Workbook for Organic Chemistry : Supplemental Problems and Solutions

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WORKBOOK FOR ORGANIC CHEMISTRY SUPPLEMENTAL PROBLEMS AND SOLUTIONS

Jerry A. Jenkins Otterbein College

W.H. Freeman and Company New York

© 2010 by W.H. Freeman and Company All rights reserved. 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 About the author vi | Acknowledgments vi | Selected concepts/reactions locator vii TIPS viii | Common abbreviations ix

v

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

1

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

11

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

15

CHAPTER 4

21

REACTION BASICS

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

27

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

49

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

55

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

65

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

69

CHAPTER 10

87

NMR

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

93

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

103

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

117

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

129

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

139

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

167

CHAPTER 17 CARBOXYLIC ACID DERIVATIVES 17.1 Reactions 177 | 17.2 Syntheses 186 | 17.3 Mechanisms 193

177

iv • Table of Contents Workbook for Organic Chemistry

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

201

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

209

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

229

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

351

ETHERS

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

413

CARBONYL CONDENSATION REACTIONS

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 ester syntheses) Brønsted-Lowry/Lewis equations Carbocation rearrangements cis-, trans- (geometric) isomers Claisen, Cope, oxy-Cope rearrangements Conformational analysis Curved arrow notation Degrees of unsaturation (units of hydrogen deficiency) Diazonium ions Diels-Alder reaction Enamines, synthesis of Enamines, reactions of Epoxides, synthesis of Epoxides, reactions of Free radical additions Free radical substitutions Hydrogens, distinguishing different Isocyanates, ketenes Kinetic isotope effects Kinetics, thermodynamics Neighboring group participation Nitriles Organometallics (Grignard, organolithium, Gilman), synthesis of Phenols Polymers Reaction coordinate diagrams Reaction types/mechanisms Resonance Thiols, (di)sulfides UV/VIS spectroscopy

18 1 5 3 14 2, 3 vi, 1 5 20 11 15 19 5 14 5 8 2 17 9 4 9 16 8 12 5 4 4 1 14 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

O

O H

H

O

right:

O H

wrong:

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

O H

O H

right:

O H

wrong: (resonance arrow)

(equilibrium arrows)

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

:H

O

O

+

H2

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

X

+

H

H X

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 OH H3O wrong H2O

-H

OH2

right

and a proton is not directly available in base. H

O OR H

O

O

OH H

ROH H) OR

+H wrong +ROH, -RO right

O H

Supplemental Problems and Solutions • ix

COMMON ABBREVIATIONS The following abbreviations and symbols are used throughout this workbook:

Ac AcOH * B: Bn Bu CA CB ' D-A or (4+2) DB DCC DIBAH DMF DMSO EAS ee equiv Et F-C [H] ~H+ HMPA HSCoA hQ H-V-Z inv L LDA mCPBA Me NAS NBS NGP NR Nu: [O] PCC Ph Pr py Ra-Ni ret rds taut THF TMS Ts TsOH TS W-K X (XS)

acetyl (CH3CO-) acetic acid chiral center or isotopic label base benzyl (PhCH2-) butyl (C4H9-) conjugate acid conjugate base heat energy Diels-Alder double bond(s) dicyclohexylcarbodiimide diisobutylaluminum hydride dimethylformamide dimethyl sulfoxide electrophilic aromatic substitution enantiomeric excess equivalent(s) ethyl (CH3CH2-) Friedel-Crafts reduction proton shift hexamethylphosphoramide coenzyme A light energy Hell-Volhard-Zelinsky reaction inversion of configuration leaving group lithium diisopropylamide m-chloroperbenzoic acid methyl (CH3-) nucleophilic acyl (or aryl) substitution N-bromosuccinimide neighboring group participation no reaction nucleophile oxidation pyridinium chlorochromate phenyl (C6H5-) propyl (C3H7-) pyridine Raney nickel retention of configuration rate determining step tautomerization tetrahydrofuran tetramethylsilane or trimethylsilyl tosyl (p-toluenesulfonyl) tosyl acid (p-toluenesulfonic acid) transition state Wolff-Kishner reduction halogen excess


PROBLEMS


CHAPTER 1 THE BASICS 1.1 Hybridization, formulas, physical properties 1. SeldaneTM is a major drug for seasonal allergies; RelenzaTM is a common antiviral.

HO

a

OH

c

OH O

HO N

2

OH N H O

b

O OH d

NH NH H2N

SeldaneTM

RelenzaTM

a. Complete the molecular formula for each. SeldaneTM: C___H___NO2 RelenzaTM: 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:

c.

b. H C O:

:O N O:

d. the conjugate base of NH2CH3

Cl e.

O

N H

f.

O

O H

zingerone (a constituent of the spice ginger)

BenadrylTM (antihistamine)

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 4. a. Which has the higher bp?

N H

or

N

OH

b. lower mp?

or catechol

HO

OH

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

a.

b.

H

H

d.

c.

H H H3C

f.

e. (CH3)3C all four carbons

H

C C C

CH3

g.

h.

H3C

H

H C C C C

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:

O

a H N d

b S

N

N

O

HN CO2H

C

N c

S

N H

N H

N cimetidine

penicillin V

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: ________

Problems • 3

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

O MeHN S O

O

NMe2 HO sumatriptan

OH 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 H

O

N

H

N

RoseremTM

H N

NH

O O

ChantixTM

Ritalin TM

ChantixTM ___________

RitalinTM ___________

a. What is the molecular formula for each? RozeremTM ___________

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 HN

N N CH3 theobromine

O

NH2

CH3 N

N H2N

N N

NH2

melamine



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

a.

O

HO2C

O N H

OH

O

O O

C CH

F c.

b. N

N NH

NH2

TamifluTM (antiviral)

HO YasminTM component (OCP)

CiproTM (antibiotic)

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

b.

Ph3P:

c.

N

+

+

AlCl3

BF3

O

+

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.

H3C O

b.

H2C CH2

1.2 Acids and bases

CH3CH2 Cl:

+

+

BF3

CH3 O CH2CH3

CH2 CH2 BF3

+

Cl

Problems • 5

c.

H3C O H

d.

:Cl Cl:

e.

+

+

+

Cl

AlCl3

+

CH3 N C S :

H3C O

:CH2 CH3

H3C CH3

AlCl4

S +

:NH3

CH3 N C NH3

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

O

a. Calculate the molecular formula:

Ha Hb C C

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.

CO2H

HO

NH O

A

B

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

1.2 Acids and bases


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 C C Ha ? a. hydroxide

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. OH

CH3NO2

HO

nitromethane

estradiol

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

Cl N H

b. The pKas for pyridinium 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

H3C C O

N

H

2. Which species has the lower pKa, H C N

O

or

H O C N ?

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

HO CH NH2

b.

O c.

d.

O

H2C

O CH3

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? H3C NH C NH2 NH

1.3 Resonance


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

O O O ozone

b.

OH H c.

d.

C C N: H

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

a.

b. N H

CH2

c.

d. O

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

Ha

O

O A

8. Either oxygen 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.

c.

b.

N

N

H

O d.

e.

f.

O

N

.

g.

.

CH3O h.

CH2

i.

H 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.

HO

N

N

S

S

luciferin

CO2H ATP, O2 luciferase -CO2

N

N

S

S

O +

HO

hv

oxyluciferin

1.3 Resonance


CHAPTER 2 ALKANES 2.1 General 1. Which compound has the highest mp? 1. n-octane

2. 2,5-dimethylhexane

4. bicyclo[2.2.2]octane

5. all have the same number of carbons and would melt at the same T

3. 2,3,4-trimethylpentane

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 compound. Although both are C20 hydrocarbon alkanes, one melts at 4200 and the other at 370. 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 s-Bu t-Bu

2. Br

4.

3.

Et

5.

n-Pr

6. F

7.

8.

i-Pr 9.

10. isohexyl iodide n-pentyl

i-Bu 12.

11.

t-Bu

n-Pr neopentyl

Cl

Give the correct IUPAC names for problems 13 – 16. 13. 2-isopropyl-4-methylheptane

2.2 Nomenclature

14. 3-(1-methylbutyl)octane

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). s-Bu

OH a.

H

Me

H

Et

H

t-Bu

b.

H

H Me

(R-OH = alkyl alcohol)

2.3 Conformational analysis, acyclic


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).

PE

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.


CHAPTER 3 CYCLOALKANES 3.1 General 1. Which compound has the highest molecular dipole moment (u)? a.

Cl

b. anti conformer of 2,3-dichlorobutane

c. C2Cl2

d. cis-1,3-dichlorocyclobutane

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.

2.

isoamyl

3.

3.2 Nomenclature

(three names!)

4.

Problems • 17

6.

5. Br

F 7.

8. Cl

t-butyl I 9.

10. neopentyl

F 11.

13.

Ph

12.

14. roof-methylhausane (!)

15.

16.

3.2 Nomenclature


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

b.

"flipped" conformer

n-Pr

i-Pr

"flipped" conformer

(if i-Pr is equatorial)

H H Me

c.

"flipped" conformer

Et

3. Which has the most negative heat of combustion ('Hcomb) in each of (a), (b), or (c)?

a. t-Bu

3.3 Conformational analysis, cyclic

t-Bu

t-Bu

Problems • 19

b.

Me

Me

Me

c. Et

s-Bu

Et

Et

s-Bu

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-tbutylcyclohexane?

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



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

O

OH

HO

OH

O

HO

OH

and 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?


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 O

OH b. OH

O

c. RCO2R + NH2R

O

d.

H

H

+

RCONHR + HOR

NMe

H2NMe

+

H2O

H e. N

+

O

Me2NOH

O

O f. OH

g.

RHC

h.

i. PhCO2H

OH

OH

CHR

+

RC

H2O

CR

OH

PhCHO

j.

k. BnOH

PhCHO

4. Reaction Basics

22 • Chapter 4 Reaction Basics

l. Ac2O

+

m. CHCl3 +

H2O

2 AcOH

KO-t-Bu

n.

+

o. Br2

2 Br

p.

Ph

q.

:CCl2 + t-BuOH + KCl

HC CH

Ph

Ph

OH

Ph

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

rx

4. Reaction Basics

Problems • 23

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

'Go

rx

b. The overall Keq for the conversion of RCOCl to RCO2R’ could be calculated from 'Go 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: (1)

Br3C

:B

H

Br3C:

+ H-B

A

-Br

(2) Br3C:

Br2C: B

Br (3) Br2C:

Br

+

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

rx

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

4. Reaction Basics


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

O

O +

H

O

O

O

H OH2

+

A

+ H2O

H OH2

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 = ____________________________________ If Keq is known, then 'Go = __________________________. 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

rx

6. Consider the following reaction:

I

+

MeOH

OMe

+

HI

A

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.

b. hausene

D

D2 / Pt

c.

D

+H

d.

+Cl H

Cl

e.

=

Br Br

Br

f.

+Br

Br Br

-Br

Cl2

g.

+

hv

H

S

Et -Cl

h.

HCl

Cl

H S

Et

-H

S

Et

Cl

4. Reaction Basics


Ph

Ph

Ph

i.

H BH2

H BH2

:B H) j.

O

O Cl

4. Reaction Basics

-BH, -Cl

H

BH2

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

a.

CH2OH

Ph

NC

vinyl

H2NH2C

t-Bu

b.

O

NH

c.

NH2

d.

O O

O

O O

O

O Ph

NH2 e.

SH

f. CH2F

H 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). a. The antidepressant fluoxetine (ProzacTM), C17H18F3NO, when treated with H2 / Ni gives a structure with molecular formula C17H30F3NO. It contains no triple bonds. How many rings are in fluoxetine?

b. CiproTM 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.)

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

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

c. styrene bromohydrin

b. propylene dichloride

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

rx

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


12. Which reaction demonstrates NEITHER regiospecificity nor stereospecificity? HF

a. trans-2-pentene

b. 1-pentene

Cl2 (XS) NaBr

Cl2

d. 1-ethylcyclopropene

c. cyclobutene

D2 / Pt

H2O

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

H

E-pinene

D-pinene

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

1.

HCl Ph

NO2 2.

3.

HI

H3O

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

5.2 Reactions

Problems • 31

NMe3

HI

5. Et

HF

F

6.

DCl

7. vinylcyclohexane

HBr

8.

H

OH

9.

O (complete)

HBr

CCl3

10.

Et 11.

DBr

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

5.2 Reactions


H,

13.

EtOH

HF

14. MeO

D

Cl

HI

15.

Cl2 / H2O

16.

1. B2D6 17. 2. H2O2, OH

Cl2

18. propylene

(XS) NaI

Et

1. Hg(OAc)2, PhOH

19. 2. NaBH4

20.

H2C C CH2 allene

5.2 Reactions

(XS) CH2I2 Zn(Cu)

Problems • 33

1. KMnO4, 21.

AcO

OH

2. HIO4

1. O3 22. 2. H3O, Zn

23.

HBr, di-t-butyl peroxide Ph

24. (E)-3-hexene

diazomethane hQ

25. cyclopentyl bromide

1. base 2. OsO4 3. NaHSO3

26. 3-methyl-1-butene

IN3

1. BD3, THF

27.

2. H2O2, OH

HO cholesterol

O

H, MeOH

28.

5.2 Reactions


29.

H

(complete)

30. styrene glycol

OH 31.

HIO4

1. H2SO4 2. KMnO4, OH

H C C O

32.

HCl

H

OR

O O

1. BH3, THF

33.

O 2. H2O2, OH

O O artemisinin (antimalarial)

1. Br2, H2O

estrone

34. 2. base HO

H

35. O

(complete)

5.2 Reactions

Problems • 35

1. 36. chlorocyclopentane

OR

2. mCPBA 3. EtOH, H

Br2, s-BuOH

37.

OH OH HIO4

38. HO

pregnenolone

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

OH

KMnO4, H

vitamin A

40.

1. OsO4 2. NaHSO3

O

3. HIO4

O O incensole acetate (found in frankincense)

41.

HCl two 1,2-shifts (complete)

5.2 Reactions


42. PhCH2Cl

43.

cyclohexene

1,1-elimination -HCl (a very strong base) +

n-BuLi

(CH3)2CI2 (1 equiv)

Cl

O

Cl

O

Zn(Cu)

O permethrin (insect repellent)

5.3 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. Br 1.

2. cyclohexyl alcohol

3. t-BuCl

t-BuF

Cl 4.

5.3 Syntheses

cyclohexyl chloride

Problems • 37

D 5.

D

O 6. cycloheptane

O

HO

OH

Cl

7. Br

H 8. allylbenzene O

9. ethylene

bromocyclopropane

10. cyclopentyl alcohol

OH 11.

OH

5.3 Syntheses


Br

O

12. O

D 13. cyclohexane Br

Br Br

14.

15. isobutane

isobutyl alcohol

Cl CO2H 16. CO2H Cl

17. t-butyl chloride

isobutylene chlorohydrin

18. O

5.3 Syntheses

O

Problems • 39

OEt

19.

20.

Br

HO

21. ethylene

OH O

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

23. cyclobutyl alcohol

O

di-t-butyl ether

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!

1.

H

5.4 Mechanisms


H 2.

H

3.

OMe

MeOH

isoprene

H

4.

I I2

5. CO2H

O 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

2. NaBH4

OH

9. H (a C11 olefin)

10.

H N

N

I2 -HI I

11.

+

H2C N N

N N

diazomethane

5.4 Mechanisms


12. The reaction of 3-bromocyclohexene with HBr yields only trans-cyclohexene dibromide, i.e., no cisproduct is formed. In contrast, 3-methylcyclohexene reacts with HBr to yield a mixture of cis- and transstereoisomers, 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 H

H

H

formaldehyde

5.4 Mechanisms

O

O

2-butanone

isobutyraldehyde

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 O

H

C

B

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


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 CH3

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

D

D

D CO2H

CO2H

D C8H17

oleic acid (a fatty acid)

5.4 Mechanisms

C8H17

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

O

O

O HO

R'

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.

O CHO _______________

_______________

_______________

A

B

C

D

5.4 Mechanisms


OH

29.

H

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 ' R N3 + 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 + A

(c) Given the observed stereochemistry, what type of mechanism does this addition reaction illustrate?

31.

H

(an olefin - complete)

32. styrene (vinylbenzene)

1. Cl2, H2O 2. base 3. dry HCl

5.4 Mechanisms

B

Problems • 47

33.

+

In

BrCCl3

CCl3

(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. O

O

O H

H

HO

H

H O B

D

C

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 1,2hydride shift and subsequent deprotonation (path b): H

H H H

H H path b

1,2-H: shift

-H path a

H

-H

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


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 Dhydrogen for deuterium atoms by the following mechanism: O Ha

O

O

:B

D OD

O

+B:

+D2O

-BH

- OD

D

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 A

O

5.4 Mechanisms

D O

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

3. phenylacetylene

1. B2H6 2. H2O2,

OH

1. OMe, HOMe 2. Cl2

4. n-BuCl

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

RC C :

5. Cl

1. Li / NH3

6.

2. HBr, di-t-butyl peroxide

7. isopropylacetylene

1. H2 / Pd(Pb) 2. BH3 3. H2O2, OH

8. 1-decyne

1. NaH 2. CH3(CH2)12Cl 3. Lindlar catalyst muscalure (pheremone for house fly)

9. 1,1-dichlorobutane

1. (XS) NaNH2 2. H3O, HgSO4

6.1 Reactions

50 • Chapter 6 Alkynes

10.

11.

Cl2, H2O

C CH

PhC CH

1. (XS) HI 2. Zn(Cu), cyclopentene

O

Cl

Cl 1. (XS) NaNH2

PCl5

12.

2. D2O OMe

13.

OMe

H C C CH2OH

1. LiNH2 (2 equiv) 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

2. acetylene

6.2 Syntheses

O

n-pentyl bromide

Problems • 51

3. vinyl chloride

4. acetylene

5. t-butylacetylene

methyl vinyl ether

(E)-3-octene

2-chloro-2,3-dimethylbutane

6.

7. propyne

8. propyne

n-propyl bromide

O 9.

C CH

O H

O H

6.2 Syntheses


10. styrene

(E)-1-phenyl-1-butene

11. diphenylacetylene

cis-1,2-diphenylcyclopropane

PhCHO

Et

Cl

12. 3-hexyne Cl Et

n-Bu 13.

Et

Cl Cl

Cl

O 14. acetylene (odor of cheddar cheese)

15. acetylene

O

disparlure (pheremone for female gypsy moth)

6.2 Syntheses

Problems • 53

O 16. 1-pentyne

2

OH

6.3 Mechanisms Br C HO

C

1. OH

O O

3.

Br

C

OH O I

O

CH3 H3O

C

C CH

O

O

OH 2.

O

[I ]

O

1. NaH 2. ethylene epoxide 3. H

OH

6.3 Mechanisms


Me2N

Me2N

O

OH 1. H3CC C:

H

4.

CH3

H H

2. H

O

H 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: R

6.3 Mechanisms

CH3

strong base R CH2

C C:

CHAPTER 7 STEREOCHEMISTRY 7.1 General 1. Which of the following molecules are chiral? O a.

Me

H

H

S

Me b.

c.

Et

CO2H O

Cl Ph

Me

Me

Ph

O

d.

e.

CO2H

f.

Cl

g.

h.

HC

C

C

C

CH

C

CH

CHCH2CO2H

CH

(an antibiotic)

Me

H

Me

Ph

i.

j.

k.

H

adamantane (an antiviral agent)

OH

Br

l.

HH

HO

Br n. Cl

m. the C2-epimer of

O N

NMe2 Ph

Ph

loperamide (ImodiumTM - antidiarrheal)

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

H N

O

S

N b.

N O

N CO2H

penicillin G

O

O

strychnine

7.1 General

56 • Chapter 7 Stereochemistry

O

O

N O

CO2Me c.

d. O O

Ph

O

O

cocaine

OCH3

aflatoxin B1

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

NH2 b.

HO

Me

Br

Br

H

H

H

Ph

H

CH2CH2NH2 c.

N

Me

(-)-norepinephrine

Ph H2N

d.

O

H

H

CO2H

O

e.

f.

Ph

HS

N H Me

NH2

H captopril (antihypertensive)

O O

OMe

g.

Me

OH

misoprostol (CytotecTM - promotes cervical ripening)

HO

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

H Et

c.

H

Et

H Me

Et

b.

Cl Me

H F Me

7.1 General

CH2OH

H

Me H

F Et Me

d.

H HO

CHO

CHO HO CH2OH H

Problems • 57

e.

CHO H OH CH3

H HO

f.

CHO

AcO

CH3

OAc

g.

h. O

O

H Cl

vinyl Cl

Cl Cl

H vinyl

H H

D-pinene (from pine resin)

O i.

Et

Et

O

O

j. O

Et Me

Me Et

5. How many “kinds” of hydrogens (enantiomeric and diastereomeric hydrogens are different!) are there in a. isohexane?

b. (R)-2-chloropentane?

6. Nomenclature. Give the complete IUPAC name for the following: Me Cl H Bn a. b. H OH vinyl

c. (S)-4-chloro-1-pentene?

c.

allyl

CH2Cl

Me d.

I H

OMe

e.

Et i-Pr

s-Bu

H allyl

s-Bu

n-Pr

vinyl H

Br

H f.

Me

Et

H Me

7.1 General


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 (-)-ephedrine

A

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. Me 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 a.

c.

b.

Me e.

d.

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? O O H O O H N N O O N vs. N O

H O

H

7.1 General


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? Me HS H

Me CO2H

NH2

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 HO N H

Ph O

Ph

O

OH

O O

HO 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 A

OH 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 H

H2N

S O

N O CO2H

HO2C

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

OH

H

1. OsO4 2. NaHSO3

Ph Me

Ph Me Br2 / H2O

Ph

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

B

a. cis-2-butene

Me

H

B

H Me

Me Me

D b.

C

D Me

C

C

D

Me

D C

7.2 Reactions and stereochemistry


c. Fumarase catalyzes the following reaction in mitochondria:

HO2C H

H

D2O

CO2H

fumarase

CO2H DO H D H 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

Cl

Ph HCl

a.

HBr

Br2

b. (Z)-3-hexene

KMnO4, H

c.

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

e.

HI

H

HF Cl

Me

f.

D2 / Ni


Problems • 63

Me g.

H, MeOH

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

k.

MeOH, H

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

Ph

Ph

racemic Ph

Ph Br



b. 2-butyne

meso-2,3-dibromobutane

OH c.

racemic OH

d. trans-2-butene

meso-glycol only

OH 5. Consider the structure

Ph

NHMe , and answer the following:

a. How many stereoisomers are possible?

b. Two of the structures are the decongestants ephedrine and pseudoephedrine: OH Ph

H N

OH Me

Me ephedrine

Ph

H N

Me

Me 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”?


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

Cl

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

5. cyclobutane

6. propane

1. Cl2, hQ 3. CuI

2. Li

4. vinyl iodide

5. HI

1. Br2, ' 2. Mg 3. phenylacetylene

7. bromobenzene

1. Li 3. n-PrBr

2. CuI 4. NBS, peroxides

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. Br

1.

O 2. O

Br 3. isopentane

4. iodobenzene

Me

5. cyclopentane

Bn

CHO 6. chlorobenzene

7. cyclohexene

8.

8.2 Syntheses

3 ways! deuteriocyclohexane

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.

O OH

2. Cl2, hv

1. CH2N2, hv

2. H2C C CH2 allene

Cl

Cl +

+

spiropentane

Cl

Cl

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 O N

G

AcO

hv D

H

O

CH2

H

H AcO

N O

H

OH

8.3 Mechanisms


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. CO2H

O

COX enzyme

O

2 O2

H H

CO2H

O

R PGG2

arachidonic acid

OH

[H] other prostaglandins

O

CO2H

O

PGH2

8.3 Mechanisms

OH

CHAPTER 9 SN1, SN2, E1, AND E2 REACTIONS 9.1 General For problems 1 – 9, circle the 1. reaction that will go faster: a. AcO

+

allyl chloride

b. AcO

+

allyl chloride

ethanol

HMPA

2. structure with the poorest leaving group: a. R-SH

b. R-NH2

c. R-OAc

d. R-OH

3. stronger nucleophile:

a.

b. Et3N

N

4. alkyl halide most reactive by an SN2 pathway: Br Br

a.

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: Me

a. Me

Me

b.

Br

Me

Br

7. approximate value of kH / kD when PhCHBrCH3, vs. PhCDBrCD3, is allowed to react with potassium tbutoxide: a. 1

b. 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

a.

methanolysis (SN)

(or diastereomer) Et

b. (R)-2-bromopentane (or enantiomer)

MeO , MeOH (SN)

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? O-t-Bu

OMe

Br

a.

Br

OH

OH

b.

Br

c.

+

OR

vs.

SR (which?) to maximize SN

Br

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

H(D) Cl

b. Cl

c.

t-BuOH

H(D) H(D)

Cl (D)H H(D)

9.1 General

KO-t-Bu

KOH MeOH

KOMe MeOH

Br

Problems • 71

12. The following reaction might be envisioned as occurring by an intramolecular SN2 process. However, kinetic evidence indicates a bimolecular mechanism. Explain. SO2 O CH3 C: TsO H

SO2 O C CH3 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. 1. n-octyl bromide + KOCH3

MeOH

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

3. potassium t-butoxide + sec-BuCl

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

5. n-hexyl iodide

KCN / DMF

methanolysis (RT)

6. Cl

refluxing EtOH

7. Br

8.

O

acetolysis (SN) Cl

9.2 Reactions


9. n-propyl bromide + Me2NH

10. isopropyl bromide + sodium t-butoxide

11. 3-iodopentane

sodium acetate / DMF

Cl NaSH (1 equiv)

12. Cl

ethanolysis (SN)

13. Cl

Br OAc / Ag 14. Ph

Et 15.

Me Cl

OEt (E)

H H Me

OMe / MeOH

16. Cl

E2

17. D

D Cl

9.2 Reactions

Problems • 73

18.

CH3 H D Br H Ph

E2

acetone

19. OH

Br

triphenylphosphine

20. 4-iodo-1-pentane

t-butyl alcohol (SN)

S

21.

methanolysis

Cl

NMe2 22.

'

I

Cl 23. conjugate base of H2Se

+

OH 24.

NHMe

Ph

PhCH2Cl (1 equiv)

ephedrine

25.

HO

NMe3

Me3O

BF4

choline

9.2 Reactions


OH

1. TsCl

26.

2.

H

OH (SN)

PhNH2

S

27.

HO

NHMe Br

28.

