 31. -H2O OH

O H HCl 32. CCO H3CCO H3CCO H CCl H 3 Cl

OR OR 1. BH3.THF 33. 2. H2O2, OH HO

5.2 Reactions 270 • Chapter 5 Alkenes and Carbocations

OH Br Br OH

1. Br , H O 34. 2 2 +

HO HO HO O a halohydrin

2. base HO -HBr

H) +H 35. -H O OH HO HO

O OH 1. OR 36. Cl 2. mCPBA 3. EtOH, H -HCl OEt trans-

Br Br

Br2 +s-BuOH 37. O -H HO

OH O OH

HIO4 38. + O HO HO H

O O OH KMnO4, H 39. OH O largest C-containing product

5.2 Reactions Solutions • 271

O H OH OH 1. OsO4 3. HIO O 40. O O 4 2. NaHSO OH O O 3 O OH O O O + O O H

HCl 1,2-R: 1,2-H: 41. + +H shift H shift Cl

Li :CH (H 2 -Cl 42. Ph CH Ph CH PhCH Ph Cl -H Cl a 1,1- (or D-) elimination to produce a carbene

H3CCH3 Cl (CH3)2CI2 Cl 43. O O OPh (1 equiv) OPh Cl Cl O Zn(Cu) O more electron-rich (nucleophilic) double bond

5.3 Syntheses

HBr 1,2-H: shift +Br 1.

OH Cl 1. H2SO4 2. HCl 2. -H2O

1. KOH, EtOH 2. HF 3. Cl F -HCl

5.3 Syntheses 272 • Chapter 5 Alkenes and Carbocations

Cl 1. OMe, MeOH 2. H / Pt 4. 2 -HCl D

1. Cl2, hv 2. NaOEt / EtOH 3. D2 / Pd 5. Cl D -HCl

Cl CO2H 1. Cl2, ' 2. KO-t-Bu 3. KMnO4, H 6. CO2H t-BuOH

Cl2 Cl 7. Cl (XS) NaBr Br Br H 1. O3 8. O 2. Zn, H

1. CH2N2, hv 2. Br2, hv 9. H2CCH2 Br

1. H2SO4 2. CH I , Zn(Cu) 10. OH 2 2 -H2O

OH 1. Hg(OAc)2, H2O 11. no rearrangement 2. NaBH4 OH

H3O 1,2-H: +H2O shift -H

O Br 1. KOMe 2. O3 12. MeOH 3. Zn, H O Br D 1. Br2, hv 2. NaOH 3. DBr 13. Br

5.3 Syntheses Solutions • 273

1. KO-t-Bu 2. HBr 14. Br t-BuOH peroxide Br

3. BH THF 1. Br2, ' 2. KOH 3. 15. Br OH 4. H2O2, OH

Cl O 1. 2. KMnO4, H CO2H 16. CO H OH 2 Cl

1. OEt, EtOH 2. Cl , H O Cl +H2O Cl 17. Cl 2 2 -H OH OH2

1. H 1,2-R: -H 18. shift

2. O3

OO3. Zn, H

1. Hg(OAc)2, EtOH 19. (not EtOH, H => rearrangement) OEt 2. NaBH4

OH 1. H3O 2. H2SO4 3. HBr, ROOR 20. Br

Cl CO2H 1. CH2N2, hv 2. Cl2, ' 3. NaOH 4. KMnO4, H 21. H2CCH2 MeOH CO2H

5.3 Syntheses 274 • Chapter 5 Alkenes and Carbocations

1. KOMe 2. H O 22. Br 3 OH MeOH 3. Hg(OAc)2, 3. H + OH O or t-BuOH O 2 4. NaBH4 2

OH 1. H SO 2. CH2I2, Zn(Cu) 23. 2 4

5.4 Mechanisms

+H 1,2-R: -H 1. shift (H H

H) +H -H 2. H H

H) Me O OMe -H 3. +H H O Me H

H +H 1,2-R: -H 4. shift (H

I I I +I2 -H 5. -I O O CO2H O C O H) O O H

5.4 Mechanisms Solutions • 275

H +H -H 6. H)

+H -H 7. H)

HgOAc OAc HgOAc Hg 1. Hg(OAc)2 O (H 8. OH OH :H HgOAc -H O 2. NaBH4 O

1,2-H: -H 9. +H shift (H

H H H N +I N N N 10. 2 -H -I I II I I

H2CNN

11. N NN H2CNN N

5.4 Mechanisms 276 • Chapter 5 Alkenes and Carbocations

Br Br Br +H anti-attack 12. Br 100% trans- H Br

+Br cis- and trans- syn- + anti- +H attack + Br 1,2-H: +Br Br shift

13. A is C14 => other aldehyde is CH2O; no. deg. unsat: C15H32 - C15H24 = H8 => H8/2 = 4 deg. hydrogenation: C15H28 - C15H24 = H4 => H4/2 = 2 DB; therefore, 2 rings are present

caryophyllene isocaryophyllene incorrect: cannot exist in cis/trans forms

OH 1. BH THF 1. Hg(OAc)2, H2O 14. 3. 2. NaBH 2. H2O2, OH 4 OH O O ozonolysis [H] + H C5H12

H / Pd 15. 2 ++ (partial) A B C D isoprene

16. no. deg. unsat: C10H22 - C10H16 = H6 => H6/2 = 3 deg.

H2 / Pt C10H16 hydrogenation: C10H22 - C10H16 = H6 => H6/2 = 3 DB C10H22 O O O 1. O3 HH + 2 HH+ 2. Zn, H O O myrcene acetone formaldehyde A

5.4 Mechanisms Solutions • 277

17. hydrogenation: C10H20 - C10H16 = H4 => H4/2 = 2 DB A = HH B A other ozonolysis product

18. a. n poly(methyl methacrylate) CO2Me Ph Ph Ph Ph Ph Ph Ph Ph Ph +H b. etc.

t-Bu O +H 19. O O O O H t-Bu t-Bu O OO etc. O

20. H2CNN H2CNN CH2 + N2

CH2 NN H2CCH2 + NN

hv 21. CH2N2 H2C: + -N2

+H 1,2-R: 1,2-R: -H 22. H) shift shift

+H 1,2-R: -H 23. (H shift

5.4 Mechanisms 278 • Chapter 5 Alkenes and Carbocations

R R' S

H3C D D 24. CH3 D CO2H -RSR' D C H 1,2-D: C8H17 8 17 shift DD DD CO2H -H H2C H) C8H17 C8H17 OAc Hg HO HgOAc

1. Hg(OAc) +H2O 25. 2 H O -H 2 +HgOAc

O H) O HgOAc HgOAc 2. NaBH O 4 -H

HgOAc AcOHg

(H +H -H 26.

R CO H 27. => trans- DB 2 R' elaidic acid

OH 28. a.

A B C b. Br2 Br A NR (deep red color of bromine persists); B or C 2 color discharged or Baeyer test:

KMnO4 KMnO4 A NR (purple color of MnO4 persists); B or C brown ppt (MnO2) forms

5.4 Mechanisms Solutions • 279

OH OH -H O 29. +H 2 2

rot'n -H

30. a., b. RNNN RN +N2 A a nitrene

c. retention of configuration suggests a concerted (or pericyclic) mechanism

+H 1,2-H: 31. 1,2-R: H shift shift

H) -H 1,2-R: shift

:B (H O O 1. Cl , H O 2. base 32. 2 2 Cl -HCl

Cl H O +Cl 3. dry HCl OH Cl +H

33. In Br CCl3 InBr + Cl3C Cl3C Br Br CCl3 etc. CCl3 + Cl3C (this mechanism is similar to the addition of HBr in the presence of peroxides)

5.4 Mechanisms 280 • Chapter 5 Alkenes and Carbocations

34. a. no. deg. unsat: C16H34 - C16H30 = H4 => H4/2 = 2 deg. hydrogenation: C16H34O - C16H30O = H4 => H4/2 = 2 DB

c. E => a cis- DB at C12; F => a trans- D B a t C 10

1. O3 OH B + C + D 12 10 A 2. Zn, H

DD DD D H H +H path a 35. H not observed -D H DD DD DD D D (H D vs. path b -H D observed 1,2-D: shift DD DD therefore, path b is favored

36. The rigidity of the bicyclic structure in the conjugate base of A prevents delocalization of the negative charge onto oxygen: such a contributing resonance structure would violate Bredt’s rule (the olefinic region cannot be planar). Loss of this stabilization prevents the carbanion from forming (pKa of A is raised relative to cyclohexanone), as required by the proposed mechanism, and therefore prevents hydrogen- deuterium exchange.

-H H O O O A

5.4 Mechanisms CHAPTER 6 ALKYNES

6.1 Reactions

1. 1. NaH 2. D2O H D

H, HgSO , PhOH OPh 2. 4

1. B2H6 Ph taut H 3. Ph CCH Ph H OH 2. H2O2, HO O

3. (XS) NaNH2 1. OMe (E2) Cl . 4. Cl 4. BH3 THF taut Cl OH H 2. Cl2 5. H O , OH 2 2 O

RC C 5. Cl + RC CH 3o R-X => elimination, not substitution!

1. Li / NH3 6. 2. HBr, ROOR (peroxide effect) Br

1. H2 / Pd(Pb) 2. BH 7. CCH 3 OH 3. H2O2, OH

1. NaH 8.

2. CH3(CH2)12Cl 3. Lindlar catalyst (cis-[H])

Cl 1. (XS) NaNH OH O 9. 2 Et CC 2. H3O, HgSO4 taut Cl

10. CCH Cl2 / H2O Cl taut Cl HO O

6.1 Reactions 282 • Chapter 6 Alkynes

Me Ph I 1. (XS) HI 2. Zn(Cu) 11. Ph .. Ph [PhCCH3] I

Cl Cl D C C C C 1. (XS) NaNH 12. 2 2. D2O

OMe OMe OMe

1. LiNH2 O 2. n-C5H11Br O 13. HC C CH2OH CC (2 equiv) (1 equiv) CC less stable anion 3. H therefore, more reactive OH CC

1. NaNH2 (1 equiv) HC CH 3. NaNH2 14. HC C n-Pr CC 2. n-Pr-I 4. Cl elimination, + CCH not alkylation 3o RX!

6.2 Syntheses

Br O 1. 1. (XS) NaNH2 2. KMnO4, H Et CCEt OH Br

1. NaH (1 equiv) 2. HC CH 3. H2 / Pd(Pb) 4. HBr, R2O2 2. n-PrBr Br

Cl 1. (XS) NaNH2 2. H, HgSO4, MeOH 3. CC O

1. NaH (1 equiv) 2. Et-I 4. n-Bu-I 5. Li / NH 4. HC CH Et CC Et CC n-Bu 3 3. NaH trans-[H]

6.2 Syntheses Solutions • 283

1. H2 / Pd(Pb) 2. HCl Cl 5. CCH 1,2-R: shift

1. Br 2 3. H / Lindlar 6. 2 2. (XS) NaNH2 cis-[H]

1. NaNH2 3. H / Pd(Pb) 7. 2 4. CH2I2 2. Me-Br Zn(Cu)

1. Li / NH 3 2. HBr, (t-Bu)2O2 8. Br or H2 / Lindlar

OH O H, HgSO4, H2O taut 9.

1. BH .THF 3 taut

2. H2O2, OH OH O H

1. Li / NH3 2. O3 O 3. Zn, H H

Et C C 1. Cl2 C 3. Et-I C 4. Li / NH 10. 3 Ph 2. (XS) NaNH2 trans-[H]

1. H / Pd(Pb) Ph Ph 2. CH N , hv 11. Ph CCPh 2 2 2 Ph Ph O 1. Na / NH3 Ph 2. O 3 H Ph 3. Zn, H

Et Cl 1. Na / NH3 Et 12. 2. CHCl3, KO-t-Bu Et => [:CCl2] Cl Et

6.2 Syntheses 284 • Chapter 6 Alkynes

1. KOH n-Bu Et 13. 2. Cl2 4. EtCl n-Bu CHCl3 Cl 3. (XS) NaNH 5. Li / NH Et base 2 3 Cl Cl

1. NaH (1 equiv) 3. H3O, HgSO4 taut 14. HC CH HC C n-pentyl 2. n-pentyl chloride OH O

1. NaNH2 (1 equiv) 4. 1-bromo-5-methyl- 2. n-decyl bromide hexane 15. HC CH n-decyl CC O 3. NaNH2 5. H2 / Lindlar 6. mCPBA

O 1. NaNH2 3. O 16. CCH 3 2 OH 2. n-Pr-I 4. Zn, H

(or 3. KMnO4, H )

6.3 Mechanisms

Br Br Br C H I I H C C O HO C O C I C O O O O 1. +I -H O O O O O OH OH O OH O OH O Br O I

H OH C C 2. C +H C -H O .. 2 +H2O:

O H OH (H -H ~H taut

6.3 Mechanisms Solutions • 285

1. -H2 2. 3. H 3. +H

H: O O H OH

O H Me O C HO Me C C C 4. 1. CCMe 2. +H

-H 5. RCHCC 2 RCHCC 2 RCHCC H H ~H B:

H H ~H H RCCC RCHCC RCHCC H

6.3 Mechanisms This page intentionally left blank CHAPTER 7 STEREOCHEMISTRY

7.1 General

1. chiral molecules: a,b,f,h, and l.

H N Bn N S 2. a. 3. b. 6. O N N O CO H 2 O O

O O N CO2Me O c. 4. d. 2. O OPh O OCH3 O

H HO Ph OH H (R)- NH2 CH2CH2NH2 3. a. HO b. Me Br c. (R)- Br H N H (R)- (S)- Me meso-

O CO H OMe Ph (R)- O 2 O O H2NH g. d. e. f. HS N Me OH HPh HMe (S)- (S)- NH2 H (S)- HO meso- (R)- (S)-

4. a. enantiomers b. enantiomers c. diastereomers d. enantiomers

e. identical f. diastereomers g. enantiomers h. diastereomers

i. enantiomers j. enantiomers

7.1 General 288 • Chapter 7 Stereochemistry

Ha H Hd Cl H He H Hc Hd f Hc b f He 5. a. 8. b. 7. c. 7. Ha H H H Cl H g b d' b He Ha' Hc' Ha Hc Hd Hg a, a'; c, c'; d, d' c, d; e, f are c, d; f, g are are enantiomeric diastereomeric diastereomeric

Cl CH3 Br 6. a. H b. HOH c. Ph Cl (R)-3-chloro-4-phenyl-1-butene (S)-1-chloro-2-propanol (R)-4-bromo-5-methyl-4-n- propyl-1-heptene

(R)-

H HMe CH (S)- 3 Et d. I e. f. Et H OCH H H 3 Me

(R)-1-iodoethyl methyl ether (3S, 4R)-3-s-butyl-4-isopropyl- meso-3,4-dimethylhexane 1,6-heptadiene

Br Cl Br Cl Br Br 7. a. 4 pairs.

(+ _ ) (+ _ ) Cl (+ _ ) Cl (+ _ )

Cl Cl Cl Cl Cl Cl b. 5. Cl Cl Cl Cl

Cl F Cl

c. 2 pairs. F (+ _ ) (+ _ )

7.1 General Solutions • 289

Et Et Et HOH HOH HOH d. 2 meso-isomers, 1 pair of enantiomers. H Cl Cl H H Cl H OH H OH HO H Et Et Et meso- meso'- (+ _ )

Me Me HCl HCl e. 2 meso-isomers. H Cl Cl H H Cl Cl H H Cl H Cl Me Me

CHO CHO HOH HOH epimerize at C4 8. a. HO H HO H HO4 H HOH

CH2OH CH2OH enantiomer of A D-xylose

(S)-

H MeHN Me MeHN Me o o b. i. = H ii. ee = +10 / +40 = 25% (+), 75% (+_ ) H OH HOH therefore, % (+) = 25% + (75% / 2) = 63% Ph Ph (R)-

N 9. ee = +68o / +170o = 40% (+), 60% (+ _ ) no. chiral carbons: 4. N therefore, % (-) = 60% / 2 = 30%

H Me CO H 10. 2 MeO (S)-

7.1 General 290 • Chapter 7 Stereochemistry

11. a. ED b. D c. ED d. ED e. D f. D

g. E h. ED

O Me O HS NH Me 12. N 13. H O CO2H H O NH2 (S)- (R)-thalidomide

AcO O OH

14. a. 11. HO O b. ee = +24o / +120o = 20% (+), 80% (+ _ ) HO O therefore, % (+) = 20% + (80% / 2) = 60% N O O O Ph H Ac Ph O O Ph

O Cl Cl meso- O CO2H 15. (!) O O A Cl HO OH B O (+_ ) 3 chiral carbons, 2 meso stereoisomers, 1 pair no. stereoisomers: 2n = 24 = 16. of enantiomers

N S H 16. a. no. stereoisomers: 23 = 8. H2N O N O O CO2H CO2H O

b. ee = +82o / +103o = 80% (+), 20% (+ _ ); therefore, % (-) = 20% / 2 = 10%

7.1 General Solutions • 291

7.2 Reactions and stereochemistry

OH Me OH H 1. OsO4 Me H 1. HO HMe = + enantiomer Ph 2. NaHSO3 HO Me Me Ph Me Ph Br syn-add'n Me Br2, H2O Me H Br H = + enantiomer Me OH Ph Me Ph OH anti-add'n

A Me H Me H Me Me H rot'n - A-B H 2. a. anti-add'n. = B BH B H A A Me Me Me Me H H cis-

C Me CD Me rot'n Me - C-C D b. syn-add'n. CD= D Me Me C Me D D D C C D Me Me trans-

OD CO2H HO2C DO H HO2C H - D2O H c. anti-add'n. DO H = rot'n D CO H DH H 2 H CO2H H CO H CO H 2 CO H D 2 2 (E)-

Ph Cl Ph Ph Ph +H Ph Cl 1,2-H: Ph Cl 3. a. Et Et shift Et

Ph ClPh Br Ph ClPh Cl Et Et Et Et meso- + +Br +Cl + Ph ClBr Ph Ph ClCl Ph Et Et Et Et chiral

diastereomers => diastereomers => 2 fractions: each E 2 fractions: 1 M + 1 E

7.2 Reactions and stereochemistry 292 • Chapter 7 Stereochemistry

H Br Et H Et b. Br2 H Et H + Br anti-add'n H Br enantiomers => 1 fraction: R Et Et HEt Br

CO2H KMnO , H c. 4 => 1 fraction: M

CO2H I

+H 1,2-H: +I Ph d. shift + enantiomers => Ph Ph Ph Ph 1 fraction: R

I H Cl

FHMe H HF e. Cl + diastereomers => 2 fractions: each E H Me Cl FH Me

D D D2 / Ni f. + diastereomers => 2 fractions: each M syn-add'n D D

Me OMe

Me Me Me H Et g. +H 1,2-H: Et +MeOH H + H H H shift H -H Me Et

H OMe

diastereomers => 2 fractions: each E

. 1. BH3 THF h. + HOH enantiomers => 1 fraction: R 2. H2O2, OH syn-add'n HOH

7.2 Reactions and stereochemistry Solutions • 293

Et OH Et OH Et OH Et OH Et Et OH H3O +H2O i. OH + Et H +H, 1,2-H: -H shift meso- chiral Et OH Et diastereomers => 2 fractions: 1 M + 1 E H2 / Pt syn-add'n H 1 fraction: E Et OH Et OH OH H 1. Hg(OAc)2, H2O H + OH H H 2. NaBH4

diastereomers => 2 fractions: each E

OH 1. OsO4 j. + diastereomers => OH OH 2 fractions: each M 2. NaHSO3 OH

OH 1. mCPBA + OH enantiomers => 1 fraction: R 2. H O 3 OH OH

O +H O +MeOH O H O OMe k. + enantiomers => -H OMe H 1 fraction: R

H Br 1. H2 / Lindlar catalyst Ph 2. Br H Ph Ph Ph 2 4. a. HPh+ syn-add'n anti-add'n Br Br H H Ph Ph Br HPh cis- racemate note: reduction of the alkyne to a trans-olefin, followed by Br2 addition, would yield the meso-dibromide (see next problem)

7.2 Reactions and stereochemistry 294 • Chapter 7 Stereochemistry

Br H 1. Li / NH3 Me 2. Br , CCl H Me b. Me Me 2 4 anti-add'n Me anti-add'n Me H H Br meso-

2. OsO 1. Na / NH3 4 H t-Bu c. HO + enantiomer (+ _ ) 3. NaHSO3 t-Bu H

H OH O O Me 1. mCPBA H 2. H3O d. Me Me H Me H Me Me Me Me H H H OH OH2 meso-glycol

2 o 5. a. 2 = 4. b. diastereomers c. cannot predict (actual [D]D = +62 )

CH2Ph CH2Ph Ph Me Ph NHMe H / Pd d. i. 2. + 2 ii. 2. H NHMe + MeHN H H NHMe H Me Me Me

(+ _ )

7.2 Reactions and stereochemistry CHAPTER 8 ALKYL HALIDES AND RADICALS

8.1 Reactions

1 2 H 1 + 1 meso Cl2, hv 7 are optically active 1. 9 fractions 1 2 are optically inactive Cl 1 2

Br Br2, hv 2. + + Br (+ - ) Br (no. H) x (reactivity) = 6 x 1.0 = 6.0 4 x 82 = 328 2 x 82 = 164 % racemic 2-bromopentane = 328/(6 + 328 + 164)*100 = 66%; therefore, % (R)- = 33%

1. Br , ' 3. 2 2. Mg 3. D2O Br MgBr D

1. conc HCl 2. Li 4. 4. OH Cl CuLi 3. CuI 2 Gilman reagent I

1. Cl2, hv 2. Li CuLi 4. 5. HI 5. 3. CuI 2 I I

1,2-R: 1,2-H: shift shift

1. Br , ' 2. Mg Ph H 6. 2 3. Br MgBr Ph CC +

Br 1. Li 3. n-PrBr Br 7. 4. NBS, R2O2 Ph2CuLi Ph 2. CuI Ph OH 6. Br2 5. KOH Ph Ph H O Br 2

8.1 Reactions 296 • Chapter 8 Alkyl Halides and Radicals

8.2 Syntheses

1. Cl2, ' 1. Cl 2. OMe / MeOH 3. NBS (-HCl) t-Bu OO Br t-Bu

Br O 1. Br2, hv 2. OH 3. O3 2. (-HBr) 4. Zn, H O

3. 1. Br2, ' 2. OEt, HOEt (-HBr) 3. HBr, ROOR Br Br

I 1. Li 3. Ph-I 4. Ph2CuLi 2. CuI

1. Br2, hv 6. Li 2. Li 5. Br , hv 5. 2 7. CuI 3. CuI Br Ph 8. PhCH2I 4. MeI

O Cl 1. Li 2. CuI 4. O3 6. H 3. I 5. Zn, H

1. H2 / Pt 1. DCl 1. NBS, R2O2 D 2. Cl2, hv 2. Mg 2. H2 / Pd 7. or or 3. Li 3. H2O 3. Li 4. D2O 4. D2O

Br 2. KO-t-Bu / t-BuOH 1. NBS, ROOR 3. NBS, ROOR 8. 4. KO-t-Bu / t-BuOH

8.2 Syntheses Solutions • 297

8.3 Mechanisms

H OR R OOR + etc. 1. O2H

-H H OO OO

Cl Cl 2. Cl Cl Cl 2 ++Cl H Cl -H Cl Cl Cl Cl

Cl Cl Cl Cl Cl Cl

D D D 3. D D D D

H CH3 NO H2C O O hv 4. AcO AcO

6-membered ring TS O H C N NO 2 H2 C OH OH AcO AcO

8.3 Mechanisms 298 • Chapter 8 Alkyl Halides and Radicals

5. a.

-HR 6. +O2 H H O R O O2 HR +O2 O O O O O OOH OH

8.3 Mechanisms CHAPTER 9 SN1, SN2, E1, AND E2 REACTIONS

9.1 General

SN2 1. faster reaction: b. AcO + Cl OAc

solvent effect: acetate in HMPA (polar aprotic) is more nucleophilic than in ethanol (polar protic); the later H-bonds to acetate, thereby dampening its nucleophilicity

2. poorest leaving group: b. 'leavability' parallels the acidity of the CA of the leaving group; :NH3 is the weakest CA (of the choices, c has the best leaving group)

Et Et 3. stronger nucleophile: a. Et N: :N Et rapid pyramidal inversion lowers nucleophilicity; Et Et such "flipping" is impossible with a b

4. most reactive by an SN2 pathway: c. least sterically crowded target carbon; note that even though a is primary, it is neopentyl-like, which generally never undergoes an SN2 reaction

SN2 Br 5. solvent that will maximize the rate of reaction: a. Et3N: + Br NEt3

polar solvents (a or b) stabilize the developing charge in the TS; the amine is more nucleophilic in DMSO (polar aprotic) than methanol (polar protic):

Et N: H H-bonding stabilizes the amine, thereby increasing G 3 O Me ' b

Me Me Me H 6. more reactive by an E2 pathway: b. H vs. H H Me Br a Br b no trans-diaxial anti-periplanar TS possible hydrogen available

Br H(D) O-t-Bu 7. approximate kH / kD: c. Ph Ph E2 a carbon-hydrogen (deuterium) bond is broken in the rate-determining-step; therefore, a primary hydrogen kinetic isotope effect (~7) is observed

9.1 General 300 • Chapter 9 SN1, SN2, E1, and E2 Reactions

8. reaction to yield the more stereochemically pure product: b. Et Br Et Br Et Et OMe MeOH +MeOH a. or SN1 -H

Et Et Et Et

either diastereomer would give the same ratio of diastereomeric ethers because of a common intermediate

OMe Br H OMe H b. or OMe SN2 S 2 Br H H N OMe (R)- (S)- (S)- (R)- stereospecific: either enantiomer gives an optically pure, but different, ether

Ph EtOH Ph +EtOH Ph 9. change in rate of reaction: b. Br H OEt Ph SN1 Ph -H Ph rds rate = k[RX] ; changing the concentration of EtOH has no effect on the rate

O-t-Bu OMe 10. a. vs. Br SN2 > E2 Br ~100% E2!

OH OH b. vs. Br Br 0 0 3 RX => ~100% E2 2 RX => SN2 + E2

OR SR c. + vs. >> Br OR Br SR

RS is a better nucleophile, and weaker base, than RO ; therefore, SN2 / E2 ratio is larger for RS

9.1 General Solutions • 301

11. expected primary hydrogen kinetic isotope effect: b. H) KO-t-Bu a. H(D) no carbon-hydrogen (deuterium) bond is Cl E2 - Hofmann broken in rds; therefore, kH / kD ~ 1

KOH b. H(D) a carbon-hydrogen (deuterium) bond is broken in rds; therefore, k / k ~ 7 Cl (H(D) E2 - Zaitsev H(D) H D

(H KOMe c. + S 2 + E2 Cl OMe N (D)H H(D) no carbon-hydrogen (deuterium) bond is broken in rds

O2 O S S2 O if intramolecular OG- 12. CH3 G- CH3 C: C TsO H TsO H G- G- unfavorable TS: Nu R L bond angle not linear

O however, O2 2 S CH S CH3 if bimolecular O 3 O - G CHOTs C: CHOTs C O2 TsO O2 TsO H H3C S H C S O H3 O G-

a linear TS is possible => < 'G ; therefore, an intermolecular reaction is kinetically favored

9.2 Reactions

KOMe, MeOH 1. Br O Me SN2 > E2

I OEt 2. E2 - Zaitsev

Cl E2 3. + O Hofmann

Br E2 4. + N 2 Hofmann

9.2 Reactions 302 • Chapter 9 SN1, SN2, E1, and E2 Reactions

KCN / DMF N: 5. I C SN2

MeOH, RT +MeOH 6. - Cl SN1 > E1 H OMe

refluxing EtOH -H 7. Zaitsev Br E1 > SN1

O -Cl O O +HOAc O 8. S 1 -H Cl N OAc racemate

Me Me Br N Me -H 9. + Me2NH N Me SN2 > E2 (H

10. Br + O-t-Bu E2

I O OAc 11. + O SN2 > E2

Cl Cl SH (1 equiv) 12. S 2 Cl N SH reactivity: 10 > 20

OEt EtOH +EtOH 13. + SN1 -H Cl OEt

Br -AgBr 1,2-H: OAc 14. SN1 shift Ph Ph Ph Ph OAc

9.2 Reactions Solutions • 303

Et H Me H Me Me H rot'n OEt, -HCl Et = Me Et 15. Cl H Et H Me Cl H E2 H Me Me Cl Me anti-periplanar TS t-Bu H Me OMe, -HCl 16. = H E2 Cl Cl trans-diaxial TS Hofmann olefin

t-Bu H D -HCl 17. = H D E2 D D H DD Cl Cl (C-H bond slightly weaker than C-D bond)

CH3 H HD Me HD rot'n -HBr D 18. = Me D Br H Br HCH H 3 Ph E2 H Ph Ph Br Ph

-Br 19. -H S 1 OH N O O O Br H H

I PPh I :PPh 3 3 + 20. SN2 O -t-Bu O -t-Bu

OH 1,2-H:

SN1 shift

MeOH +MeOH 21. SCl S S SOMe SN1 -H

N -I I 22. I N intra- SN2

9.2 Reactions 304 • Chapter 9 SN1, SN2, E1, and E2 Reactions

Cl 23. HSe + S 2 >> E2 N SeH

OH H OH Me Ph Cl (1 equiv) 24. N N Ph Ph Me Ph SN2

Me O Me Me3N -Me2O, -H Me3N Me BF4 25. OH Me O SN2

OTs H OH 1. TsCl 2. OH 26. ret SN2, inv H H OH

Me S 27. Me :NH2 S + :N SN2 H

Me HO HO F NHMe better leaving group N F 28. + Br S 2 OH (1 equiv) N OH OH OH

refluxing MeOH rigidity of bicyclic structure prevents formation of a planar carbocation 29. NR! backside attack impossible Bredt's rule precludes double bond at bridgehead Cl

Ph IF ISe SePh (XS) (vinyl halides unreactive under S 2 conditions) 30. N SN2

O O EtBr (1 equiv) Br 31. H2N NEt H N 2 2 NEt3 O SN2 O more nucleophilic nitrogen

9.2 Reactions Solutions • 305

Br 1. NaNH2 2. 32. Ph C CH Ph C C: Ph C CH + E2 > SN2

MeOH 1,2-R: -H 33. E1 shift I (H

OTs 34. + SCN: NR! (aryl tosylates unreactive under SN conditions)

Br OAc HOAc, RT 1,2-R: +HOAc 35. SN1 shift -H

1. MeI 2. refluxing EtOH -H 36. S S SN2 -Me2S, E1

O O

O-t-Bu 37. E2 O O Br Hofmann olefin

(XS) MeI 38. :N N: Me NNMe2 I SN2

anti- H Br Me OEt 39. a. O O O E2 - Zaitsev O H H syn- H Br H Me OEt b. O O O E2 - Hofmann O H H

9.2 Reactions 306 • Chapter 9 SN1, SN2, E1, and E2 Reactions

9.3 Syntheses

Ph Ph Ph Ph H O, Ag 1,2-H: +H O 1. 2 2 Ph Ph Ph SN1 shift -H Ph Br OH Ph NaOMe, MeOH

SN2 Ph OMe LDA or KO-t-Bu Ph E2 - Hofmann Ph

MeOH, ' -H 2. E1

(not OMe! => E2 - Hofmann olefin)

D OEt 3. = H Br H -DBr, E2 HD Br H Hofmann olefin (not HOEt, '! => E1 - Zaitsev olefin)

Ph H Me H Me Me H rot'n KOEt, EtOH Ph 4. = H Ph Me Ph HI I Me H E2 Me Me Me I H

Et I EtOH, ' 1,2-R: -H 5. E1 shift

OPh Br PhOH, AgNO3 1,2-R: +PhOH, -H 6. + SN1 + E1 shift or, -H

O O

Ph OH +PhCO2H 7. Cl OPh SN1 -H

9.3 Syntheses Solutions • 307

Br 1. KO-t-Bu, t-BuOH 2. HBr, ROOR 8. E2 - Hofmann Br

1. Br , hv 2. EtOH, RT 9. 2 EtO Br SN1 (not OEt! => 100% E2)

O 1. MeOH, ' 2. O3 10. E1 3. Zn, HCl Br O

CO2H 1. KO-t-Bu, t-BuOH 2. KMnO4, H E2 - Hofmann CO2H

Ph 1. NBS, peroxide Ph 3. HBr, peroxide Ph 11. Ph 2. NaOMe (E2) Ph 4. KO-t-Bu (E2 - Hofmann) Ph

OH OTs 1. KOH, EtOH 2. H3O 3. TsCl 12. Ph Ph E2 Ph Ph OTs

3. BH THF 1. Br2, ' 2. KOMe 3. 13. E2 Br 4. H O , OH 2 2 OH

SOCl 2 O 14. OH S O Cl -HCl -SO2 Cl

[SN2 reactivity slow (neopentyl-like); avoid SN1 (rearrangement)]

HS S S -Br 1. Br2, CCl4 Br 2. SH 15. H2CCH2 Br base, SN2 Br S SN2 S

9.3 Syntheses 308 • Chapter 9 SN1, SN2, E1, and E2 Reactions

C: 1. HCl CuLi C 4. vinyl chloride 5. Cl2 16. 2. Li 2 3. CuI 6. (XS) NaNH2

H taut OH 7. BH3.THF O 8. H2O2, OH

1. Li 2. H O 3. Br , hv 17. 2 2 I Li H Br (SN or E not possible)

H D MeOH, ' -H D -Br, E1 H D DH D 18. Br D D D = H H OMe / MeOH H -DBr, E2 Br trans-diaxial elim D

9.4 Mechanisms

Br SH -Br S Br -HBr S 1. S 2 S 2 N Br N

HOAc 1,2-R: +HOAc 2. = SN1 shift -H Cl OAc HOAc

CH CH3 CH3 3 N CN: N CH3 N CH3 SN2 -CH3Br 3. C Br SN2 :N CBr N:

9.4 Mechanisms Solutions • 309

Cl HO Ph NEt2 -Cl 4. Ph N Ph SN2SN2 EtEt OH Et2N:

O H O intra-S 2 O inter-SN2 H 5. N O double inv = net ret O inv inv O Cl H OH OH O HO

(in conc OH, product is O , formed by initial inter-SN2) H

H (H S S S I DMF -H 6. -I SH SN1

- NH -Br 7. O N SN2 N H O O

Br H(H

H H H OH O H O O H Cl HO -Cl -H 8. Cl intra-SN2 H trans-A with inv possible B H Cl O Cl H H HO H B intra-Walden inv not possible

H cis-A OH H HO OH therefore, inter-SN2 with inv H

Cl Cl rds S +H2O repeat 9. S S S -H -Cl Cl OH OH OH Cl OH2

9.4 Mechanisms 310 • Chapter 9 SN1, SN2, E1, and E2 Reactions

9. (cont.) similarly,

Cl rds Cl +H2O repeat N: N N N -Cl -H Cl OH OH OH Cl OH2

H AcOH H OAc intra-S 2 (NGP) therefore, +HOAc 10. OTs N OTs not possible inter-SN2: O -TsOH OAc I O OAc trans-enantiomer only vs. AcOH OTs OAc OAc NGP +HOAc H O + -TsO -H O OAc OAc II O O racemate

H Ac O (H O O -H O OAc OTs Ac O NGP 11. O 40% AcH) - OTs O OAc -H O O O H Ac 60%

X X OAc NGP O +HOAc 12. vs. O rds -H O I II O NGP not possible - H Ac therefore, no kinetic enhancement

H H O O +EtOH NGP, rds EtO 13. OH k -H Cl less strained intermediate => < 'G vs. therefore, k >> k' H O NGP, rds H +EtOH O EtO OH Cl k' -H

9.4 Mechanisms Solutions • 311

H Ac O OTs OTs H OAc

- OTs 14. vs. +HOAc -H I II ret NGP not carbocation possible stabilized by S electrons

TsO NGP + OAc 15. = H = AcO - OTs OTs OAc H OAc

CH2 NR"2 CH2 NR"2 CH2 NR"2 O dil OH O -R'O O 16. RNA: -H, NGP O OH O O O O R'O P R'O P P O O + OH O O OO OH CH NR" 2 2 CH2 NR"2 CH2 NR"2 O O O 2' 3'

OH O HO (H O O H) OH O OH P P P O O or O O(H O O O O O 2'-phosphate 3'-phosphate

CH2 NR"2 O DNA:

O NGP not possible; therefore, more stable in dil base R'O P O O

' -H 17. Cl + racemate -Cl (H (H

1,2 -R: -H shift

9.4 Mechanisms 312 • Chapter 9 SN1, SN2, E1, and E2 Reactions

O adenine 2- O adenine HO6P2 O P O O 18. O O O P O OH OH OH 2- HO P O 6 2 O (H adenine O -H O O O PO O OH - HO P OPO = PPi O O O

OPP OPP -BH 19. a. . . . -OPP . OPP (H :B

OH +H b. rot'n -H2O

+H2O -H

OH2 OH c. coupling mechanism: bond formed OPP . +I-PP OPP -H, -OPP . .