+ OH

F

(1 equiv)

OH epinephrine

29.

refluxing MeOH Cl

I

F

(XS) NaSePh

30.

O 31.

H2N

NEt2

O NovocaineTM

32.

Ph C CH

1. NaNH2 2. cyclohexyl bromide

MeOH (E1)

33. I

9.2 Reactions

EtBr (1 equiv)

Problems • 75

34.

KSCN

Ph OTs

Br acetolysis

35. RT

1. MeI 36.

S 2. refluxing EtOH

O

KO-t-Bu / t-BuOH

37. O Br

AromasinTM (an aromatase inhibitor used in breast cancer therapy)

38.

N

(XS) MeI

N

paraquat (an herbicide)

39. H Br Me a.

O O

EtO EtOH

H H Me Br

EtO O

b. O

EtOH

H

9.2 Reactions


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

Ph 1. Ph

Ph Br

OH

Ph Ph OCH3 Ph Ph

t-Bu Br 2.

t-Bu

3. Br H

D

H

Me

Ph 4.

Me H

H I

Ph

Me H

Me

Et

I

5.

OPh 6.

9.3 Syntheses

Br

+

Problems • 77

O O

Cl

7.

8.

Ph

Br

Br

9.

EtO

O 10. O

Br

CO2H CO2H

11.

Ph

Ph

Ph

Ph

OTs 12.

Ph Ph

OTs

13. OH

9.3 Syntheses


14.

OH

Cl

S 15. ethylene S

H

via an alkyne

16.

O

17. I

Br

D

D

D

18. Br H

D

D

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

1. Br

9.4 Mechanisms

NaSH / HCO3

S

Problems • 79

acetolysis

2. Cl

OAc

Me N Me

Br

3.

C N

Me N C

Cl

NEt2

4.

OH

Ph

Ph

NEt2

5.

H

N

OH

O dil OH O

H

O O

Cl

OH (Note: retention of configuration!)

I

6.

S

DMF

vinyl

SH

7. n-butyl bromide +

O N O

H

pyridine N-oxide

9.4 Mechanisms


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

S

Cl

HO H2O

Cl

N

Cl

HO

S

N

OH

OH

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

OAc

HOAc

OAc a racemate

I

9.4 Mechanisms

II

Problems • 81

O OTs

11.

O

OAc

HOAc

O

OAc

+

60%

40%

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

X O

I

II

13. Cl

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:

OTs

HOAc

AcO

NaOAc

9.4 Mechanisms


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 RO P O CH2 O O

NR"2

O RO P O CH2 O O

O O R'O P O

3'

O O R'O P O DNA

O RO P O CH2 O O dil OH, H2O

NR"2

OH

RNA

O O O P O

O RO P O CH2 O O

NR"2 +

2'

OH

OH

O O P O O

a 3'-phosphate

a 2'-phosphate

_______________________________________________________

'

17.

Cl

racemic camphene + HCl

camphene hydrochloride

18.

O O P O adenine O O O P O O OH OH O P OH O ATP

9.4 Mechanisms

adenine O O O P O O

NR"2

+ OH cAMP

PPi

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 (DMAPP) and isopentenyl pyrophosphate (I-PP) to form geranyl pyrophosphate (G-PP): O-PP

O-PP a.

H2O

base

+

O-PP

OH

DMA-PP

I-PP G-PP O O PP = P O P OH (-OPP is a good leaving group) O O

geraniol (a monoterpene)

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 O-PP

I-PP O-PP

F-PP

(a C20-diterpene)

[O] H3O

x2

x2 OH

triterpenes (C30) (e.g., squalene => cholesterol)

A

tetraterpenes (C40) (e.g., lycopene, E-carotene)

H

OH vitamin A (retinol)

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

9.4 Mechanisms


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. :H

O-PP

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.

S + H2N CO2H methionine

O O P O adenine O O O P O OH OH O O P OH O ATP

S

adenine O

H2N CO2H

OH

SAM OH HO

HO

epinephrine

9.4 Mechanisms

OH

NH2

norepinephrine

Problems • 85

21.

Ph2C N N

22. PhCH2Cl

+

23. HCCl3 + KI

TsOH

EtOH -N2

:P(OMe)3

KOH, H2O

PhCH2

HCCl2I

O P(OMe)2

Ph2CHOEt

+

MeCl

(Note: reaction does NOT occur in the absence of KOH!)

9.4 Mechanisms


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).

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

4. C10H14O: IR (3200 cm-1). 1H 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).

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

16. C7H14O2: IR (1610 cm-1). 1H 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 17. C11H12O2: IR (1705 cm-1). 1H 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 ? CH3 H 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-dichloro2-fluoropropane?

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

10. NMR


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

CO2H

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. b. What is the maximum number of 13C chemical shifts that would be expected for the C8H17 alkyl side chain? 25. Treatment of 2,3-dibromo-2,3-dimethylbutane with SbF5 (a very strong Lewis acid) in SO2 at -600 yields SbF6- and a substance whose 1H 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?

(i-Pr-O)2

10. NMR

O

O

P

P (O-i-Pr)2

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.)

Cl

PPh3 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


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.) Ha

Hb

H

H2SO4

H

=

homotropylium ion

COT

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.

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

G 0.6

10. NMR

0.4

0.2

0.0

-0.2

-0.4

-0.6

-0.8 ppm

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

+

MeO2C

CO2Me

retro D-A

4.

' 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) O 8.

O

+

N N

N Ph

O Cookson's dienophile

11.1 Reactions

94 • Chapter 11 Conjugated Systems

9.

+

N H

10. cyclohexene

(Z)-1,2-diphenylethene

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

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

C CH

12.

O

13.

intra D-A

1. vinyl chloride 2. KO-t-Bu

1.

CO2Me

14. 1,3-cyclohexadiene 2. O3

3. Zn, H

O

15.

11.1 Reactions

' (retro 4+2)

Problems • 95

1. ' (retro D-A)

16.

2. cis-1,2-diphenylethylene

CO2H

+

17.

'

HO2C fumaric acid

O 18.

+

estrone

MeO O

19.

..

HO2C OAc

CO2H '

'

+ OAc

CO2Me

20.

CO2Me

O 21.

N

'

SO2 +

O2S

11.1 Reactions


O base

22.

4+2

C19H24O2 MeO NMe3

I

Br

MeO R

CHO

Si Me2

23.

OMe 24.

N

+ Me3SiO

R

R' 4+2 R

Danishefsky's diene

11.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. D 1. cyclohexane

via a conjugated diene

Br

2. cyclohexane

bicyclo[2.2.2]octane

Me 3. cyclohexene Me

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

H

Z

R

'

Z = C, O

R

C Z H

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

'

CO2Et

+ OH

11.2 Syntheses


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 redpurple color. Explain. O O

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 a.

+

O

'

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 O CO2Me

Me

'

+ O

H CO2Me A

O H occidentalol

11.3 Mechanisms

OH


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

R R R

O

O

'

8.

O tautomerize

9.

O

' (ene reaction) (see 11.2, 7)

NC

Ts N

N N

CN

Ts N

hQ

-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 OH

N NH

H N

N

N

OH

N N

H N

CO2H

CO2H

CO2H

CO2H

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


CHAPTER 12 AROMATICS 12.1 General 1. Circle the compounds that would be expected to have aromatic character. O B

a.

b.

c.

d.

O

N N N N

H e.

N

O

f.

O N

H j.

i.

h.

g.

H

N B

H B N H

N

O N B

H k.

HN O

H

N H

SbF5

l. carbocation in the reaction of

Cl 2 Li

m. product in the reaction of

2 Li

2 MeLi

n. product in the reaction of

+

2 MeH

+

Br Zn

o. product in the reaction of

-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 H

N

N CH3

purine

ZofranTM (antiemetic)

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

a.

O

b.

O

c.

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

c.

b.

d.

6. Which would undergo an SN1 reaction most readily? O

O a.

b.

O c.

Cl Cl

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

Cl

N N

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. antibonding

0 bonding

12.1 General

Problems • 105

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

c.

b.

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.

2. o-methylphenol

fuming H2SO4

HONO2 / H2SO4

12.2 Reactions

106 • Chapter 12 Aromatics

..

PH2

H2SO4, SO3

3.

Cl2 / Fe

4. N H

5. benzene

1. PhCH2CH2Br, AlCl3 2. NBS, R2O2 3. KOMe, MeOH

Br2 / CCl4

6.

Br2, Fe

NBS, R2O2

Se 7.

8.

Br2, FeBr3

Ph N O

9.

ICl, Fe

Me

SH

Cl2, FeCl3

Cl2, BF3

10. N

12.2 Reactions

Problems • 107

OH 11.

2 Cl

Cl

H2SO4

+ formaldehyde

Cl (C13H6Cl6 - hexachlorophene, a disinfectant)

O O

12.

N N

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

picric acid

13.

F CN 14.

NH3

Cl

NMe2 1. MeLi 15.

2. H Br

Cl 1. HNO3, H2SO4

16.

2. NaOMe, MeOH CF3

17. toluene

+

O

H -H2O (a bicyclic C13 compound)

12.2 Reactions


OH 18. H BHT (C15H22O - a food preservative)

Br2 / Fe

19. O

O

D-pyrone

20.

F3C

1. fuming HNO3 (x2!)

Br

2. i-Pr2NH Trifluralin BTM (a pre-emergent herbicide)

21. anisole

1. MeI, AlCl3 2. NBS, ROOR 3. KOH (flavor in licorice) partial 1 H 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 I2

HO

CO2H NH2 tyrosine

12.2 Reactions

HO

I CO2H

O

catalyst I

I thyroxine

NH2

Problems • 109

1. [O]

23. N

3. [H]

2. Cl2, BF3 N O

N (completecontrast to 12.2, 10)

pyridine N-oxide

12.3 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. 1. benzene Cl

D

D

2. benzene D

3. benzene

4. benzene

D

D

1,2-diphenylethane

5. Cl

12.3 Syntheses


6. PhH

7. O

o-nitrobenzoic acid

8. benzene

9. benzene

10. NHMe

N

11. benzene CO2H ibuprofen

12. acetone, phenol

12.3 Syntheses

HO

OH

HO

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 OCH2CO2

Cl 2,4-D

2,4,5-T

Br O

14. NO2

NH

OH

TylenolTM

H2N

O CO2H

H2N

15. p-hydroxybenzoic acid O

O NEt2

O

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.

1. styrene

H Ph

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


O 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

OAc

OH 1. OH 2. Dry Ice

CO2H

CO2H

3. H salicylic acid

aspirin

Br 5.

Br2 / AlCl3

+

isobutylene

O 1. R C X , AlCl3 6. 2. HBr R

12.4 Mechanisms

Problems • 113

O 7.

O

OH

AlCl3

R

R O

Cl

Cl

Cl

Cl

NaOH

8.

Cl

O

Cl

Cl

O

Cl

dioxin

BF3

9. 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 CO / HCl H

CHO

Cl A

11. toluene

H2O2 CF3SO3H

OH (a convenient way to substitute an hydroxyl group onto an aromatic ring)

12.4 Mechanisms


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 aminocontaining 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? Cl N Cl

1. dye-NH2

N N

2. cotton-OH

Cl

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 Cl3C

H

H

H2SO4

+ 2 chlorobenzene

Cl

Cl CCl3 DDT

14.

12.4 Mechanisms

1. BF3 +

Cl

2. H

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 O CH3 SnCl4 CH2Cl Merrifield resin

poly(stryene)

1. Br2

17.

C7H8Br2

2. :B

C7H7Br

3. H2O O ditropyl ether

Me 18.

H Ph

OTs H Me

HOAc SN (acetolysis) (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


Cl

CHCl3

19. N H

OR

N

H 20.

OMe 21.

OMe H3PO4 / H2SO4

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 1. H2SO4

3.

2. H3O HCl

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

1. NaH 5. 1-hexen-3-ol 2.

O 6.

Ph

O MeO S OMe O

1. PhMgCl 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

2. H

9. Ph

O

1. LiAlH4

O

2. H H2 / Pt

Me 10.

H Me

POCl3

OH D

pyridine

H

HO 1. NaOH 11.

O NMe

2. CH3I (1 equiv)

HO morphine

codeine

OH

OH Jones reagent

12. HO

13.1 Reactions

HO

Problems • 119

1. Br2, H2O 2. Me3SiCl

13.

3. Li 4. acetone 5. H3O

O BnO

14.

1. (XS) MeLi OBn 2. H

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

O

O

1. NaBH4

OAc

16.

2. H O cortisone acetate

1. LiAlH4 2. H OH

H

17. HO

18.

(pinacol rx)

H

Ph OH OH

O

O 1. NaBH4

aromatase

19.

2. H O

HO andostenedione

estrone

estradiol

13.1 Reactions


OH

O

1. SOCl2, Et2O 2. Mg

20.

3. H patchouli alcohol (used as a fragrance)

O Ph

OEt

1. n-PrMgCl

21.

2. H

N Me

DemerolTM (narcotic analgesic)

O

OCH3

1. toluene, ' 22.

+ TMSO

2. H

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

4.

OH

H

H

OH

racemic s-BuCl

*OH H (* = 18O)

5. n-butane HO OH

6. cyclohexane

7. vinyl chloride

8. n-hexyl alcohol

O

1,3,5-hexatriene

C N

13.2 Syntheses


D CO2H 9.

10. p-chlorophenol

p-hydroxybenzaldehyde (via a Grignard)

OH

O

11. HO

HO estradiol

OH

12.

O 13. n-butane

OH

O OH OH

Cl

14.

O OH 15.

13.2 Syntheses

Problems • 123

O

OH 16.

O

Cl

OH

OH 17.

OH

HO

C

CH

18. HO

HO

OH

OH C CH

19. Me

HO

O

estradiol O

HO

OH C CH

O

EnovidTM constituents (OCPs)

20. Br

13.2 Syntheses


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. B

A

O

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 H2SO4

OH OH

(a cyclic ether)

1. NaBH4

2.

2. H2SO4 retinal

3. cycloheptene glycol

O

H

H

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

13.3 Mechanisms

Problems • 125

Ph

Ph

H OH

4.

O

OH

H

5.

+

OH

H

6. glycerol

H O

H

7. OH

OH (an aldehyde)

2. 1. H3PO4

8. CO2H

O

MgCl

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

13.3 Mechanisms


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 H

O

O

H3O O H

O O

O

OCH3

OCH3

O H A

aflatoxin B1

10.

O

CH3 H Br HO H CH3

CH3 H Br H Br CH3

HBr

+

CH3 Br H H Br CH3

racemate

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

11.

or OH

HBr

only trans-product is formed!

OH

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

H

OH CO2H

CO2H O

13.3 Mechanisms

1. [H]

CO2H

2. (-H2O)

NH2 valine

Problems • 127

H2SO4

13. 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 O H H2C CD2 D DH2C D NOT H3C OH OH

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:

O

OH

- H2O

taut

OH

OH

H

+H - H2O

OH

H ~ H:

H

-H

O

H

cyclohexene glycol

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

D H

D O B

a. Which pathway is consistent with this observation?

b. Suggest a preparation of A from cyclohexene.

13.3 Mechanisms


O

OH

H

16.

(

OH

Cl

= 13C)

O

1. Li 2. acetone C4H8 +

17. 3. H

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

OH

Cl

D 18.

D

:PPh3

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) 2. TsCl

1. O

3. KOAc, 18-crown-6

2. benzyl phenyl ether

(XS) HI

3. phenyl mercaptan + Me3S I

OH KOH

4. Br

5.

1. HF

O

2. PCC

6. 2-isopropyloxirane

NaCN MeOH

1. styrene epoxide 7. PhLi

2. H2SO4

OMe 8.

1. HI (1 equiv) 2. CrO3, H 3. NaBD4 4. H

14.1 Reactions

130 • Chapter 14 Ethers

1. PhCO3H 9.

2. PhOH, H

H, MeOH

10. O

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

F 2.

1. ClCH2COCl

F

AlCl3

3. H

N N

F 5. OH

F

4. base

N

6. H

N N N flutriazole

1. mCPBA 12.

2. MeNH2

Ph

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 a.

CO2H

'

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 O

'

H

Claisen

CO2H O

HO2C prephenic acid

chorismic 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


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.

O

Claisen

Cope

'

'

OH ~H

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

O

HO

'

tautomerize

H

Predict the oxy-Cope product for the reaction below: OH '

SH 18.

[O]

2 H2N

CO2H cystine [crystallization in kidneys can lead to one type of calculi (stone)]

cysteine

OH 19.

SH

HS

[O]

OH dithiothreitol

14.1 Reactions

C4H8O2S2

Problems • 133

14.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. 1. 3-methylpentane

2. cyclohexene

3-methoxy-3-methylpentane

trans-cyclohexene glycol

OTs 3.

4. propylene

S 2 diallyldisulfide (found in garlic)

O 5. cyclohexane H

6. cyclohexene oxide

via an epoxide

cyclohexane

14.2 Syntheses


O 7.

O N

HS CO2H

N 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

CHO

1. styrene epoxide

1. LDA 2. methyloxirane

allyl alcohol

2. H

H

3. O

14.3 Mechanisms

OH

Problems • 135

O

O

O

OH

OH, MeOH

4.

O

OH

O

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

O

H

O

H HO

O

H O H

H

O

H

H

H O H

O

O H H

O

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. HO O

H

H H2O

O O

H

O

HO

H

O

H

H

H

O

O

H

H

H

O

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):

H H HO

O squalene epoxide

lanosterol

14.3 Mechanisms


7. Biochemical hydroxylation of aromatic compounds proceeds via arene oxides, which subsequently undergo ring opening to form phenols: H O

O

H

OH

OH (a) - H

+H

cytochrome P450

benzene oxide

A (b)

tautomerization

1,2-H: shift

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

OH

O 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 8.

R

OH R'

1.

O

OH

Cl

R'

R

2. BF3, Et2O

R'

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

14.3 Mechanisms

Problems • 137

9.

N H

PhO O

DBN

S

N H

PhO

S

O

N

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:

N

N

H H A

N

N

A

DBN

14.3 Mechanisms


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 OH

1.

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

H

O

11-cis-retinal

O

(a protein)

OH

rhodopsin

H3O

5. Ph

OH 6.

1. PCC 2. H3O 3. HOEt, H

O 7.

Cl

1. KO-t-Bu / t-BuOH 2. HCl

15.1 Reactions

140 • Chapter 15 Aldehydes and Ketones

8. methyl n-propyl ketone

1. NaBD4 2. H 3. H2SO4 (E1)

1. KMnO4 2. semicarbazide

9.

3. H2 / Pt (XS)

HO vitamin D

O CH 10.

OMe O

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

O 1. Ph3P O

11. p-nitrobenzaldehyde 2. H3O

a fluorescent "spy dust" ingredient

12.

O

1. Ph3P-CHOCH3 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 14.

1. CH2I2 / Zn (Cu) 2. H3O

O

3. Ph3P=CHC=CH2 OMe

15. cyclopropanecarbaldehyde hydrazone

1. H3O 2. EtMgI 3. H

1. H3O 16. acetophenone diphenyl ketal

O

2. H2NOH

H3O

CHO

17. a heterocycle

18.

1. Ph3P 2. n-BuLi

O

O

Br

O 19.

O

3. butanal 4. H3O

H3O O

20. t-butylacetylene

1. H3O, Hg2+ 2. hydrazine, OH

15.1 Reactions


1. Br2, H2O 2. O

,H

21. 3. Li 4. ethylene oxide 5. H3O

1. HONO2 / H2SO4 22. CHO

O

2. H2NOH, H nitrofuroxime (used in treating urinary tract infections)

23.

Et

H3O

O O

multistriatin (European elm bark beetle pheremone)

OH / ROH

24. 2-oxopropanal

(intra-Cannizzaro)

CHO 1. N2D4, OD, D2O

25. OMe

2. HI

OH vanillin

CN

CO2H O

26.

OH

O

Ph H

OH

CO2H O H3O

OH

+

OH OH

OH HCN

+

??

OH

This reaction, with the release of the very toxic HCN, provides a defense mechanism for millipedes.

15.1 Reactions

Problems • 143

O 27.

H3O O frontalin (insect pheremone)

MeOH, H

28. 3-oxobutanal

C6H12O3

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

1H

O

O

OMe

H3O

29.

O 30.

H3O O

safrole (odor of sassafras)

NH2 31.

H

1. 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


O

O

H

H

OH

33.

OAc 1. LiAlH4

OAc

34.

2. H3O

HO

O

HO

O

35.

H3O

O O F flunisolide (anti-inflammatory in allergy medication)

36. 1,2-cycloheptanedione

(XS) hydroxylamine

heptoxime (used in quantitative determination of Ni)

H3O

37.

O

O

O

O

paraformaldehyde

38. Chain degradation of a hexose: OH

OH

O

1. NH2OH, H H

OH

OH

15.1 Reactions

OH

2. Ac2O (dehydrates an oxime)

Problems • 145

O Me N N

N H

39.

mild acid O O

O tadalafil (CialisTM)

40.

H N

O

Me2N

S

H N

mild acid

MeNH2 +

NO2 ranitidine (ZantacTM - antiulcerative)

N N N H3O

41. N H

S

olanzapine (ZyprexaTM - antipsychotic)

1. PCC

42.

2. Me2CuLi

OH 3. H3O citral H N O

43.

O

H3O

O

F paroxetine (PaxilTM - antidepressant)

15.1 Reactions


44. Ammonia is produced in the mitochondria primarily by the oxidation of glutamate to produce an imine, which is subsequently hydrolyzed: NH2 O2C

[O]

H3O

CO2

+ NH4

glutamate

D-ketoglutarate

O O O 45.

H3O

4 steps

HO

O diosgenin (from Mexican yams)

progesterone

OH 1. PCC 2. MeLi

46. 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

H

O

O dehydrostrychninone

15.1 Reactions

2. H2 / Lindlar catalyst

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 R

+

H

SH

SH

R

S

H

S

R

n-BuLi

S

1. R'X

S

2. H3O

Li

a dithiane

O R

R'

Corey-Seebach reagent

Predict the products of the following reactions: 1. HS(CH2)3SH, H 2. MeLi

a. benzaldehyde

3. EtI 4. H3O

S

1. n-BuLi 2. n-decyl bromide

S

3. H3O

b.

1. HS c. acetaldehyde

SH , H

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

H2N

CO2H

serine

H2N

CO2H

+

H

H

glycine

(cont. on next page)

15.1 Reactions


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.

H2N FH4 +

H N

N

H3O

N

CH2O

FH4 +

N

H O

catabolism

N

CH2O

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 50. Cl

O

NH2

H -H2O 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

OCH3

OH

2. H

D

3. cyclopentanone

CO2H

O

Cl

4.

5. benzaldehyde

D

Ph

CO2H O

15.2 Syntheses


O 6. Ph

O CH2Cl

C

O C

Ph

O

OH OH OH

O OMe

O 7.

O

OH OMe

O OMe

OH OH

O OH

8.

9. MVK (methyl vinyl ketone)

15.2 Syntheses

via a Wittig

O OMe

Problems • 151

10.

via a Wittig

OH

O

O

O 11.

H + Br

O HO

CHO

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

NR2 (H

N

n-BuLi

-H O

How could this observation be used to form

CHO

NR2

OH Ph

from acetone and benzaldehyde?

CHO

13. O O

O 14.

H

15.2 Syntheses


OH

OH C CH

15.

O

O major component in OCPs

O 16.

Ph , benzaldehyde

Ph

Et Ph

R

R tamoxifen (NolvadexTM - antiestrogen)

(R = -OCH2CH2NMe2)

O 17. 1-butene

18.

CO2Me CO2Me

O 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. OH HO

Br

HO

O O

OMe N

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 Ph

H2 / Ra-Ni

Ph

PhCH2Ph (hydrogenolysis)

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

O 22.

O

via an enamine

H

N O

15.2 Syntheses


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. 1.

H, H218O

O

18O

H3O

2.

H2N

N

CHO

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

4.

+ hydrazine

H

N N

H

H

Cl 5.

OMe / HOMe

OH

O H

15.3 Mechanisms

OMe MeO

Problems • 155

O

O

R H2NOH, H

6.

R

R

N O

R

HO

O

O

7.

O

H

OH

O

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 stateadministered 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 H2N

O

H

O N

N

H3O

OH

HO

OH

H OH tetrodotoxin

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

1. SN2 +

CH3I

Me2

O 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. O

A

15.3 Mechanisms


O

OH HCl

10.

Cl O

OH

O

11.

H3O acetaldehyde

O

OH H3O

12. O

O

diazomethane

13. cyclohexanone

Ph N 14. N H Ph

15.3 Mechanisms

cycloheptanone

H

NHPh + NHPh

acetaldehyde

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 O

17.

H +

OH

18. 2 phenol

HO

OH

+

O

NH2R

OH

acetone

OH

H

NHR

OH

C15H16O2 bisphenol A (a starting material in the synthesis of LexanTM)

O 19.

H

n-PrNH2

Et N

15.3 Mechanisms


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 2. CH2O

1. NaH OH

O MOM

2. ClCH2OCH3

HO

3. H3O

OH

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 21.

O

H3O

O

OH SCH3

MeSH

22.

0o O

OH

O 23.

CN

H2O, OH NC

OH

15.3 Mechanisms

H

Problems • 159

O

OEt

O

1. MeLi 24.

2. H3O

E-vetivone

25.

H H

O

O

Et N

EtNH2 H

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

PhCHO

+

PhNH2

1.

2.

Ph N Ph

Ph

OMe (Danishefsky's diene)

N O

3. H

Ph

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


b. the Vilsmeier reaction (outline a mechanism for both steps)

Me

Me N

POCl3

H

Me

Me N

O

OH

O2PCl2 C

1. phenol

Cl

CHO

2. H3O

H

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: CN

2 PhCHO

O

OH

Ph

Ph

benzoin

formaldehyde, H

NH

29.

N

OH

HO O

30. OH

OH

OH

E-D-ribose

OH

HO H

O OH

OH OH

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: H N

NH2 HO2C arginine

O

H3O

NH

H2N

NH2

NH2 +

NH2

NH2

HO2C

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

O

Ph

N O

RNH2

O

OH O CO2H

H O 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. O R C O

(CH2)nCH3

O

H3O O O

P OR' 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 Ph

O Ph

+

RCO3H

Ph

O Ph

15.3 Mechanisms


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

O

Ph

OH

KOH, EtOH Ph

Ph

CO2 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 Ph

O

1.