. OPP (H conversion of A to vitamin A:

OH +H OH

A H H (H -H OH vitamin A

9.4 Mechanisms Solutions • 313

OPP - OPP + OPP d. F-PP PPO N-PP

:H OPP H) [H] -H-OPP . . squalene . - OPP . OPP OPP F-PP N-PP

CO2H H3C SNH CH3 O 2 O S adenine 20. PP PO adenine O O O SN2 H2N

OHOH CO2H OHOH SAM OH OH HO NH 2 CH3 HO N CH3 S adenine H HO O SN2 H2N HO epinephrine CO2H OHOH

+H -N2 +HOEt Ph 21. Ph2 CNN: Ph2CH N2 Ph2C OEt -H Ph H) OTs

Cl -Cl Cl OMe Ph CH O 22. 2 Ph P CH3 Ph P + CH Cl S 2 OMe 3 N (OMe) SN2 :P(OMe)3 2 O

OH Cl -Cl +I 23. Cl3C(H Cl2C: Cl2C: Cl2CI Cl2CI -H HOH H I

note: HCCl3 + I HCCl2I via S N 2

9.4 Mechanisms This page intentionally left blank CHAPTER 10 NMR

Cl O OH 1. 2. 3. 4. Cl O OH OH

O F O 5. O 6. Br O 7. Cl 8. O F

Br Cl O Br 9. 10. 11. 12. Cl OH

O 13. Br 14. 15. 16. H HO J => trans- F OH O

O 17. O H

Ha Cl 19 18. Ha and Hb are diastereomeric protons; F (I = 1/2) Hb CH3 therefore, max multiplicity for H = doublet x doublet x doublet = 8 lines Fb a or b

Ha Ha Br Me Br Br 19. vs. Me Br Me Me Ha Hb A B

methylene protons are identical Ha => doublet; Hb => doublet => singlet (appears as a multiplet)

10. NMR 316 • Chapter 10 NMR

Ha Ja,b = 16 Hz

Ja,c = 8 Hz Hb Hc Hc

3 Br 20. 1 2 multiplicity: 5 lines (pentuplet) Ha

Cl F Hc Ha 21. a. highest field proton is Ha Hb Cl Ja,F => doublet

Hb (Hc) Jb,c

Jb,F

b. Hb and Hc are diastereomeric: Jb,c ~ Jb,F => triplet

Hb Hc Ha H and H are diastereomeric and independently couple with H ; if J J , 22. Ph CO2H b c a a,b a,c Ha would appear as a doublet x doublet = 4 lines (assuming no coupling through nitrogen) NH2

O 23. O

1 2 3 5 6 4 24. a. Ha (lowest field proton): doublet x doublet => 4 lines 8 7 b. C and C are diastereomeric carbons; therefore, 8 chemical shifts 7 8 Hb DO Hc Ha

Br Br Br 25. -SbF5Br Me Me Me Me Br SbF5 all methyls are equivalent therefore, appear as a singlet

31 26. P (I = 1/2), nP = 2; therefore, 2nI + 1 = 3 (triplet) Ha Ha (i-Pr-O) (O-i-Pr)2 2 PP OO

10. NMR Solutions • 317

Ha 27. a. amplitude: signal at G -16.1 is highest amplitude because most molecules PPh3 JH,P (66%) contain Pt with I = 0 (no further spin-spin coupling with Ha Cl Pt Ha is observed, i.e., J = 0) H,Pt PPh3

31 multiplicity: P (I = 1/2), so JH,? = JH,P > 0, nP = 2 therefore, 2nI + 1 = triplet -16.1

Ha JH,Pt b. amplitude: signals at G -13.6 and -19.6 arise from fewer (34%) JH,P molecules containing 195Pt (I = 1/2) per above

multiplicity: Ha now undergoes spin-spin coupling with both P and Pt to give a doublet of triplets (JH,Pt >> JH,P) -13.6 -19.6

c. Ha is very highly shielded, essentially a hydride, because of the polarization of the G+ G- Pt H Pt-H bond (much higher electron density around Ha than typically encountered in a C-H bonds)

Ha Ha Hb J Because of magnetic anisotropy of a,c Bo the aromatic ring current, Ha experiences diamagnetic (and H paramagnetic) lines b Ja,b 28. Hc of force relative to applied field Bo. H c Therefore, Ha is more shielded (and Hb deshielded) than normally observed in hydrocarbon protons on sp3 carbons. pseudo-quartet (same for Hb) 5.5 ppm!

5.0 (Hb) 0 0.5 (Ha)

H H 29. multiplicities: H 11B H vs. H 10B H H H

11B (I = 3/2) 10B (I = 3) therefore, 2nI + 1 = quartet therefore, 2nI + 1 = septet (higher amplitude) (lower amplitude)

relative amplitudes of quartet/septet reflect the natural abundance of 11B/10B = 80%/20%

10. NMR This page intentionally left blank CHAPTER 11 CONJUGATED SYSTEMS

11.1 Reactions

+H +Br 1. Br (1,4-addition)

D D 2. DCl (1 equiv) 1,4-add'n most stable carbocation D D Cl +Cl

CO2Me CO Me D-A 2 CO2Me 3. = = CO2Me MeO2C CO2Me

retro-D-A 4. = O O O

HBr 5. +H H H Cl Cl

+Cl +Cl H H 1,2-add'n 1,4-add'n product of thermodynamic control - more conjugated system than 1,4-adduct, therefore, more stable

11.1 Reactions 320 • Chapter 11 Conjugated Systems

' 6. + retro-D-A

DBr (1 equiv), ROOR 7. +DBr 1,4-add'n -Br D Br Br Br Br

O O O D-A O 8. N N N N (4 + 2) N Ph NPh O O

H H N N D-A 9. Ph Ph Ph Ph

1. NBS, R O Ph Ph 2 2 3. 10. = 2. KOMe (E2)

Me 1. retro-D-A 2. 11. 2 Me ' Me Me (E)-

intra-D-A 12. =

11.1 Reactions Solutions • 321

O O O 1. D-A 13. 2. KO-t-Bu E2, -HCl Cl Cl

1. D-A 2. O 14. 3 3. Zn, H CO2Me CO2Me CO2Me OH

O 15. ' + retro-D-A

1. retro-D-A 2. 16. 2 Ph ' Ph Ph Ph cis-

CO2H

HO2C CO H 17. = 2 '

s-cis-diene CO2H

O O

(4 + 2) 18. O MeO O MeO

11.1 Reactions 322 • Chapter 11 Conjugated Systems

AcO AcO

CO H CO2H 19. = 2 ' CO H ' 2 + OAc CO2H OAc HO2C CO2H

CO2Me (4 + 2) CO Me 20. 2 CO Me 2 CO2Me

O O O N ' N (4 + 2) N 21. - :SO2

O2S

O O O H) B: base, E2 22. (4 + 2) -NMe3 MeO MeO MeO NMe I 3 C19H24O2

MeO MeO

R R Si D-A Si 23. Br Me2 Br Me2 H CHO

OMe OMe R' R' N (4 + 2) N 24. + R Me SiO RR Me SiO 3 3 R

11.1 Reactions Solutions • 323

11.2 Syntheses

1. Cl2, hv 3. NBS 4. KOMe (E2) 5. DBr 1. 2. KOMe (E2) ROOR Br 1,4-add'n Br

1. Cl2, hv 2. KOMe 5. ethylene 6. H2 / Pt 2. CH 2 D-A 3. NBS, R2O2 H2C 4. KOMe

1. NBS, peroxides 3. Me 3.

2. KO-t-Bu, t-BuOH 4. H2 / Ni Me

1. NBS, R2O2 3. 2-butyne 4. 2. KO-t-Bu (E2) CH3 D-A

H3C

O O (4 +2) 5. A O

Me Me N N O (4 + 2) 6. NMe + O = O O O O O O O

H CO Et O 2 CO2Et ' 7. + = H CO Et O OH 2 H

11.2 Syntheses 324 • Chapter 11 Conjugated Systems

11.3 Mechanisms

O O O O > pH 8.5 1. O -H (H O OH HO HO - sp3 carbon prevents conjugation - > conjugation results in a 'red shift;' from one ring to the other two; absorption occurs at a longer wavelength, therefore, absorption occurs at shorter moving into the VIS, and the molecule, wavelengths (UV in this case) therefore, is "colored"

' ' 2. + retro-D-A D-A

O O ' (4 + 2) 3. a. O O

O O

O O O

' ' 3. b.

4. a. longest Omax: n S* N

b. low probability A = H c d H (molar absorptivity) only ~ 10 - 100 for n S* transitions (vs. > 10,000 for S S* transitions)

c. no non-bonding electrons in the CA of pyridine; therefore, no n S* transition N H

11.3 Mechanisms Solutions • 325

+H -H O 5. 2 OH OH2 OH2 G 1.70 H HG 4.10 -H +H2O = OH OH OH G 2.25 2

HG 1.79 HG 5.45 B

O O O O Me O retro-D-A 6. = ' D-A -CO O 2 O R H CO2Me R O O CO2Me R = -CO Me 2 A B

R inter-D-A intra-D-A 7. = R R R R R

O O O D-A 8. =

(H CH2 H CH2 O O O taut ' 9. ene-reaction

Ts Ts N N N hv 10. NTs NTs CN N = - :N N: CN CN CN

11.3 Mechanisms 326 • Chapter 11 Conjugated Systems

OH OH N OH N OH [O] NH N N 11. H H N N [H] N b CO2H CO2H c site of redox CO2H CO2H bilirubin (red) biliverdin (green)

Increased conjugation promotes a ‘red shift’ in Omax, causing the color of pigments to move toward the green-blue end of the VIS spectrum:

x biliverdin is conjugated from Ca to Cb, whereas bilirubin’s conjugation is disrupted (as a consequence of reduction) at Cc x biliverdin, therefore, absorbs at longer wavelengths (red) than bilirubin; alternatively, biliverdin is transparent to shorter wavelengths (green).

12. a. b. c.

(4 + 4) (2 + 2) (4 + 2) thermally forbidden thermally allowed therefore, most likely to occur

11.3 Mechanisms CHAPTER 12 AROMATICS

12.1 General

1. The following compounds obey the Hückel (4n +2) rule and would be expected to have aromatic character: = lone pairs of electrons are in a p orbital (other lone pairs are in sp2 orbitals)

H N N H B H b. d. N f. O i. j. N N N N B B O H N H H

O OH

HN taut N k. (6 S electrons) O N HO N H

2 Li 2- m. product: 2 Li (6 S electrons)

H H 2 MeLi n. product: 2 Li (10 S electrons) H) (H -2 CH4 H3C: H H :CH3

Br ZnBr Zn -ZnBr o. product: 2

Br Br both rings are aromatic P >> 0!

SbF5 Note: l. carbocation from the reaction of (8 S electrons = 4n+ 2) Cl -SbF5Cl

12.1 General 328 • Chapter 12 Aromatics

2. largest P: a.

etc. G+ G-

P both rings are aromatic

note: flow of electrons in either direction in b or c would result in one ring being aromatic and the other anti-aromatic, thereby lessening the benefit of charge separation and lowering P.

O N N N N 3. H3C N N most basic (localized) H least basic (delocalized, part N of aromatic ring current) CH3

4. least stable: b. O O

anti-aromatic a and c have corresponding aromatic, and therefore stabilizing, contributing resonance structures

-H 5. most acidic: d. etc. (H

aromatic CB

loss of a proton from a, b, or c would produce a resonance-stabilized, but not aromatic, CB

O O O -Cl aromatic (4n + 2) 6. most likely to undergo an SN1 reaction: a. contributing structure Cl stabilizes carbocation

most basic more basic (localized) O OH OH H N N +H 7. a. b. c. N N O O O

aromatic more basic (localized)

12.1 General Solutions • 329

_ -Cl 8. = 0 Cl +

aromatic - all protons are equivalent all bonding MOs are filled

O O O O 9. largest molecular dipole moment: b. vs. d.

aromatic contributor aromatic contributor with longer charge with shorter charge separation (>d) separation P = H d -aromaticity promotes charge separation in b and d (but not a or c), thereby increasing H -charge separation distance (d) is greater in b than d

O OH OH OH OH BF4 H HBF4 10. Ph Ph Ph Ph Ph Ph Ph Ph Ph Ph

aromatic cyclopropenium moiety

H H 11. +H +Cl Cl aromatic note: does not undergo a 1,2-H: shift!

(10)

Hb ~ G7 (4) Ha ~ G-1 12. Ha Hb 1H NMR: G10 G0

magnetic anisotropy causes the four Ha protons to be highly shielded (above TMS) and the ten Hb protons to be deshielded (into the aromatic region)

12.1 General 330 • Chapter 12 Aromatics

12.2 Reactions

N N H fuming sulfuric acid H 1. + o-isomer O O HO3S

OH OH no! (avoid 1,2,3-subst'n) Me Me HONO / H SO 2. 2 2 4

NO2

o-, p-director H m-director PH2 PH PH H SO , SO 2 3 3. 2 4 3 +H

SO3H

(H (H (H -H Cl N Cl N Cl N Cl N H H H Cl2 / Fe H 4. vs. H H N H Cl Cl more important set of contributing resonance structures N N H H

1. AlCl 5. Ph 3 Ph 1,2-H:- Ph PhH Ph Br shift F-C alkylation

CH2 2. NBS, R2O2 Ph 3. KOMe Ph Br E2

Br2 / CCl4 Br Br Br Br , Fe 6. 2 Br Br Br

NBS, R O 2 2 +

12.2 Reactions Solutions • 331

Se Se G+ G- -FeX3Cl Me Me 7. ICl [I ] + + p-isomer EAS FeX3 I

N N N O Br2, FeBr3 O O 8. + Br Br

SH SH Cl2, FeCl3 9. + p-isomer Cl

HCl HCl HCl

N N N Cl2, BF3 least important 10. N (H (H (H Cl -H Cl Cl Cl N N N N more important set of contributing resonance structures

OH OH O OH OH Cl Cl +H EAS OH 11. HH HH HH+ Cl Cl +H HO Cl Cl Cl OH OH OH Cl Cl Cl EAS -H2O OH2 Cl Cl Cl Cl Cl Cl Cl Cl

O 1. -CO2, -N2 2. D-A 12. + NN: benzyne

12.2 Reactions 332 • Chapter 12 Aromatics

Me OH Me

O2N NO2 NO2 Me 13. + Me Me O2N OH Me NO2 NO a S-complex 2

F NH2 FNH3 CN CN CN :NH3 -HF 14. nucleophilic add'n - elim aromatic subst'n mechanism Cl Cl Cl

NMe2 NMe2 NMe2 NMe2 1. MeLi (-HBr) + MeLi 2. H 15. NAS, Me Me Br benzyne mechanism

Cl Cl OMe

NO2 NO2 1. HNO3, H2SO4 2. NaOMe, MeOH 16. EAS NAS, add'n - elim mechanism CF 3 CF3 CF3

+H 17. O OH OH EAS OH

EAS +H

-H2O

OH OH OH

+H EAS 18. + again

12.2 Reactions Solutions • 333

Br Br Br Br , Fe -H 19. 2 H) H) OO EAS O O O O OO most stable intermediate

Br Br N O NNO 1. fuming HNO 2 2 2. i-Pr NH O2NNO2 20. 3 2 EAS NAS, -HBr

CF3 CF3 CF3

OMe OMe OMe OMe

Ha 1. MeI, AlCl3 2. NBS, ROOR 3. KOH 21. SN2 Hb

Br OH 1H NMR: aromatic a-b quartet suggests p-subst'n

I I 22. HO 2 HO catalyst thyroxine CO2H CO2H H2N I H2N

H H H Cl Cl Cl 3-subst'n N N N O O 2. Cl2, BF3 O 23. N H) Cl H) Cl H) Cl Cl O -H + 4-subst'n N 3. [H] (or 2-subst'n) N N N NCl O O O best resonance structure

12.2 Reactions 334 • Chapter 12 Aromatics

12.3 Syntheses

1. CH3I, AlCl3 2. Cl2, Fe 3. H2 /Pt 1. high T, P F-C Cl Cl Br Li H DD DD DD 1. Br2, Fe 3. Li 4. H2O 2.

2. dil D2SO4 (EAS) D D D

Cl MgCl D

1. Cl2, AlCl3 2. Mg 3. D2O 3. (x2) Et2O

Cl MgCl D

O OH HH

1. PhCH2COCl 2. H2 / Ra-Ni followed by 4. Ph AlCl3 Ph hydrogenation Ph hydrogenolysis

1. Cl2, BF3 3. KOMe, MeOH 5. 2. NBS, ROOR Cl E2 Cl

1. cyclohexyl chloride 2. NBS 3. NaOH 6. AlCl 3 ROOR Br E2

1. O3 2. Zn, HCl 4. SOCl2 7. 3. O [O] 2 CO2H 5. AlCl3 (F-C) O

12.3 Syntheses Solutions • 335

Cl CO2H 1. 2. HONO , H SO 3. KMnO , H 8. 2 2 4 4 AlCl3 NO2 NO2

Cl 1. Cl2, Fe 2. PhLi PhLi 3. H 9. -HCl Ph 4. MeI, AlCl3 Ph

1. Br2, FeBr3 2. NaNH2 10. NHMe NHMe :NH NHMe Br 3

~H N N (H Me Me

O O Cl 1. EtCl, BF3 2. 3. H , Ra-Ni 11. 2 AlCl Et 3 Et Et

OH O OH OH 1. H2SO4 2. PhH, H 12. HO OH EAS +H, -H2O t-Bu

3. , H EAS HO Ph HO Ph HO EAS t-Bu (see 12.2, 18)

Cl O O CO2 Cl Cl Cl 13. 1. NaOH 2. Cl CO2 (Cl)H NAS (Cl)H SN2 (Cl)H Cl Cl Cl

12.3 Syntheses 336 • Chapter 12 Aromatics

Br O OH OO OH

1. OH (NAS) 2. a. SnCl2, HCl 3. 14. O b. neutralize (1 equiv) NO -HOAc 2 NO2 NH2 NH O CO H CO H CO H 2 2 CO2 2

1. HONO2 2. KOH 4. SnCl2, HCl 15. H SO Cl 2 4 NO 3. 5. neutralize NH 2 NO2 2 OH OH (1 equiv) O O

12.4 Mechanisms

H EAS 1. Ph -H Ph (H Ph Ph

O O OH H) O O +H O EAS -H 2.

H) OH OH

NMe2

O AlCl3 AlCl AlCl3 O O 3 NMe2 3. Cl Cl Cl Cl Cl Cl EAS AlCl Cl O 3 O NMe2 NMe2 Cl

EAS -Cl -AlCl4

Me2N NMe2 Me2N C Cl O O

OCO OH O O O OH (H CO CO2H 1. OH 2. CO2 O -H 2 3. H 4. C -H O

12.4 Mechanisms Solutions • 337

[Br ] Br Br

Br2, AlCl3 -H 5. + (H

O 1. AlCl3 6. RX RCO: -AlCl3X Ph EAS Ph -H Ph 2. HBr R R +H RCO: H) HO R O H) O H

-H3O ~H EAS

(H R H2OR HO R

AlCl 3 AlCl3 AlCl3 H O O O O O O OH ~H 7. R R (H R R :O C R O O

HO Cl Cl Cl O Cl O -HCl -Cl Cl 8. Cl Cl NAS NAS Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl HO

ClO Cl Cl O Cl -Cl -HCl Cl O Cl NAS Cl O Cl NAS Cl

-BF3Br 9. Br

MeO BF3 MeO MeO

(H -H EAS

MeO MeO

12.4 Mechanisms 338 • Chapter 12 Aromatics

O O H EAS -H 10. :C O: H CO: H) H H

O H +H O H EAS -H 11. H O H O HO -H2O H) H HO OH

Cl Cl Cl O Cotton

N N 1. NAS N N -HCl N N 2. Cotton-OH N N 12. Cl -2 HCl Cotton Cl N Cl N Cl HN N Cl HN N O H2N NAS x2 Dye Dye Dye H2NDye

O OH OH Cl OH +H 13. Cl C H Cl 3 Cl3C H Cl3C H EAS CCl3 +H H Cl -H2O Cl Cl Cl EAS CCl3 CCl3

1. BF3 H 14. Cl -BF3Cl EAS BF3 2. H (H -H 1,2-R: H shift

12.4 Mechanisms Solutions • 339

(H H EAS +H -H 15. H -H -H O O OH OH 2 E1 H OH H OH OH OH OH

OCl-SnCl5 O -H, EAS 16. O

SnCl4 +SnCl4 O +Cl -MeOSnCl4

SN1-like SnCl Cl O 4

Br Br 1. +Br +Br 2. :B 17. 2 Br -Br -HBr Br Br Br Br H) C7H7Br

C7H8Br2

3. H2O, -Br O OH +H2O -H -H SN1 aromatic

Me Me H H OTS HOTs 18. = rot'n -OTs Ph H Me OTs Ph H NGP Me H Me Me Me H HOAc Me H

Ph Me H +HOAc -H +HOAc Me Ph Me H H OAc -H Me H Me H Me H OAc HOAc enantiomers

12.4 Mechanisms 340 • Chapter 12 Aromatics

Cl OR -Cl 19. Cl3C(H Cl2C: Cl2C: + Cl N N Cl H H Cl Cl -Cl -H

N N H

+H 1,2-R: shift H 20. H

-H 1,2-R: shift (H

O O O +H -H2O 21. HO P OH HO P OH2 HO P OMe OH OH OH

O EAS -H H) (HO)2 P OMe OMe H2O3P

12.4 Mechanisms CHAPTER 13 ALCOHOLS

13.1 Reactions O OH Cl 1. NaBH 4 3. PCl3 1. 4. KO-t-Bu Ph 2. NH4Cl Ph Ph Hofmann, E2 Ph OH 1. H / Pd 2. H SO 2 2 4 Ph Ph E1

OH O 1. i-PrMgBr 2. 2. Br + Et3COH Hofmann

OH 1. H SO 3. 2 4 2. H3O E1 OH

HCl, SN1 1,2-H: +Cl +H, -H2O shift Cl

O O O 1. TsCl 4. 2. NaOAc -HCl SN2 OH OTs O O Me OH O O 5. 1. NaH 2. Me OSO3Me

-OSO3Me, SN2

O O O 1. PhMgCl PhMgCl 2. H 6. Ph O Ph Ph Ph Ph Ph3COH -n-BuO Ph not isolable O O 1. LiAlH4 LiAlH4 2. H PhCH2OH -n-BuO Ph H Ph H H not isolable O O OH 1. Li 2. 3. H 7. CH3I CH3Li -LiI

13.1 Reactions 342 • Chapter 13 Alcohols

3. BH .THF 8. 1. HBr 2. LDA, E2 3 HO Hofmann OH Br 4. H2O2, HO

O OH

1. NaBH CO2Me CO2Me 4

9. 2. H OH O2CPh OPh OH 1. LiAlH4 + PhCH2OH O 2. H OH OH CO2Me H2 / Pt

O2CPh Me O OPOCl2 H HOH POCl HOPCl2 Me 10. 3 rot'n H E2 py Me Me D Me D Me Me D -HOPOCl2 Me H -HCl H H D :py

HO O O Me

1. OH 2. Me-I 11. O O O S 2 NMe NMe N NMe

HO HO HO

OH OH OH O 12. CrO3, H O

HO HO HO HO

O 1. Br2, H2O 2. Me3SiCl 4. 13. OH OSiMe3 OH

Br 3. Li Li 5. H3O OH

O O O O 1. MeLi O 14. O O -OBn MeLi -OBn BnO OBn BnO BnO PhPh O 2. H MeLi OH

13.1 Reactions Solutions • 343

CO2H O 15. 1. LiAlH4 2. (XS) HBr Br S 1 HO O N HO

O OH O OAc HO OAc 16. 1. NaBH4 OH 2. H HO OH O HO

1. LiAlH4 + EtOH 2. H HO

OH OH O +H 17. 1,2-R: -H OH -H2O O shift H

18. +H 1,2-R: Ph -H -H2O shift Ph OH OH O Ph Ph H O (H O

O OH

1. NaBH4 19. 2. H HO HO

OH MgCl O O O HO 1. SOCl , Et O 20. 2 2 3. H 2. Mg

O O O HO Ph Ph Ph OEt Ph 21. 1. n-PrMgCl n-PrMgCl 2. H N -OEt N N N Me Me Me H Me

13.1 Reactions 344 • Chapter 13 Alcohols

OCH O 3 OMe O OMe O 22. 1. ' 2. H D-A TMSO taut Si O O

13.2 Syntheses O D 5. NaBD4 1. H2 / Pt 3. H3O 6. H 1. O 2. H2SO4 4. CrO3, H 7. POCl3 / py

O 3. CH O O OH Li 2 6. i-PrLi 1. HBr 4. H 2. H 7. H 2. Li 5. PCC [O] 8. Na2Cr2O7 [O]

O 1. H3O 3. PhMgCl OH 3.

2. CrO3, H 4. H Ph

OH OTs 4. 1. TsCl (ret) 2. OH (SN2 - inv) H

H H OH Me HCl +Cl Et racemic H SN1 Cl *OH H * 1. PCl3 (inv) 2. OH (SN2 - inv) (* = 18O) Cl double inv => net ret H

1. Br2, hv 3. CH O 5. H SO 5. 2 2 4 6. H3O 2. Mg MgBr 4. H E2 HO OH

Br O 1. Cl2, ' OMe 2. KOMe (E2) 4. KOH (SN2) 6. MeLi 7. H SO 6. 2 4

3. NBS, R2O2 5. Jones reagent E1

13.2 Syntheses Solutions • 345

O 1. Mg 3. H2SO4 7. Cl 2. H E1

1. TsCl N 8. OH OTs 2. KCN / DMF C

SN2

1. LiAlH4 3. SOCl2 or PCl3 4. Mg CO2H 9. OH Cl D 2. H 5. D2O

OH O TMS O TMS OH OH 1. TMS-Cl 2. Mg 10. 4. H3O 5. PCC (protect) 3. CH2O (deprotect) [O] Cl Cl CH2O CH OH 2 HO

OH Li O

1. PCl (not HCl) 11. 3 4. CH3CHO

2. Me3SiCl 5. H3O HO 3. Li TMSO 6. PCC HO

1. BH .THF 3 4. Li 2. H2O2, OH 5. CH O 12. 2 Br OH 3. PBr3 6. H

1. Br2, hv 3. BH .THF 13. 3 5. K2Cr2O7 O 2. KO-t-Bu / t-BuOH [O] 4. H2O2, OH OH (Zaitsev E2) OH

1. AlCl3 14. 2. H2 / Pd Cl O F-C hydrogenation, then O hydrogenolysis

CO2H 3. KMnO , H OH 5. LiAlH4 4 OH 6. H CO2H

13.2 Syntheses 346 • Chapter 13 Alcohols

O O 3. B2H6 6. MeLi OH 1. HCl (SN1) 4. H O , OH 15. 2 2 H 7. H Me 2. LDA (E2) 5. PCC 8. PCC

H OH OTMS OH 1. Me3SiCl 3. O 16. O 5. KMnO4, H Cl MgCl 2. Mg 4. H3O

HO O

OH O OH 1. H2SO4 (E1) 2. O3 4. MeMgI 17. 3. Zn, H 5. NH Cl O 4 (weak acid to OH avoid dehydration)

1. TMS-Cl OH OH 4. CrO3, H C CH 2. BH3.THF 18. 5. LiC CH

HO 3. H2O2, OH TMSO 6. H3O HO

O OH CH OH C 1. NaOH HC CLi 2. MeI (S 2) 4. 19. N 5. H 3. CrO , H HO 3 MeO MeO

O O CH OH C C CH

1. TMS-Cl 3. H3O

2. LiC CH 4. CrO3, H HO O O TMS

1. Mg 3. H 20. -H2O 2. CH2O Br O OH2

-H 1,2-R: shift

13.2 Syntheses Solutions • 347

1. NaBD OH 4 SbF5 21. etc. 2. H (- OH) O D D D B deuterium is 'equilibrated' among all five cyclopentenium carbons

13.3 Mechanisms

+H 1. -H OH -H2O O O OH O H) H H

1. NaBH4 2. 2. +H 2. H2SO4 -H2O 1. H: OH HO 2. H

-H

(H O 1729 cm-1 O OH OH 3. +H 1,2-R: H -H H -H2O H shift OH G 9.5 (d)

Ph Ph Ph +H 1,2-H: Ph 4. OH H -H shift OH -H2O OH O (H H O

+H 1,2-R: -H 5. OH -H2O shift

1,2-R: -H shift (H

13.3 Mechanisms 348 • Chapter 13 Alcohols

OH H HH (H +H H 6. 1,2-H: H ~H H H E2 H OH OH -H2O shift OH OH OH O OH O -H3O H 2 O

H +H 7. 1,2-H: H 1,2-H: H -H O 2 shift H shift, OH OH OH O (H -H O

Ph 1. H PO Ph 8. 3 4 -H2O EAS +H -H CO2H OH2 C O O O 2. MgCl

-HBr 4. HBr (SN1) 3. H3O

Br OH 5. Me2NH NMe2

+H +H2O 9. O O O O O O HO OH2 O H H2O H O O H O ~H O taut HO H H H O -H O O O OCH3 O H) O O H H H H

Br Br Me H Br H Me +H -H O 10. Me H Me 2 +Br Br Br H HO H H HMe Br Me = H Br Me H Me Br H Me OH2 Me (+) Br Br - Me +Br H Me H Br H = Me Br H Br Me

13.3 Mechanisms Solutions • 349

Br Br Br +H NGP 11. OH -H2O +Br trans- anti- OH2 Br Br Br +H Br SN2 OH2 OH +Br, -H O 2 Br cis- gauche- Br NGP not possible trans-dibromide O H H O O H OH +H O 12. OH O (H 1,2-R: shift CO H O 2 CO H 2 CO2H -H CO2H CO H OH 2 taut CO2H 2. -H2O OH 1. [H] O OH E1 CO2H H +H -H 13. +H H -H O OH 2 +H -H H H no energy benefit to -H 1,2-H: shift 1,2-H: shift

pathway (1) - E2 HO D O H CCD H D O 14. 2 2 + H, taut -HOH or DOH + OH OH H D H OH HCD2H H3CD NOT formed pathway (2) - H) O D H D O O + D 1,2-H: or D: + pinacol H H D H O(H shift, -H HCD2H DH2CD +H, -H2O formed -- therefore, pathway (2) is favored

D H -DOH H, taut 15. a. dehydration-tautomerization: A D NOT B OH vs. O D D pinacol-like: A +H 1,2-D: shift -H D D B -H2O OH O(H

-- therefore, the pinacol-like pathway is favored

13.3 Mechanisms 350 • Chapter 13 Alcohols

OH OH 1. OsO4 3. CrO , H O 4. NaBD 15. b. 3 4 D 2. NaHSO3 OH O 5. H D OH A

H O OH OH OH +H 1,2-H: 16. OH -H shift OH OH OH O(H H O

alternatively, H H O OH OH taut taut OH ~H (H ~H (H O O

O Cl Cl 1. Li 3. +2H 17. + + 2. O -H2O aromatic C4H8 carbocation

Cl O PPh3 Cl SN2 SN2 - Ph3PO 18. Ph3P: CCl3 Ph3PCCl3 D D -Cl Cl -HCCl3 O D D (H D D

OH 19. OH +H 2 -H2O

H -H

13.3 Mechanisms CHAPTER 14 ETHERS

14.1 Reactions

1. HBr 3. KOAc 1. 2. TsCl O Br Br S 2 Br SN1 OH OTs N OAc H

Ph OH HI I 2. O HI + NR! H SN1-like

Ph -Me2S Ph Me -H 3. S S S S 2 S Ph Me H N H

(H O O KOH 4. -HBr Br

O H 1. HF F 5. 2. PCC F S 1 O N OH [O] H

NaCN / MeOH 6. :N C: N S 2 C O N OH

O O 1. PhLi Ph Ph 2. H 7. Ph Ph :Ph E1 Ph

H O OH CH3 O OH 1. HI 2. [O] 3. - 4. [H] 8. SN2 D I

1. PhCO H 2. H +PhOH, -H 9. 3 SN1 O O OPh H OH

14.1 Reactions 352 • Chapter 14 Ethers

H Me O OMe H, MeOH 10. H H O OH H trans-diaxial ring opening determines regioselectivity

F F 2. MgI F 1. ClCH2COCl 11. F AlCl 3 3. H OH Cl O F Cl F F 5. 4. base

6. H F OH O F N N N N

N N

Me NH2 Ph Me 1. mCPBA 2. 12. Ph Ph O Me NHMe OH

OH O O OH H 1. NaOH 3. ' 13. taut 2. RX

O H CO2H O 14. a. ' CO2H OBn OBn

O (H HO CO H CO2H O 2 ' HO +H -H 14. b. -H O O 2 -CO2 CO H CO H 2 O 2 HO2C O HO2C O

14.1 Reactions Solutions • 353

' 15.

O O OH O ' ' 16. taut

OH OH O ' taut 17.

CO2H H2N [O] 18. H2N S S NH2 HO2C SH CO2H

OH HO 19. HS [O] S SH S OH HO

14.2 Syntheses

Br OMe 1. Br2, hv 2. HOMe 1. SN1 not OMe ( => 100% E2!)

OH 1. mCPBA 2. H3O 2. O OH

MgBr O OTs O 1. NBS, ROOR 4. H 3. 3. 2. Mg 5. TsCl

14.2 Syntheses 354 • Chapter 14 Ethers

2. H N NH 1. NBS, ROOR 2 2 5. I 4. 2 S 3. OH, H O [O] S Br 2 SH 4. H

Li O 1. Cl2, ' 3. O 5. PCC 5. OH H 2. Li 4. H [O]

O 1. NaBH 2. H2SO4 3. H2 / Pd 6. O 4 H -H2O H :H

O O 2. thiourea O 7. 1. HCl 3. OH N Cl N HS N conjugate addition 4. H HO2C HO2C HO2C

Br Br SH SH S 1. NBS, R2O2 2. HBr 3. thiourea 6. Br S 8. Br 2 [O] CO2H CO2H 4. OH CO2H 5. H CO2H CO2H

14.3 Mechanisms

O BF3 BF3 O O ~H: H -BF3 1. Ph BF3 Ph PhCH2CHO Ph Ph H H O BF3

CH2 1. LDA 2. (H N 2. H OH O 2 O H

3. -H

OH OH OH O H

-- see 14.3, 6 for an even more impressive polycyclization!

14.3 Mechanisms Solutions • 355

OOO HO O OO 4. OH HO

H) OMe HO O O OH HO O O O

H HO O O H H O O O 5. SN2, ~H O H O O O H H H H H H HO O ~H ~H O product O O H H H

H H 6.