O OH

Ph 2. H

15.3 Mechanisms

MgX

Ph Ph

Problems • 163

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

OH H

36. R

R

R R

OH HO base

37.

acid

OTs

30% HClO4

38. O

HO

However (!), H2SO4

39. O

OH

15.3 Mechanisms


And, lastly, a steroid dienone – phenol rearrangement: HO

HO O

O

OH

O

OH

O OH H

40. O prednisone (anti-inflammatory)

Problems 41 and 42 illustrate the Favorskii-type rearrangement. O

CO2R

Cl

CO2R

OR

41.

( = 13C tag)

+

HOR 50%

50%

Br O

CO2R

OR

42.

HOR

Br

OMe H3O

43.

+ NMe2 O

15.3 Mechanisms

O

NMe2

OH

Problems • 165

N

R N R

Br2 / H2O

44.

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 N

O C H

HS

CO2R

H2N

O

O

H

+

N

CO2H O

penicillamine

RO2C

S HN 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

NH2

N H

O

NH H

N

strychnine

H phe HO OH

HO OH

15.3 Mechanisms


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

HO

H3O

Br

OH

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

H

O2

O

H

O

O

O

OH 2

OH

49. An impressive biomimetic conversion in Johnson’s (Stanford) total synthesis of progesterone (see 19.3, 16 for the final stage): O 1.

O CF3CO2H

O

H

O

O

O

O

H 2. aq K2CO3

H

H

H

H

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.

1.

KMnO4, H

N N nicotine

niacin 1. NaOH

2. phenylacetic acid

2. Me3O

BF4

1. NaBH4

3. 3-oxobutanoic acid

2. H

Ph

O

Ph

1. H3O

4. O

2. CrO3, H

O Ph

1. NaOH (1 equiv) 5. J-bromobutyric acid

2. ' 3. LiAlH4 4. H3O

OH

1. KOH 2. acrylic acid (propenoic acid)

6. 3. BH3 4. H3O

7. benzyl chloride

1. NaCN 2. PhMgCl 3. H3O

O 8.

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

16.1 Reactions

168 • Chapter 16 Carboxylic Acids

9.

N Ph

CO2H

OH

Ph

1. (XS) PhLi 2. H

OH

fexofenadine (AllegraTM - antihistaminic)

OAc

1. LiAlH4

CO2H

2. H

10.

aspirin

OH 1. KCN, H 2. H3O 11. 3. BH3 4. H3O

O testosterone

12.

Ph

CO2H

NMe2

1. EtLi 2. H

Ph

13.

methadone

CO2H

1. H3O 2. PCC

CO2 +

14. A reaction in the biosynthesis of the amino acid leucine: CO2H CO2H HO

1. [O] 2. (-CO2)

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

CO2H

1. OH

3. Jones reagent 4. '

2. H O O

Ph

OH ecgonine

cocaine

tropinone

OH 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 N H

.

COOH CH2 CH2 COOH

=

16.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. 1. benzoic acid

2. propylene

PhCH2CO2H

pentanedioic acid

16.2 Syntheses


O 3.

acetone + CO2

OH

O 4. o-chloroacetophenone CO2H

CO2H

5. styrene

ibuprofen (MotrinTM - antipyretic)

CO2Na 6. 1-butanol sodium valproate (used in the treatment of epilepsy)

CO2H 7.

O OH

CO2H O

HO

8. benzyl bromide

16.2 Syntheses

via a nitrile

CHO

O

Problems • 171

N N

NC 9.

O NC O

10.

pentanedioic acid

O 11. RCO2H

RCH2

R'

O 12. 3-oxobutanoic acid

13. ethanol

OH Ph

butanedioic acid (succinic acid)

CO2H 14. CO2H

16.2 Syntheses


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. O 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 HO + R R OH 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 R

O

CO2H

O

+

H3O,

O

'

R

H

+ CO2 +

N

N

8. Strecker also developed a synthesis of amino acids: O R

H

+

R'NH2 +

CN

H3O H

R' N

O OH R

16.3 Mechanisms


9. The biosynthesis of the amino acid phenylalanine involves an acid-catalyzed decarboxylation of prephenic acid: CO2H

HO

H

CO2H

CO2H

-CO2

O

CO2H

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 OH

-H2O

O O + NH2CHRCO2H

OH O

O N

O

O

O + CO2 + RCHO

ninhydrin

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

H H NH2

N R

[H] [O]

NAD+

O NH2

O

OH CO2

N R

NADH CO2

pyruvate

NADH

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. O O

O CO2

OAA

16.3 Mechanisms

O O O + O P O P O P O guanosine O OH O GTP

CO2 CO2 +

O + GDP O P OH O PEP

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 Dketocarboxylic 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 S

R N

O

PDC CO2

H

thiamine

+

CO2

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.’) H

O

R

CO2H OH

N H

N

NH

histidine

NH2

NH2

PLP -CO2

N

NH

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


CHAPTER 17 CARBOXYLIC ACID DERIVATIVES 17.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate. O 1.

OEt

Cl

Et2NH (1 equiv) / pyridine

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 6. N,N-diisopropylpropionamide

H3O

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

O 8.

MeOH, H

O O

1. LiAlH4 2. H

9.

3. Ac2O CO2H

17.1 Reactions

178 • Chapter 17 Carboxylic Acid Derivatives

saponification

10. H-caprolactone

O NH 11.

S

Bn

H2O, OH

N O

CO2H

penicillin G

O OR

Et 12.

+ urea OR

Et O

VeronalTM (a barbiturate, sedative) O

N

1. OH 2. H

O

13.

3. Jones reagent 4. ' Ph

O cocaine

O

O

OH 14. HO

..

SCoA NH2CH2CO2H (glycine) OH

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

N 15.

glycocholate (a major bile salt)

1. H3O 2. CH2N2

N O

O strychnine

17.1 Reactions

Problems • 179

OH OEt

16.

1. H3O (a lactone) 2. PhMgCl 3. H

O O O

1. LiAlH4

17.

2. H O

O

Spanish fly

H 18.

1. phosgene

N

N

2. LiAlH4 3. H

H

protein O O P F Me

19.

OH

SarinTM (a cholinesterase inhibitor) H

CO2H 1. SOCl2

N 20.

2. diethylamine N H

LSD lysergic acid O S Cl O

21.

O +

H2N

N H OrinaseTM (for diabetes)

O NH

22.

H2O, OH

S O

O

saccharin

17.1 Reactions


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

Al2O3, '

24. acetic acid

H2C

(-H2O)

O 25.

C

O

aniline

ketene

O

MeO

+

OMe

[ethylene glycol]n

H

n

dimethyl phthalate

DacronTM

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

OH

bisphenol A OH 27.

Me N C O

+

methyl isocyanate (active ingredient in the insecticide SevinTM)

1. Li 2. CuI

O O

28.

3. benzoyl chloride 4. H3O

Cl

O

NHMe

29. F3C

fluoxetine (ProzacTM - antidepressant)

17.1 Reactions

1. propionic anhydride 2. LiAlH4 3. H

Problems • 181

HO

O O

O O

NH3

30.

simvastatin (ZocorTM - antilipemic) O O

C CH 1. saponification

31.

2. H3O N OH Ortho TriCyclinTM (OCPs) O MeO

1. LiAlH4

N H

32.

2. H3O

HO capsaicin (active agent in cayenne pepper)

H2N

O S O

1. H2O,

33.

N

N CF3

OH

2. SOCl2 3. urea

celecoxib (CelebrexTM - anti-inflammatory)

F S HO

O O

34.

O

H3O

F O F fluticasone propionate (FlonaseTM - anti-inflammatory)

17.1 Reactions


1. PhLi

CO2Na

35.

2. H3O

MeO

naproxen sodium (AleveTM - anti-inflammatory)

O

36.

N

HN

O S NO

N

H3O

N

N

OEt

sildenafil (ViagraTM - treatment of ED)

1. base, '

CO2H

Br

37.

2.

NH2

O NH

OH

38. n

H-caprolactam

39.

Nylon 6 (a polyamide)

1. ethyl chlorocarbonate

NH N

2. Et2NH diethylcarbamazine (anthelmintic)

O

OEt N N

40.

H3O CO2

Cl loratadine (ClaritinTM - antihistaminic)

17.1 Reactions

Problems • 183

NH 1. ethyl benzoate 2. LiAlH4

41.

3. H

Cl Cl sertraline (ZoloftTM - antidepressant) OMe 1. SOCl2

CO2H 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 H3O

43.

CF3 O

N H dutasteride (AvodartTM - treatment of BPH)

CO2H 44.

O

N H

H2N

H3O OMe

O Ph aspartame

O N

N

exhaustive hydrolysis

N

45. N

O

Cl

O N N

zopiclone (LunestaTM - sedative, hypnotic)

17.1 Reactions


O 46.

1. NaBH4 CO2H 2. H

2 H2O + C6H8O4

pyruvic acid

47. Chain elongation of a tetrose sugar: OH

O H

OH

OH

1. HCN, CN 2. H3O 3. SOCl2 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-butoxycarbonyl). R O O R R' N + H2N OH H O Cl H2N CO2H CO2H H2N O R' C A B A-B 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


O H

O glucose

53.

+ glucose

OH tuliposide (found in tulip bulbs)

N

tulipaline (a J-lactone) (produced when bulbs are damaged - a fungicide)

O N

54.

1. SOCl2 2. ammonia

N CO2H

F

3. SOCl2

O LevaquinTM (antibacterial)

OH 55. S

O

N

N N H

S

N

H3O '

CO2 + MeNH3 + H3N

O O meloxicam (MetacamTM - anti-inflammatory)

+

(gives a positive Tollens' test)

17.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. 1. sec-butyl acetate O

O

O 2.

R

O NH2

R

H O

R

17.2 Syntheses

??

S

CH3

Problems • 187

O

O N

NH

3.

4. O

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

O 6.

H

OH

CO2H 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 NHCH3 NCH3 NCH 3

Cl

Cl

CH3 O N Cl

O

O

Cl

H NCH3

Cl Cl

O

CH3 O N Cl

O

H3C N NH2 N

Cl

ValiumTM

10. 3-oxohexanedioic acid

O O

11. methyl benzoate

17.2 Syntheses

O

methyl phenylacetate

Problems • 189 12. Following is an outline for the synthesis of fluoxetine (ProzacTM). Supply reagents for each step. O

O NaO

O

N O

N O

F3C

F3C

O

N H CH3

O

N H

OEt O

F3C

F3C

NH2

O

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

O

O 5. Cl

O

F3C

HO

NMe2

Cl

NMe2

NMe2

OEt F3C

mechanism?

O

NMe2

Me N O

OEt

O

Me N

H

F3C ProzacTM

mechanism:

17.2 Syntheses


Et

HO2C

15.

ibuprofen

O

O

O

16.

OH CO2H

O

OH

17.

O

Cl

CONEt2

18.

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. NH2

HO

N H serotonin (5-HT)

H N

MeO

O

H N

O H

N H 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

O

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 O catalyst CHO (- CO)

Br

A O Cl

Cl O

CN

B

Nylon 66

CN H2N

NH2 C

21. Name the following polymer and devise a synthesis for it. Remember, appropriate starting monomer. Why?

OH

OH

OH

OH

is not an

OH

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

CH3O

O

CH3O

O

O

N R'O R, R' = H (morphine)

N O

OH N

O

hydrocodone (a component of VicodinTM)

oxycodone (HCl salt = OxyContinTM, 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


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

O

H3O

O N

O

N O

codeinone

neopinone

O2C

N

N

CO2H OH

N NMe2

CO2H O

17.2 Syntheses

OH

H

N 23.

oxycodone

AmbienTM (sedative)

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

2.

(show location of the labeled oxygen)

1. (XS) RMgX

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

4.

TsOH O

an alkene

+

?? (show location of the labeled oxygen)

Et

O 5. ethyl 5-oxohexanoate

PhMgCl (1 equiv)

O Ph

17.3 Mechanisms


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

Ph N C S PITC

H2N

+

S

Ph

O OH

N H

O

N

alanine residue

+ H2N

O

O

1. Me3O BF4

O

NH2

2. H3O

OMe

7.

NH

C PTH-alanine

B

A

H

CO2H

valine

8. The Swern oxidation: a. "activation" of DMSO step:

Me

O S

O

O

Cl

Cl

+ Me

DMSO

b. oxidation step: H OH + R R

- CO2 - CO

NR3

oxalyl chloride a chlorosulfonium salt

17.3 Mechanisms

O R

R

Problems • 195

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

S

+

Cl N

H

OH

R

R

O R

NR3

O N-chlorosuccinimide

R

a chlorosulfonium salt

10. The biosynthesis of pyrimidine bases, e.g., uracil, begins with the formation of dihydoorotic acid. Formulate a mechanism. O O H2N

O P O O OH

CO2H +

H N

H2N CO2H

carbamoyl phosphate

O

H O

aspartic acid

N N H

CO2H

O

dihydroorotic acid

N H 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 3. CO2

1. POBr3 CF3

N H

CF3

N

CF3

4. H 5.

CF3

N

(- 600) Li

6. H3O HO N H

N

7. [H]

CF3

CF3 mefloquine

17.3 Mechanisms


12. Acid halides react with diazomethane to give diazomethyl ketones, which, like diazomethane, decompose to give carbenes. O R

O

CH2N2

..

Cl

R

O

hQ

C N N H

R

..C

H

+ N2

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 [D]D +480

1. H2 / Pd 2. Wolff-Kishner =

twistane [D]D +4340

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

O

R

S

17.3 Mechanisms

R' N H

1. N C Br 2. H3O

R O + O

H N

H2N O

R'

Problems • 197

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

OH N

(Baeyer-Villager oxid)

H

H2O2, H

N O

O

pseudostrychnine

O

O

N

O

N H

H3O N

N O

O

O

O A

strychnone

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

Ac

N

N Ac2O, pyridine

N O

Me

CO2H

N OAc O

17.3 Mechanisms


16. The final step in Sheehan’s (MIT) total synthesis of penicillin V involved the formation of a strained Elactam. 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 N H CO2 N H

O

N C N

S

OPh N H

DCC

O

N O

CO2

S

CO2

salt of penicillin V

O 17. H2N

OH

J-aminobutyric acid (GABA)

1. PCl3 (2 equiv), H3PO3 (1 equiv) 2. H2O

OH ONa O P OH P OH O pH 4.3 OH OH O P OH O P OH H2N H2N OH OH TM (a bisphosphonic acid) Fosamax (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 O

,H O

OH

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 2. NaBH4

NH2 20. N H

+

CO2Me

3. H

N H

N

O

17.3 Mechanisms


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 2. KO-t-Bu / t-BuOH

O 11. O

3. LiMe2Cu 4. H3O

12. t-butyl methyl ketone

1. Br2, H 2. (CN)2CH: 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.

+ H

H (pKa 15)

NaOEt HOEt (C8H10)

18.1 Reactions

Problems • 203

O 1. LDA 16. 2. Cl2, H

Br

1. HCl (1,4-addition) 2. O

17. isoprene

3. H3O,

O '

O 1. Cl2, H 18. 1. Cl2, H 2. KO-t-Bu

2. OH (SN2) 3. KMnO4

3. H3O 4. KMnO4

Cl

19.

CH2(CO2Me)2

1. KOH, EtOH

LiH (XS)

2. '

Cl

91%

95%

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

O 20. O

O

1. base 2.

tautomerize

O

O

O

OH

O

coumadin (WarfarinTM - an anticoagulant)

18.1 Reactions


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)


butanedial

O NH

4. dimethyl malonate O

O

N H

seconal (a sedative)


2-benzylbutanoic acid

O 6. dimethyl malonate


+

+

styrene

Ph

methyl acrylate

Br

(acrylic acid is propenoic acid)


18.2 Syntheses

OH

s-Bu-CONH2

Br

Problems • 205

O

O OH

9. ethyl acetoacetate O

O

O O


O

O

Cl

Cl



s-butyl methyl ketone

2-methylbutanoate

O 13. cyclopentanone OAc

O 14. cyclopentanol

CO2H

O

O

15. 2,4-pentanedione

18.2 Syntheses


16. methyl acetoacetate

2,4-pentanedione

via an organoselenium cmpd

17. 3-pentanone

ethyl vinyl ketone

Cl

O

O

O

18. chlorobenzene N H

WellbutrinTM (an antidepressant)

OH 19.

O ,

OH

O

N H

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 20. dimethyl malonate O

N

N Na O

sodium pentothal (used to induce pre-surgical anesthesia in combination with sedatives)

21.

CN

Ph

Ph N

NC

N

O 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 O

H

1. OH

OH

OH

H

(a central reaction in glycolysis catalyzed by the enzyme TIM, triose isomerase)

O 1. EtO, HOEt 2. ethyl acetoacetate

O 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 H

O

O OMe

5.

MeOH O O

O

18.3 Mechanisms


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 HN

O NH

CO2, ATP

N

HO2C

S pyruvic acid

A

biotin

OH H 8.

O O

O O O P O OH

H

O 9.

18.3 Mechanisms

+

Me3Si

O

O

OH

CO2H

S

HO2C

HN

O

Cl

Me3N

O TMS

CO2H CO2H OAA

CHAPTER 19 CARBONYL CONDENSATION REACTIONS 19.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. benzaldehyde + acetophenone

1. OH, (-H2O) 2. NaCH(CO2R)2 3. H3O

H

2. isobutryaldehyde

O 3.

+

OEt, EtOH

CHO

O O OMe, MeOH

4.

OMe

O

5. dimethyl heptanedioate

1. OMe

CO2 +

2. H3O, '

O OH, ROH

6.

(conj. add'n + retro-aldol)

O OH 7. O

8. propionaldehyde

cis-jasmone (a perfume) 1. BrCH2CO2Me, t-BuO 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 10.

+

base

ethyl vinyl ketone

O (complete)

11. benzaldehyde

+

OH

CH3NO2

OMe, MeOH (retro-Claisen)

12.

CO2Me O

13. cyclopentanone

1. NaOEt / EtOH (-H2O) 2. NH2NHPh

1.

O

2.

14. MeO

15. acetone

OMe, MeOH O Cl

+

C5H6O

3. H3O

O

OH

1. CH2(CO2Me)2, OMe

(aldol)

2. H3O, ' C6H10O

19.1 Reactions

CO2

C8H12O2

Problems • 211

CHO 16.

+

OH (-H2O)

3-pentanone

CHO

OH

OH retro-aldol

O

17. OH

H

OH

18.

1. methyl vinyl ketone

N

2. H3O 3. NaOH (aldol)

19. acetaldehyde + (XS) formaldehyde

OH

C(CH2OH)4 -- pentaerythritol O PhCHO, H 20.

CHO 21.

+

CH2(CO2Et)2

1. base 2. H

OH

C12H10O4 O CHO 22.

KOH

+ CHO O

(a pentacyclic dione)

19.1 Reactions


23. A reaction in the biosynthesis of the amino acid leucine: O O

SCoA (aldol)

1.

CO2H

2. hydrolysis

24. Trans-resveratrol, isolable from red wine, has been implicated as a cardioprotective and can be synthesized as follows: MeO MeO

CHO 1.

+

CN

OEt + CO2

2. H3O OMe

HO

OH

trans-resveratrol (82%)

O

3. BBr3, RT (hydrolysis)

OH

1. H3O

25.

2. KO-t-Bu (retro-aldol)

26. Forward and retro-aldol-like reactions that occur in plants: O

CO2

O

retro-aldol

CO2

SCoA O2C

HO CO2

-

O2C

CO2

succinate

19.1 Reactions

H

glyoxylate

hydrolysis

malate

Problems • 213

O O OEt / EtOH

+

27.

(Robinson annulation)

O

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 B

A

29. The Krebs Cycle begins with an aldol-like condensation of a thioester (acetyl coenzyme A) with oxaloacetate, followed by hydrolysis:

O

O SCoA

+

O2C

H3O

CO2

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

[O]

H3O SCoA

(conj. add'n)

HSCoA (retro-Claisen)

O +

SCoA

19.1 Reactions


31. Excessive accumulation of acetyl CoA can lead to metabolic ketosis by the following pathway: H3O Claisen A O 2

[H]

- CO2

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 O

Diels-Alder

2. a. LiAlH4 b. H

EtO

C A

B

(complete)

19.1 Reactions

HO H

1. H2 / Ni (1 equiv)

C B

EtO

1. H3O (=> a ketone) 2. methyl vinyl ketone, base (Robinson annulation)

H

OH

Problems • 215

b. Construction of the D ring began as follows. Fill in the bracketed structure and outline the mechanism for step 3. R

X , t-BuO

1.

C O

R 3. mild acid

C H

2. a. EtO C C MgX

CO2Et

b. H

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

C

D 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. O

O

O 1.

C

D

RO

OR

O

O C

H

OAc

O

O

D

OR H

RO

O

2. I2, OH 3. OH, ' 4. KOAc

O

O C

D H

19.1 Reactions


e. The final four steps yielded cortisone. Deduce the structure of cortisone acetate. OAc O

O C

1. HCN 2. (-H2O)

D

3. KMnO4, OH (- HCN)

H

OH O

O

OH

4. H3O

O cortisone

cortisone acetate

___________________________________________________

CO2Me

33.

1.

CO2Me

OMe, MeOH

2. H3O, '

O

O H

34.

NO2

CH3NO2 NaOMe / MeOH

N H C8H7N2O5 Na

H N

O

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: O O

Et3N

longifolene

19.1 Reactions

Problems • 217

19.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. 1.

?

+

via an aldol

?

1,3-diphenyl-1-propanol

O via an enamine

2. cyclohexanone

via an aldol 3.

O

CHO

? OH

4.

via an aldol

?

O

5. diethyl ketone

6. acetone O

CO2R

O

7.

CO2R

CO2R O

19.2 Syntheses


8. cyclohexanone

4-benzyl-1,3-cyclohexadione

9. 1-pentene CHO

10. phenylacetaldehyde

PhCH(CH2OH)2

O 11.

?

+

via a Robinson annulation

?

O

O 12.

13. cyclohexanone, acetone

via a Wittig* O

*Why not via an aldol?

O

O via a Robinson annulation

14. O O

19.2 Syntheses

Problems • 219

O

15.

H

O 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. OH

O

O +

2.

CH2N2

+

N2

O NaOEt / EtOH 3. O

O

O OEt

19.3 Mechanisms


O

O

1. CH3NH2

OCH3

4. 2 methyl acrylate 2.

5. p-chlorobenzaldehyde

+

OMe

CHBr3

N

OH

OH Cl CO2

CO2Et

1. NaOEt, EtOH, MVK 2. H2O, OH

O

3. '

6.

O

O

O 1. acetone, OEt

H

7.

2. H

geranial

O

(a step in the commercial synthesis of vitamin A)

O

O

O

OEt 8.

19.3 Mechanisms

OEt

+ PhCO2Et

Ph

OEt

+ ethyl acetate

Problems • 221

OH

O +

9.

O

OH

methyl vinyl ketone

CHO H

O

O HCO2 +

10. The anaerobic breakdown of glucose (glycolysis) involves the following isomerization and retro-aldol: HO O

HO

H

OH OH

O

OH

OH

O

aldolase

OH H

+

OH

OH

OH

OH

OH

OH

O

D-D-fructose

D-D-glucose

R2N

OH

CO2R

R2N [Br ]

11.

R1

CO2R

CO2R

' %r

(from NBS) NH

1

NH

12. The following illustrates the Stobbe reaction. Hint: a key intermediate is a J-lactone. CO2R

O +

CO2R

base Ph

Ph

Ph Ph

CO2R CO2

19.3 Mechanisms


O O

1. OMe / MeOH

13. O

O O

O

OR

O

14.

2. H

O CO2R

CO2R

EtO

O EtO

O

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 3. H2O

O

O

O

Et

Et

O

O base

CO2Me

18.

CO2Me

O 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 R

HO

O HO

humulone: R = i-Bu

O

H, H2O '

R HO O

OH

cis- and trans-isohumulones

cohumulone: R = i-Pr adhumulone: R = s-Bu

19.3 Mechanisms


21. Woodward’s (Harvard) total synthesis of the alkaloid strychnine included the following steps:

N H

NMe2

1. CH2O, Me2NH

OMe

OMe

H

OMe

N H

OMe 92% 2. MeI 3. NaCN, DMF

CO2Et N

CN 4. LiAlH4, THF

OMe N H

OMe N H

5. ??

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. O P OH O O

OH

O

HO 2.

1. hydrolysis

H OH

CO2 phosphoenolpyruvate (PEP)

an enol 3. - H2O O OH

phenylalanine, tyrosine

HO

CO2H HO

19.3 Mechanisms

4. H

HO2C O

OH

O

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 CO2H

H

NH2

2 HO2C ALA

N H

H3N

porphobilinogen

24. Several steps in Sheehan’s (MIT) total synthesis of penicillin V are shown below. CO2H

1. ClCH2COCl

2. Ac2O N

valine

Cl

O

O

O

SH N H

O

NH

NH2

Me

O

CO2H

3.

CO2Me

isomerization

SH, OMe MeOH

O 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 SR

+

SR O

CO2

+

O

O SR

O

19.3 Mechanisms


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 (1)

O retro-aldol

E D

O

O +

H

R' R N

O

1. thiamine

(2) OH

2. R

H

R

+

O

S

O O

(CB of thiamine) OH

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 3

SCoA

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 HMG-CoA

OH

P O

ATP

a reductase CO2

O PP

CO2

O P O OH O O P O P O O OH

mevalonate P =

PP =

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. P O

O PP - CO2

CO2

O PP

O PP

I-PP

DMA-PP

O N H

OH N

OH

O OH

F LipitorTM

19.3 Mechanisms


CHAPTER 20 AMINES 20.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate.

1. phthalimide

1. base 2. ClCH(CO2Et)2 3. base 4. isopropyl chloride 5. H3O

alanine

1. (XS) CH3I 2. Ag2O, H2O, ' 3. OsO4 2. 2-methyl-2-hexylamine 4. NaHSO3 5. (XS) COCl2 6. NH3 C9H18N2O4 - MiltownTM 1. (XS) MeI 2. Ag2O, H2O 3. N-ethylcyclohexylamine

3. '

1. (XS) MeI 2. Ag2O, H2O

4. N

3. '

H coniine (toxin in hemlock, killed Socrates)

HO 1. (XS) CH3I

5. O

2. Ag2O, H2O, ' N

HO morphine

Me 6.

Ph2N

H

H

Me

1. H2O2 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 H NMe2 H

9.

Hofmann elimination

Cope elimination

H 10. 3-pentanone + dimethylamine + formaldehyde

(Mannich rx)

NH2 N

11.