Me Me HO H O H Me Me H H

-H Me H Me two 1,2-H: shifts followed by HO HO two 1,2-R: shifts

H H D O DOH DOH O OH H H D 7. D 1,2-D: shift -H (H path (b)

D OH OH H observed -D H methyl group determines path (a) direction of ring not observed opening

14.3 Mechanisms 356 • Chapter 14 Ethers

O(H O R OH R OH R OH 2. 1,2-R: R -BF R 8. R' R' R' 3 BF3 shift R' ~H R' O O BF3 OBF3 OBF3 OH H

N S Cl PhO S -DBN-H N 9. H (H PhO H O N CH2 O N O O O O DBN: Cl

N S N S -Cl PhO PhO H H ~H O N O N O O O O H DBN: DBN H

14.3 Mechanisms CHAPTER 15 ALDEHYDES AND KETONES

15.1 Reactions

1. CrO3, H 1. OH N NH2 2. H2NNH2, H

O O OBr 1. Ph3P O PPh3 3. 2. 2. MeLi a Wittig ylid a vinyl ether

O H O acetal H O O H O 3. 3 3 H Ph + O Ph O OH OH OH

-H2O 4. + opsin-NH2

opsin HO HN

O O OH H3O 5. HO Ph Ph hemiketal

O OH CHO OH OEt 6. 1. PCC 2. H3O H 3. H 3. HOEt, H OH O O

O O O Cl 1. KO-t-Bu / t-BuOH 2. HCl 7. E2 conj. add'n Cl

O DOH D 1. NaBD4 3. H2SO4 8. 2. H E1

15.1 Reactions 358 • Chapter 15 Aldehydes and Ketones

1. KMnO4 [O] 3. H / Pt 9. 2 O [H] HO 2. N HN H2NNH NH2 NH NH NH2 (-H2O) NH O 2 O

O O O

CH 1. HO OH , H 3. Ph3P=CMe2 10. O H OMe o H 2. DIBAH, -78 4. H3O O O

O O CHO 1. Ph3P 11. O H O N 2. H3O 2 O2N

OH O

1. Ph3P=CHOCH3 O CH3 2. H O O H O 12. O 3 3 H CH (+H2O) 3 (-MeOH)

1. Ph3P: OR O 3. H 4. H3O 13. RO-CH2-X Ph3P-CHOR 2. MeLi (-HOR) H

OMe OMe O H H H 1. CH2I2 2. H O H 3. Ph P=CHC=CH 14. O O 3 3 2 Zn (Cu) H H H H OMe OMe O

H O OH 1. H3O 2. EtMgI 15. NNH2 H 3. H3O

O PhO OPh Ph 1. H3O 2. H2NOH N 16. Ph Me Ph Me (-2 PhOH) (-H2O) Me OH

15.1 Reactions Solutions • 359

OH O O O OH CHO H O OH H 17. 3 OH

O 1. Ph P O O O 3 O 3. H O 18. O O 4. H3O Br PPh3 2. n-BuLi

acetal H hemiacetal H O O OH O O O H3O H3O H3O 19. + 2 OO OO HH OO etc. HH H

1. H O, Hg2- 20. CCH 3 2. H2NNH2 W-K +H O, taut 2 OH N O NH2

OH O Br 2. , H 1. Br2, H2O O O 21. 3. Li Li OH OO 5. H3O 4. O

HO O

1. HONO , H SO 2. H NOH, H 22. 2 2 4 2 N CHO CHO OH O O2N O O2N O note position of EAS!

H3O Et H O 23. Et O O 3 Et HO O OH O HO HO

O O O OH O ~H H OH 1,2-H: shift O O 24. OH H O H OH O

15.1 Reactions 360 • Chapter 15 Aldehydes and Ketones

D CHO H C D CHD2

1. D2NND2, OD, D2O 2. HI 25. +MeI W-K SN2-like OMe OMe OH OH OH OH

CO2H CN CO2H O O Ph O OH O 26. H3O HO CN OH H OH + HCN + Ph H OH OH Ph H OH OH

OH O O H3O H O 27. 3 O HO HO OH O

OO 1715 cm-1 O O MeOH, H 28. O H H H 3.4 HH4.9 G2.2 2.8

OMe OOOMe HO HO O OMe HO HO OH HO O O OH 29. H3O -H2O -HOMe

O OH O H3O 30. + HH O OH

NH2 1. H 31. H 2. H2 / Pt CO2H -H O N CO H N CO2H O 2 2 H

HO O O O OH OH OH

1. LiAlH4 32. + EtOH + MeOH MeO 2. H3O O MeO

15.1 Reactions Solutions • 361

OO OO O 1,2-H: shift O HO O 33. ~H H H HH OH intra-Cannizzaro OH H H H OH H O

O O OH 1. LiAlH O 4 -H2O 34. O EtOH + OH 2. H3O O

HO O HO O O HO HO OH O O H3O OH 35. +

O O F F

OH O N H2NOH 36. O N OH

O 37. H3O OOOO n HH acetal

OH OH OH O OH OH N OH OH N 1. NH2OH, H 2. Ac2O C 38. H H -H O OH OH OH OH OH OH 2 OH OH OH OH OH a hexose an unstable cyanohydrin H

OH OH O -HCN a pentose

15.1 Reactions 362 • Chapter 15 Aldehydes and Ketones

O O Me Me O N mild acid N 39. N + HH N N N H H O O

O acetal OH O OH

HH H H H OH O O NN mild acid 2 Me N N ~H N 40. 2 S NHMe NH2Me OH2 H NO2 H NO H NO2 -H 2 H H -MeNH2 O N O -H N O(H Me2N S

NO 2 NO2

N O N NH N NH2 N 2 N H3O CO2H 41. H3O + N N N N H H H S S S H

O O 1. PCC 2. LiMe2Cu 42. OH [O] H H 3. H3O

H H N N O O O H O OOH 43. 3 + HH O acetal OH

F F

H NH2 N O [O] H O 44. 3 O2CCO2 O2CCO2 HO2CCO2H + NH4

15.1 Reactions Solutions • 363

O O

O ketal H3O 45. OH OH HO HO

OH O OH Me 1. PCC 2. MeLi 46. EtO EtO 3. H3O EtO EtO O

O N O N 1. HC CNa / THF 47. OH O 2. H / Lindlar catalyst N 2 N H H O O note: addition to ketone, not amide carbonyls

O O 1. HS(CH2)3SH, H S Ph 3. EtI S Ph 4. H O 48. a. 3 Ph H Ph 2. MeLi S SN2 S Et

S S 1. n-BuLi O 3. H O b. 3 S H S 2. CH3(CH2)9Br SN2

S O O 1. HS SH, H S 3. O c. H S 2. NaNH2 S O O or 4. H3O

-H2O HO

15.1 Reactions 364 • Chapter 15 Aldehydes and Ketones

H H

H2N NN H2N NN +H O N 2 N 49. H N H N -H O H O N 2 O HN + nitrogen analog of OH an acetal H O formaldehyde O HN HN CO2H CO2H

CO2H CO2H tetrahydrofolic acid

N N N N N N H 50. NH2 -H O Cl O 2 Cl N

XanaxTM - (anxiolytic)

15.2 Syntheses

1. H O O HO CN HO CO H 3 3. CN, HCN 4. H3O 2 1. 2. CrO3, H OH O NH NH2 1. Hg(OAc)2, H2O 3. KMnO4 4. NH3 5. H2 / Pt

-H2O 2. NaBH4

OCH 3 OH O OH 1. HI 2. Cr O 2- 3. NaBD4 2. 2 7 H H D -CH3I 4. H

3. H2NNH2, OH W-K

15.2 Syntheses Solutions • 365

1. HCN, CN OH 2. H O 3. O 3 4. H2 / Pd CO2H CO2H CN 3. H2SO4 (-H2O)

O OH Cl 1. NaBD4 3. SOCl2 4. D D 2. H SNi - ret

O OH O 1. HCN, CN 3. CrO , H H CO2H 3 CO2H 5. 2. H3O

O O O O 1. OH Cl 2. PCC 3. ethylene glycol, H O 6. Ph OH O S 2 Ph Ph Ph N [O] (1 equiv) H O (aldehyde more reactive than ketone)

OH 1. LiAlH4 OH 2. H OO OH O 1. NaBH4 7. OMe OMe 2. H OH O H2 / Pd OMe

OO 1. LiAlH4 OH 2. H 7. (cont.) OMe OH 3. H2 / Pd O 1. ethylene glycol, H (protect ketone) OH 2. LiAlH4 3. H3O O

1. BH3.THF OH 3. PCC H 4. Ph3P 8.

2. H2O2, OH

O OOH 1. H3O 2. H O MeOH, H O 9. conj. add'n OH OMe

15.2 Syntheses 366 • Chapter 15 Aldehydes and Ketones

O 1. MeOH, H 2. PCC 10. OH OH H MeO OMe MeO OMe O

Ph3P 4. H3O 3. MeO OMe O

O HO H MgBr H 1. OH, H 3. 11. O O HO Br O 2. Mg 4. H3O

CHO

NR2 1. H2NNR2 NR2 2. n-BuLi NR2 3. PhCHO N 12. O N N H O OOH Ph 4. H3O Ph

CHO O O CHO 1. HO OH 2. KMnO , OH 3. CrO , H 13. O 4 O 3 H or 2. a. OsO4 HO 4. H3O O b. NaHSO3 OH O

O O

H 1. LiMe2Cu H 2. H2NNH2, OH 14. W-K

OH O HO CCH

1. ethylene glycol, H 3. HC CLi 15. O 2. K2Cr2O7 4. H O O 3 O O

15.2 Syntheses Solutions • 367

O OH Et 1. EtMgBr 3. HBr H Et Li 16. 2. H 4. Li O 5. R Ph Et Ph Et Ph -H2O 6. H3O Ph OH Ph R R

OH O O 1. NBS, R2O2 3. PCC [O] 4. Me2CuLi 17. 2. OH (SN2) 5. H

1. DIBAH, -780 3. Ph P=CH 18. 3 2 CO2Me CHO 2. H CO2Me CHO

1. PCC 2. O3 19. HO 3. Zn, H H OH O O

H OH O 1. Me SiCl 2. Ph P O 20. 3 TMS 3 TMS Br Br 3. MeLi PPh3 O O CO2H 5. H (deprotect) TMS 4. H 6. CrO3, H

O SH SH H2 / Ra-Ni 21. Ph Ph SS PhCH2Ph H Ph Ph

O 1. HNMe 2. O 2 3 H 22. O NMe2 N -H2O 3. Zn, H O

15.2 Syntheses 368 • Chapter 15 Aldehydes and Ketones

15.3 Mechanisms

H OH (H O O 2 O O HO OH2 H2OOH +H +H2O ~H -H2O -H 1.

= 18O O +H 2. ~H -H H

N N OH2 N OH NO(H H 2 NH2 H H HH

H H H 3. ~H O O O OH2 OOH CA of a OOH 2 hemicetal H O -H OH H (H HOH O

H O OH O H OH 4. -H conj. add'n N N H NH2 H taut HN H NH H H NH2 2 H H2NNH2 ~H -H -H2O OH2 N N N N N N H H (H H H

Cl Cl OMe -Cl OMe 5. O OMe OMe O H O Me Me O O Me Me O O OH O H) OMe H OO OH O OH2 O ~H -H2O RR 6. RR R R R R H N HN 2 N O H2NOH OH OH H O H H) R R RR R R ~H -H3O R R ~H N O OH N O N O N O OH2 H O H H

15.3 Mechanisms Solutions • 369

H H O O 7. OOH O O OH OH OH OH OH O O -H O O (H

(H O H O H O H O H O H O 8. O O O

OH OH OH OH OH O OH O HO H OH O OH O -H OH OH H3O HO OH hydrolysis OH OH OH

O O O O H S O S O 9. C + S H

H H H O O O O H OH

10. -H (H Cl Cl Cl Cl O O O O OH H H

O O O O 11. ~H H O OH O O H2O O H H H2O

H) O O O -H O H + + H O taut O H)

15.3 Mechanisms 370 • Chapter 15 Aldehydes and Ketones

OH OH OH OH H H H 12. OH2 O O O O ~H OH2 OH2 OH OH (H -H3O ~H -MeOH OH O O OH OH

O O O N N 1,2-R: shift 13. +N2 CH2 NN

Ph Ph Ph Ph N N N N 14. OH2 OH2 N N NHPh NHPh Ph H Ph H ~H Ph H O (H NHPh N -H O + OH H H NHPh NHPh

H H Me Me Me 1. mCPBA O 2. O rotate 15. PPh3 Me Me PPh3 H Me H PPh3 Me PPh H 3 Me -OPPh H 3 O O Me Me H H Me H Me H

O O O P(OMe) 3 -OP(OMe)3 16. P(OMe)3

:P(OMe)3

15.3 Mechanisms Solutions • 371

HO OH HO HO OH O 2 :NH2R 17. +H O -H2O O OH OH OH OH OH OH HO HO H) NHR NHR O -H O

OH OH OH OH

O OH OH HO +H EAS 18. OH OH H) -H HO +H HO OH OH HO OH -H, EAS -H2O

O Et O Et ~H 19. H N H N taut O H H H2N 2 H

Et Et (H Et Et -H - OH ~H N N N H N OH O

1. -H 2. S 2 OO 20. a. 2 O N O O Williamson Cl (H :H Cl Cl ether synthesis Cl MOM

b. 3. O O O O HO OH H + OH H OH H O O 2 H3O O +H2O + MeOH H2COMe H H OH -H

15.3 Mechanisms 372 • Chapter 15 Aldehydes and Ketones

Br Br Br OH2 H3O OH 21. 2 Br H O O (H -HBr -H Br

H O HS Me O Me O OH S (H SMe (H ~H ~H 22. SMe conj. add'n taut

O O OH OH

HO H O C N -HCN 23. H H NC H + CN O OH O CN O (H :C N

:Me O OEt O OEt H O OEt 1. MeLi 2 24. 2. H -H2O 1,2-add'n

(H O O OH2 OEt -H OEt ~H OEt +H2O -HOEt H OH2

H Et Et Et H O N H H HO N CHO EtNH CHO ~H H N 25. 2 H O -H2O (H

Et Et Et N N 2O NH ~H -H -H2O

OMe H OMe OMe Ph Ph Ph Ph N 2. D-A N 3. H N -H N 26. O -MeOH Ph O Ph O Ph O Ph TMS TMS (H

15.3 Mechanisms Solutions • 373

O O Cl O Cl2C OH ~H Cl C 27. a. Cl2C(H Cl2C: H D-elim H Cl HO O OH O O Cl O C C H) C SN2 -Cl H 2. H H -HCl O H

Me O O NH-Cl Me OPCl +Cl Me OPCl Me H b. Me N C N C N C Cl Cl O2PCl2 O Me H Me H Me Cl Cl Cl Cl OPCl Cl

OH OH OH Cl (H HMe1. -H NMe2 CN NMe2 Cl Me Cl OH OH -Cl CHO 2. H3O NMe2 hydrolysis

O O OH H OH ~H O 28. Ph H Ph H Ph O Ph H CN CN Ph Ph :C N CN

O OH O OH ~H - CN Ph H Ph Ph CN Ph

O H

NH HH N OH ~H N OH2 29. H OH OH OH -H2O

N N N -H (H EAS OH OH OH

15.3 Mechanisms 374 • Chapter 15 Aldehydes and Ketones

HO HO H HO H OH OH H O +H O E- O 30. OH bottom-side hemiacetal OH OH OH OH attack OH OH

HO HO H top-side O -H O D- attack OH OH OH OH OH OH

NH H NH2 H OH NH2 H 2 H 2 OH +H2O 2 NNH NNH2 N 31. HO2C HO2C HO2C NH2 NH2 NH2 NH2 NH2 H NH O O (H 2 -H N ~H HO C NH NH2 + 2 H 2 HO2C H2NNH2 NH2

Ph Ph R R H) O O O O H N HN O O -H 32. O Ph Ph :NH2R O O H H CO2H CO2H CO2H O H NHR HRH HO H O: N +H O +H2O 2 O -H -H O Ph H O Ph CO2H OH CO2H OH2 R N Ph (H Ph R N R N +H Ph -H3O OH O OH O OH O CO2H CO2H CO2H

OH2 OH2 O +H O O RO C H H ~H 33. 2 H H H O PO3R' O (H H OH O O RO2C -H (CH2)nCH3 H O + H H PO3R'

15.3 Mechanisms Solutions • 375

O H OH OH O 1,2-R: shift Ph Ph Ph 34. Ph O Ph O RCO2H O O O O Ph R R O H + R O (H H O (H -H O Ph Ph Ph O Ph O

OO OO OO OOH 1,2-R: ~H 35. a. Ph Ph HO Ph HO Ph OPh shift Ph hP OH Ph

OH OH O O O 1. 1,2-R: 2. H b. O O OH O O shift, ~H OH OH

OO O O 1. O 1,2-R: O 2. H c. OH PhPh shift O Ph Ph Ph Ph Ph Ph

O OH OH OH OH

+H 1,2-R: -H 36. R shift (H R RR RR RR R R

O (H :B O OH OH OH OH +H 1,2-R: -H 37. -HOTs shift (H

OTs

H) +H 1,2-R: -H 38. shift O HO HO HO HO

15.3 Mechanisms 376 • Chapter 15 Aldehydes and Ketones

+H 1,2-R: 39. shift O HO HO HO

-H 1,2-R: HO H) shift HO OH

O O O

. +H . . 40. O . HO . HO . 1,2-R: shift O HO -H 1,2-R: . HO O H) . . O shift . . HO .

HO OH O O O O O OR H OR RO H) Cl -HCl 41. RO H) OR O H H) OR OR

Br RO O O O -HBr OR -Br 42. Br CO2R Br Br (H OR

OMe H +H 1,2-R: -H 43. -MeOH shift (H

O NMe O NMe2 O O 2 NMe2 NMe2 A

15.3 Mechanisms Solutions • 377

OH OH 43. (cont.) 2 2 NMe O 2 O O NMe ~H A NMe2 2

O NMe -H + H 2 NMe2 OH O H O(H H R R R R N N N N -Br ~H 44. OH (H OH 2 O Br Br Br 2 Br Br H R (H N -H N R -H N R -Br CHO CHO CHO Br

O O O CO R CO R CO2R +H 2 ~H 2 45. N N N H OH NH OH2 O 2 NH2 O H NH2 O S S CO2H O CO H HS H 2 CO2H -H2O O O O CO2R CO2R -H H CO2R N H N N N (H N NH2 CO2H CO H S O S 2 O O S CO2H

46. H +H ~H H2N: NH2 NH N O N N H H OH H OH2 H HO HO

HO -H O HO HO 2 OH NH NH -H NH N N N H H HO HO HO OH OH HO

15.3 Mechanisms 378 • Chapter 15 Aldehydes and Ketones

H H S SH SH2 O Br Br +H O Br Br ~H Br Br -H Br Br 47. 3 Br Br Br OH2 Br Br O Br -H2S Br Br OH2 (H H) OH2 HO OH HO OH +H O Br Br -H Br Br 3 Br Br Br Br

H O OH H) OH H OH ~H O 48. OO OO O H O O H -H 1,2-H: O shift OH 2

O O 1. +H !! O 49. -H2O OH (H O O :B O O O O O O 2. +H2O H) OH2 OH -H

15.3 Mechanisms CHAPTER 16 CARBOXYLIC ACIDS

16.1 Reactions

CO2H KMnO , H 1. N 4 N N

1. OH O 2. -Me2O O 2. Ph CO2H Ph O Ph Me BF S 2 O 4 N O

OO OH O 1. NaBH4 3. OH OH 2. H

O O PhO Ph 4. 1. H3O H 2. CrO3, H OH OO 3

Ph acetal

O E O 1. OH 2. ' O 3. LiAlH4 5. J OH O O D OH Br SN2 Br 4. H3O OH

OH O O 1. OH 3. BH 6. 3 O OH 2. CO2H 4. H3O (conj. add'n) OH

1. NaCN 2. PhMgX 3. H3O Ph 7. PhCH2Cl Ph C N Ph C N Ph SN2 Ph O

O O 1. SOCl2 N 0 N 8. C 2. DIBAH, -78 3. H3O NH2 C -H2O H H

16.1 Reactions 380 • Chapter 16 Carboxylic Acids

1. (XS) PhLi 2. H -H O 9. 2 CO2H O OH O O HO Ph Ph Ph

OO OH 1. LiAlH4 10. + OH OH CO2H 2. H

11. 1. HCN, CN 2. H3O 3. BH3

O conj. add'n O O O CN CO2H 4. H3O OH

O O Et O Et CO2H 1. EtLi Ph NMe Ph NMe2 2. H 12. 2 Ph NMe2 Ph -H2O Ph Ph

1. H O OO 3 CO H 2. PCC 13. CO2H 2 conj. add'n HO [O] HOH

H) OH OO O taut + CO2 H H HO

labile carboxyl CO2H 1. [O] CO2H 14. D 2. -CO2 E CO2H CO2H CO2H HO O O

N N N CO2H CO2H 3. CrO , H 4. ' 15. 3 [O] -CO H 2

OH O O

16.1 Reactions Solutions • 381

OH O OMe OMe CO2H CO2 CO CO H 1. OH (2 equiv) 2. MeI (1 equiv) 2 2 16. 3. H SN2 more stable anion, therefore, less reactive