HONO N H

O

H2O uracil

cytosine

12.

1. Li

Ph Cl

2. CO2 3. H

4. SOCl2 5. NH3 6. Br2, OH, H2O phentermine (a diet drug)

20.1 Reactions

Problems • 231

OH 13.

Br2, OH

NH2

Me

NH3 +

H2O

O

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.

2. Ag2O, H2O, '

MeO

3. H EffexorTM (antidepressant)

Ph OH

17.

(gives a positive DNP test)

1. SOCl2 2. NH3 3. Br2, OH, H2O

O OH

4.

I (2 equiv) DetrolTM (treatment of urinary incontinence)

18.

CO2H

1. a. Br2, PBr3 b. H2O 2. KO-t-Bu 3. HCN, CN 4. H2 / Pt pregabalin (LyricaTM - first treatment approved for fibromyalgia)

20.1 Reactions


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 2. Cl2, Fe 3. H2O, OH

H N 22.

O

4. I-Cl, Fe 5. HONO 6. H3PO2

1. NaH 2. H3O 3. NaBH4

CO2Me 23.

O

+

N

N

4. HBr (SN1) 5. ' (-HBr)

O

HO

H

1. 24.

O

OH NH

HO

nicotine

NaBH3CN, EtOH 2. HCl 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

NH2

2. cyclohexane

3. isopentane

3-methyl-1-butene

4. 3-methyl-1-butene

isopropylamine

OMe

OMe MeO

(via a Hofmann elimination)

OMe

MeO

OMe OMe

MeO

OMe

5.

NH2

O

H

NHPh

mescaline (from peyote cactus)

6. p-nitrotoluene

7. p-nitrotoluene

p-nitrobenzylamine

p-nitroaniline

20.2 Syntheses


O 8.

9.

NH2

HO3S

O

NO2

HO3S

N N

NEt2

methyl orange

N CO2Me

O

N

10.

OCOPh cocaine

11. toluene

tropinone

O

2,6-dichlorotoluene

O 12. p-nitroaniline

HO

N H

acetaminophen (TylenolTM)

O 13. 1-nitro-2,6-dimethylbenzene, ethylene oxide, diethylamine

N H

lidocaine

14. benzene

20.2 Syntheses

anisole

NEt2

Problems • 235

NO2

15.

N N

NH2

butter yellow

16. benzyl methyl ketone

H N

Ph

methamphetamine

Me

Me

H

Br

N

N O 17.

S

S S

O O

S

O

OH

O

OH

tiotropium bromide (SpirevaTM - bronchodilator)

N

OH

OH

, benzene

18.

NO2

N

para-red dye

D 19. benzene D

OH

OH

20. NHAc

propofol (intravenous anesthetic)

20.2 Syntheses


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 1. butanamide + phenol + Br2

N

OH

OPh O

O 2.

1. Cl2, H2O, OH

CO2H

2. H

NH2

NH O

anthranilic acid

CO2H

1. HONO 2. adjust to pH 8

NH2

3. 1,3-butadiene

3.

O 4.

1. HCN, CN 2. H2 / Pt

O

3. NaNO2, HCl

O 5. cyclohexanone

1. NH2OH 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 1. PhMgX

HO

O N

Ph 1. (XS) MeI N

2. Ag2O, H2O, '

2. H

CH3O

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

NaN3 SN2

O

H3O

'

NH3

H

- 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

HO HN3

O

N3 '

NH

20.3 Mechanisms


9. The Fischer indole synthesis involves an isomerization known as a [3,3] sigmatropic rearrangement, shown by the arrows below: O H2SO4 PhNH-NH2 + Ph N Ph H N N Ph H 2-phenylindole H tautomerization

[3,3] H

N N H

N

Ph

H

N H

Ph

Outline a mechanism for conversion of the intermediate in brackets to the indole product.

10.

11. acetophenone

CH2O NH2

+

NH

H

CH2O

+

NH3

O

H 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

O

N 1. [H]

CHO + H2NMe + CHO

2. esterification

CO2 CO2

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. N + acetonitrile H

O O CN

O N H

b. MeO

H2SO4

t-Bu

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


15. The Corey-Link reaction (step 2) may be used to prepare D-amino acids: O

OH

1. LiCCl3 R

H

R

H

N3

2. base

CCl3

R

3. NaN3, MeOH

CO2Me

H

NH2

4. H2 / Pd R

5. hydrolysis

H

CO2H

a. Outline a mechanism for steps 2 and 3.

b. Account for the product in a mechanistically similar reaction:

1. :CCl3

O

2.

HO

NH2

O

O

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 N

N

O

[H]

NH

N

H

N

H C C

[O] Ph

H

DeprenylTM

20.3 Mechanisms

H N

R N

O NH

N reduced flavin H

O

flavin cofactor

R N

O

H C

H oxidized drug

O

C H H C N Ph

Ph

O NH

N

N

H C C

H N

SOLUTIONS TO PROBLEMS

CHAPTER 1 THE BASICS 1.1 Hybridization, formulas, physical properties 1. a. SeldaneTM: C32H42NO2

RelenzaTM: C12H20N4O7 c

HO

OH

a

OH O

HO

b.

O

N

b

OH

OH N H

NH

O

2

d

NH H2N

c. a: sp3 – sp3; b: sp3 – sp2; c: sp2 – sp2 2. a. :C C:

d. SeldaneTM oxygens: sp3; nitrogen d: sp2

b. H C O:

c. O N O

d. the conjugate base of :NH2CH3 -H H3C N

Cl e.

N H

O

O

O

f.

H

H

3. a.

H

:C

sp2

H

b.

=> bond angle ~ 1200

H

H

=

CH2

a linear HCH bond angle implies sp hybridization; therefore, each lone electron lies in an unhybridized p orbital with spins aligned (Hund's rule) G+ N H

4. a. higher bp:

H N G-

b. lower mp: catechol O G+ H O GH

this isomer is capable of intermolecular H-bonding, thereby increasing intermolecular attractive forces and raising its bp relative to the other amine

5. a. no

b. no

6. a. CHCl3 H C

Cl Cl

d. yes

e. yes

f. no

g. yes

F vs.

P

C Cl Cl Cl

N

H H P H3C

h. yes c. SO2

b. CH3NO2

P

Cl

c. yes

catechol, unlike hydroquinone, can undergo intramolecular H-bonding, which decreases intermolecular attractions and results in lowering its mp relative to hydroquinone

vs.

O N

CH3

P

O

O C O linear, P = 0

vs.

S O O bent, P > 0

permanent charge separation (> H)

7. a. penicillin V: C16H18N2O4S b. arrow a: sp2

b: sp3

cimetidine: C10H16N6S c: sp



O c. Ph

O N H

N H 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; O N S H N O

cimetidine: 8. N: C

HN

N S

N

N H

O

N 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 N

O S MeHN O

O O

NMe2

9. a. RozeremTM: C16H21NO2

HO

ChantixTM: C13H13N3

b. lone pairs: RozeremTM: 5; O

OH

RitalinTM: C14H20NO2

ChantixTM: 3;

RitalinTM: 4.

N

N H

O

H

H N

NH N

10. lone pairs: theobromine: 8; O HN O

O

melamine: 6. NH2

CH3 N

N

N N CH3

11. a. alkene, amide, amine, ester, ether

H2N

N N

NH2

b. alkene, amine, arene, carboxylic acid, halide, ketone

c. alcohol, alkyne, arene.


O

Solutions • 245

1.2 Acids and bases 1. strongest base in ammonia: H2N

(H

H2N

:H

3. a.

O

b. Ph3P:

BF3

O

N

O

BH3

N BH3

CH3 O CH2CH3

CH3CH2 Cl

+

Cl

LA

H2C CH2

BF3

LB

LA

H3C O

AlCl3

O

Ph3P BF3

LB

c.

nitrogen is more electron-releasing than oxygen

AlCl3

4. a. CH3 O

b.

H2

+

H2N

2. stronger base: (CH3)2NH

c.

amide anion - the CB of ammonia

CH2 CH2 BF3

:CH2 CH3

(H

+

H3C O

H3C CH3

LA LB (also Bronsted-Lowry acid-base, respectively) d.

Cl Cl

AlCl3

LB

LA

Cl

+

AlCl4

S e.

CH3 N C S

:NH3

LA

CH3 N C

LB

pKa ~ 35 O 5. a. C20H28O

H) NH2

b. C

pKa ~ 16 O H +

NH2

C

C

C

H WB

NH3

WA

SA

H

c. Keq 4)

7. lowest pKa: b.

a. ~ 16

8. quantitative rx: b.

b. ~ 5

R C C (H pKa 22

c. ~ 16

d. ~ 10

e. ~ 38

NH2

Keq >> 1

SB

a. hydroxide (Keq 4)

1.3 Resonance CH2 1.

O

CH3 C O sp2

p*

H p* sp3 per

* not VSEPR; electrons may resonate if housed in a p orbital

1.3 Resonance

N sp2

Solutions • 247

-H

H) O C N:

2. lower pKa: H-O-CN

O C N:

O C N:

negative charge delocalization => more stable CB -H

H) C N:

HO C NH2 H

HO C NH2 H

3. a. 3.

negative charge localized on C

:C N:

HO C NH2 H

localized charge

b. 1.

saturated, no p orbital O

O

O

O

O

c. 5.

O

O

O

O

O

OCH3

OCH3

OCH3

OCH3

OCH3

CH2

CH2

CH2

CH2

CH2

d. 5.

NH 4.

Me

most basic +H

N H

NH2

H Me N H

NH

H

NH

NH2

Me

vs.

N H

NH2 H

vs.

Me

NH

N H

NH2

no important resonance contributions H 4 resonance structures => a more stable CA

Me

N H

H

NH Me

NH2

N H

H N 5. a. O O O

O O O oxygen 'octetted,' closer charge separation

H

NH NH2

Me

NH

N H

NH2

H N

b. carbon 'octetted,' additional S bond

1.3 Resonance


OH H

H

C C N: H negative charge is borne by more electronegative atom

C C N:

c.

OH

H

d. carbon 'octetted,' additional S bond

O 6. a. 2. N H

N H

O

O

b. 3.

CH2 c. 1.

d. 3. O H

(Ha

O

O

7.

O

O

-Ha O

A

O

O

O

O

O

CB of A is stabilized by charge delocalization over three nuclei and, therefore, more easily formed => pKa of Ha is lowered O O

O +H

8.

OH

H

charge localized

H

vs. O

H

O

OH

H OH

charge delocalized, => > stability therefore, more favored CA species

9. a. 5.

NMe2

NMe2

NMe2

NMe2

H

H

H

NMe2

b. 5. H

c. 4. N

1.3 Resonance

N

N

N

H

not N nitrogen 'sextetted'

Solutions • 249

O d. 2.

O

O

e. 3. O

O

f. 4. N

N

CH2

CH2

N

CH2

N

CH2

g. 4.

OCH3

OCH3

OCH3

OCH3

h. 4. H

H

Cl

Cl

H

H

Cl

Cl

i. 2. O

O

O

O

O

O

vs.

10. A

B these additional resonance structures increase both H and d (P = H x d); therefore, PB > PA

N

N

S

S

O

N

N

S

S

O

11. O

O CB of oxyluciferin

1.3 Resonance


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)

3. 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

a. 2,3-dimethylpentane: 6.

Ha H Hc e H f Hb

Ha b. 2,4-dimethylpentane: 3. Hb

Hd Hb

Hc

Hc

c. 3-ethylpentane: 3.

d. 2,2,4-trimethylpentane: 4.

Ha

Hd

Ha

Hb

Hc

He Ha e. 2,5,5-trimethylheptane: 7.

Hb

Hd

Hd

Hf

Hb f. 4-ethyl-3,3,5-trimethylheptane: 10.

Hc

He

Hg

Ha

Hg

Hi

Hf Hc

Hh

Hj

2.1 General


2.2 Nomenclature I NO2 1.

1

2.

5 9

1

7

Br

3-nitro-4-ethyl-2,2,5-trimethylheptane

7-bromo-2-iodo-3-ethyl-5,6-dimethylnonane

7 4

3.

4. 1

3

8 1

4-ethyl-3,3,5-trimethylheptane

5-ethyl-3,5-dimethyloctane 8

4

5.

7

6.

F

1

4-fluoro-2-methyl-2-phenylheptane

5-ethyl-3,4-dimethyloctane

5

4

7.

9 (1)

1

5

1

8.

8

3

(3) 1

2,3,7-trimethyl-4-n-propyloctane

5-(1,2-dimethylpropyl)nonane

(choose path with more branching) 1

I

10.

9. 9

1-iodo-4-methylpentane

2,3-dimethyl-4-n-propylnonane

1

11.

7 4

(1)

10 (3)

12.

4 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

7

5-ethyl-4-methyldecane

3

15.

16.

10 1

diethylpentane (3,3- not necessary!)

3,7-diethyl-2,2,8-trimethyldecane

OH

17. a.

b.

Br

c.

O


1.

3.7 kcal/mol

Br H

3.7 kcal/mol -2.0 (2 x 1.0 kcal/mol) 1.7 kcal/mol

H H

PE

HH 1.0 kcal/mol

rot'n

H H

2

2. a.

=

Me

3

1

H b.

=

2

Me

OH 3. a.

Me H

H Me

H

OH

= H

H

largest R-groups are anti H

H Et Ph

H

H Me

H

isobutyl alcohol

b.

gauche: dihedral angle ~ 60 0

s-Bu H Et t-Bu H Ph

H =

Ph

1

7

H 4-t-butyl-3-methyl-5-phenylheptane



H

PE

H

4.

Me

H Me

Me

dihedral angle only 600 between both methyls

rot'n about C2-3 bond

5. intramolecular hydrogen-bond stabilizes a nearly eclipsed conformer for FCH2CH2OH

H FO vs. H H

HH

P >> 0


F H H

H F P~0

H

CHAPTER 3 CYCLOALKANES 3.1 General P! 1. highest molecular dipole moment: d.

Cl

a.

Cl

Cl b.

Cl

Cl

H

H

H

H

Cl

c. Cl C C Cl

P=0 Cl

Cl

Cl

2. constitutional isomers for a. dichlorocyclopentane: 3.

Cl

Cl Cl 1,1-

1,3-

1,2-

b. C6H12 that contain a cyclopropyl ring: 6.

Cl 3. cis/trans stereoisomers for a. dichlorocyclopentane: 2 pairs.

Cl Cl

+

trans-

Cl

1,2-

+

1,3-

Ph Ph

Ph +

b. diphenylcyclohexane: 3 pairs.

cis-

cis-

+

Ph

trans-

+

1,2-

1,3-

1,4-

Cl

Cl

cis-

Ph

Ph Cl

Cl

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

Hg

4. different kinds (constitutional and geometric) of hydrogens in a. 1-ethyl-1-methylcyclopropane: 5.

Hf

Hc

Ha

b. allylcyclobutane: 9.

Ha

Hb

Ha Hc Hf Hb Hd

d. chlorocyclopentane: 5. Ha Hb

Hc

Hd

Cl He

Hi He

He Hd Ha and Hb are cis- and trans- to methyl, and so are 'different'

c. methylcyclobutane: 6. He

Hh

Hc

Hb Hd

e. vinylcyclopentane: 8. Ha Hb

Hf

He

Hc

Hd

Hg Hh

3.1 General


H H H 5. least strained: a.

b.

H H

H transno strain

H

H

H d.

c. ring strain prevents alkene from being planar violates Bredt's rule

H

H

cisdiaxial strain

angle strain

Me Me

6.

H

H Me

Me

exo-

endostereoisomers

H

Me H

Me +

7. only 2.

=

=

H

Me

Me identical structures (for now!)

3.2 Nomenclature 4

1

1.

1

4

2.

2

1-cylclopropyl-3-methylbutane 4-s-butyl-1-ethyl-2-n-propylcyclohexane

1

(2)

3.

4. t-pentylcyclopentane (1,1-dimethylpropyl)cyclopentane 2-cyclopentyl-2-methylbutane

3.2 Nomenclature

(1)

4

7

4-(2-cyclohexylethyl)-3-methylheptane

Solutions • 257

5.

6. Br trans-1-bromo-2-s-butylcyclopentane cis-1-isopentyl-5-n-propylcyclodecane

Cl 7.

F (4)

(1) (2)

1

8.

6 3

(2-chloro-1-methylbutyl)cycloheptane 1-fluoro-6-t-butyl-3-vinylcyclooctane 1 (2)

9.

10.

I

(1)

6 3

9

cis-1-allyl-2-isobutylcyclohexane

5-iodo-3-(1-cyclobutylethyl)-6-ethyl2,2,8,8-tetramethylnonane 6 1

F

2

Ph

11.

12. 7

5

trans-1-fluoro-3-phenylcyclohexane 2,6,6-trimethylbicyclo[3.1.1]heptane

6

13.

5

7

14.

10

1

4

2 2

1

9

5-methylbicyclo[2.1.0]pentane

7-allylbicyclo[4.3.1]decane

2

15.

(1)

16.

(4)

9

1

8 7

2,9,9-trimethylbicyclo[5.2.0]nonane

trans-1-(2,3-dimethylbutyl)-2-n-propylcycloheptane

3.2 Nomenclature


3.3 Conformational analysis, cyclic H

H

OH

H

Me

1. most stable conformer:

Me

O H H

H

menthol

H

neomenthol Ph

2. a.

Ph Ph H

degenerate structures (same energy) therefore, Keq = 1

Ph

H H

H

b.

H

H H > 1,3-diaxial interactions => less stable conformer therefore, Keq < 1

H

H H c.

H

Me

Et

H Et

Me larger ethyl group in more stable equatorial position therefore, Keq > 1

3. most negative 'Hcomb (=> least stable):

a.

t-Bu

t-Bu

(most stable, therefore, least negative 'Hcomb)

least stable

Me

Me

t-Bu

Me

Me Me

b. Me

> dimethyl repulsion therefore, least stable

c.

least stable


(most stable)

Solutions • 259

4. least negative 'Hcomb (=> most stable): a.

= all alkyl groups are equatorial therefore, most stable

b. same compound!

H 5. most stable conformer for:

H

Cl trans-

t-Bu

vs.

H

t-Bu

:B

t-Bu

H

(-HCl)

Cl

cis-

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 6.

OH

F

vs.

OH

F

F

trans-

cistwist-boat stabilized by intramolecular hydrogenbonding; not possible in chair conformer (or in trans-isomer)

HO O

HO

7. a.

OH

HO OH OH

O

HO

HO

OH

HO 2

1

b. configuration 2 is less stable (one substituent must be axial) and would burn with a more negative 'Hcomb

H

Me Me

8.

Me

Me

5.4 kcal/mol -0.9 (2 Me-H 1,3-diaxial strain interactions) -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.

1

2

3

4

5

6

7

8



b. for conformational chair-chair flipping, Keq = 1 for configurations 1, 3, and 5: 3e/3a

2

1 3e/3a

3

4e/2a

3a/3e

4 5e/1a

3e/3a

5

6

7

8

3e/3a

4e/2a

4e/2a

6e/0a

c. least stable: 1.

three 1,3-diaxial steric interactions exist between two methyl groups (only one such interaction exists in configurations 3 and 5)

1

d. least likely to flip: 8.

8 all methyls are equatorial


all axial!

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 rx

A

B

+3 kcal/mol

B

C

+7

A

C

+10 kcal/mol

TS 'G for rds

intermediate -'Go / RT b. Keq = e

3. a. 'Go 'Go =

c. rate = k[conc. term(s)] = koe

-'G / RT

rx

'G for rds

4. a.

B b. 'Go = -RT ln Keq

'Go

-'G / RT Keq = e = e -2,500 / (2)(300)

Br3C:

Keq = 1.6 x 10-2

A rx

'Go = +2.5 kcal/mol

4. Reaction Basics


5. a. type of reaction: rearrangement;

mechanism: polar / ionic. H

O

O

O

H) OH2

b.

H

O

O

O +

H)

A

B

H2O

c. Keq = [B] / [A] = 75% / 25% = 3.0;

H OH2

'Go = -RT ln Keq.

d. nucleophiles: A, H2O, and B.

e. TS

intermediate 'Go 'G0 = rx

6.

I

fast

rds

+MeOH, -H

-I

7.

OMe

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 1. a.

1

3

b.

H Ph

Cl

8

H H

= H

H

1 2

H Ph

5

H

trans-5-phenyl-2-pentene

(E)-3-chloromethyl-4-s-butyl-2methyl-3-octene 1 3

c.

9

3

1

9

d. 7 6

3-n-propyl-1-nonene

2. a. (Z)-

b. (E)-

6-methylbicyclo[5.2.0]-3,8-nonadiene

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-

OMe

cis, transOMe

5. most stable carbocation: H N

O

6. a. no. deg. unsat: C17H36 - C17H20(18+3-1) = H16 => H16/2 = 8 deg. hydrogenation: C17H30F3NO - C17H18F3NO = H12 => H12/2 = 6 DB no. rings = 8 - 6 DB = 2.

F3C

fluoxetine HN

b. no. deg. unsat: C17H36 - C17H16(18+1-3) = H20 => H20/2 = 10 deg. no. DB = 10 - 4 rings = 6.

N

N OH

F O

O

CiproTM

5.1 General


N OH

c. no. deg. unsat: C28H58 - C28H34(35-1) = H24 => H24/2 = 12 deg. no. rings = 12 - 5 DB -1 TB(= 2 DB!) = 5.

C C

O

RU 486 O

d. no. deg. unsat: C17H36 - C17H10(14-0-4) = H26 => H26/2 = 13 deg. no. DB = 13 - 3 rings = 10.

O O S O

rofecoxib F

e. no. deg. unsat: C19H40 - C19H16(17+2-3) = H24 => H24/2 = 12 deg. no. rings = 12 - 8 DB = 4.

N

Cl

O S

N H O

N

O

floxacillin

O HO

H N

f. no. deg. unsat: C19H40 - C19H20(20+1-1) = H20 => H20/2 = 10 deg. hydrogenation: C19H32FNO3 - C19H20FNO3 = H12 => H12/2 = 6 DB no. rings = 10 - 6 DB = 4.

O

O

PaxilTM

F

7. number of stereoisomers for 2,4-hexadiene: 3;

O

for 2-chloro-2,4-hexadiene: 4. Cl Cl Cl Cl

Cl 8. a.

Cl

=>

b.

propylene dichloride (note: no double bond)

(Z)-3-methyl-2-phenyl-2-hexene Br

OH

OH c.

d.

, NOT Br styrene bromohydrin

5.1 General

OH

O e.

OH trans-cyclohexene glycol

isobutylene epoxide

Solutions • 265

+H

9.

1

1,2-R: shift

1

2

+Cl

2

3

H 'G0 'G0 =

1,2-H: shift

Cl

3

rx

+H

10. a.

CA

H

H H

most important contributing resonance structure

H

H 1,2-H:

+H

b.

shift most stable intermediate: benzylic carbocation

H

H ~H: 11.

H

H rigidity of the carbon skeleton prevents carbocation from being planar

HF

12. neither regiospecificity nor stereospecificity: a.

F + F

Br Cl2

b.

Cl

regiospecific

(XS) NaBr Cl c.

Cl2 / H2O

d.

anti-add'n

D2 / Pt syn-add'n

OH stereospecific

D D stereospecific

5.1 General


H +H

13.

E-pinene

-H

H

D-pinene more highly substituted double bond => more stable olefin therefore, Keq >> 1

5.2 Reactions HCl

1. Ph

+H

+Cl

1,2-H: shift

Ph

Ph

Ph

Cl

O NO2

HI

2.

NO2

N O

I

+H

3.

+H2O

1,2-R: shift

OH

-H Br 1. Cl2, '

4.

3. Br2

2. KOMe

Cl

MeOH

NMe3 5.

NMe3

HI

Br

CHCl3

NMe3

1,2-H:

NMe3

shift Et

6.

F

HF

Et

Et

Et

I

F

F

F F

7.

5.2 Reactions

DCl

D

1,2-H: shift

D

Cl

D

Solutions • 267

HBr

8.

1,2-R:

again! Br

shift

+H

9.

-H O H

O H HBr

CCl3

10.

O

O (H

CCl3

CCl3 Br

Et

Et

Et

Br D

DBr

11.

D H

Et D

+

H anti-add'n

2. Br2 / hv

1. H2 / Pd

12.

Et

Br

Br D

= H

H syn-add'n

Br

OEt +EtOH

+H

13.

-H H

HF

14. MeO

H

F

Cl

MeO

D

HI

H

F

D

+ OMe

D

D

syn-

15.

OMe

Cl

anti-

Cl

Cl

I Cl 16.

Cl2 / H2O OH

17.

1. B2D6

2. H2O2, OH D BD2

D OH

5.2 Reactions


Cl2

18.

G+

I

Cl

Cl

G+

I Et 19.

20. H2C C CH2

Et

Et 1. Hg(OAc)2, PhOH

2. NaBH4

OPh

OPh HgOAc

(XS) CH2I2 Zn(Cu)

spiropentane CHO

1. KMnO4, OH

21. AcO

2. HIO4

AcO

OH OH

AcO CHO

1. O3 22.

O

2. H3O, Zn H

O

HBr

23. Ph

Ph

Br

+HBr -Br

Br H note: opposite regioselectivity than HBr without peroxide and no rearrangement

O O

Ph

Et H2C N2 24.

+

hv

N2

Et transOH

25.

26.

Br

2. OsO4

1. base

OH

-HBr

I N3

3. NaHSO3

I

I N N3

N N

5.2 Reactions

Solutions • 269

.

1. BD3 THF

27.

2. H2O2, OH

HO

O 28.

O

O

+H

HO

D

OH

O

+MeOH H

H

OMe

-H

H +H

29.

-H H)

O

OH HIO4

30.

O H

+

OH

H

H

OH

OH 1. H2SO4

31.

2. KMnO4, OH

-H2O OH

O

H C C O

32.

HCl

H3C C O

H3C C O

H

OR 33.

H3C

Cl

Cl

OR

.

1. BH3 THF 2. H2O2, OH HO

5.2 Reactions


OH

Br Br

1. Br2, H2O

34. HO

OH

+ HO

HO

O a halohydrin 2. base -HBr

HO

H) +H

35. O

-H HO

H O

HO

OH

O 36.

Cl

1. OR

3. EtOH, H

2. mCPBA

-HCl

trans-

Br 37.

OEt

Br +s-BuOH -H

Br2

O

HO

OH O OH HIO4

38.

+ O HO

HO

H

O OH 39.