CH2CH2N(CH3)2 COO CH3NHSO2CH2 17. H CH2 a salt! CH2 N COOH H

16.2 Syntheses

CO H 2 1. LiAlH 4 OH 3. TsCl CN 1. 5. H3O CO2H

2. H 4. KCN (SN2)

1. NBS, ROOR O O 3. NaCN 2. 4. H3O 2. HBr, ROOR Br Br SN2 CN CN HO OH

O OH O OO O 1. H3O 2. CrO , H 3. OH 3 ' OH OH conj. add'n Jones reagent -CO2

O O O HO O 1. OH , H 3. CO2 4. Cl 2. Mg 4. H3O MgCl CO2H

O 1. 5. Cl O 2. H2NNH2, OH AlCl3 W-K CN CO2H 3. HBr 4. KCN 5. H3O

16.2 Syntheses 382 • Chapter 16 Carboxylic Acids

1. PCC 6. OH n-Pr 3. HCl n-Pr 2. n-PrLi O Cl n-Pr 5. NaOH, H2O n-Pr 4. NaCN CO2Na CN

CO2HO 1. H3O OO O 3. ' (-CO2) 4. BH3 O 7. OH HO C OH 2. CrO3, H 2 CO2H 5. H HO

o 8. Ph Br 1. KCN (SN2) 2. DIBAH, -78 N 3. H3O O Ph CN Ph Ph H hydrolysis H

N NH 2 N N H o CHO CN 1. DIBAH, -78 3. NH NH , H CHO 9. O O 2 2 CN 2. H3O CHO -H2O -H2O O O

1. KMnO4, H HO C CO2H 10. 2 2. ' HO OH -2 CO2 CO2H CO2H OO

1. LiAlH4 3. SOCl2 or HCl 5. R'MgX 11. RCO H RCH2OH 2 RCH2-CN RCH2COR' 2. H3O 4. NaCN 6. H3O 5. H3O or 6. R'Li

7. H3O

OO 1. BH O OOH 3 3. HO OH , H O 5. PhMgBr 12. O O OH OH Ph 2. H3O 4. PCC H 6. H3O

O 1. H SO (E2) Br 2 4 3. (XS) KCN NC 4. H O 13. OH 3 HO OH 2. Br , CCl Br CN 2 4 O

16.2 Syntheses Solutions • 383

O O O OH

1. Cl 2. CH Cl 3. H2 / Ni 14. 3

AlCl3 AlCl3 hydrogenation

CO2H a benzylic alcohol 4. KMnO , H 4 hydrogenolysis

CO2H

1. NaNH2 (1 equiv) 15. HC CH HC C 3. KMnO4, H HO 2. n-pentyl chloride O

OH O O OH 1. NaH (2 equiv) 2. Et-I 3. H 16. OH O (1 equiv) O O Et Et more reactive anion

16.3 Mechanisms O O O H +H O 1. CO H -H 2 O (H O H

H) OH O O OH CO2H taut C 2. O ' -CO n-pentyl n-pentyl 2 O n-pentyl tautomerization to E-ketocarboxylic acid facilitates decarboxylation

O H)O O Cl O CCl HO Cl 2 CO2 +OH -H -Cl 3.

Cl Cl Cl Cl

16.3 Mechanisms 384 • Chapter 16 Carboxylic Acids

O +H +CO +H2O -H 4. CO CO2H O(H CO OH2 H

O OH H) HO OH O HO ~H 5. R OH RN HO H N N 2 H RH O O -H2O HO -H R N R N (H

O O H OH OH2 +H ~H 6. O H O O CO2H H O (H O H O O O Me Me O -H +MeOH Me -H2O O O O

O O O

OO NH2 ON -CO 7. 2 N OH taut ON RCO2H (H -H2O R O RH RH O N O H N dimerize 2 H3O + hydrolysis N -2 H2O NH2 O -RCHO

O H OH OH2 +H ~H -H2O R' H 8. RH R'HN R'N N R H H R H R R'NH2 R' O NC R' H N H O H OH 3 N C H N R hydrolysis R

16.3 Mechanisms Solutions • 385

O CO H (H 2 O -H 9. HO CO H +H 2 H2O CO H 2 -CO 2 CO2H O O O

O O O(H O CO2H O H H2NCHR 1. -CO2 10. O N R N R -H2O 2. taut

O O O -RCHO H2O O O O O N -H -H2O blue dye O + H2N (H O O O O

H O2C O C O 2 11. R N H) OH OH + RHN H H NH 2 NH2 O O NADH NAD+

OO O O O CO2 -CO2 -GDP 12. OPOP OCO2 O CO2 OH O PO3H GTP

R' R' R' R'

-H S -CO S 13. N S S R N 2 R N R R N O (H (H CO2 O OH O O HOH H) R' OH R' O -H N S + S R H R N (H O H

16.3 Mechanisms 386 • Chapter 16 Carboxylic Acids

O R' H O HN NH R' -H O 2 -CO2 R OH H 14. PLP + histidine R OH

N R' = N H NH H N H H HN R' +H R OH H O PLP + histamine 3 imine hydrolysis N H

O O H O intermolecular hydrogen bonding forms a 15. 2 OH O tight dimeric complex in nonpolar solvents H O MW = 60 MW = 120

16.3 Mechanisms CHAPTER 17 CARBOXYLIC ACID DERIVATIVES

17.1 Reactions O O Et NH (1 equiv) 1. OEt 2 OEt Cl Et N -HCl 2 more O O reactive than ester

OO O O 2. + MeNH Me O 2 N + OH H O O

O (XS) EtOH, H OEt 3. + OH transesterification

O O O -HCl -HCl 4. + OO OH OH Cl Cl HO O Cl

OO 1. PCl3 OO 2. LiAlH(O-t-Bu)3 OO 5. HO OH (or SOCl2) Cl Cl 3. H HH

O H3O 6. CO2H + NH N hydrolysis 2

O O MgBr MgBr Ph or O 1. 1. 7. OH - OPh 2. H3O -H2O

not isolable O O Me O MeOH, H MeOH, H MeO OMe 8. O NAS CO H O 2 OO

1. LiAlH4 3. 9. O 2. H NAS CO2H OH OO

17.1 Reactions 388 • Chapter 17 Carboxylic Acid Derivatives

O O O OH, H O 2 HO 10. O saponification H D

H NH2 Bn N S S OH, H O CO2 O C 11. 2 + 2 O N N O H CO2H CO2 O O Et Et OR H2N - 2 HOR NH 12. + O Et Et H N OR 2 O N O O H

O O N N N O 1. OH 3. CrO3, H 13. PhCO2H + OH 2. H + 4. '(-CO2) MeOH OPh OH O O

O O OH OH SCoA NH

NH2CH2CO2H CO 14. A B 2 -HSCoA HO OH HO OH H H

N N N 1. H3O 2. CH2N2 15. N N (diazomethane) N H H O O O O

O OH O OCH3

OH O OH OH 1. H O 3 O 2. PhMgCl Ph or 16. OEt O Ph -HOEt 3. H HO Ph -H O 2 Ph a J-lactone

17.1 Reactions Solutions • 389

O O O 1. LiAlH4 OH O 17. = 2. H OH HO OH O O

O O H O N Cl N 18. 1. Cl Cl 2. LiAlH N NN 4 NN N -HCl H -HCl H 3. H

protein protein O OH O O 19. OPF P -HF Me OMe

O O H H H CO2H Cl NEt2

N 1. SOCl N 2. HNEt2 N 20. 2 -HCl

N N N H H H lysergic acid diethylamide O O O O -HCl 21. SCl + H2NNH S N N H H O n-Bu O tosyl chloride more nucleophilic nitrogen

O O

OH, H2O O 22. NH + NH3 S SO3 O O

OO O O 23. HO NH2 HO N + HOAc 1 equiv H

O O ~H OH taut 24. H2CCO H2CC H2C NHPh NH2Ph NHPh :NH2Ph

17.1 Reactions 390 • Chapter 17 Carboxylic Acid Derivatives

H 25. dimethyl phthalate + HO OH O transesterification O O O O O O O O O a polyester O O

26. HO OH + Cl Cl - n HCl

OOO O

O O O

O O

O OH O O H OH C O NMe O NMe O N Me H N ~H taut 27. Me

O O O O LiCu Ph Cl O 1. Li 3. Ph 28. O O 2. CuI 2 4. H3O Cl Gilman reagent

O O NHMe Me 1. 2O N 29. 2. LiAlH4 F C OH 3 - O 3. H O Me ON

F3C

17.1 Reactions Solutions • 391

HO O

O H2N OH O O :NH3 O OH 30. O NH2 + OH

O OH CCH CCH 1. OH, H O 31. 2 2. H3O + HOAc + NH OH N O 3 OH

O MeO 1. LiAlH MeO N 4 N 32. H H HO 2. H3O HO

O O O H2N H2N N S 1. H2O, OH H S O 2. SOCl2 O 33. N N 3. O N N CF3 CF3 H2N NH2

F HO S O O HO OH HO O O H O 34. 3 O F F O + HSCH2F + O OH F F

O CO Na O 2 1. PhLi 2. H3O 35. O -H O Ph MeO Ph 2 MeO

17.1 Reactions 392 • Chapter 17 Carboxylic Acid Derivatives

O O

O HN N O HN N N H N S H3O N H HO S 36. N N N O Me N + O N Me OEt H OEt

O O O NH 1. base O -Br 2. 2 37. N Br CO H O H 2 SN2 OH Br O O OH O NH NH 38. ~H NH2 O2CNH2 O2CN ~H H O O O H NH N etc. O2CN N H H O

O O O NH 1. Cl OEt NOEt 2. Et NH 39. 2 N NEt2 N N -HCl -HOEt N

O OEt OOH H N N N

H O -CO 40. N 3 N 2 N

Cl Cl Cl

O HH NH N Ph NPh O

1. Ph OEt 2. LiAlH4 41. -HOEt 3. H

Cl Cl Cl Cl Cl Cl

17.1 Reactions Solutions • 393

H2N S

OMe OMe O N OMe O H O CO2H N 1. SOCl2 2. CO H 42. Cl 2 S OMe OMe -HCl OMe N O CO2H

O CF3 O N OH H CF3 H N H3O 3 43. + CF3 CO2H O N CF3 H H3N

O CO2H O CO2H H N H O 3 OH 44. N 3 H2N H OMe + + MeOH H NCOH Ph O 3 2 Ph aspartic acid phenylalanine

O Cl CO2H CO H N N N N 2 N exhaustive Cl 45. N + N hydrolysis N N N H OO O H a carbinolamine N -CO2 H N Cl N NCO2H N + + H N N H2N O O OH HO O OO 1. NaBH4 H 46. CO2H + - 2 H O 2. H OOHHO 2 O O

OH O OH OH OH OH OH HO 1. HCN 2. H3O Cl 4. LiAlH(t-BuO)3 47. H CN OH CN 3. SOCl OH OH OH OH 2 OH OH O 5. H HO a tetrose a cyanohydrin O H a pentose

17.1 Reactions 394 • Chapter 17 Carboxylic Acid Derivatives

R O O R 48. a. -HCl :NH2 CO2H O Cl O N CO2H H

(H O R O O R O +H -H b. N N O N H OH O N OH H H OR' H H H OR' + R O O R O -CO N 2 H) N H2N H OH O N H OH H OR' OR' H

O OH

49. SCoA + -HSCoA CO2 NMe3 transesterification O CO2 O NMe3

CH3(CH2)14CO2H, H 50. O -H2O HO CH3(CH2)14 O

O O

HN H3O HO 51. NH4 +CO2 + O N H N H 3

O O O OH O 1. ATP 2. H2N 52. P OH OH O OR' N O -HPO4R' (AMP) H NAS O 2. HSCoA SCoA -HPO4R'

17.1 Reactions Solutions • 395

+H -H 53. O O HO-glucose O + OO O transesterification O O glucose H glucose H) H

N O N O N O 1. SOCl2 3. SOCl 54. N N N N 2 N N 2. NH 3 O F CO H F F C 2 N O O NH2 O

OH O N OH O OO N N S H3O OH taut OH 55. H H3N + N NHMe S S NHMe SO3H SO H OO OH 3 H OH O NHMe NHMe NHMe -CO2 H taut

SO3H SO H 3 OH SO3H O H3O MeNH + (aldehyde gives a positive Tollens' test) 3 H SO3H

17.2 Syntheses

O 1. H3O -HOAc 3. Li 5. MeOH, H 1. O 2. PCl3 4. CO Cl 2 CO2 MeO O

O O O 1. H2O, OH 3. LiAlH(O-t-Bu)3 2. R NH 2 R Cl R H 2. SOCl2 4. H O 1. H3O2. SOCl2 O or 2. MeLi (2 equiv) O R OH R Me R OH 3. LiMe2Cu 3. H, -H2O O O 1. LiAlH NH 4 NH 3. Ac2O N 3. 2. H -HOAc

17.2 Syntheses 396 • Chapter 17 Carboxylic Acid Derivatives

3. CrO , H 1. BH THF 3 3. [O] 5. AlCl3, -HCl 4. F-C acylation 2. H2O2, OH O Cl O H 4. PCl3 O

CO2H 1. D-A 2. Ac2O, or O 5. C CO2H C ' H2SO4, -H2O HO2C O CO2H O

HO H 1. BH3 OH 6. HO H H O O 2. H O O HO or 1. OH O 2. a. LiAlH4 H CO2HO b. H3O

O O O O 1. H3O 3. LiAlH(O-t-Bu)3 5. HO OH Ph 7. Ph NH2 Ph Cl Ph H O 2. SOCl2 4. H H

O O O OO 1. AgNO3, EtOH 2. MeLi 3. H 8. H OH Me (Tollens' reagent) (2 equiv) -H2O

O O O NHCH3 Ac O NCH3 NCH3 9. 2 Ph Cl , AlCl3 O Cl Cl F-C acylation Cl

CH3 H O N O NCH3 Cl Cl O Cl O H O, OH Cl Cl 2 -HCl, NAS

CH3 H3C O O N N

NH2 NH (S 2) O H 3 N Cl Cl N -HCl -H2O

17.2 Syntheses Solutions • 397

O O HO CO2H 1. ' (-CO2) 2. NaBH4 H 10. O CO2H CO2H CO2H -H O 3. H 2 O

O 1. LiAlH4 2. H 4. KCN (SN2) O 6. CH2N2 O 11. Ph OMe Ph OTs Ph Ph 3. TsCl 5. H3O -N2 OH OCH3

O O F C F NaO N 3 ON 12. O nucleophilic aromatic subst'n O F3C

OH, H2O O ONOEt H Cl OEt ONH2 O F C 3 F3C

1. LiAlH4

:H H H H H H H N N -EtO N H +H: N +AlH3 OEt H O O O OAlH3 H O N H :H CH 2- 3 +H: N - OAlH3 CH2 F3C 2. H3O

O 1. NBS Br 2. Li O2C 4. EtOH, H EtO 13. Ph N N N N peroxide Ph Ph Ph 3. CO2

O ONMe2 HO NMe2 Cl NMe2 2. a. NaBH4 1. Me2NH b. H 3. PCl3 14. conj. add'n

Me O ON ONMe2 H F3C O 5. Cl OEt 4. F C F3C 3 6. OH, H2O SN2 -CO2 (continued on next page)

17.2 Syntheses 398 • Chapter 17 Carboxylic Acid Derivatives

14. (cont.) Mechanism for step 5: Me Cl O NMe2 CH3 ONOEt -Cl CH +Cl N 3 Cl OEt NAS O F3C -CH3Cl F3C EtO O O

O Et O Br 1. Cl 2. NBS, R2O2 15. Et AlCl 3 3. H2 / Ni (or W-K)

HO C 2 5. CO2 BrMg 4. Mg

6. H

O O O O

1. KCN, HCN 2. H3O 3. BH3 16. OH CN CO2H 4. H

O 5. HO O O OH , 7. MeLi O O 6. PCC H OH 8. H3O O OH H

O O O 1. KMnO4, H HO 2. ' 17. OH OH -CO OO 2 O O 3. OH OH, H OH OH H O 4. LiAlH4 OO O OH 5. H3O

O O 1. Li Cl CO2H 2. CO2 4. SOCl 5. Et NH NEt 18. 2 Cl 2 2 3. H -HCl

O H HO NH2 O NH2 O N 1. KOH Me Me Ac 19. 3. 2O more 2. MeI -HOAc N N nucleophilic N H H than NAS H

17.2 Syntheses Solutions • 399

Br CN O O H2 / Pt (XS) HBr (XS) NaCN / DMF 20. high pressure/temp SN2 Br O CN 1. H O 2. SOCl Cl CN 3 2 Cl O B H / Ni 2 H N CN 2 NH or 1. LiAlH 2 4 C 2. H

21. Poly(vinyl alcohol). Vinyl alcohol is unstable and rapidly tautomerizes to acetaldehyde: O taut OH H

1. H OAc OAc OAc OAc OAc OAc OAc etc. HOAc, H, Hg2+ 2. H3O HC CH or saponification OH OH OH OAc OAc OAc poly(vinyl alcohol) poly(vinyl acetate)

HO O CH3O

1. NaOH 2. CH3-I c. 22. a. O O O N N SN2 N

HO HO HO 1. H / Pd morphine codeine 2 2. PCC O CH3O OO b. O (XS) O O O NAS O N N

O O heroin hydrocodone

CH3O CH3O CH3O CH3O

~H ~H +H2O d. O O O O taut taut -H N N N N H (H OH H) O O O O H codeinone H neopinone oxycodone

17.2 Syntheses 400 • Chapter 17 Carboxylic Acid Derivatives

17.3 Mechanisms

H O O (H H) HO OH2 HO O OOH OH2 +H O ~H O H O H 1. O O =

-H -H O O -label appears in both carboxyl OH OH oxygens - but NOT in alcohol oxygen HO + HO

O O O O 1. (XS) RMgX 2. OCl O R O + RR R R R :R :R :R 2. H 2. H

t-BuOH R3COH OH O -H O O 3. 2 D-lactone does NOT form because of ring strain O OH O O O -H2O O -H2O O = O intermolecular OH intramolecular O O condensation HO2C condensation O O much less ring strain