5.2 Reactions

KMnO4, H

O

OH O largest C-containing product

Solutions • 271

O OH 1. OsO4

O

40.

OH O

OH

O

O

O

3. HIO4

O

2. NaHSO3

O

H

O

OH

O

O

+

O

H O HCl

41.

1,2-R: shift

+H

1,2-H: H shift

Cl

+

Li (H 42. Ph CH Cl

:CH2 -H

Ph CH Cl

-Cl

Ph

PhCH

a 1,1- (or D-) elimination to produce a carbene

H3C 43.

Cl

O

Cl

O

OPh

(CH3)2CI2

Cl

(1 equiv) Zn(Cu)

Cl

CH3 O

OPh

O

more electron-rich (nucleophilic) double bond

5.3 Syntheses Br HBr

1.

OH 2.

3.

1. H2SO4

+Br

1,2-H: shift

Cl

2. HCl

-H2O

Cl

1. KOH, EtOH -HCl

2. HF

F

5.3 Syntheses


Cl 1. OMe, MeOH

4.

2. H2 / Pt

-HCl D 1. Cl2, hv

5.

3. D2 / Pd

2. NaOEt / EtOH

Cl

D

-HCl Cl 1. Cl2, '

6.

CO2H

3. KMnO4, H

2. KO-t-Bu

CO2H

t-BuOH

Cl

Cl2 7.

Cl

(XS) NaBr

Br Br H

1. O3

8.

O

2. Zn, H

1. CH2N2, hv

9. H2C CH2

10.

OH

2. Br2, hv

1. H2SO4

Br

2. CH2I2, Zn(Cu)

-H2O OH

1. Hg(OAc)2, H2O

no rearrangement

11.

2. NaBH4 OH H3O

+H2O

1,2-H: shift

-H O

Br 12.

2. O3

1. KOMe MeOH

3. Zn, H O

13.

1. Br2, hv

Br

2. NaOH

3. DBr

D Br

5.3 Syntheses

Solutions • 273

1. KO-t-Bu

14.

2. HBr

t-BuOH

Br

1. Br2, '

15.

Br

peroxide

Br

.

3. BH3 THF

2. KOH

OH

4. H2O2, OH Cl 1.

O

2. KMnO4, H

16.

CO2H CO2H

OH Cl

17.

Cl

1. OEt, EtOH

Cl

2. Cl2, H2O

-H OH2

18.

1. H

1,2-R: shift

Cl

+H2O OH

-H

2. O3 O

19.

1. Hg(OAc)2, EtOH

1. H3O

3. Zn, H

(not EtOH, H => rearrangement) OEt

2. NaBH4

20.

O

OH

2. H2SO4

3. HBr, ROOR

Br

Cl 21. H2C CH2

1. CH2N2, hv

2. Cl2, '

3. NaOH MeOH

4. KMnO4, H

CO2H CO2H

5.3 Syntheses


22.

Br

1. KOMe

2. H3O

OH

MeOH

+

OH

3. H

O

3. Hg(OAc)2, t-BuOH 4. NaBH4

or

2

OH

2. CH2I2, Zn(Cu)

1. H2SO4

23.

O 2

5.4 Mechanisms

+H

1.

-H

1,2-R: shift

(H

H

H) -H

+H

2.

H

H

H)

Me O

3.

OMe -H

+H H Me

O

H

H +H

4.

+I2

5.

-H

1,2-R: shift

(H

I

I

I

-H

-I CO2H

5.4 Mechanisms

O C O H

O

O H)

O

O

Solutions • 275

H +H

6.

-H H)

H +H

7.

-H H)

HgOAc

Hg

OAc

HgOAc O (H

1. Hg(OAc)2

8. OH

OH

:H -H

HgOAc O

O

-H

1,2-H: shift

+H

9.

10.

2. NaBH4

H N

+I2

H N

(H

H N

N

-H

-I I I

I

I

I

H2C N N 11. H2C N N

N

N

N N

5.4 Mechanisms


Br

Br 12.

Br

anti-attack

+H

Br 100% trans-

Br

H

+H

+Br syn- + antiattack

+

cis- and trans-

Br +Br

1,2-H: shift

Br

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

OH

incorrect: cannot exist in cis/trans forms

isocaryophyllene

.

14. O

1. BH3 THF

1. Hg(OAc)2, H2O

2. H2O2, OH

2. NaBH4

O +

OH

[H]

ozonolysis H

C5H12

H2 / Pd (partial)

15.

A isoprene

+

+

D

C

B

16. no. deg. unsat: C10H22 - C10H16 = H6 => H6/2 = 3 deg. C10H16

H2 / Pt

hydrogenation: C10H22 - C10H16 = H6 => H6/2 = 3 DB C10H22 1. O3

O +

2. Zn, H myrcene

5.4 Mechanisms

acetone

O

O 2

H

H

+

formaldehyde

H

H O

A

O

Solutions • 277

O 17. hydrogenation: C10H20 - C10H16 = H4 => H4/2 = 2 DB B A

18. a. n Ph

Ph

+H

H H other ozonolysis product

poly(methyl methacrylate)

CO2Me

b.

A =

Ph

Ph

Ph

Ph

Ph

Ph

Ph etc.

H

O 19.

+H

O

O O

etc.

20.

23.

CH2

H2C N N

CH2

22.

t-Bu t-Bu O

O

O

H2C N N

21. CH2N2

O

O

H

t-Bu

N N

hv

H2C CH2

+

+

N2

N N

H2C: +

-N2

+H

+H

1,2-R:

1,2-R:

shift

shift

1,2-R: shift

-H

H)

(H

-H

5.4 Mechanisms


R

R'

S

H3C

24.

D

D

D

CH3

CO2H

D

-RSR'

D

D

D

-H

CO2H

C8H17 Hg

OAc

HO

1. Hg(OAc)2

+H2O

H2O

-H

O 2. NaBH4

O

D

H2C H)

C8H17

25.

1,2-D: shift

C8H17

C8H17

HgOAc

+HgOAc H) O

HgOAc

HgOAc -H

HgOAc AcO Hg

(H +H

26.

-H

R 27.

CO2H

=> trans- DB R'

elaidic acid

OH 28. a.

A

B

b.

A

Br2

C

NR (deep red color of bromine persists);

B or C

or Baeyer test: KMnO4 A NR (purple color of MnO4 persists);

B or C

5.4 Mechanisms

Br2

KMnO4

color discharged

brown ppt (MnO2) forms

Solutions • 279

OH

29.

OH2

+H

-H2O

rot'n

-H (H

30. a., b. R N N N

+

R N

A

N2

a nitrene

c. retention of configuration suggests a concerted (or pericyclic) mechanism

1,2-H: shift

+H

31.

H

1,2-R: shift H) 1,2-R: shift

-H

(H

:B

O

O 2. base

1. Cl2, H2O

32.

-HCl

Cl Cl

O

H

+Cl

3. dry HCl

OH

33. In

Br

CCl3

Cl

InBr + Cl3C

+H

Cl3C Br

etc.

CCl3

Br CCl3

+ Cl3C

(this mechanism is similar to the addition of HBr in the presence of peroxides)

5.4 Mechanisms


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- DB at C10 1. O3 12

D

10

D

D

A

35. D

D

D

vs.

+

D

D H

D

C

2. Zn, H

D

+H

+

B

OH

H

path a

H H

not observed

-D D

D

(H D

path b

D

D D

-H

observed

1,2-D: shift D

D

D

D

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 hydrogendeuterium exchange.

-H H

O

A

5.4 Mechanisms

O

O

CHAPTER 6 ALKYNES 6.1 Reactions 1.

1. NaH

H

D OPh

H, HgSO4, PhOH

2.

3.

2. D2O

Ph

1. B2H6

Ph C CH

2. H2O2, HO

1. OMe (E2)

Cl

4.

taut

H

H

Ph

OH

O 3. (XS) NaNH2 4. BH3.THF

Cl Cl

2. Cl2

OH

taut

H

5. H2O2, OH

O

RC C

5.

+

Cl

RC CH

1. Li / NH3

6.

3o R-X => elimination, not substitution!

2. HBr, ROOR Br

(peroxide effect)

7.

1. H2 / Pd(Pb)

C CH

2. BH3 OH

3. H2O2, OH 1. NaH

8.

2. CH3(CH2)12Cl 3. Lindlar catalyst (cis-[H])

Cl

9.

1. (XS) NaNH2

Et C C

2. H3O, HgSO4

OH

O

taut

Cl

10.

C CH

Cl

Cl2 / H2O HO

taut Cl O

6.1 Reactions


Me 11.

I

1. (XS) HI

Ph

Ph

Cl

..

2. Zn(Cu)

[PhCCH3]

I

Cl C

1. (XS) NaNH2

12.

C

OMe 1. LiNH2

O C C

(2 equiv)

C

C

D

2. D2O

OMe

13. HC C CH2OH

Ph

OMe

2. n-C5H11Br

O C C

(1 equiv) less stable anion therefore, more reactive

3. H OH C C

1. NaNH2 (1 equiv)

14. HC CH

HC C

3. NaNH2

n-Pr

C C

2. n-Pr-I

4. +

C C H

Cl

elimination, 3o RX!

not alkylation

6.2 Syntheses Br 1. (XS) NaNH2

1.

Et C C Et

O

2. KMnO4, H

OH

Br 1. NaH (1 equiv)

3. H2 / Pd(Pb)

2. HC CH

4. HBr, R2O2 Br

2. n-PrBr

3.

Cl

1. (XS) NaNH2

1. NaH (1 equiv) 2. Et-I 4. HC CH

Et C C 3. NaH

6.2 Syntheses

C C

4. n-Bu-I

2. H, HgSO4, MeOH

Et C C

n-Bu

5. Li / NH3 trans-[H]

O

Solutions • 283

5.

C CH

1. H2 / Pd(Pb)

2. HCl

Cl

1,2-R: shift

1. Br2

3. H2 / Lindlar

6.

cis-[H]

2. (XS) NaNH2

7.

1. NaNH2

3. H2 / Pd(Pb)

4. CH2I2 Zn(Cu)

2. Me-Br 1. Li / NH3

2. HBr, (t-Bu)2O2

8.

Br

or H2 / Lindlar OH

H, HgSO4, H2O

O

taut

9. 1. BH3.THF

taut OH

2. H2O2, OH

O H

1. Li / NH3

2. O3 O

3. Zn, H

10.

1. Cl2

C

H

C C

3. Et-I

C

2. (XS) NaNH2

11. Ph C C Ph

Et 4. Li / NH3

1. H2 / Pd(Pb)

Ph

Ph

2. CH2N2, hv O

Ph

1. Na / NH3 Ph

12.

Ph

trans-[H]

2. O3

Ph Ph

H

3. Zn, H Et

1. Na / NH3 Et

2. CHCl3, KO-t-Bu => [:CCl2]

Et Cl Cl Et

6.2 Syntheses


1. KOH 2. Cl2

13.

Cl 3. (XS) NaNH2

14.

HC CH

n-Bu Et

n-Bu

4. EtCl

CHCl3 Et

5. Li / NH3

1. NaH (1 equiv) n-pentyl

HC C

3. H3O, HgSO4

base

O

OH

15. HC CH

n-decyl C C 3. NaNH2

4. 1-bromo-5-methylhexane O 5. H2 / Lindlar 6. mCPBA

1. NaNH2

C CH

16.

Cl

taut

2. n-pentyl chloride

1. NaNH2 (1 equiv) 2. n-decyl bromide

Cl

O

3. O3 2 4. Zn, H (or 3. KMnO4, H )

2. n-Pr-I

OH

6.3 Mechanisms Br C HO

H

I

O C

C O

1.

Br C

H O

I

OH

O

O

O

-H

O

O

O

O

C

O

+I O

OH

Br C I

OH

O

O

Br

OH O I

H OH 2.

C C

+H

C C .. +H2O:

-H2O

O

H -H ~H taut

6.3 Mechanisms

OH (H

Solutions • 285

1. -H2

H

3.

2. O

H:

H

O 4.

5.

3. +H

1.

R C C CH2 H

O

O

C C Me

-H

C C

H

Me

OH

HO

Me C C

2. +H

R C C CH2

R C C CH H ~H

B: H R C C C H

~H

H R C C CH

H R C C CH

6.3 Mechanisms


CHAPTER 7 STEREOCHEMISTRY 7.1 General 1. chiral molecules: a, b, f, h, and l.

H N

Bn 2. a. 3.

N

S b. 6.

N

O O

N

CO2H

O

O

O

O

N O

CO2Me c. 4.

d. 2. O

O

Ph

O

OCH3

O

HO OH 3. a.

HO

NH2

Ph

(R)-

H

CH2CH2NH2

Me Br Br H Me (S)meso-

c. (R)-

b. H

(R)-

H

N

O Ph d.

(R)-

H2N H H Ph (S)NH2 meso-

CO2H

O

OMe

O

O e.

f. (S)H

HS

N H

g.

Me

Me (S)-

OH

HO (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


Ha Hc

Hd He

5. a. 8. Ha'

Hb

Hc'

c. 7.

Ha

Ha Hc Hd Hg

CH3 OH

H

b.

Ph

Cl

Hb He

Hg

c, d; f, g are diastereomeric

c, d; e, f are diastereomeric

Cl

H

Hd Hf

Hc

b. 7.

Hd'

a, a'; c, c'; d, d' are enantiomeric

6. a.

HbHe Hf

Cl

Br

c.

Cl (R)-3-chloro-4-phenyl-1-butene

(S)-1-chloro-2-propanol

(R)-4-bromo-5-methyl-4-npropyl-1-heptene

(R)H CH3 d.

I H

H

(S)e.

OCH3

f.

H

Me

Et

H

Et

Me

(R)-1-iodoethyl methyl ether

(3S, 4R)-3-s-butyl-4-isopropyl1,6-heptadiene

Br Cl

Br

meso-3,4-dimethylhexane

Br

Cl

Br

7. a. 4 pairs. _) Cl (+

_) (+

Cl

Cl Cl

_) (+

Cl

b. 5. Cl Cl

Cl

Cl

F

Cl

c. 2 pairs. _) (+

7.1 General

Cl

_) (+

F

Cl (+ _)

Cl Cl

Solutions • 289

Et

Et H H H

d. 2 meso-isomers, 1 pair of enantiomers.

OH Cl OH

e. 2 meso-isomers.

Et

meso-

meso'-

H Cl Cl H

Cl Cl Cl Cl

8. a.

CHO OH H 4 H CH2OH

H H HO

OH Cl H Et _) (+

Me Cl H H Cl Me

Me

H HO HO

OH H OH

Et

Me H H H H

H Cl H

Et

epimerize at C4

enantiomer of A

CHO H OH HO H H OH CH2OH D-xylose

(S)MeHN

H b. i.

MeHN H

Me OH

=

H OH

H

Ph

Me

Ph

_) 9. ee = +68o / +170o = 40% (+), 60% (+ therefore, % (-) = 60% / 2 = 30%

_) ii. ee = +10o / +40o = 25% (+), 75% (+ therefore, % (+) = 25% + (75% / 2) = 63%

(R)-

no. chiral carbons: 4.

N

N

OH H

Me CO2H

10. MeO

(S)-

7.1 General


11. a. ED

b. D

g. E

c. ED

d. ED

e. D

f. D

h. ED

Me

O

O

HS

NH

N

12.

13.

O

H

O (R)-thalidomide

NH2

AcO

HO

14. a. 11. N H

Ph O

Me CO2H

H

O

OH

O O

Ph

HO O O Ac O Ph

(S)-

_) b. ee = +24o / +120o = 20% (+), 80% (+ therefore, % (+) = 20% + (80% / 2) = 60%

O

Cl

Cl (!)

15.

O

Cl

O

meso-

CO2H

O

O A

Cl O

HO

no. stereoisomers: 2n = 24 = 16.

S

N H O

H2N

N

O

O CO2H

CO2H

O

_); therefore, % (-) = 20% / 2 = 10% b. ee = +82o / +103o = 80% (+), 20% (+

7.1 General

OH

_) (+

3 chiral carbons, 2 meso stereoisomers, 1 pair of enantiomers

16. a. no. stereoisomers: 23 = 8.

B

Solutions • 291

7.2 Reactions and stereochemistry OH Me

H

1. Ph

Me HO

1. OsO4 2. NaHSO3

Me

Ph

OH H Me HO Me Ph

H =

Me

+

enantiomer

syn-add'n

Br

Me Me

Br2, H2O Ph

H

Br Me

=

Me

H OH

+

enantiomer

Ph

OH anti-add'n

A 2. a. anti-add'n.

Me B

Me

H H

=

B

Me

H

Me B

rot'n Me

- A-B

A

H A Me

Me

H

H Me

H

H cis-

C

Me b. syn-add'n.

C Me

D D

C =

Me

C

c. anti-add'n.

Ph 3. a.

Et

rot'n

Me C

D Me

Me - C-C

D

D D

D

C

Me trans-

Me

OD

CO2H DO H D H CO2H

Cl Ph

D

DO =

D

H

HO2C

rot'n

H CO2H

H

CO2H

+H

Ph

HO2C - D2O

H

H

CO2H

CO2H

D

Cl Ph

Et

1,2-H:

Ph

shift

Et

H

(E)-

Cl Ph

Ph Cl Ph

Br

Ph Cl Ph

Cl

Et

Et

Et

Et meso-

Ph ClBr

Ph

+ Cl Cl Ph

Et

Et

Et

+

+Br

diastereomers => 2 fractions: each E

+Cl

Ph Et chiral

diastereomers => 2 fractions: 1 M + 1 E



Br

H b.

Et

Br2

H

anti-add'n

Et

H

H H

Et

+

Et

Br

Br

Et H

Br

enantiomers => 1 fraction: R

Et

CO2H KMnO4, H

c.

=> 1 fraction: M CO2H I 1,2-H:

+H

d.

Ph

Ph

+I +

shift

Ph enantiomers => 1 fraction: R

Ph

Ph

I

H H

e.

F Cl

HF

Cl

H Me +

diastereomers => 2 fractions: each E H

Me H

F Me

D

D2 / Ni

f.

Me +H H

diastereomers => 2 fractions: each M

D

Me H

H

D +

syn-add'n

g.

Cl

D

Me

1,2-H: H

Et H

shift

Me

OMe

H

Et

+MeOH -H

+ Me

Et

H

OMe

diastereomers => 2 fractions: each E

h.

1. BH3.THF + 2. H2O2, OH syn-add'n


H

OH

H

OH


Solutions • 293

Et

Et

OH

Et

H3O

i.

H

Et

OH

OH

+H2O -H

+H, 1,2-H: shift

Et Et OH

OH Et

+

mesoEt

OH

chiral

diastereomers => 2 fractions: 1 M + 1 E

Et

H2 / Pt

OH

H

syn-add'n

1 fraction: E

Et

Et

OH

1. Hg(OAc)2, H2O H

OH

OH H +

H OH

H

2. NaBH4 diastereomers => 2 fractions: each E

OH

1. OsO4

j.

OH

+

diastereomers => 2 fractions: each M

OH

2. NaHSO3

OH OH

1. mCPBA

+ OH

O

+H

k.

O

O

+MeOH

Ph

syn-add'n

OMe

H


O

H

H

2. Br2 H H

Ph

+

Br

Ph

Br Br

Ph

Ph cis-


H

Br Ph

anti-add'n

H

OMe

+

-H

4. a. Ph


OH OH

2. H3O

H

Ph

racemate note: reduction of the alkyne to a trans-olefin, followed by Br2 addition, would yield the meso-dibromide (see next problem)



Br b.

Me

H

1. Li / NH3

Me

anti-add'n

Me

H

2. Br2, CCl4

Me

Me

anti-add'n H

Me H

Br meso-

OH 2. OsO4

1. Na / NH3

c.

H HO

3. NaHSO3

d.

Me

Me

H O H

O

1. mCPBA

H Me

Me

t-Bu

2. H3O Me

H

t-Bu

H

OH H

Me Me

b. diastereomers

Ph

Me

d. i. 2.

Ph

NHMe

+ H

NHMe

H

Me

OH meso-glycol

c. cannot predict (actual [D]D = +62o)

H2 / Pd

ii. 2.

CH2Ph H NHMe Me

+ _) (+


Me H

H OH2

5. a. 22 = 4.

_) (+

+ enantiomer

CH2Ph MeHN H Me

CHAPTER 8 ALKYL HALIDES AND RADICALS 8.1 Reactions 1

2

1 + 1 meso

H

Cl2, hv

1. 1

1

Cl

7 are optically active 2 are optically inactive

9 fractions

2 Br Br2, hv

2.

+ (no. H) x (reactivity) =

+

(+ -)

Br

Br 2 x 82 = 164

4 x 82 = 328

6 x 1.0 = 6.0

% racemic 2-bromopentane = 328/(6 + 328 + 164)*100 = 66%; therefore, % (R)- = 33%

1. Br2, '

3.

2. Mg

3. D2O MgBr

Br

4.

D

2. Li

1. conc HCl OH

Cl 3. CuI

1. Cl2, hv 2. Li

CuLi

3. CuI

2

4. CuLi 2 Gilman reagent

I

4.

5. HI

5.

I

I H 1,2-R: shift

6.

1. Br2, '

Br

2. Mg

Br

1. Li

7. 2. CuI

MgBr

1,2-H: shift 3. Ph

3. n-PrBr Ph2CuLi

H

Ph C C

Br

4. NBS, R2O2

Ph

+

Ph

OH 6. Br2

Ph Br

H2O

5. KOH Ph

8.1 Reactions


8.2 Syntheses 1. Cl2, '

1.

3. NBS

2. OMe / MeOH (-HCl)

Cl

t-Bu O O t-Bu

Br 1. Br2, hv

2.

2. OH

1. Br2, '

3.

O

3. O3

(-HBr)

Br

O

4. Zn, H

2. OEt, HOEt (-HBr)

Br

3. HBr, ROOR

Br

I 4.

1. Li Ph2CuLi

3. Ph-I

2. CuI

1. Br2, hv 2. Li

5.

6. Li 7. CuI

5. Br2, hv Br

3. CuI 4. MeI

Cl 6.

7.

O

1. Li 2. CuI 3.

1. H2 / Pt 2. Cl2, hv

4. O3

H

5. Zn, H

I

1. DCl 2. Mg or

3. Li 4. D2O

or 3. H2O

1. NBS, ROOR

1. NBS, R2O2 2. H2 / Pd 3. Li 4. D2O

Br

2. KO-t-Bu / t-BuOH 3. NBS, ROOR

8. 4. KO-t-Bu / t-BuOH

8.2 Syntheses

Ph

8. PhCH2I

D

Solutions • 297

8.3 Mechanisms H

O R

R O O R

etc.

+

1.

O2H H

-H O O O O

Cl

Cl Cl

2.

Cl2

+

H -H

Cl

+ Cl

Cl

Cl

Cl

Cl

Cl Cl Cl

Cl Cl

Cl Cl

D D

D

D

D

D

D

3. D

CH3 N O O 4.

H2 C

O

hv

AcO

H

AcO 6-membered ring TS H2C N OH AcO

O N O

H2 C OH AcO

8.3 Mechanisms


5. a.

b.

-HR

6. H

+O2 O

H R

O

O2

H R +O2

O

O O

O OOH

8.3 Mechanisms

O OH

CHAPTER 9 SN1, SN2, E1, AND E2 REACTIONS 9.1 General 1. faster reaction: b. AcO

SN2

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)

3. stronger nucleophile: a.

Et :N Et Et

Et Et N: Et b

rapid pyramidal inversion lowers nucleophilicity; such "flipping" is impossible with a

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 S N2 reaction

5. solvent that will maximize the rate of reaction: a.

Et3N: +

Br NEt3

SN2

Br

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): H

Et3N:

H-bonding stabilizes the amine, thereby increasing 'G

O Me b

Me

Me

H

Me 6. more reactive by an E2 pathway: b.

H

H

vs. Me

H

Br b anti-periplanar TS possible

a Br no trans-diaxial hydrogen available

Br 7. approximate kH / kD: c.

Ph

H(D)

O-t-Bu E2

Ph

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


8. reaction to yield the more stereochemically pure product: b. Et

Br

Et

Et

Br

or

a. Et

Et

MeOH

+MeOH

SN1

-H

Et

OMe

Et

Et

either diastereomer would give the same ratio of diastereomeric ethers because of a common intermediate H

b.

OMe

OMe SN2

Br

Br or

H (S)-

(R)-

H

OMe SN2

H (S)-

OMe (R)-

stereospecific: either enantiomer gives an optically pure, but different, ether

9. change in rate of reaction: b.

Ph Br Ph

EtOH

Ph

SN1 rds

Ph

H

+EtOH

Ph

-H

Ph

OEt

rate = k[RX] ; changing the concentration of EtOH has no effect on the rate

O-t-Bu

10. a.

~100% E2!

Br

OH

b.

OMe

vs.

Br

vs.

SN2 > E2

OH

Br

Br 30 RX => ~100% E2

OR

c. Br

+ OR

20 RX => SN2 + E2

SR

vs. 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 H(D) a. Cl E2 - Hofmann

KOH

H(D) (H(D)

b. Cl

E2 - Zaitsev

(H c.

no carbon-hydrogen (deuterium) bond is broken in rds; therefore, kH / kD ~ 1

H(D)

KOMe Cl

(D)H

O2 S 12. C: TsO H

OMe

H(D)

+

SN2 + E2

no carbon-hydrogen (deuterium) bond is broken in rds O2 S

O GG- CH3 C TsO H

if intramolecular

O CH3

a carbon-hydrogen (deuterium) bond is broken in rds; therefore, kH / kD ~ 7

Gunfavorable TS: Nu O2 S

however, if bimolecular

O

R

GL bond angle not linear O2 S

CH3

C: O2 S TsO H H3C O

CHOTs

TsO

O

CH3

GC H

C H3 O G-

O2 S

CHOTs

a linear TS is possible => < 'G ; therefore, an intermolecular reaction is kinetically favored

9.2 Reactions KOMe, MeOH

Br

1.

SN2 > E2 I

O

Me

OEt

2.

E2 - Zaitsev Cl 3.

E2

O

+

Hofmann

Br 4.

+

E2

N 2

Hofmann

9.2 Reactions


KCN / DMF

I

5.

C

N:

SN2

+MeOH

MeOH, RT

6.

Cl

-H

SN1 > E1

OMe

refluxing EtOH

7.

-Cl

O

Br

10.

O

Br

-H

Me2NH

+

+

+

Me N Me

-H

E2 O

I

OAc racemate

Me N Me (H

SN2 > E2

O-t-Bu

11.

O

+HOAc

O

SN1

Cl

9.

Zaitsev

E1 > SN1

Br

8.

-H

OAc

O

SN2 > E2 Cl

Cl SH (1 equiv) 12.

SN2

Cl reactivity:

10 >

SH

20 OEt

13.

EtOH

+EtOH

SN1

-H

+

Cl

OEt

Br 14. Ph

9.2 Reactions

-AgBr

1,2-H:

SN1

shift

Ph

OAc Ph

Ph

OAc

Solutions • 303

Et 15.

Me Cl

Me

H H

=

H

H

Cl

rot'n

Me

H Et

Me

H

Me

Cl

H

E2 Hofmann olefin t-Bu

D

-HCl

HD

= D

Cl

Cl

18.

Me

OMe, -HCl Cl trans-diaxial TS

D

H

Et

t-Bu

H

H Br

E2

H

=

17.

Me

OEt, -HCl

Cl anti-periplanar TS Me

16.

Me

Et

E2

H

D

D

(C-H bond slightly weaker than C-D bond)

CH3 D H Ph

H =

Br

H

D

rot'n

H CH3

H

Ph

Me Ph

Me D

E2

H Ph

Br

-Br

19. OH

Br

-H

SN1

O H

O H

O

PPh3 I

I :PPh3 20.

D

-HBr

+

SN2

O -t-Bu

O -t-Bu

OH 1,2-H: SN1

MeOH

21.

S

Cl

N 22.

shift

I

+MeOH S

SN1

-I

S

-H

S

OMe

I N

intra- SN2

9.2 Reactions


Cl 23. HSe

+

OH

SN2 >> E2

H N

24. Ph

Ph

25.

O Me Me

OTs

ret

HO

Me

HO better leaving group

(1 equiv)

+

Me :N H

Me N

F

SN2

OH OH

refluxing MeOH 29.

NR!

Cl

F

rigidity of bicyclic structure prevents formation of a planar carbocation backside attack impossible Bredt's rule precludes double bond at bridgehead

I

SePh (XS)

30.

Se

Ph (vinyl halides unreactive under SN2 conditions)

SN2

O 31. H2N O

9.2 Reactions

BF4

F

Br

OH

I

Me

OH

S

SN2

OH

O

H

SN2, inv

:NH2

+

28.

Me3N

2. OH

H

NHMe

Ph

SN2

H

S

OH Me N

-Me2O, -H

1. TsCl

26.

27.

Ph

SN2

OH

OH

Cl (1 equiv)

Me

Me Me3N

SeH

NEt2

EtBr (1 equiv)

O H2N

SN2 more nucleophilic nitrogen

O

Br NEt3

Solutions • 305

1. NaNH2

32. Ph C CH

Ph C C:

Br

2.

Ph C CH E2 > SN2

MeOH E1

33.

1,2-R:

+

-H

shift (H

I OTs S C N:

+

34.

NR! (aryl tosylates unreactive under SN conditions)

OAc

Br 1,2-R:

HOAc, RT SN1

35.

1. MeI S

36.

shift

S

SN2

+HOAc -H

-H

2. refluxing EtOH -Me2S, E1 O

O

O-t-Bu

37.

E2

O

O

Br

Hofmann olefin

38. :N

N:

antiH Br 39.

Me O

a.

(XS) MeI

N Me

OEt

H

2I

O

E2 - Zaitsev

O

O H

synH Br b.

Me O

O H

Me N

SN2

H OEt

O

E2 - Hofmann

O H

9.2 Reactions


9.3 Syntheses Ph 1.

Ph

H2O, Ag Ph

Ph

SN1

Br

Ph

1,2-H:

OH

Ph OMe

Ph

E2 - Hofmann

Ph

-H

Ph

NaOMe, MeOH SN2 LDA or KO-t-Bu

Ph

shift

Ph

+H2O

Ph

Br MeOH, '

2.

-H

E1 (not OMe! => E2 - Hofmann olefin) D

3.

H

= Br H

-DBr, E2

H

D

OEt

Br

H

Hofmann olefin

(not HOEt, '! => E1 - Zaitsev olefin) Ph 4.

Me H

Me

H I

=

Me

I Ph

H

H

H rot'n Me

Me

Me

I

5.

Ph

Me

KOEt, EtOH

H

E2

Me

H

I

EtOH, '

1,2-R:

E1

shift

Et -H

OPh 6.

Br

PhOH, AgNO3

1,2-R:

+PhOH, -H

SN1 + E1

shift

or, -H

+

O 7.

9.3 Syntheses

Cl

Ph

O OH

SN1

Ph

+PhCO2H -H

O

Ph

Solutions • 307

Br 8.

1. KO-t-Bu, t-BuOH

2. HBr, ROOR

E2 - Hofmann

Br

2. EtOH, RT

1. Br2, hv

9.

Br

EtO

SN1 (not OEt! => 100% E2)

O 2. O3

1. MeOH, '

10.

E1

3. Zn, HCl

O

Br 1. KO-t-Bu, t-BuOH

CO2H

2. KMnO4, H

E2 - Hofmann

11.

CO2H

Ph

1. NBS, peroxide

Ph

3. HBr, peroxide

Ph

Ph

2. NaOMe (E2)

Ph

4. KO-t-Bu (E2 - Hofmann)

Ph

1. KOH, EtOH

12. Ph

Ph

E2

OTs

OH

SOCl2

O

-HCl

Ph

.

3. BH3 THF

E2

Br

14.

OTs

3. TsCl

Ph

2. KOMe

1. Br2, '

13.

OH

2. H3O

4. H2O2, OH

S O

Cl

OH

Cl -SO2

[SN2 reactivity slow (neopentyl-like); avoid SN1 (rearrangement)]

15. H2C CH2

1. Br2, CCl4

Br

2. Br

HS

S SH

base, SN2

Br

S

-Br SN2

S S

9.3 Syntheses


1. HCl 2. Li

16.

3. CuI

C

CuLi 4. vinyl chloride

5. Cl2

2

6. (XS) NaNH2 H

.

7. BH3 THF

OH

taut

O 8. H2O2, OH

1. Li

17.

3. Br2, hv

2. H2O

I Li (SN or E not possible)

H

Br

H

MeOH, '

D H

-Br, E1 D

H

D

-H

D

D

18. D

Br D

D

=

H

OMe / MeOH

H

-DBr, E2

H Br trans-diaxial elim

D

9.4 Mechanisms Br 1.

SH Br

-HBr SN2

2.

S

-Br

S

SN2

Br

HOAc

1,2-R:

SN1

shift

-H

Cl

CH3 N CH3 3. :N C Br

9.4 Mechanisms

+HOAc

=

OAc

HOAc

SN2

CH3 N CH3 C

-CH3Br Br

N:

SN2

CH3 N C N:

C:

Solutions • 309

Cl

HO

-Cl

4.

Ph

N

SN2

Et

Et2N: O

H

O

Ph

SN2

Et

O

intra-SN2

5.

NEt2

Ph

O

H

inter-SN2

O

OH

O

inv

inv H

Cl

double inv = net ret

OH

OH HO

O

(in conc OH, product is

O , formed by initial inter-SN2) H

DMF

I

6.

7.

O

SN2

N O

N O

H

H

Br

(H H

H O

H HO

N H

-

-Br

H

H

Cl

OH

O

Cl

O

-H

-Cl

intra-SN2 with inv possible

H trans-A

H O

Cl HO

H

B

HO therefore, inter-SN2 with inv

rds

Cl

OH H

S

+H2O -H

-Cl Cl

intra-Walden inv not possible

H

OH

S

B

Cl

H

H H cis-A

9. Cl

S

-H

-I SN1

SH

8.

(H S

H S

repeat

S Cl

OH

S OH

OH

OH2

9.4 Mechanisms


9. (cont.) similarly, Cl

Cl

rds

N:

+H2O

N

-Cl

repeat

N

-H

Cl

N OH

OH

OH

Cl OH2 AcOH H

H OTs intra-SN2 (NGP) not possible O

10. O

I

therefore, inter-SN2:

OTs

OAc

+HOAc -TsOH

OAc

OAc

trans-enantiomer only

vs.

AcOH

OTs NGP

H II

-TsO

O

O

OAc

+HOAc

O

OAc +

-H

OAc

O

OAc

racemate

H OTs

O

11.

NGP

O

Ac

O

X

-H

rds H

NGP not possible therefore, no kinetic enhancement H O

+EtOH

NGP, rds

9.4 Mechanisms

EtO

OH

less strained intermediate => < 'G therefore, k >> k'

vs.

Cl

O

Ac

-H

k

H O

O

OAc

+HOAc

O

II

Cl

O 60%

NGP

O

NGP, rds

OAc

-H

Ac

X

I

O

Ac O

O

vs.

13.

OAc 40%

H)

H

H O

O

-H

O

- OTs

12.

(H O Ac

k'

O

H

+EtOH -H

EtO

OH

Solutions • 311

H OTs

Ac

H

OTs

14.

O

OAc

- OTs

vs.

+HOAc -H

II

I NGP not possible

ret

carbocation stabilized by S electrons TsO + OAc

NGP 15.

= OTs

CH2

NR"2

O

dil OH

16. RNA:

H

OAc

OAc

CH2

AcO

=

H

- OTs

NR"2

O

-R'O

CH2

NR"2

O

-H, NGP O O R'O P O O

O OH R'O P O O CH2

NR"2

O

CH2

2'

O

O + OH

NR"2

CH2

HO (H O

O P O O O

or

O H) OH P O O (H O

DNA:

O

O R'O P O O

17.

Cl

O

O

OH

NR"2

O

OH P O O O

2'-phosphate CH2

O

3'

OH

O P

3'-phosphate

NR"2

NGP not possible; therefore, more stable in dil base

-H

'

+

-Cl

(H 1,2 -R: shift

racemate

(H -H

9.4 Mechanisms


2-

18.

HO6P2

O O O P O

adenine

adenine

O

O OH

O 2-

OH

O P O HO6P2 O (H O

OH

adenine -H

O O O P

O

O O - HO P O P O O O

OH

O OPP

OPP

. .

19. a.

OH

b.

..

-BH -OPP

(H :B

+H

= PPi

OPP

rot'n

-H2O

+H2O -H OH2

OH c. coupling mechanism: OPP

.

+I-PP

. conversion of A to vitamin A:

..

-H, -OPP OPP (H

OH

OH

+H

A

H

H

(H vitamin A

9.4 Mechanisms

-H

bond formed OPP

OH

Solutions • 313

OPP

+ OPP

- OPP

d.

PPO

F-PP

N-PP :H

OPP

. .

H)

..

[H] - OPP

OPP F-PP

-H-OPP

squalene

OPP

N-PP

CO2H H3C

S

O O 20. PP O P O

CH3 S

NH2 adenine O

H2N

SN2

OH

adenine O

CO2H

OH

OH

OH

SAM OH HO

NH2

OH

CH3 S

HO

N CH3 H

adenine

HO

SN2

O

H2N

HO

epinephrine CO2H

+H

21. Ph2 C N N:

OH

OH

-N2

Ph2CH N2

Ph2C

H) OTs

Cl Ph CH2 22.

-Cl

:P(OMe)3

Ph

SN2

Cl Cl2C:

OH 23.

Cl3C (H

Cl O P CH3 (OMe)2

-Cl

Cl2C:

-H

Ph

OMe P OMe O

SN2

+I

I

I

Ph

Ph

-H

note: HCCl3 +

+HOEt

Cl2C I H OH

OEt

+

CH3Cl

Cl2C I H

HCCl2I via SN2

9.4 Mechanisms


CHAPTER 10 NMR

Cl 1.

OH

O

2.

3. Cl

4.

O 5.

OH

OH

O

F O

O

6. Br

O

8.

Cl

7. F

O

Br Cl 9.

O

10.

Cl

Br

11.

12.

OH O

O Br

13.

14.

HO

J => trans-

H

16.

15. F

OH

O

O 17.

O H

Fa 18.

Ha Hb

Cl CH3

Ha and Hb are diastereomeric protons; 19F (I = 1/2) therefore, max multiplicity for Ha or b = doublet x doublet x doublet = 8 lines

Fb

Br

Ha

Br

Me

19.

Ha

Br

vs. Me

Ha A

Br

methylene protons are identical => singlet

Me

Hb B

Me

Ha => doublet; Hb => doublet (appears as a multiplet)

10. NMR


Ha

Ja,b = 16 Hz Ja,c = 8 Hz

Hb Hc 1

20.

2

Hc 3

Br

multiplicity: 5 lines (pentuplet)

Ha

Cl 21. a.

F Ha

Hc Hb

Cl

highest field proton is Ha Ja,F => doublet Hb (Hc)

Jb,c Jb,F Jb,c ~ ~ Jb,F => triplet

b. Hb and Hc are diastereomeric:

22.

Hb Hc H a CO2H Ph NH2

Hb and Hc are diastereomeric and independently couple with Ha; if Ja,b Ja,c, Ha would appear as a doublet x doublet = 4 lines (assuming no coupling through nitrogen)

O 23.

O 1 2

3 5

24. a. Ha (lowest field proton): doublet x doublet => 4 lines Hb DO

Hc Ha

-SbF5Br

25. Br

Br

Me Me

Br

Me Me

SbF5 all methyls are equivalent therefore, appear as a singlet

26.

31P

(I = 1/2), nP = 2; therefore, 2nI + 1 = 3 (triplet) (i-Pr-O)2

10. NMR

8 7

b. C7 and C8 are diastereomeric carbons; therefore, 8 chemical shifts

Br

6

4

Ha

Ha

P O

P O

(O-i-Pr)2

Solutions • 317

Ha 27. a. amplitude: signal at G -16.1 is highest amplitude because most molecules (66%) contain Pt with I = 0 (no further spin-spin coupling with Ha is observed, i.e., JH,Pt = 0)

Cl

JH,P

PPh3 Pt Ha PPh3

multiplicity: 31P (I = 1/2), so JH,? = JH,P > 0, nP = 2 therefore, 2nI + 1 = triplet -16.1 Ha

JH,Pt JH,P per above

b. amplitude: signals at G -13.6 and -19.6 arise from fewer (34%) molecules containing 195Pt (I = 1/2) 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

G+ Pt

c. Ha is very highly shielded, essentially a hydride, because of the polarization of the Pt-H bond (much higher electron density around Ha than typically encountered in C-H bonds)

GHa

Ha

Ha

Hb

Bo Hc Hc

28.

Ja,c

Because of magnetic anisotropy of the aromatic ring current, Ha experiences diamagnetic (and Hb paramagnetic) lines of force relative to applied field Bo. Therefore, Ha is more shielded (and Hb deshielded) than normally observed in hydrocarbon protons on sp3 carbons.

Ja,b

pseudo-quartet (same for Hb)

5.5 ppm!

0 0.5 (Ha)

5.0 (Hb)

H 29. multiplicities:

H

11B

H H

vs.

H

H 11B

(I = 3/2) therefore, 2nI + 1 = quartet (higher amplitude)

10B

H

H 10B

(I = 3) therefore, 2nI + 1 = septet (lower amplitude)

relative amplitudes of quartet/septet reflect the natural abundance of 11B/10B = 80%/20%

10. NMR


CHAPTER 11 CONJUGATED SYSTEMS 11.1 Reactions H +H

1.

+Br

Br

(1,4-addition)

D

D DCl (1 equiv)

2.

1,4-add'n most stable carbocation D

D +Cl

Cl

CO2Me 3.

CO2Me

D-A

=

CO2Me =

MeO2C

CO2Me

CO2Me

retro-D-A

4.

= O

O

O

HBr 5.

H

H

+H

Cl

Cl +Cl

+Cl H

H 1,2-add'n product of thermodynamic control more conjugated system than 1,4-adduct, therefore, more stable

1,4-add'n

11.1 Reactions


'

6.

+

retro-D-A

DBr (1 equiv), ROOR

7.

+DBr

1,4-add'n

-Br

Br

Br

O

O

O N N N

8.

D-A

N

(4 + 2)

N

Ph

O

H

D

Br

O N Ph

O H

N Ph

9.

N

D-A Ph Ph

Ph

Ph

1. NBS, R2O2 10.

Ph

3.

= 2. KOMe (E2)

1. retro-D-A 11.

'

Me

2 Me

12.

11.1 Reactions

=

2.

Me Me

(E)-

intra-D-A

Br

Solutions • 321

O

O

O

1. D-A

13.

2. KO-t-Bu E2, -HCl Cl

Cl

H 1. D-A

14.

O

2. O3 3. Zn, H

CO2Me

CO2Me

CO2Me O

H

O O '

15.

+

retro-D-A

1. retro-D-A

16.

'

2

Ph

2. Ph Ph

Ph cis-

CO2H 17.

HO2C

=

CO2H '

s-cis-diene

CO2H

O

O

(4 + 2) 18. MeO

O

MeO

O

11.1 Reactions


AcO

19.

AcO CO2H

= OAc

'

CO2H

'

CO2H

+ CO2H

HO2C

CO2H

CO2Me

20.

(4 + 2)

CO2Me

CO2Me

CO2Me

O

O

O

N

N

'

21.

OAc

(4 + 2)

N

- :SO2 O2S

O B:

O

O

H) base, E2

22.

(4 + 2)

-NMe3 MeO

MeO NMe3 I

MeO C19H24O2

MeO R 23.

MeO

Si Me2

Br

D-A

R

Si Br Me2

H

CHO

O OMe

OMe N

24.

R'

+ Me3SiO

11.1 Reactions

R

R

(4 + 2)

N

R' R

Me3SiO

R

Solutions • 323

11.2 Syntheses D 1. Cl2, hv

3. NBS

2. KOMe (E2)

ROOR

4. KOMe (E2)

1.

5. DBr 1,4-add'n

Br

Br

1. Cl2, hv 2. KOMe

2.

5. ethylene CH2

3. NBS, R2O2 4. KOMe

6. H2 / Pt

D-A

H2C

Me

1. NBS, peroxides

3.

2. KO-t-Bu, t-BuOH

4. H2 / Ni

3.

Me 1. NBS, R2O2

4.

3. 2-butyne CH3

2. KO-t-Bu (E2)

D-A

H3C

O O 5.

(4 +2)

A

O

Me

Me N

O 6.

O

NMe +

= O

H

O

O

O

H

O +

O

O

O

7.

CO2Et

N

(4 + 2)

= H

CO2Et O

'

CO2Et OH

11.2 Syntheses


11.3 Mechanisms O

O

O

O

> pH 8.5

1. O

O

-H

(H OH

HO

HO

- > conjugation results in a 'red shift;' absorption occurs at a longer wavelength, moving into the VIS, and the molecule, therefore, is "colored"

- sp3 carbon prevents conjugation from one ring to the other two; therefore, absorption occurs at shorter wavelengths (UV in this case)

'

2.

'

+

retro-D-A

D-A

O

O '

3. a.

(4 + 2)

O

O O

O

O

O

'

3. b.

longest Omax: n

4. a.

O

'

S*

N b. low probability

c.

A = H c d

H (molar absorptivity) only ~ 10 - 100 for n S* transitions) (vs. > 10,000 for S

no non-bonding electrons in the CA of pyridine; therefore, no n N H

11.3 Mechanisms

S* transitions

S* transition

Solutions • 325

5.

+H

-H2O

OH

OH2 G 1.70 H

OH2

H G 4.10

OH =

-H

+H2O

OH2

OH G 2.25 H G 1.79 H G 5.45 B

O O

6.

O

O

retro-D-A

D-A

-CO2

R

O

O A

R

intra-D-A R R

R

O

R R

O

O

8.

B

inter-D-A

=

D-A

=

H CH2

(H C H2 O

O

O '

taut

9.

ene-reaction

Ts

Ts

N 10. Ts N

O

H CO2Me

R

R = -CO2Me

7.

Me

'

=

O CO2Me

O

N N

hv - :N N:

Ts N

=

N CN CN CN

CN

11.3 Mechanisms


a OH N NH

11.

H N

OH

OH

N [O]

N

[H]

CO2H c site of redox

N

H N

N b

CO2H CO2H

OH

CO2H biliverdin (green)

bilirubin (red)

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.

c.

b. (4 + 4)

(2 + 2) thermally forbidden

11.3 Mechanisms

(4 + 2) thermally allowed therefore, most likely to occur

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)

N N N N

d.

b. O

O

f.

H

N B

N H

N B

H H

OH taut

HN O

j.

i.

H

O k.

N

H B

N

N H

HO

2 Li

m. product:

2-

2 Li

(6 S electrons)

H

n. product:

H

(10 S electrons)

2 Li

-2 CH4

(H

H

:CH3

Br

ZnBr Zn

o. product:

H

2 MeLi

H) H3C:

(6 S electrons) N

-ZnBr2

Br

Br

both rings are aromatic P >> 0!

SbF5

Note: l. carbocation from the reaction of Cl

-SbF5Cl

(8 S electrons = 4n+ 2)

12.1 General


2. largest P: a. G+

etc.

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

3. N H

N

N

H3C least basic (delocalized, part of aromatic ring current)

4. least stable: b.

O

most basic (localized)

N CH3

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

6. most likely to undergo an SN1 reaction: a. Cl

most basic O H 7. a.

N more basic (localized)

12.1 General

more basic (localized) OH

OH N

+H

N

c.

b. O

aromatic (4n + 2) contributing structure stabilizes carbocation

O

O aromatic

N

Solutions • 329

_ -Cl

8.

=

0 +

Cl aromatic - all protons are equivalent

O

all bonding MOs are filled

O O vs.

9. largest molecular dipole moment: b.

O

d.

aromatic contributor with longer charge separation (>d)

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 H 10. Ph

OH

OH

aromatic contributor with shorter charge separation

OH

BF4

OH

HBF4

Ph

Ph

Ph

Ph

Ph

Ph

Ph

Ph

Ph

aromatic cyclopropenium moiety

H

H

+H

11.

+Cl Cl

note: does not undergo a 1,2-H: shift!

aromatic

(10) Hb ~ G7 12.

Ha

(4)

Ha ~ G-1

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


12.2 Reactions N H

1.

N H

fuming sulfuric acid O

HO3S

OH

OH Me

o-isomer

no! (avoid 1,2,3-subst'n)

Me

HONO2 / H2SO4

2.

+

O

NO2 H PH2

o-, p-director PH2 3.

H2SO4, SO3

m-director PH3

+H SO3H

(H

(H Cl2 / Fe

4.

N H vs.

N H

Cl H

N H

Cl

1. AlCl3 Br

N H

-H

Cl

Cl

N H

more important set of contributing resonance structures

Cl N H

1,2-H:shift

Ph

Cl

H

N H

5. Ph

(H

PhH

Ph

Ph

F-C alkylation

CH2

2. NBS, R2O2 3. KOMe

Ph

Br

Ph

E2

Br Br2 / CCl4 Br

Br Br

6.

Br2, Fe Br

Br NBS, R2O2

12.2 Reactions

+

Br

Solutions • 331

G+ G7. I Cl

Se

-FeX3Cl

[I ] +

Se

Me EAS

FeX3

N 8.

O

N

Br2, FeBr3

N

O

p-isomer

O

+ Br

SH

Cl2, FeCl3

9.

+

I

Br

SH

Me

+

p-isomer

Cl

H

Cl

H

N

H

N least important

Cl2, BF3

10.

Cl

N

(H

Cl

N (H

(H

Cl

Cl

Cl

N

Cl

-H N

N

N

more important set of contributing resonance structures

OH

OH O 11. H

OH

+H H

H

OH

H

H HO

H

EAS Cl

Cl

Cl Cl

OH

Cl

OH

EAS

+

Cl

Cl

Cl

Cl

OH

OH

Cl

-H2O

Cl

+H

Cl Cl

OH2

Cl

Cl

Cl

Cl

Cl

O 12.

O

1. -CO2, -N2

+

2. D-A

N N: benzyne

12.2 Reactions


Me

Me

OH O2N

13.

NO2

+ Me

NO2

Me

NO2

a S-complex

F

F

CN

add'n - elim mechanism Cl

Cl

NMe2

NMe2

NMe2 1. MeLi (-HBr) 15.

Cl

Cl

NO2

2. NaOMe, MeOH NAS, add'n - elim mechanism

CF3

CF3

+H

Me

OMe NO2

EAS

O

2. H Me

1. HNO3, H2SO4

16.

NMe2

+ MeLi

NAS, benzyne mechanism

Br

17.

CN

-HF

nucleophilic aromatic subst'n

Cl

Me NO2

NH2

NH3 CN

:NH3

14.

OH

O2N

Me

CF3

O H

OH EAS OH +H

EAS

-H2O

OH 18.

12.2 Reactions

+H

+

OH

OH

EAS again

Solutions • 333

Br Br2, Fe

19. O

H) O

EAS

O

Br H) O

O

Br

-H O

O

O

most stable intermediate

Br

Br O2N

1. fuming HNO3

20.

N NO2

NO 2

NAS, -HBr

EAS CF3

CF3

CF3

OMe

OMe

OMe

OMe Ha

3. KOH

2. NBS, ROOR

1. MeI, AlCl3

21.

O2N

2. i-Pr2NH

SN2

Hb

Br

OH 1H

NMR: aromatic a-b quartet suggests p-subst'n

I 22.

I2

HO

HO

catalyst

CO2H H2N

I

H

N O

H)

H

Cl

N O

2. Cl2, BF3

23.

H2N

H

Cl 3-subst'n

Cl

N O Cl

thyroxine

CO2H

H)

Cl

N O H)

Cl

Cl -H

4-subst'n (or 2-subst'n)

N O

N O best resonance structure

N O

3. [H]

+ N

N

Cl

12.2 Reactions


12.3 Syntheses 2. Cl2, Fe

1. CH3I, AlCl3

1.

3. H2 /Pt

F-C Br 2.

high T, P

Cl D

1. Br2, Fe

Cl

Li D

3. Li

H

D

D

D

4. H2O

D

2. dil D2SO4 (EAS) D

D

3. D2O

2. Mg

1. Cl2, AlCl3

Et2O

(x2)

D

MgCl

Cl

O 4.

D

MgCl

Cl 3.

D

OH

1. PhCH2COCl

H followed by

2. H2 / Ra-Ni Ph

AlCl3

Ph

hydrogenation

Ph

hydrogenolysis

Br 1. Cl2, BF3

3. KOMe, MeOH

5. 2. NBS, ROOR

1. cyclohexyl chloride 6.