H O +H O OEt -H 4. + O O H Et Et O (H 30 carbon

OOEt OEt OOEt O O O PhMgCl - OEt 5. O Ph O Ph O Ph (1 equiv) Ph: ketone more reactive than ester S H S S O H PhN PhN Ph NH2 ~H NH +H 6. Ph NCSH2N R N NH N HO H NR NR H O O HN H H R S S R = CO H 2 Ph N NH -H Ph ~H + N NH O H2NCO2H O H) HN H R

17.3 Mechanisms Solutions • 401

Me Me O O O OMe OMe 7. 1. Me3O BF4 2. H3O Me NH NH2 NH 2 2 -Me2O NH2 H2O OH2 OMe OMe ~H -NH4 NH3 O O (H O O O Me Cl -Cl O 8. a. SO +Cl ClCl S Cl SCl ++CO :C O: Me 2 Cl O

R H :NR3 :NR O b. 3 -NHR R (H -NHR (H 3 3 R O RR -Cl R OS -Me2S SCl

O O O R2CHOH, NR3 9. S Cl N SCl N [O] RR O O (mechanism per 17.3, 8)

OH O O 2- -HPO4 ~H HO 10. PO3H HO H2NO NAS H2N H2N CO H N CO H N CO2H 2 H 2 H H2N O H O O CO2H HO OH H) OOH2 HN -H O ~H 3 HN H2N

O N CO2H H O N CO2H OCON 2H H H

O O POBr2 Br Br OPOBr2 Br Br2PBr 1. POBr +Br 11. O 3 -HO2PBr2

N CF3 -Br N CF3 N CF3 N CF H H 3 CF3 (H CF3 N CF3 CF3 O 2. Li CO2H 3. CO2 0 7. [H] 5. C5H4NLi, -60 mefloquine 4. H 6. H O N CF3 3 N CF3 CF3 CF3

17.3 Mechanisms 402 • Chapter 17 Carboxylic Acid Derivatives

O O O -Cl -H NN: NN: 12. a. R Cl R H R H(H :C N N: H H :CH2N2

1. SOCl2 hQ b. -N2 2. CH2N2 CO2H C C O H NN: O H O

2. W-K 1. H2 / Pd hydrogenolysis of cyclopropyl bond O

cleavage here R H OR' H OR' H H N N 1. BrCN N -MeSCN N 13. N N -Br O ORH ORH O R' S S Br CN C N O R H 2. H3O O + H3N N iminium hydrolysis O R' (see 17.3, 7 for mechanism)

a carbinolamine OH O H H O N N N O H H2O2, H 14. N N (see 15.3, 34) N

O O O O O O A +H O N O H H N O N HO O -H O 2 -H H N N N H O O O O O O

17.3 Mechanisms Solutions • 403

Ac N OO O O O -pyH O 15. -HOAc O O N CO2H O H) O O O O Ac :py OO N O -CO2 - OAc O = Me N OAc O O O

S S O O N +H O N +H 16. H H N N O O H H CO2 CO2 R dicyclohexylurea RN CNR R = cyclohexyl N NHR H H + N PhO S S H -H O N N O O H) CO2 CO2 O OH ~H :P OH 17. HPOH R = -CH2CH2NH2 OH taut OH nucleophilic form O H H) O O O OH (H 1.:PCl3 +Cl :P(OH)3 RO PCl2 P OH -HCl RO -POCl2 RCl R PCl2 Cl O (H Cl Cl -HCl OH OH (H OH -H O H O POH O -H R P(OH) :PCl3 OH 2 2. +3H2O R PCl3 :PCl3 P(OH) 2 P(OH)2 R P(OH)2 H N O POH -3 HCl 2 O O O OH

O O O +H Me CO: 18. O O + an acylium ion OMe OMe OMe H

-H

O O O OH O O H) O O Me CO: Me Me

17.3 Mechanisms 404 • Chapter 17 Carboxylic Acid Derivatives

19. a. Carboxylate as a nucleophile: O O O O

O O ~H O + HOAc O O OH O O H H 18O-label appears in salicylate

b. Carboxylate as a base: O O O O H H O OH ~H O + HOAc O O OH O O no label! HO c. Therefore, pathway b is preferred.

:H HOH H NH2 1. -H O NC 20. 2 CO2Me N CO2Me N H H H OR

2. NaBH4, HOR O OMe 3. H N -HOMe H) N O N N H H

17.3 Mechanisms CHAPTER 18 CARBONYL Į-SUBSTITUTION REACTION AND ENOLATES

18.1 Reactions

1. LDA 2. n-PrBr 1. O O O

H O OMe 1. OMe O OMe 2. Ph Br 2. CO2Me OO MeOH OO O Ph O OO CO2H ' 3. H3O Ph -CO 2 Ph O O O O O O O 2. OEt 1. LDA EtO OEt 4. H O 3. OEt Et 3 NAS 3. EtI CO Et ' Et O 2

O O O CO Et O 2 1. base CO Et 3. H3O, ' -CO2 4. Ph 2 Ph OH Ph CO2Et 2. (PhCO) O -CO2 2 CO2Et

HO C NC CO2Et 2 Cl NO2 H CN CO2H 1. OEt NC 2. H O, ' 5. NC-CH2-CO2Et 3 CCO2Et N -Cl -CO2 nucleophilic aromatic subst'n NO2 NO2

H H NaCH(CO2R)2 6. H Me HC Me SN2 (RO2C)2 OTs H cis- trans-

O O O OO O 1. H3O 3. NaH 4. PhCH2Cl 5. H 7. 2. CrO , H 2 equiv 1 equiv 3 O O O Ph more reactive enolate Ph

18.1 Reactions 406 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

EtO C H CO Et CO2Et 1. OEt 2 2. 2 3. H O, ' 8. 3 O CO2Et EtO C EtO2C -CO OH 2 O 2 OH O

O O O O 1. (XS) OEt 3. H O, O 9. OEt -Br 3 ' 2. Br(CH ) Br -CO2 EtO O 2 4 CO2Et Br

O O Ph O O 1. LDA Se 2. PhSeBr -PhSeOH 4. MeOH 10. O H 3. H2O2 H OMe

O O O O O O 1. Br , H O O 11. 2 3. LiMe2Cu 4. H3O O O 2. KO-t-Bu O

O O O O 1. Br , H 12. 2 2. (CN)2CH: 3. H3O, ' OH SN2 -CO2 Br CH(CN)2 O

O O EtO C EtO C H 2 1. OEt 2 2. n-Pr Cl EtO2C H 13. 3. H3O, ' EtO C EtO2C 2 EtO2C NAS -2 CO2 O

O O O O 1. a. Br2, OH (-CHBr3) OH 2. PCl3, Br2 Cl 3. MeOH OMe 14. Br Br haloform rx H-V-Z rx b. H

H OEt H +EtOH H -H O 15. 2 H O -EtO OH

18.1 Reactions Solutions • 407

O O 1. LDA O Cl 16. O 2. Cl2, H -Br Br -H

OEt 1. HCl 2. 17. OO 3. H3O, ' Cl conj. add'n EtO2C -Cl -CO2 O O

O O O O O OH O 3. KMnO 1. Cl , H 1. Cl , H 18. 4 2 2 3. H3O 2. OH 2. E2 4. [O] O

Cl -Cl C(CO2Me)2 CO2Me 1. KOH 19. -Cl Cl CO2 LiH CO2Me EtOH Cl 2. ' :CH(CO2Me)2 1. SOCl2 O 3. mCPBA O 2. Me2NH CH2(CO2Me)2 LiH O

NMe2 NMe2

OO OO OO base 20. Ph O taut conj. add'n O O OPhO

18.2 Syntheses

O O O O 1. PCl , Cl 3 2 3. KCN 1. 4. H3O OH OH OH OH 2. H2O Cl SN2 CN CO2H H-V-Z rx

18.2 Syntheses 408 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

(CO Me) CO2Me 1. OMe, MeOH 2 2 CO2H 3. H3O -H O 2. 2 CO2Me 2. Br OH -CO2 OH OH OO

O O MeO2CCO2Me CO2Me 1. OMe 3. H O 3. 3 Cl 5. LiAlH(O-t-Bu)3 H CO2Me -CO2 Cl 2. Br CO2Me CO2Me H 4. PCl5 6. H O O

O 1. OMe O CO Me 2 2. 2-chloropentane CO Me H N NH 4. 2 5. 2 2 NH CO2Me (urea) 3. OMe CO2Me 4. allyl chloride -2 MeOH O N O H

O CO2Me 1. OMe Et CO2Me 3. OMe Et CO2Me 5. 5. H3O, ' OH CO2Me 2. EtI H CO2Me Bn CO Me -CO 4. PhCH2Br 2 2 Ph

. O 1. BH3 THF 3. CrO3, H 5. (MeO C) CHNa 6. 2 2 OH Cl 2. H2O2, OH 4. SOCl 2 NAS

OH OH O 7. NaBH4 6. H3O, ' O OO CH(CO2Me)2 8. H (-H2O) -CO2

3. H O (-CO ) 3 2 OH CO2Me 1. OMe CO Me 4. LiAlH Br 7. 2 4 6. HBr CO2Me O OH Br (CO2Me)2 5. H SN2 2. OMe

O O

1. NaOEt, HOEt 3. H2NNH2 4. NH3 8. OH, ROH Et Et 2. MeI Et OO H NO O O O O W-K 2

18.2 Syntheses Solutions • 409

O O O OO 1. OEt O 3. H O, 9. 3 ' OH Et OO -CO O O Et OEt 2 O 2. O O EtO OEt

O O OO 1. KOEt / EtOH 3. H 10. O Et -HOEt O O 2. Cl OH Et O OOHtransesterification

O O 1. OEt 4. NaBH 11. 3. H3O, ' O 4 Cl Et O O 2. -CO2 5. H EtO2C O O O 6. PCl3 Cl

O 1. OEt O O O 2. EtI Me 5. H3O, ' 6. I , OH 12. 2 O Et Et -CO haloform rx O O 3. OEt Et 2 4. MeI O O

+ HCI3

O O O O 1. Cl2, H 3. H3O 4. Ac2O 13. 2. KO-t-Bu (E2) OH OAc [or 3. HCl, 4. NaOAc (SN2)]

OH O O 3. KCN 1. Jones reagent Br CO H 14. 2 2. Br2, H 4. H3O

[or 3. ethylene glycol, H, 4. Mg, 5. CO2, 6. H ]

OO OO(H O taut O 15. -H +H

18.2 Syntheses 410 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

O OO 1. OMe OO 3. H3O 16. Me OO ' O O 2. Me O O O -CO2

O O O 1. LDA 3. H2O2 O 17. -PhSeOH 2. PhSeBr Se Se Ph Ph O OOH OH OH 4. H3O 6. CH2O, H 5. H2 / Pt O O

Cl O Cl O Cl O Cl O 1. NH2 Cl 2. Cl2, H 3. 18. AlCl3 Cl -HCl N-t-Bu H

OH CO2H CO H O O 2 1. KOH 3. Br , PCl Br 19. 2 3

2. 4. H2O CO2H H-V-Z rx O 5. SOCl2 ON O N 6. H2N-i-Pr H 7. LiAlH4 H OH 8. H Br

9. OH ((SN2)

S S S Na 1. NaOMe / MeOH O O CO Me 2 2. EtI 5. H N NH HN NH HN N 20. MeO OMe 2 2 6. NaO-t-Bu CO Me 2 3. NaOMe / MeOH thiourea O O (1 equiv) O O 4. 2-iodopentane -2 MeOH

NC NC H NC CN 1. NaNH2 MeN(CH2CH2Cl)2 NMe intra-SN2 21. Ph NMe Ph S 2 (-Cl , ) -H -Cl Ph N Cl O O 2. OH, ROH EtO 3. EtOH, H O NMe NMe Ph Ph

18.2 Syntheses Solutions • 411

18.3 Mechanisms

O O H O H OH H HO H +H (H H 1. -H +H -H O

OH OH OH OH OH O H OH O (H OH H

O O O O OO O OEt O 1. OEt 2. O -OEt 2. OEt O O Et -H Et O O O O O

IH II O O O O I2 I2 O O -H I O 3. PhCO2 + OH OH Ph Ph Ph Ph Ph Ph OH I Ph O II H) O Ph I3CH + PhCO2 I3C: I3C Ph O OH

(H O OH OH O O +H -H +H -H 4. H -H (H +H -H +H H H H achiral

O O O O +MeO H) OMe 5. + MeO -MeO O O MeO O O

H O H H O O taut O ~H -H O O 6. O 3 +H OH OH2 ~H O H H) H)

18.3 Mechanisms 412 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates

O O H) O O H O taut HO N NH -H OAA 7. CO2H + HO2C HO C HO C 2 H 2 biotin S CO2H

OH H OH OH OH H O H 2- H 8. taut -HPO4 -H H taut H O O O OH O (H O PO H O PO3H 3

H Me3N: O O H) Si TMS taut SN2 O -H O 9. Si Cl -Cl

18.3 Mechanisms CHAPTER 19 CARBONYL CONDENSATION REACTIONS

19.1 Reactions CO R O RO2C 2 1. OH 2. :CH(CO2R)2 O 1. PhCHO + PhCOCH3 Ph Ph -H2O Michael Ph Ph aldol HO2C O 3. H3O -CO2 Ph Ph O OH O H H H 2. H H taut aldol OOH

O O O OEt O 3. t-Bu-CHO t-Bu or Knoevenagel t-Bu O O OOH O

O O O OMe OMe 4. OMe - OMe -H O O O

CO2Me CO2Me CO2Me 1. OMe - OMe 2. H O 5. 3 OMe OMe Dieckmann O -CO2 O O O

O O O OH, ROH 6. conj. add'n retro-aldol OH O

O O HO OH 7. -H2O -H

OEt OEt O Et O Et

19.1 Reactions 414 • Chapter 19 Carbonyl Condensation Reactions

O H 1. BrCHCO2Me -Br O 8. CO2Me CO2Me O O Br O H 2. H3O, ' taut -CO2 H OH H) OO O O O O 1. base O NAS 9. MeO CHO MeO MeO CO2 O O (H CO2H 2. acid CO2 OO MeO MeO -CH3CH2CO2H

OCH3 OCH3 OCH3 base 10. -H2O Michael O O aldol (Robinson annulation) O O O OH OH 11. O NCH O NCH PhCHO NO NO2 2 3 2 2 Ph 2 or Ph

O OMe, MeOH 12. MeO OMe CO2Me retro-Claisen CO Me MeO 2 O O O

O O O O 1. NaOEt -H O 2. NH NHPh 13. 2 2 conj. add'n aldol OH NHNHPh

O O CO2Me O O O O 1. OMe -Cl 3. H O, ' 14. 3 aldol 2. Cl Cl O O -CO2 -H2O O O O OH OH

O O O O O OH 1. :CH(CO2Me)2 -MeO 15. O 2. H3O Michael -CO OMe O 2 O mesityl oxide CO2Me CO2Me dimedone

19.1 Reactions Solutions • 415

HO CHO CHO OH -2 H2O 16. O O O O CHO HO HO

OH OH H OH O H OH 17. retro-aldol taut O + HO HO OH O H OH O H (H H O O

1. 2. H3O 3. NaOH 18. N N O O O

O O O O O HOCH HO + mixed again 2 again 19. H H H H H H aldol CH OH CH OH 2 2 OH O OH O HO H H H OH HCO2 +C(CH2OH)4 mixed Cannizzaro

OH O O OH O

~H Ph H -H 20. taut H -H2O mixed aldol HO Ph Ph

OH CHO CO2Et CO2Et CO2Et 21. 1. :CH(CO2Et)2 2. H CO Et O OH 2 -H2O O -HOEt OH OEt OO H transesterification

O O O CHO OHC KOH again 22. + + -2 H O CHO 2 OHC aldol O O O

OH O O OH O O 1. 2. H O 23. + 3 OH SCoA SCoA CO2H CO2H CO2H

19.1 Reactions 416 • Chapter 19 Carbonyl Condensation Reactions

OMe MeO CHO CN OMe 1. OEt 2. H3O 24. + MeO

OMe HO OMe CN O (H OMe O OMe -CO2 MeO MeO -H2O HO OMe OMe H

O H OH O O 1. H O 25. 3 2. KO-t-Bu taut conj. add'n retro-aldol H H

O (H O O

CO2 CO2 O taut O C O 2 26. retro-aldol CO2 O CO2 CO H 2 O HO O H aldol HO H) H O hydro- CO2 O O CO2 O lysis CO2 O2C SCoA SCoA

O O O O O C OEt taut 27. C Michael EtO (H O O O O O O

-H2O aldol

O O OH

O H H) O O O OH HOH taut ~H 28. OH OH OH OH OH OH aldolase OH OH OH

19.1 Reactions Solutions • 417

OH CO2H O O CO2 CO H3O 29. 2 + O C HO CO H O2C SCoA 2 SCoA 2

O CO2H

O OH O OO H H3O [O] 30. SCoA SCoA SCoA S O O OH H) CoA + taut SCoA SCoA SCoA retro-Claisen

O O OH O [H] O Claisen H3O OH O C 31. SCoA -CO CoAS O HO O 2 A B

O O HO H 1. H / Ni (1 equiv) H 32. a. D-A 2 + O OH H H EtO 2. a. LiAlH4 EtO O b. H EtO a vinyl ether 1. H3O HO HO HO H H H aldol 2. MVK, base OH OH OH H -H2O H H O Michael O O O

R X R R b. 1. , t-BuO 3. +H C O 2. a. EtO CCMgX C -H O C OH OEt 2 C OEt b. H

R R R taut +H2O H C OEt -H C C C C CO2Et OEt OH

19.1 Reactions 418 • Chapter 19 Carbonyl Condensation Reactions

O O O O 1. a. OsO4 2. KIO c. OH OH 4 b. NaHSO3 OTs OTs OTs O O O O 3. OMe -OTs -H H OTs

O O I O I O O 1. RO2CCO2R O 2. I2, OH O d. O

OR CO2R CO2R 3. OH O I O O I I I O O O OH ' O C OH C O -CO ,-I O 2 O O

~H taut O O O I I OAc O O O -CO2 taut C O O 4. KOAc SN2

OAc OAc OH NC O O OAc 1. HCN O CN 2. -H O O e. 2

3. KMnO4, OH O OAc NC OAc O OH O O (H 4. H3O cortisone -HCN OH

O cortisone acetate

O O CO2Me O 33. 1. - OMe 2. H3O, ' Dieckmann OMe -CO2 CO2Me

19.1 Reactions Solutions • 419

HOMe O OH O OH O Na N N OMe H O -H O 34. O2NCH2 O2NCH2 H) (H O2N NO2 NO2

O O O O

O Et3N O +H 35. O O -H Michael taut

19.2 Syntheses

O O O OH 1. H3O 2. H / Pt 1. + 2 Ph H Ph Ph Ph Ph Ph -H2O

O O N 1. H 2. N 2. + N -H2O H OO O 3. H3O

O O O H OH, ROH H 3. H H H OH O O

OH H O O O 4. O 3 -H2O

O O HO 1. OH (aldol) 2. H / Pd 3. H SO (E1) 5. 2 2 4 -H2O 4. H2 / Pd

19.2 Syntheses 420 • Chapter 19 Carbonyl Condensation Reactions

1. LDA O O HO O O 2. MeI O 5. Ph P: O 6. 4. OH 3 3. Cl2, H Cl H (protect) Cl 6. MeLi PPh3 O O 7. acetone - via a Wittig, not a mixed 8. H3O aldol! O Wittig

O O O CO2R 1. NaCH2CO2R OR 2. OR, HOR 7. CO2R Claisen OR Dieckmann CO2R O O

O O O O 1. Cl2, H 3. H3O 8. 5. NaH (2 equiv) 2. KO-t-Bu (E2) 4. CrO , H 3 O O O O 7. H 6. PhCH Br Ph Ph 2 (1 equiv) O O

O 1. BH .THF N O 3 4. N , H 2. H2O2, OH H H 5. Cl 9. H H -H O 3. PCC 2 6. H3O

OH OH O 1. CH2O, OH O 2. NaBH4 OH 10. H mixed aldol H 3. H

O O O OR, ROH OR, ROH 11. + Michael -H O, aldol O O O O 2 O

O O O 1. O 3 3. KOH, EtOH 12. -H2O 2. Zn, H aldol O OH

19.2 Syntheses Solutions • 421

1. Br , H O 2 5. O 2. ethylene glycol, H O O 13. PPh3 3. Ph P: (S 2) O 3 N 6. H3O 4. n-BuLi - the reaction of cyclohexanone with acetone via an aldol would yield four possible products!