AlCl3

E2

Cl

2. NBS

Cl

3. NaOH

ROOR

Br

1. O3 2. Zn, HCl

E2

4. SOCl2

7. 3. O2 [O]

CO2H

5. AlCl3 (F-C) O

12.3 Syntheses

H

Solutions • 335

Cl

1.

8.

2. HONO2, H2SO4

AlCl3

NO2

Cl

1. Cl2, Fe

9.

2. PhLi

NO2

3. H

PhLi

-HCl

Ph

Ph

4. MeI, AlCl3

2. NaNH2

1. Br2, FeBr3

10.

CO2H

3. KMnO4, H

NHMe

Br

NHMe

NHMe

:NH3

~H N Me

N (H Me

O O Cl 2.

1. EtCl, BF3

11.

AlCl3

Et

O

1. H2SO4

3. H2, Ra-Ni

OH

Et

Et

OH

OH

HO

12.

OH

2. PhH, H

EAS

+H, -H2O

t-Bu HO

Ph

3.

,H

HO

Cl

NAS

O Cl

1. NaOH Cl

HO

O Cl

(Cl)H

EAS

EAS (see 12.2, 18)

t-Bu

13.

Ph

2. Cl

CO2 SN2

(Cl)H Cl

CO2 Cl

(Cl)H Cl

12.3 Syntheses


Br

OH

O 2. a. SnCl2, HCl

1. OH (NAS)

14.

OH

(1 equiv) -HOAc

NH

3.

b. neutralize NO2

O

O

NH2

NO2

O

O CO2H

CO2H

4. SnCl2, HCl

2. KOH

1. HONO2 15.

CO2H

CO2

H2SO4 OH

Cl

3.

NO2

5. neutralize

NO2 (1 equiv)

OH

NH2 O

O

12.4 Mechanisms H EAS

Ph

1.

-H

Ph (H Ph

Ph

O OH

H)

O

+H

2.

O

O EAS

O

-H H)

OH

OH

NMe2 O 3. Cl

O AlCl3

AlCl3 Cl

Cl

O AlCl3

Cl

Cl

NMe2

Cl

O

EAS

Cl

NMe2 EAS

Me2N

NMe2

NMe2

C O

AlCl3

Cl -AlCl4

-Cl

Me2N

O

Cl

O

O C O OH 4.

O

O 1. OH -H

12.4 Mechanisms

2. CO2

OH

O (H C O

O

-H

CO2

3. H

CO2H

Solutions • 337

[Br ]

Br

Br Br2, AlCl3

5.

-H

+

(H

O 6.

1. AlCl3 R

X

R C O:

-AlCl3X Ph

EAS

Ph R

R C O:

H)

O

H2O

7.

R

R EAS

HO R (H

R

AlCl3 O O

O

HO

~H

R

AlCl3

+H

O H

H)

-H3O

O

2. HBr

Ph R

-H

AlCl3 O

H O

R

OH

~H

R

(H :O C R

R

O

O

HO Cl

Cl

8. Cl

Cl

Cl Cl

-HCl NAS

O O

Cl

O

Cl

Cl

Cl

Cl

Cl

NAS

Cl

Cl

-Cl NAS

Cl

Cl

-Cl

Cl

Cl HO

O

O

Cl

Cl Cl

Cl

Cl -HCl

Cl

O

Cl

NAS

Cl

-BF3Br

9.

Br BF3

MeO

MeO

MeO (H

-H MeO

EAS MeO

12.4 Mechanisms


O :C O:

10.

EAS

H C O:

O

H

-H

H)

H

H

+H

11. H O O H

H

O

O H

H

N

H)

Cl 1. NAS

N

-H

N

O Cotton

Cl -HCl

N

N

2. Cotton-OH

N

Cl Cl

N

Cl

HO

OH

Cl 12.

EAS

H O -H2O

N

H2N

Cl

HN N Dye

Dye

Cl

-2 HCl NAS x2

N HN N Dye

N O

Cotton

H2N Dye

O

OH

+H 13. Cl3C

H

Cl3C

Cl

OH

H

Cl3C

OH Cl

H

EAS

CCl3 +H -H2O

Cl

H Cl

Cl

Cl EAS

CCl3

CCl3

1. BF3

14. Cl

H

-BF3Cl

EAS

BF3 -H

(H

2. H 1,2-R: shift

12.4 Mechanisms

H

Solutions • 339

15.

H O

H OH

16.

O

+H

-H

-H2O

OH

H OH

OH

-SnCl5

Cl

(H

EAS

E1

H

OH

O

-H

OH

OH

-H, EAS

O

SnCl4

+SnCl4

O -MeOSnCl4

+Cl

SN1-like Cl

O

Br

1. +Br2

17.

Br

Br

2. :B

+Br

-Br

Br Br

SnCl4

-HBr

Br Br

C7H7Br

H) C7H8Br2

3. H2O, -Br

O

OH

+H2O SN1

-H

-H

aromatic

Me 18.

H Ph

OTs H

=

rot'n

Ph

Me

Me

OTS

H

OTs

H Me Me

H

Ph H

Me Me

H HOAc

-H

-OTs NGP

Me

+HOAc

H

Me H Me H OAc

Me Me

H

Me Ph Me

H HOAc H +HOAc -H

OAc H

enantiomers

12.4 Mechanisms


Cl

OR 19. Cl3C (H

-Cl

Cl2C: Cl

Cl2C:

+

Cl

N H

N H

Cl

Cl

-H N

N H

+H 20.

-Cl

1,2-R: shift

H

H

1,2-R: shift

-H (H

21.

O HO P OH OH O (HO)2 P

+H

O HO P OH2 OH

OMe

-H

-H2O

H) H2O3P

12.4 Mechanisms

O HO P OH

OMe

EAS OMe

CHAPTER 13 ALCOHOLS 13.1 Reactions O

OH

1. NaBH4 1. Ph

Cl 3. PCl3

2. NH4Cl

Ph

2. H2SO4

Ph

Ph

E1

Ph

OH

Hofmann, E2

Ph

OH

1. H2 / Pd

O

1. i-PrMgBr

2.

4. KO-t-Bu

2.

Br

+

Hofmann

Et3COH

OH 1. H2SO4

3.

2. H3O

E1

OH

HCl, SN1

1,2-H: shift

+H, -H2O

O

Cl

O

O

1. TsCl

4.

+Cl

2. NaOAc

-HCl

OH

SN2

OTs

O O

OH

O

1. NaH

5.

O

2. Me OSO3Me

Me

-OSO3Me, SN2 O 6.

Ph

O

O

O

1. PhMgCl

PhMgCl Ph

-n-BuO

Ph

Ph

Ph

Ph

2. H

Ph3COH

not isolable O

1. LiAlH4 Ph

-n-BuO

O

LiAlH4 H

Ph

not isolable

H

2. H

PhCH2OH

H

O O 7. CH3I

1. Li -LiI

CH3Li

2.

OH 3. H

13.1 Reactions


1. HBr

8.

3. BH3.THF

2. LDA, E2

OH

HO

Hofmann

Br

4. H2O2, HO OH

O

CO2Me

1. NaBH4

CO2Me 9.

2. H O

OH O2CPh

Ph

OH

1. LiAlH4

O

+ PhCH2OH OH

2. H

OH

CO2Me

H2 / Pt

O2CPh O O PCl2

Me 10.

H Me

OH D H

POCl3 py -HCl

H Me

rot'n

H Me

D Me

D H

H

H

1. OH

Me

-HOPOCl2

O

SN2

NMe

NMe

HO

HO

OH

HO

OH

OH

HO

O

O

CrO3, H

12.

D

O

2. Me-I

NMe

Me

:py

Me

O

O

Me

E2

O

HO 11.

OPOCl2

HO

HO

HO O

1. Br2, H2O

13.

OH

2. Me3SiCl

Br

3. Li

OSiMe3

4.

Li

5. H3O

OH OH

O 14. O

O

O

1. MeLi BnO

Ph

O

-OBn OBn

BnO

-OBn

O

BnO

Ph

O OH

13.1 Reactions

O

MeLi

2. H

MeLi

Solutions • 343

CO2H

15.

O

1. LiAlH4

2. (XS) HBr SN1

O

HO

O

HO

OH

OAc

O

Br

OAc

HO 1. NaBH4

16.

OH 2. H O

OH

HO

HO 1. LiAlH4

+ EtOH HO

2. H

OH

OH +H

17.

18.

Ph OH OH

1,2-R:

-H2O

OH

O

O

shift

H

+H

1,2-R:

-H2O

shift

Ph

O

-H

-H Ph

Ph

O (H

H

O

O

OH 1. NaBH4

19.

2. H

HO

OH

O

O

HO

MgCl

O

HO

1. SOCl2, Et2O

20.

3. H

2. Mg

O

N Me

Ph

n-PrMgCl

1. n-PrMgCl -OEt N Me

HO

Ph

Ph

OEt 21.

O

O

Ph

2. H N Me

N H

Me

13.1 Reactions


O

OCH3

OMe O

OMe O

1. '

22.

2. H

D-A

TMSO

taut

Si O

O

13.2 Syntheses O 1.

O

OH

1. H2 / Pt

3. H3O

2. H2SO4

4. CrO3, H

7. POCl3 / py

1. HBr

O

3. CH2O 4. H

Li

2. 2. Li

H

5. PCC [O]

O

1. H3O

3.

OH

OTs

H

OH Ph

2.

H

OH (SN2 - inv)

H

OH

Me

HCl

+Cl

Et H

1. PCl3 (inv)

Cl 2.

Cl

1. Br2, hv

1. Cl2, ' 2. KOMe (E2) 3. NBS, R2O2

13.2 Syntheses

*OH

MgBr 4. H

*OH

(SN2 - inv)

5. H2SO4 OH

4. KOH (SN2) 5. Jones reagent

6. H3O

E2

O

Br

(* = 18O)

H

double inv => net ret

3. CH2O

5. 2. Mg

racemic

H

SN1

6.

8. Na2Cr2O7 [O]

4. H

1. TsCl (ret)

O

6. i-PrLi 7. H

3. PhMgCl

2. CrO3, H

4.

D

5. NaBD4 6. H

HO

O 6. MeLi

Me 7. H2SO4 E1

Solutions • 345

O

1. Mg 7.

3. H2SO4

Cl 2.

E1

H O 1. TsCl

OH

8.

2. KCN / DMF

OTs

C

N

SN2

CO2H

9.

1. LiAlH4 2. H

Cl

O TMS

OH 1. TMS-Cl

10.

O TMS 2. Mg

(protect)

3. CH2O

HO

CH2O

H

O

O 4. CH3CHO

2. Me3SiCl 3. Li

5. H3O 6. PCC

TMSO

HO

4. Li 5. CH2O

12. Br

3. PBr3

OH

6. H

1. Br2, hv

3. BH3.THF

2. KO-t-Bu / t-BuOH (Zaitsev E2)

4. H2O2, OH

O

[O] CH2OH

1. PCl3 (not HCl)

1. BH3.THF 2. H2O2, OH

Cl

5. PCC

Li

11.

14.

OH

OH

(deprotect)

OH

D

5. D2O

4. H3O

Cl

Cl

13.

4. Mg

3. SOCl2 or PCl3 OH

5. K2Cr2O7 OH

1. AlCl3

2. H2 / Pd

F-C

hydrogenation, then hydrogenolysis

O OH OH

5. LiAlH4

CO2H

6. H

CO2H

[O]

O OH

3. KMnO4, H

13.2 Syntheses


15.

O

3. B2H6 4. H2O2, OH

OH 1. HCl (SN1) 2. LDA (E2)

O 6. MeLi 7. H

H

5. PCC

Me

8. PCC

H OH 16.

Cl

1. Me3SiCl

OTMS MgCl

2. Mg

3.

OH

4. H3O

OH 17.

HO

O

O

OH

2. O3

1. H2SO4 (E1)

4. MeMgI

3. Zn, H

5. NH4Cl (weak acid to avoid dehydration)

O

OH

1. TMS-Cl 2. BH3.THF

18. HO

3. H2O2, OH

5. H

CH

MeO

MeO

O

OH

C CH

C CH

3. H3O

1. TMS-Cl 2. LiC CH

C

4. HC CLi

O

HO

OH

O

3. CrO3, H

HO

C CH

HO

6. H3O

1. NaOH 2. MeI (SN2)

OH

OH

4. CrO3, H 5. LiC CH

TMSO

OH

19.

O

5. KMnO4, H

O

4. CrO3, H

O TMS

1. Mg

3. H

20.

O

-H2O

2. CH2O Br O

OH2 1,2-R: shift

-H H)

13.2 Syntheses

Solutions • 347

1. NaBD4

OH

SbF5

21. O B

etc.

(- OH)

2. H

D

D

D

deuterium is 'equilibrated' among all five cyclopentenium carbons

13.3 Mechanisms 1.

+H OH

-H

-H2O

OH

O H

H)

1. NaBH4

2. +H

2. H2SO4

-H2O

H

2.

O

O

1. H: O

HO

H

2. H -H H) O (H OH 3. OH

+H

OH

-H2O

H

H

-H

O H

shift G 9.5 (d)

Ph 4.

1,2-R:

1729 cm-1

OH

+H

OH

-H2O

Ph

H

OH

Ph 1,2-H:

H O (H

shift

Ph

-H

O

(H +H 5. OH

-H2O

1,2-R:

-H

shift 1,2-R: shift

-H (H

13.3 Mechanisms


OH

H

6. OH

-H2O

OH

OH

OH

OH

Ph

+H

CO2H

(H H

~H

O H

E2

H

-H3O

OH2 O

H

1,2-H: shift, -H

H O (H

Ph 1. H3PO4

8.

OH

1,2-H: shift

-H2O

OH

H H

1,2-H: shift

O

H

+H

7.

H

H H OH

+H

-H2O

H O

EAS

OH2

-H

C

O

O

O 2.

-HBr

4. HBr (SN1)

3. H3O

Br

9.

+H

O

O O

O H

O

O OH2 O H

HO

H2O

~H

O taut

O H O O H

10. H

Me

Me

HO H

-H

OCH3

O H)

Br

Br

H HO

+H2O

O

O

H

OH

5. Me2NH

NMe2

MgCl

+H H

H

Me Me

OH2

-H2O

H O O

H

H

H H

+Br

Me Me

Br

Br H

Br

Me =

Br H

Me

+Br H

H Br

Me Me

H Br Me

Br

Br

13.3 Mechanisms

Me

H

Br

O O H

(+) -

Me =

H Br

Br H Me

Solutions • 349

Br

Br

Br

+H

11.

NGP -H2O

OH

+Br

anti- OH2

trans-

Br Br

Br +H

OH2

OH cis-

gauche-

Br

SN2 +Br, -H2O NGP not possible

Br

Br trans-dibromide

O

O

OH

12.

H

+H

O

OH

CO2H

H O (H

CO2H CO2H O

O CO2H

OH

CO2H

2. -H2O

OH

H

-H

CO2H

taut

O

1,2-R: shift

1. [H]

OH

E1

CO2H

H 13. OH

+H

-H

-H2O

+H

-H

no energy benefit to 1,2-H: shift

14. H2C CD2 OH OH

H

+H H H

-H

pathway (1) - E2

HO

D

H

D

+

-HOH or DOH

H

D

H

O

O

H, taut

OH

1,2-H: shift

H

+

CD2H

H3C

D

NOT formed pathway (2) pinacol +H, -H2O

H) O H

D

H +

H

D

H

D D

1,2-H: or D:

O (H

shift, -H

O

O +

H

CD2H

DH2C D formed

-- therefore, pathway (2) is favored

15. a. dehydration-tautomerization: A

D

H

H, taut

D O

OH

vs. pinacol-like: A

-DOH

+H -H2O

D D OH

NOT B

D 1,2-D: shift

D O (H

-H

B

-- therefore, the pinacol-like pathway is favored

13.3 Mechanisms


15. b.

1. OsO4

OH

2. NaHSO3

OH

O

3. CrO3, H

O

OH 4. NaBD4

D D

5. H A

O

OH

OH

+H

16. OH

OH

alternatively,

H

H O

H OH

taut

OH (H

OH

H 1,2-H: shift

~H

OH

-H OH O (H

O

OH

taut

O (H

~H

O

O

Cl

Cl

1. Li

3. +2H

17. 2.

+

+

-H2O

O

aromatic carbocation

Cl 18.

Ph3P:

CCl3 Cl

SN2

Ph3P CCl3

-Cl

O

SN2

O PPh 3

-HCCl3

Cl - Ph3P O

D D

D

OH

OH2

+H

H -H

13.3 Mechanisms

D D

(H

D

19.

OH

-H2O

C4H8

CHAPTER 14 ETHERS 14.1 Reactions 1.

1. HBr O H

SN1

3.

Ph

O H

2.

OH

3. KOAc

2. TsCl

Br

OTs

HI

S

-Me2S SN2

S

H

O

Ph

SN2

Br

OAc

OH

I

HI

+

SN1-like

Ph

Br

Me

-H

S H

Ph

S

NR!

Me

(H O KOH

4.

-HBr Br O H

5.

1. HF

F

SN1

2. PCC

F O

[O]

OH

H 6.

NaCN / MeOH

:N C: O

C OH

O

O

1. PhLi

7. Ph

N

SN2

Ph

Ph

E1

:Ph H O CH 3

8.

Ph

2. H Ph

OH

1. HI

O

2. [O]

OH 3. - 4. [H]

SN2

D

I

9.

1. PhCO3H

2. H O

+PhOH, -H O H

SN1 OPh OH

14.1 Reactions


H

O

Me OMe H, MeOH

10.

H H OH

O H

trans-diaxial ring opening determines regioselectivity F

F

MgI

2.

F

1. ClCH2COCl

11.

F

AlCl3

3. H

Cl

OH

F

O

Cl

F

F 5. 6. H

OH

F

4. base O

F

N N

N N N

N Me NH2 12.

Ph

1. mCPBA Ph

Me

2.

Ph

NHMe

O Me OH

OH

O

O

1. NaOH

OH H

3. '

13.

taut

2. RX

O

H

CO2H

14. a.

'

O CO2H

OBn

OBn O

HO

CO2H

14. b.

'

HO

O O HO2C

14.1 Reactions

+H -H2O CO2H

(H O

CO2H

-H -CO2 HO2C

CO2H

O O

Solutions • 353

'

15.

O

O

O '

16.

OH

'

taut H

OH

OH

O

'

17.

taut

O

H2N

CO2H

[O]

18.

H2N

SH

HO2C

S

NH2

CO2H OH

19.

S

HO

[O]

HS

S S

SH HO

OH

14.2 Syntheses Br

1. Br2, hv

OMe

2. HOMe

1.

SN1 not OMe ( => 100% E2!)

2.

1. mCPBA

OH

2. H3O

O

OH

O

MgBr 3.

1. NBS, ROOR 2. Mg

O 3.

OTs 4. H 5. TsCl

14.2 Syntheses


S 2. H2N

1. NBS, ROOR

4.

3. OH, H2O 4. H

Br

Li

1. Cl2, ' 5. 2. Li

6.

SH

O

3.

OH

O

S

O H

[O]

-H2O

H

O 1. HCl

N

conjugate addition

O

2. thiourea 3. OH

O Cl

4. H HO2C

HO2C

1. NBS, R2O2

Br

CO2H

N

HS

N

HO2C

8.

S

3. H2 / Pd

2. H2SO4

H

:H

7.

5. I2 [O]

5. PCC

4. H

1. NaBH4

O

NH2

2. HBr

Br

Br

3. thiourea

SH

S S

6. Br2 [O]

4. OH CO2H 5. H

CO2H

SH

CO2H

CO2H

14.3 Mechanisms O

O BF3

BF3

1. Ph

2.

O

Ph Ph

O

CH2 (H

N

H

H

~H:

BF3

H

Ph

2. H O

PhCH2CHO

O BF 3

1. LDA 2

-BF3

OH

H

H) -H

3. O H

OH

OH

-- see 14.3, 6 for an even more impressive polycyclization!

14.3 Mechanisms

OH

Solutions • 355

O

O

HO

O

O

O

O

OH

4. HO

H) OMe HO

O

O

HO

HO

O

O

O

H O

O

O

5.

OH

O

H

SN2, ~H H

O

H

O O H H HO ~H

product

O

H

H

H

O

H

O

~H

O H

O

H

H

O

H

H

6. Me

HO

H O

Me

H Me

Me

H

H

-H Me

H

HO

D 7.

Me

HO

H O

D

OH

D

H

two 1,2-H: shifts followed by two 1,2-R: shifts

O

OH

H

H 1,2-D: shift path (b) D

methyl group determines direction of ring opening

(H

D

-H

OH

OH H

OH D

-D path (a)

H

observed

not observed

14.3 Mechanisms


R

R

OH 2.

R'

8.

OH

R'

BF3

O

R

OH

R'

O BF3

1,2-R: shift

O (H R -BF

O

R' ~H O BF3

R' OH

R

3

O BF3 H

9.

N H

PhO

S

O

-DBN-H

(H

N

O

CH2

O

Cl

N

O

O

DBN:

O

S

N H

PhO

Cl S

N H

PhO O

~H N

O

O

S

N H

PhO O

O

N H

DBN: DBN H

14.3 Mechanisms

-Cl O

CHAPTER 15 ALDEHYDES AND KETONES 15.1 Reactions 1. CrO3, H OH

1.

N 2. H2NNH2, H

NH2

O O

O

1. Ph3P

Br

PPh3

O

3.

2. 2. MeLi a vinyl ether

a Wittig ylid

O

3. O

acetal

O

H3O

Ph

O

O

H H3O

H

+

Ph

OH

OH

OH

+

4.

H

O

H

O

H3O

5.

HO

hemiketal

CHO

1. PCC

2. H3O

H OH

OH

3. H

Cl

1. KO-t-Bu / t-BuOH

OEt

3. HOEt, H O

O

O 7.

Ph

O

OH 6.

N opsin

O

OH Ph

-H2O

opsin-NH2

O

O 2. HCl

E2

conj. add'n Cl

O 1. NaBD4 8. 2. H

D

OH

3. H2SO4

D

E1

15.1 Reactions


[O]

1. KMnO4 9. 2.

HO

3. H2 / Pt

O

H2NNH

NH2

HN NH

NH2

(-H2O)

NH2

O

O

O

O

O CH

10.

[H]

N NH

OMe

1. HO

OH , H

2. DIBAH,

-78o

3. Ph3P=CMe2

O H

H

4. H3O

O

O O CHO 11.

1. Ph3P

O O

2. H3O

O2N

H O2N OH

1. Ph3P=CHOCH3

O

12.

O CH3 2. H3O

O

O CH3

(+H2O)

H3O

H

(-MeOH)

O 1. Ph3P:

Ph3P-CHOR

13. RO-CH2-X

3.

OR

H

H 1. CH2I2

O

14.

O

OMe

H

H

2. H3O

O

H H

Zn (Cu) H

OMe

3. Ph3P=CHC=CH2

H

OMe

H

O

H

O N NH2

15.

OH 2. EtMgI

1. H3O

H 3. H3O

16.

PhO

OPh

Ph

Me

15.1 Reactions

H

(-HOR)

2. MeLi

OMe

O

4. H3O

O

1. H3O (-2 PhOH)

Ph

Me

2. H2NOH

Ph

(-H2O)

Me

N OH

Solutions • 359

O 17.

CHO

O

H3O

OH

OH O

H

OH

OH O 18.

1. Ph3P

O

O

O

O

Br

3. H

2. n-BuLi

acetal O 19.

O

H3O

O

O

C CH

20.

H O

O

O

PPh3

hemiacetal OH H3O H

O

1. H3O, Hg2-

H O

O O

2

etc.

H

H

H

W-K

OH

N NH 2

O

O

OH Br

1. Br2, H2O

21.

O

2. H2NNH2

+H2O, taut

O

H3O

+

H

O

4. H3O

2.

,H

O

O

3. Li Li OH

O

5. H3O

2. H2NOH, H

1. HONO2, H2SO4 CHO

O

4. O

O

HO

22.

O

O2N

CHO

O

O2N

N OH

O

note position of EAS!

23.

Et

H3O

O

Et

O

OH

H

24. O

O

OH H

OH

Et

HO O

HO

O

O

H3O

O

OH

O 1,2-H: shift

O H

HO

OH

~H

O O

15.1 Reactions


H

CHO

D C D

CHD2

1. D2NND2, OD, D2O

25.

2. HI

W-K

OMe

OMe

OH

CN O

O OH

OH

CO2H Ph

O

H3O

H

OH

OH

OH

HO

CN

Ph

H

+

OH OH

O HCN

+

Ph

H

O

O

OH

OH

O

O

H3O

27. O

H3O

HO

HO

OH O

MeI

OH

OH

CO2H 26.

+

SN2-like

1715 cm-1 O

O

O

O

MeOH, H

28.

H H G2.2 O

O

OMe

HO

HO

O

OMe

H H 4.9 2.8 HO

O

H 3.4

HO

OH

H3O

29.

OMe OH

HO -H2O -HOMe

O 30.

OH

H3O

O +

O

H

OH

H

NH2 31.

H

1. H

CO2H

O

N

-H2O

O

O

2. H2 / Pt CO2H

N H HO

O

CO2H

OH OH

OH 1. LiAlH4

32.

+ MeO MeO

15.1 Reactions

2. H3O O

EtOH

+

MeOH

Solutions • 361

O

O

O

O

33. H

H

H

OH

H

O

1,2-H: shift

OH

O

H

intra-Cannizzaro

HO

~H

OH

H

O

H

O

H

O O O

34.

OH

1. LiAlH4 EtOH

O

-H2O

OH

+

2. H3O O

HO

HO

O O

HO

HO O

35.

O OH OH

H3O

O

O +

O F

F

O

OH N

H2NOH

36. O

N OH O

37.

O

OH

O O acetal

OH

O

OH OH a hexose

n

OH H

38. OH

O

H3O

OH

N

1. NH2OH, H

H

H

OH

OH

H OH

OH

OH

2. Ac2O -H2O

OH OH

OH

OH H

OH

OH O a pentose

C OH

N

OH

an unstable cyanohydrin -HCN

15.1 Reactions


O

O N

39.

Me

N

mild acid

N

N H

H

H

O

O

OH

acetal

O

OH

O

40. Me2N

+

N

N H

O

O

Me

S

H N

H N

H

OH2 NO2

S

O

H H O N NH2Me

~H

H

NO2

H

-H

H N

O

Me2N

H OH2 N NHMe

mild acid

H N

-H

NO2

NO2 -MeNH2

O (H NO2

N N

N

41. N H

NH2 O

H3O

N H

S

N

OH

S

O

O

O O

N H

S

O

2. LiMe2Cu

H 3. H3O

O

H3O

O

OH +

H

H

OH

acetal

F

F

NH2

N

H

[O]

15.1 Reactions

+

H N

43.