O O O O O 1. OR, HOR 3. , OR OR, -H O 14. 2 2. MeI Michael aldol O O O O O

O OH 1. O3 H 3. H O -H2O O 15. H 2. Zn, H aldol O H H

O O O O O OH, HOR aldol 16. 3 aldol, -H2O conj. add'n -H2O O

19.3 Mechanisms

O OH O OH OH

1. +H -H taut taut retro-aldol HO H) O O O O O O OH (H -H ~H +H -H2O aldol OH2 OH

O O 1,2-R: shift O 2. +N2 NN: :CH2 NN:

19.3 Mechanisms 422 • Chapter 19 Carbonyl Condensation Reactions

O O O H) OEt O retro- 3. O OEt CO2Et Claisen O OEt O OEt H) OMe OMe 4. OMe 1. conj. add'n 1. again N O O NH O Me Me OMe :NH2Me CO2Me O O O O OMe 2. OMe OMe MeO - OMe -H Dieckmann O N N Me Me

O O -Br 5. Cl Cl Cl O H CBr2 Br Br OH Br HO CHBr3 Br3C: OH OH Cl Cl O -HBr O -Br O HO Br

CO2Et CO2Et CO2Et CO2Et 1. OEt, aldol -H O 6. 2 O O O O O Michael HO O (H

H) OH O O 2. H2O, OH

3. ' O O -CO2 O

O O OH O O O H 1. 7. H -H2O 2. H aldol O O (H -H

OO O O OO retro- Claisen 8. OEt + EtO Claisen OEt EtO EtO Ph O - OEt OEt Ph

19.3 Mechanisms Solutions • 423

OH OH O O O O 9. OH, Michael O aldol CHO CHO HO OH HO O O (H H O O ~H - OH

-HCO2 RO (H

HO HO HO HO H (H O OH OH O 10. H taut H OH OH OH taut OH OH O O OH (H OH OH OH OH OH OH O H A O (H H

D-D-fructose

OH OH OH -H HO HO H O OH retro- H (H OH O HO + aldol O O HO H ~H H O (H OH taut OH OH A OH

R NCOR O O R2NCO2R R2NCO2R 2 2 R2N OR R2N OR 11. (H -H Br Br (H -Br Br Br N -H N NH NH NH

RO C Ph O 2 Ph OR O base Ph O - OR -H Ph CO2R 12. O Ph O RO C Ph Ph RO C Ph CO2 2 2 (H CO2R :B

O O O 1. OMe 13. O

O O O H) OMe O

19.3 Mechanisms 424 • Chapter 19 Carbonyl Condensation Reactions

OH OH 2. H -H taut 13. OH O (cont.) O O (H O O

O O OR OR CO2R retro- ~H Dieck- 14. O H OR OR O mann O Claisen OR CO2R OR O OOR O O O O

EtO OEt EtO EtO 15. -H O O O EtO O EtO O H) O O O EtO (H O O

O O

1. a. O3 2. OH 16. b. Zn, H O O O O

HO (H aldol -H2O

O O H) OH

NH, H 1. +H 17. -H N O -H2O N N OH (H H A - a dienamine

2. EtI (SN2) 2. EtI (SN2)

N 3. H O A 3 O O 3. H3O N A Et I Et Et IEt

O O O HB O

:B taut 18. Michael Michael

OMe CO2Me CO2Me MeO O O

19.3 Mechanisms Solutions • 425

O O O O 19. 1. NaH 2. H2O

MeO (H O MeO HO MeO MeO O O O OH O OH O OH O R' R' R' R taut, +H2O R retro-aldol- R 20. OH like O HO O R' = H H O O(H HO OH HO R' HO R' H) OR' O O (H O O OH O aldol- like R' R R' R' R HO taut R -H2O HO O OH HO O O (H R' O H O OH R' R' H

NMe2 :N C: N CN Me I 2. -I 3. -NMe3 21. a. Mannich b. R R R N SN2 N SN2 N H H H CN O O H NH2 N 4. LiAlH4 5. H OEt , H c. R R R CO2Et N [H] N -H O N H H 2 H

H O OH OH (H H OH (H 1. H O O 22. PEP 3 OH HO2C H -H PO HO C 2 4 2 2. aldol OOHOH OH 3. -H2O, taut O O H) H) O OH CO2H CO2H -H2O 4. H HO2C HO CO2H HO HO OH taut aldol O OOHO HO HO H HO H

CO2H CO2H CO HCOH CO2H 2 2 CO2H CO2H CO2H H, -H2O 23. H) +H -H O aldol O -H2O (H N H H N O NH N H N 3 2 H N H H 3 H3N H 3 H3N

19.3 Mechanisms 426 • Chapter 19 Carbonyl Condensation Reactions

O O O O AcO H) O O -HOAc O 24. a. O(H O -HOAc NO N NH O Cl H) Cl O O O O Cl ~H -H -Cl O O O H) N N N H Cl

O O O O +MeOH OMe b. HS O HS O HS O OMe - OMe HS N N N N taut MeO (H H O

OO O O OO -CO2 - SR 25. OSR RS SR SR

R' R' R' O H R' R' R N S R N S - R N S 1. HR 26. S 2. R N S R N OH OH O (H -H + HOH HO R O O H) O O O (H -H H R H H OH

O O OH O OH O Claisen aldol- partial 27. a. SCoA SCoA SCoA SCoA like -HSCoA O(H SCoA hydrolysis O CO2 H SCoA O

PO H O 3 2- -CO2 2- 2- P2O6H P2O6H ~H P O H b. O O O 2 6 -HPO 2- 4 (H O H O

19.3 Mechanisms CHAPTER 20 AMINES

20.1 Reactions

O O O 1. base CO2Et (CO Et) 1. NH 3. base 2 2 N H N 2. ClCH(CO2Et)2 CO Et 4. 2 Cl O S 2 N O O O CO2H CO2H 5. H O phthalic acid + 5. 3 N CO H NH 2 3 -CO2 O Cl O NH2 OH 1. (XS) CH I OH O 3 3. OsO4 O Cl 2. 5. (XS) COCl2 2. Ag O, H O, 2 2 ' 4. NaHSO3 -2 HCl O

O NH2 O O 6. NH3 O -2 HCl NH2

Me Me Me N N N H 1. (XS) MeI OH 3. ' Me 3. + H2CCH2 2. Ag2O, H2O -H2O

1. (XS) MeI OH 4. 3. ' N 2. Ag2O, H2O -H O H N 2 N

HO HO HO

OH 1. (XS) CH3I ' 5. O O O N 2. Ag2O, H2O N -H2O N

HO HO HO

H NPh Ph2N O 2 Me 1. H2O2 6. Me Me H 2. ' H HMe H) + Ph NO Cope Et 2 Et Et Me (cis-elimination) Me

20.1 Reactions 428 • Chapter 20 Amines

O O 1. SOCl 2 3. ', -N 7. CO2H 2 NNN NNN NCO 2. NaN3 Curtius O H2O -CO2 NH2 N OH H

OH OH OH OH 1. (XS) MeI OH OH taut 8. 2. Ag2O, H2O, ' NHMe HO HO O

D H D Hofmann elimination anti-periplanar H H 9. NMe 2 NMe3 H D D Cope elimination syn- (H

N O

O H +H N -H 10. Me2NH + HH N N -H2O H OH O O taut OH

NH2 N N OH O HONO H O 11. N N 2 N taut HN -N O N O N 2 O N O N H H H H

1. Li 12. Ph 2. CO2 4. SOCl 6. Br , OH Cl Ph 2 Ph NH2 2 CO2H Ph 3. H NH2 5. NH3 O H2O

OH OH Br , OH O NH 2 13. 2 NH3 + Me Me NH2 H2O Me H O acarbinolamine

20.1 Reactions Solutions • 429

1. NaNH 2 3. PBr 4. Me NH 14. Ph CHOH Ph2CH O 3 Ph2CH Br 2 Ph CH NMe 2 O O 2 O 2 2. O O

N O

1. HBr 2. O NH 15. + ROOR Br NH2 NH 3. H2NNH2 O H Me2N (H O OH O 1. MeI 3. H 16. 2. Ag O, H O, ' pinacol-like MeO 2 2 rearrangement MeO MeO

Ph Ph

OH NH2 1. SOCl2 3. Br2, OH 17. O 2. NH O OH 3 OH H2O Ph Ph 2 I 4. N NH2

OH OH

CO H CO2 N CO2H 2 1. a. Br , PBr 2 3 C H2N CO2H b. H2O (H-V-Z) 3. HCN, CN 4. H / Pt 18. 2 2. KO-t-Bu (E2) conj. add'n

Br Br Br NH2 NH N 1. Br2 2 2. NaNO 2 3. KI I 19. 2 HCl -N Br Br 2 Br

CN CN CN CN 1. Cl2 2. NaNH 3. KNO , H CO2H 20. 2 2 5. H3O FeCl3 (NAS via 4. CuCN benzyne) Cl NH 2 CN CO2H

CO2H CO2H CO2H 1. KMnO , H 4 3. Fe, HCl 5. HBF 21. 4 2. fuming nitric 4. NaNO , HCl 2 F acid NO2 N2

20.1 Reactions 430 • Chapter 20 Amines

Cl 1. Br2, Fe Cl Cl NHAc 2. Cl2, Fe NH2 4. ICl, Fe N2 22. 5. H3PO2 3. H O, OH 5. HONO 2 Br Br I Br I

CO2Me O 1. NaH 23. N 2. H O N 3 O -H2 O N -OMe N O N -CO Me Me 2 Br Me O NHMe N 5. ' 3. NaBH4 Me -HBr 4. HBr N N H)N N Me

HO O HO H 1. H OH reductive 24. O NC O OH NH NaBH3CN amination N

HO HO HO

O OH 2. HCl N H Cl HO

20.2 Syntheses

1. Br2, hv 2. Mg 5. SOCl2 NH2 7. Br2, OH 1. CO2H NH 3. CO 6. NH 2 2 3 O H O 4. H 2

Cl CN 1. Cl2, ' 2. KCN 3. LiAlH4 2. NH2 SN2 4. H

4. potassium 1. Br , hv 3. 2 3. HBr phthalimide 6. (XS) MeI 2. KOH (E2) R2O2 Br 5. H2O, OH NH2 7. Ag2O, H2O, '

20.2 Syntheses Solutions • 431

1. KMnO4 2. SOCl2 4. OH 4. Br2, OH NH2 H 3. NH3 NH O H2O 2 O

OMe OMe OMe 1. (XS) MeI OMe MeO OMe 2. Ag O, H O, MeO OMe MeO OMe MeO OMe 2 2 ' 5. H NPh 5. 2

3. O3 NaBH3CN NH2 4. Zn, H OH H NPh NHPh

N K

1. NBS, ROOR 2. 6. Br O NH2 O2N O2N 3. H2O, OH O2N

O CO2H 7. 1. KMnO4, H 2. PCl3 NH2 NH2 4. Br2, OH O N 2 3. NH3 O2N O2N H2O O2N

1. (XS) CH I 3 3. O3 5. HO OH O 8. O NH2 2. Ag2O, H2O 4. Zn, H H O

1. SnCl2, H HO SNO 3. PhNEt2 9. 3 2 HO3SN2 HO3SNNNEt2 2. NaNO2, HCl

N O CO Me N O 2 N 4. MeI 1. H O 10. 3 OH 3. ' 5. Ag2O, H2O, ' 2. CrO3, H -CO2 6. repeat 4. and 5. O Ph O O (double Hofmann) O

1. HONO , 2 Cl Cl Cl Cl H SO 3. SnCl , HCl Cl Cl 11. 2 4 2 5. H3PO2

2. Cl2, Fe 4. NaNO2, HCl NO2 N N

20.2 Syntheses 432 • Chapter 20 Amines

NH2 1. Ac2O NHAc NHAc 12. 3. KNO2, H

O2N 2. SnCl2, HCl H2N 4. H2O HO

NHEt2 1. CrO , H 13. OH 3 Cl NEt O Et2N Et2N 2 2. SOCl O O 2 -HCl NH Fe, H

NO2 NH2

NH OMe 1. HONO2, 2 OH H SO 3. NaNO , H 5. KOH 14. 2 4 2

2. Fe, HCl 4. H2O 6. MeI

NO2 N2 1. SnCl2, HCl 3. PhNH2 15. NN NH2 2. KNO2, HCl -H EAS

H2NMe (-H2O) 16. [H] O N NHMe NaBH3CN, H Me

(alternatively, 1. H2NMe, 2. H2 / Pd)

H H N N N Br O O 1. mCPBA 2. (XS) MeBr 17. S S SN2 S O S O S O S OH OH OH O O O

1. HONO2, H2SO4 NHAc 2. SnCl2, HCl 4. HONO2, H2SO4 NH2 18.

3. Ac2O 5. OH, H2O O2N OH NNO N 2 7. N2 6. NaNO2, HCl OH

O2N

20.2 Syntheses Solutions • 433

NO 2 NO2 3. Fe, HCl 1. HONO2, 4. KNO2, HCl 19. 2. DCl H2SO4 EAS DD5. H3PO2 DD

OH OH OH OH

1. propylene, H 2. H O 3. KNO2, HCl 20. 3 F-C alkylation -HOAc 4. H3PO2 NHAc NHAc NH3

20.3 Mechanisms

O O O H OH Br Br 1. -HBr Br N N N H (H -Br O N C ~H O H OPh NC NCO taut O H) Ph HO Ph

O O OH CO 1. Cl 2 2 -H 2. NH NCl OH, H2O N O Cl -Cl O O CO H 2 CO2 CO -CO2 OH 2 2. H NH2 N CO2 taut H NCO

O CO2H CO2H 3. 1. HONO 2. pH 8 O -H -CO NH2 NN NN 2 -N2

3. D-A

20.3 Mechanisms 434 • Chapter 20 Amines

O HO CN HO NH2 1. HCN 2. H2 3. NaNO2 4. CN Pt HCl H) O O HO HO N2 -H -N2

OH OH2 OH2 O N N 1. NH2OH 2. H -H2O 5. N -H2O O OH OH2 taut -H +H2O NH N N

MeO MeO MeO XMg 1. PhMgX 6. O XMg O XMg O NMe NMe NMe

MeO MeO MeO

MeO MeO MeO Ph Ph HO 1. NMe XMg O NMe XMg O 2. H NMe

MeO MeO MeO

MeO MeO Ph Ph Hofmann: HO 1. (XS) MeI HO NMe NMe 2 2. Ag2O, H2O, '

MeO MeO

Cl SN2, -Cl ' N 7. NNN NNN NNN -N2

20.3 Mechanisms Solutions • 435

H) N3 OH O O HO H O NNN NNN NNN -N 8. 2 N ~H NH '

Ph +H (H 9. ~H N NH2 N N Ph NH Ph H H H Ph H 2 N H N H N ~H H H H H) Ph Ph -NH4 N N NH3 H H

10. +H ~H EAS -H NH 2 NH2 -H2O NH NH NH H H H2C H H OH O

O (H O H O O 11. -H CH2O, H Ph CH3 taut Ph CH2 Mannich Ph NH2 Ph N CH O + NH H 2 3 H2CNH2 H)O O H)O O O O O -H Ph , -H CH2O, H Ph N Ph Ph N Ph Ph N Ph 3 (again) (again) H CH2

taut acetone dicarboxylate H)O OH OH CHO CO2 ~H N ~H - OH CO -H (H N N 2 12. H CHO H2NMe O OH OH O N N N O2CCO2 CO2 CO -2 CO2 -H 2 - OH N O(H taut O2C OH O O CO2

20.3 Mechanisms 436 • Chapter 20 Amines

OH2 +H OH2 -H 13. a. NCMe N NHAc NCMe Me taut

O O +H b. -HOAc O O NOMeC H O +H2O, -H OMe NOMeC t-Bu N taut H OH2

O O H O Ph NH2 1. OH 2. CH I 14. Ph S Cl Ph S N Ph S NPh 3 -HCl Ph S 2 O O O N CH3 O 3. H2O, OH CH3 PhSO3 + N Ph S N H O Ph

(H :B O O O R 2. -HCl Cl 3. -Cl 3. -HCl R O 15. a. R CCl2 N3 Cl NAS Cl R Cl N3 OMe NNN O H) Me

H H N 1. :CCl Cl O N b. O 3 2. -HCl -HCl O -Cl Cl O Cl O O OH H NH H)

R H RH RH N N O N N O N N O -H 16. NH NH +H NH N N N H) H O C O OCH H C C N N H C H C H CC Ph Ph H H N H Ph

20.3 Mechanisms