O2C

N

CO2H

H

[O]

H N

44.

NH2 H3O N H

1. PCC

42.

N

CO2

O2C

CO2

O H3O

HO2C

CO2H

+

NH4

Solutions • 363

O

O ketal

O

H3O

45.

OH OH

HO

HO

OH

46.

O 2. MeLi

1. PCC EtO

EtO

EtO

OH Me

3. H3O

O

EtO

O

N

O

N 1. HC CNa / THF

OH

47. O N


O

1. HS(CH2)3SH, H

S

2. MeLi

S

1. n-BuLi

S

2. CH3(CH2)9Br SN2

S

Ph

SN2

S

Et

O

4. H3O Ph

O

3. H3O

1. HS H

3. EtI

S

S b.

c.

Ph

H

O

H

O note: addition to ketone, not amide carbonyls

O

48. a. Ph

N

H

S

H

SH, H

S

S

O

3.

O S

2. NaNH2 O

S

O

or -H2O

4. H3O HO

15.1 Reactions


N

H2N 49.

H

H N

H2N +H2O

N

H

N O

N N

-H2O

N

H N N H

O

+ H

nitrogen analog of an acetal

O

H formaldehyde

O HN

HN

O HN

CO2H

CO2H tetrahydrofolic acid

CO2H

N

N

N

N

N

-H2O

O

Cl

N

H

NH2

50.

CO2H

N

Cl

XanaxTM - (anxiolytic)

15.2 Syntheses O

1. H3O 1.

HO

3. CN, HCN

CN

HO

4. H3O

CO2H

2. CrO3, H 1. Hg(OAc)2, H2O

OH

O 3. KMnO4

2.

H

1. HI -CH3I

OH H

2. Cr2O72-

O

3. NaBD4 4. H

3. H2NNH2, OH W-K

15.2 Syntheses

NH2 5. H2 / Pt

-H2O

2. NaBH4

OCH3

NH

4. NH3

OH D

Solutions • 365

O

3.

OH

1. HCN, CN

CN

O

2. H3O

OH

1. NaBD4

Cl

3. SOCl2

D

2. H

D

SNi - ret

OH

O 1. HCN, CN

H

CO2H

3. H2SO4 (-H2O)

4.

5.

4. H2 / Pd

CO2H

O 3. CrO3, H

CO2H

CO2H

2. H3O O Cl

6. Ph

O

1. OH

OH

Ph

SN2

[O]

O

O

2. PCC

3. ethylene glycol, H

O

Ph

O

Ph (1 equiv)

H

O

(aldehyde more reactive than ketone)

OH

1. LiAlH4 O 7.

OH

2. H

O

1. NaBH4 OMe

OH

O

OH

O

OMe

2. H

H2 / Pd

OMe O

1. LiAlH4 2. H

O

7. (cont.)

OH

OMe

OH 3. H2 / Pd

O

1. ethylene glycol, H (protect ketone) OH 2. LiAlH4 3. H3O O

.

1. BH3 THF

8.

OH

3. PCC

H

4. Ph3P

2. H2O2, OH

O 9.

1. H3O conj. add'n

O

OH

2. H

O

MeOH, H OH

O OMe

15.2 Syntheses


O 2. PCC

1. MeOH, H

10.

H

OH

OH

MeO

O

MeO

OMe 3.

4. H3O MeO

O

OMe

Ph3P

OMe

O 1. HO

H 11. Br

O

H O

OH, H

MgBr

3. HO

O

2. Mg

4. H3O CHO

12.

1. H2NNR2

O

H

N

NR2

NR2 2. n-BuLi

NR2

N

N

3. PhCHO

O O

OH

4. H3O

Ph

Ph

CHO 13.

O

O 1. HO

OH

2. KMnO4, OH

O

or 2. a. OsO4 b. NaHSO3

H

O

3. CrO3, H 4. H3O

HO

O

OH

O

O

O 14.

CHO

H

1. LiMe2Cu

H

2. H2NNH2, OH W-K

O

OH

HO C CH

1. ethylene glycol, H 15. O

15.2 Syntheses

2. K2Cr2O7

3. HC CLi O

4. H3O O

O

Solutions • 367

O

OH 1. EtMgBr

H

16.

Et 3. HBr

Et

2. H

Li

4. Li

O 5. R

Ph

Ph

Et

Ph

Et 6. H3O

-H2O Ph

OH Ph R

R

OH

1. NBS, R2O2

O

3. PCC [O]

O

4. Me2CuLi

17. 2. OH (SN2)

18.

5. H

1. DIBAH, -780 CO2Me CO2Me

3. Ph3P=CH2 CHO CHO

2. H

O 2. O3

1. PCC

H

19.

O

3. Zn, H OH

H

O

O

H OH

O

1. Me3SiCl

20. Br

TMS

Br CO2H

O

5. H (deprotect)

2. Ph3P

O

3. MeLi

PPh3

TMS

O 4.

TMS

H

6. CrO3, H

O 21. Ph

22.

SH Ph

SH

H

O

1. HNMe2 -H2O

S

S

Ph

Ph

H2 / Ra-Ni

NMe2

PhCH2Ph

2. O3 3. Zn, H

O H

N O

15.2 Syntheses


15.3 Mechanisms H

O

O

OH2

OH2

+H2O

+H

1.

HO

H2O

~H

(H

O

O H

O

-H2O

-H

= 18O

O +H

2. N

-H

~H N H

H

OH2

N H

H

H

OH2

N H

O (H

H

O

OH2

O

O

O OH

O H

4.

OH

conj. add'n

H

~H CA of a hemicetal

H

HO

-H

O H

Cl

N N H H H

-H2O

Cl

O

N N H H

OMe

-Cl OMe

H

OMe

O

O O

Me O

O

H

O

6. R

OH

O

O

R H2N

R H2N OH

R HN

R

R N O

15.3 Mechanisms

-H3O

R

-H2O

R R

N O OH2

~H

R N O OH H

H R

R N

O H

Me

H) OMe

R

R OH

OH H)

R

O OH2 O

~H

Me

Me

OH O

~H

OH2

N N H (H

OMe

OH

OH

NH2

N N H

O H

O (H

H

H

HN

-H

5.

OH2

-H

taut

NH2 H NH2

H2N NH2

NH2

H

3. O

H

O

H

R N

O H

O ~H

Solutions • 369

H

H O

O 7.

O O OH

OH

OH OH

O

O

O (H

O (H

O H

H O

OH

O

-H

O

O H

O H

H

O

8.

OH

O

O

O OH

OH

OH

OH

OH

O HO

OH

O O H

OH

OH OH OH HO

-H

O

H3O

OH

hydrolysis

OH

OH

OH

O 9.

H

O

H

O S

O

O C S H

O H

O

O H

O S

+

O H

OH -H

10.

(H O

O

O O

O

H

Cl

Cl

H

H

O

11. H

Cl

Cl

O

OH

O

O

H2O

O

~H

O

H O

H2O O

H)

O -H H

+

O

taut

O H

O O

+

H)

O

H

15.3 Mechanisms


OH

OH

OH H

OH

H

H

O

O

12.

OH2 O

OH2 (H

-H3O O

O

OH2

~H

OH

OH

~H

-MeOH

OH

OH

O

O

O 13.

N N

O H

O

1,2-R: shift

+

CH2 N N

Ph

Ph N

Ph

N

N

Ph N OH2

14. N

N H

Ph

H

N2

NHPh

Ph

OH2

NHPh ~H

O

O

-H

O

Me

Me Me

H

H

Me

H 2.

O Me

-OPPh3

Me PPh3

rotate

H

PPh3

Me

P(OMe)3

Me O

O H

Me

O

15.3 Mechanisms

NHPh

PPh3

H

:P(OMe)3

NHPh

PPh3

H

Me

16.

OH

H

Me 1. mCPBA

O

NHPh +

H

15.

(H

Ph H N

H

O P(OMe)3

Me

-OP(OMe)3

H

H

Solutions • 371

HO

HO

OH

O OH

O

+H

17. OH

OH HO O OH

O

-H2O OH HO

H)

OH

OH

OH

HO

EAS

OH

OH

NHR

O

-H OH

:NH2R

O

OH

NHR

OH

+H

18.

HO

OH2

OH H) -H

HO HO

OH

OH -H, EAS

O

Et

19.

H

O

N H2

H2N Et

Et

+H HO

-H2O

Et

~H H

(H

-H

N H

taut

Et

N

~H N

H N

OH

1. -H2

20. a.

O

Cl

O

(H :H

O

Cl

O

O

2. SN2 Williamson ether synthesis

Cl

O H Et

- OH

N

OH

O MOM

Cl

3.

b.

O

O H

OH

H

HO

O

OH

O H

O

H

+

MeOH

H3O

O OH

OH

H2O +H2O

+

H2C O Me

-H

15.3 Mechanisms


Br 21.

OH2

Br

Br

H3O

OH2

Br

H O (H

O

O

O

HS Me

Me S

22.

-HBr

H

OH

O (H

(H

~H

O

C N

23. O

OH

OH

O

HO H

-HCN

H

OH

(H

SMe

~H taut

SMe

conj. add'n O

-H

Br

H + CN

O :C N

CN

NC

H

O

:Me O

OEt O

1. MeLi

24.

OEt

H2O

2. H

OEt

-H2O

1,2-add'n O (H OEt H

O -H -HOEt

H CHO

OH2 OEt

~H

OH2

Et O

CHO

EtNH2

25.

N

-H2O

H

~H

H

OEt

+H2O

Et N H

H

Et HO

H N

O

(H Et N

Et N

-H

-H2O

H2O

Et N

~H

H)

OMe N

26. O TMS

Ph

15.3 Mechanisms

H

OMe Ph

N

2. D-A O TMS

Ph Ph

3. H

OMe N

O

Ph Ph

(H

-H -MeOH

N O

Ph Ph

Solutions • 373

O OH

27. a. Cl2C (H

O Cl2C

Cl2C:

D-elim

Cl Cl C H

~H

H

Cl

H

b. Me

Me N

O C

OH H

2. H

H

O C

O O P Cl N C Cl Me H Me

-Cl

O

Cl O P Cl Cl

O

HO

O

Cl -HCl

H)

O

O SN2

C H

O Me O P Cl N C H Cl Me Cl

+Cl

Me

-Cl

H N C

Me

Cl

O2PCl2

Cl

OH

OH

OH H

Me

(H

1.

C N

-H

NMe2

OH

-Cl

NMe2

Me

Cl

Cl

Cl OH CHO

2. H3O

NMe2

hydrolysis

O 28.

Ph

O Ph

H

OH

Ph

O

O

Ph

Ph

CN O

OH O

H

Ph

CN

:C N

NH H

OH

~H

H

H Ph CN

O OH - CN

Ph

~H H

Ph CN Ph

H N

H

OH H

29.

N

~H

OH2

OH

OH

N

OH

N OH

-H

-H2O

N (H OH

EAS

OH

15.3 Mechanisms


HO 30.

HO

OH

O

H O

+H OH hemiacetal

OH

OH

HO

H O

-H OH OH

OH

NH2

H N

31. HO2C

NH2

H NH

H N

HO2C

NH2

HO2C

+

H2N

32.

-H NH2

O :NH2R

O H

OH R N

Ph O H CO2H

OH O CO2H

R

H

O CO2H (H

R N

+H

Ph

O

+H H

H

O

Ph

OH O CO2H

OH O CO2H

O

Ph

R N

-H3O

OH2 33. RO2C

O

O

OH2 Ph

N

OH2 NH2 NH2

~H

-H

-H

H

Ph

H2O:

+H2O

O O CO2H

O (H NH2 NH2

R HN

H

+H -H

H N H

Ph

NHR H

HO

H N

O CO2H

CO2H

O

NH2

R H) O N H

O O

OH

HO2C

HO2C

Ph

Ph O

OH

+H2O

NH2 O

OH

NH2

OH2 NH2

NH2 NH2

H

top-side attack

DOH OH

OH

H O

bottom-side attack

OH

HO O

HO

OH E-

OH2 O

H

H

~H

H

PO3R' OH RO2C

+

O PO3R'

15.3 Mechanisms

O H

(CH2)nCH3

-H

H O

O (H

H

H

Solutions • 375

H

O Ph

34.

OH

Ph

Ph

O R

O

O

Ph O

O

(H

OH O

Ph

O R

35. a.

O

Ph

O

Ph

HO Ph

OH

OH 1.

b.

O 1.

Ph

Ph

O

O

O

Ph

Ph

O OH

1,2-R:

O

O (H

:B

Ph

OH

O

OH

O OH

O

shift

R

2. H

2. H Ph

Ph

OH

R

Ph

Ph

OH

OH 1,2-R:

R

O

OH

O

Ph

OH R

shift R

O

~H

Ph

Ph

shift, ~H

+H R

37.

HO

shift

1,2-R:

OH

36.

O

Ph

O

O

O O

O

c.

Ph

(H

O Ph

O

1,2-R:

RCO2H +

-H

Ph

OH

O

O

O

O

1,2-R: shift

R

O Ph

O

O H

Ph

H

O

-H R

(H

R

R

R

OH

OH

OH

1,2-R:

+H -HOTs

shift

-H (H

OTs H) 38. O

1,2-R:

+H HO

HO

shift

-H HO

HO

15.3 Mechanisms


+H

1,2-R:

39. HO

O

shift

HO

-H

HO

1,2-R: HO

shift

H)

HO

OH

O

40. O

. .

HO O

O +H HO

. .

HO

-H

. . ..

HO H) O

O

HO

. .

1,2-R: shift O

. .

1,2-R: shift

HO

OH O O O

O 41. RO

H)

Cl

O

OR

-HCl

H) OR H

H) OR

Br

RO O

42.

(H

O OR

O

O -HBr

Br

H

RO

OR

OR

-Br

CO2R

Br

Br OR

OMe H +H

43.

1,2-R: shift

-MeOH O

NMe2

-H (H

O

NMe2

O A

15.3 Mechanisms

NMe2

O

NMe2

Solutions • 377

OH2

43. (cont.) NMe2

O

OH2 O

O

A

~H

NMe2

NMe2 O +

-H

NMe2

H

NMe2

OH

R

N

N

-Br

44.

R

N R CHO

O

~H

N

Br H R N CHO Br

-Br

N

NH2

O H HS

CO2R

~H

N

OH NH 2 S

O

CO2R N

CO2R

-H

H

N

N CO2H

S

O

O

H2N:

N H

CO2R

N (H

N S

O

NH2

N H

HO

NH2

CO2H

S

CO2H

~H

O

H

CO2H

O H

+H

H

46.

S

NH2

-H2O

O

O

OH2

O

CO2H

H

CO2H

-H

O CO2R

+H

O (H

Br

O CO2R H O

H R N

OH2

(H N R CHO

-H

O (H

R

N

Br OH2

Br Br

45.

O H

NH

N H

OH

HO

HO

HO

HO

OH2 -H2O

OH NH

NH -H N H

N

N H

NH

HO HO

HO OH

OH

HO

15.3 Mechanisms


S 47.

H

Br

SH

Br

Br

Br

+H3O

Br

Br OH2

O H

SH2 Br

~H

Br

OH2 Br

O

Br

Br

-H

Br

-H2S

Br (H

Br

Br

Br OH2

H) HO

OH

Br Br

O H

Br

O

OH

+H3O

Br

Br

Br

H) O

H

~H

OH O

O O

Br

OH

H

48.

HO

-H

Br

O

H

H

O O

O

-H 1,2-H: shift

O OH

2

O O 1. +H

49.

!!

O

-H2O (H

OH O

OO

O

:B

O

O 2. +H2O H)

15.3 Mechanisms

OH

-H

O O OH2

CHAPTER 16 CARBOXYLIC ACIDS 16.1 Reactions

1.

CO2H

KMnO4, H

N

N

N

1. OH

CO2H

2. Ph

O

Ph

O

3.

Ph 4.

O

O

O

O H

3

O

2. CrO3, H

OH

acetal

O

E D

1. NaCN SN2

O

SN2

O (conj. add'n) OH

Ph

C N

3. LiAlH4

CO2H

4. H3O

Ph

C N Ph

OH

3. H3O

NH2

-H2O

C

N

2. DIBAH, -780

C

N H

Ph

Ph O

O 1. SOCl2

OH

O

3. BH3

2. PhMgX

OH

4. H3O

O

1. OH 2.

O

2. '

O Br

6.

7. PhCH2Cl

O

1. OH

OH

OH

8.

Me

OH

1. H3O O

O

Ph

O

2. H

Ph

Ph

5. J Br

OH

1. NaBH4 OH

SN2

BF4

O O

2. -Me2O

O

3. H3O

O H

16.1 Reactions


1. (XS) PhLi

9.

2. H

CO2H

O Ph

O

O

O

-H2O

+

conj. add'n

O

OH

OH

2. H

1. HCN, CN

11.

Ph

OH

1. LiAlH4

10. CO2H

O

OH Ph

HO

3. BH3

2. H3O O

O

CN

CO2H

O

4. H3O

OH

12. Ph

CO2H

O Ph

1. EtLi

NMe2

Et NMe2

1. H3O

CO2H

-H2O

O

CO2H 14. CO2H HO

N 3. CrO3, H H OH

16.1 Reactions

CO2

OH

labile carboxyl CO2H 2. -CO2 D E CO2H O

CO2H

[O]

H)

O

CO2H O

N 4. ' -CO2

O

O

H

N

CO2H 15.

H

O

O +

H

1. [O]

NMe2

OH

taut H

O

[O]

HO

Et

Ph

2. PCC

CO2H

conj. add'n

O Ph

2. H

Ph

Ph

13.

O

O

Solutions • 381

O

OH CO2H 16.

OMe

OMe CO2

1. OH (2 equiv)

CO2

2. MeI (1 equiv)

CO2H

3. H

SN2 more stable anion, therefore, less reactive

17.

CH2CH2N(CH3)2 COO H CH2 CH2 N COOH H

CH3NHSO2CH2

a salt!

16.2 Syntheses CO2H 1.

1. LiAlH4

OH 3. TsCl

2. H

1. NBS, ROOR

3. NaCN Br

2. HBr, ROOR

O

Br

OH

SN2

conj. add'n

CN

1. HO

OH , H

4. 2. Mg

O

' OH

Jones reagent

O

OH

O

2. CrO3, H

O

O

HO

CN

O OH

O

4. H3O

O

1. H3O

OH

Cl

CO2H

4. KCN (SN2)

2.

3.

CN 5. H O 3

-CO2

O

O 3. CO2 4. H3O

MgCl

CO2H

O 1. Cl

5.

O

2. H2NNH2, OH W-K

AlCl3

CN

CO2H 5. H3O

3. HBr 4. KCN

16.2 Syntheses


1. PCC

OH

6.

n-Pr

2. n-PrLi

3. HCl

O

n-Pr

Cl n-Pr

5. NaOH, H2O

CO2Na CO2H

O

7.

8.

Ph

O

2. CrO3, H

1. KCN (SN2)

Br

4. NaCN

CN

1. H3O OH

n-Pr

HO2C

Ph

O

O 4. BH3

3. ' (-CO2) OH

CO2H

2. DIBAH, -78o

CN

N

Ph

9.

CN

O

1. DIBAH,

CHO O

2. H3O

CN

1. KMnO4, H

10.

-H2O

1. LiAlH4

RCH2OH

2. H3O

H

N N

-H2O

O

CO2H

HO2C

O

Ph

CHO

3. NH2NH2, H

2. '

O

HO

-2 CO2

CO2H CO2H

11. RCO2H

3. H3O

N NH2

H

CHO

HO

hydrolysis

H

-78o

5. H

O

3. SOCl2 or HCl

OH O

O

5. R'MgX RCH2-CN

RCH2COR' 6. H3O

4. NaCN

or

5. H3O 6. R'Li 7. H3O

O

O

12.

1. BH3 OH

3. HO OH

2. H3O

1. H2SO4 (E2) 13.

O

Br

16.2 Syntheses

Br

O

O O

O

5. PhMgBr

OH Ph

H 6. H3O

4. PCC

3. (XS) KCN

OH 2. Br2, CCl4

OH , H

O

NC

4. H3O CN

HO

OH O

Solutions • 383

O

O 1.

14.

O

Cl

OH

2. CH3Cl

3. H2 / Ni

AlCl3

hydrogenation

AlCl3 CO2H

a benzylic alcohol 4. KMnO4, H

hydrogenolysis

CO2H 15.

HC CH

1. NaNH2 (1 equiv)

3. KMnO4, H

HC C

HO

2. n-pentyl chloride O O

O

OH

2. Et-I

1. NaH (2 equiv)

16.

OH

O

(1 equiv)

O

OH

3. H O

Et

Et

more reactive anion

16.3 Mechanisms O +H

1.

O O

CO2H

O

H O (H

-H

O

H

H)

OH

O CO2H

O C

taut

2. n-pentyl

OH '

O

n-pentyl

-CO2 n-pentyl

O tautomerization to E-ketocarboxylic acid facilitates decarboxylation

O

H)O

Cl

HO

Cl

O2C

Cl +OH

3.

O

CO2 -Cl

-H Cl

Cl

Cl

Cl

16.3 Mechanisms


+H

4.

+CO

O 5.

R

HO ~H

OH

OH

R

OH

H)

O

CO2H

O

HO

N H

H2N

R

-H

O (H H

OH2

C O

HO

O

+H2O

C O

N R

-H2O

O

HO

N

H -H

N

R

(H

O H

O

OH2

OH +H

6.

O

CO2H

-H

O

7.

R

+MeOH

O N -H2O

R N

O

O (H

-CO2

N R

R

H R'NH2

+H

OH R'HN H

H

R' N

R'N H

OH

H3O hydrolysis

N R

OH2

H

hydrolysis -RCHO

O

-H2O

R

R'

H N

H H

R' N H

O

H2N

NH2

O

R

16.3 Mechanisms

R

~H

taut

H

+

-2 H2O

H

OH

O

dimerize

N

8.

-H2O

Me

O O

O

O

O

H O

O

O

CO2H

Me

O

O

NH2

O

(H O

Me

O

H

O O

~H

O H

R

C N

N C

R

H3O

Solutions • 385

O 9.

CO2H

HO

H2O

-CO2

O

O

N

O

O (H R

-H2O

O N

O

O

O N

-H

O + H2N

O

O

O

O2C

H

O

O2C

H) OH

H NH2

R N

H NH2

NADH

NAD+

O

O

R N

+

O

O

-CO2

CO2

CO2

R' 13.

OH

H

O

O

O 12.

H2O

O

-H2O

(H

R N

R -RCHO

O

O

11.

H

1. -CO2 2. taut

O

blue dye

CO2H

O

CO2H H2N CHR

O

10.

-H

CO2H

O

O

(H

O

+H

CO2H

O O O P O P OH O GTP

-H

R N

O

+

H OH -H

R N

S

S

R N

O

OH O

H)

R'

R' H

-CO2 O

CO2 O

R'

S

R N

S

(H

O

PO3H

R'

R'

S

CO2

-GDP

(H

OH

S

R N H

O

(H

16.3 Mechanisms


O

R'

14. PLP + histidine

R' =

NH

N

H

-H2O

H N

R

H

O

R' -CO2

OH

H H

O

16.3 Mechanisms

+H

H

O H

MW = 60

H

H

N H

O OH

H

H3O

O 2

OH

N R' OH

imine hydrolysis

15.

R N

N H R

PLP + histamine

N

O

MW = 120

intermolecular hydrogen bonding forms a tight dimeric complex in nonpolar solvents

CHAPTER 17 CARBOXYLIC ACID DERIVATIVES 17.1 Reactions O 1.

Cl

2.

OEt

Et2N

-HCl

more O reactive than ester O

O

Et2NH (1 equiv)

OEt

O

O

O +

O

MeNH2

O N H

Me

+

OH

O

O (XS) EtOH, H

O

3.

OEt

OH

+

transesterification O O

4.

+ OH

OH

O

Cl

O

O

-HCl HO

Cl

O

1. PCl3

O

O

-HCl

O

O

Cl

O

2. LiAlH(O-t-Bu)3

O

5. HO

(or SOCl2)

OH O

6.

Cl

H

3. H

H

H3O N

CO2H

hydrolysis

+

NH2

O

O O

7.

Cl

Ph

1.

MgBr

MgBr

1.

or OH

2. H3O

- OPh

-H2O

not isolable

O 8.

O MeOH, H

O O

NAS

Me O CO2H

MeO

MeOH, H

O O

1. LiAlH4

9.

3.

O

O O

NAS

2. H CO2H

OMe

OH

O

O

17.1 Reactions


O O

O

OH, H2O

10.

HO

saponification

H N

Bn 11.

O

O

D

H

NH2 S

CO2

OH, H2O

N

O

S

O2C

+

H

N

CO2H

CO2 O

O OR

Et 12.

H2N

+ Et

Et

- 2 HOR O

H2N

OR O

O

N

1. OH

O

Ph

O

N 3. CrO3, H

OH

PhCO2H + + MeOH

2. H

N H

O

N O

13.

NH

Et O

4. '(-CO2) O

OH

O O

OH

O

OH SCoA

14.

NH NH2CH2CO2H

A

HO

CO2

B -HSCoA

OH

H

N 15.

2. CH2N2 N H O

O

O O

O

OH OH

OH O OEt

1. H3O

O

-HOEt

O

OCH3

OH

2. PhMgCl 3. H

a J-lactone

17.1 Reactions

OH

(diazomethane)

N H

O

H

N

N

1. H3O

N

16.

HO

HO

Ph

or

Ph

-H2O

Ph Ph

Solutions • 389

O

O

O

1. LiAlH4

17.

OH OH

2. H

O O

18.

N

1. Cl

N H

Cl N H

Cl

-HCl

OH

O N

N

2. LiAlH4

N

-HCl

N

protein OH

O

O

P

-HF

O

Me O

O H

H

H

CO2H N

20.

N

3. H

protein

O O P F Me

19.

HO

O

O H

O

=

NEt2

Cl N

1. SOCl2

N

2. HNEt2 -HCl

21.

N H lysergic acid diethylamide

N

N H

H

O S Cl O tosyl chloride

O + H2N

NH

n-Bu more nucleophilic nitrogen

OH, H2O

NH

O

S O

O

+

O HO

N H

1 equiv

:NH2Ph

NH3

O O

NH2

24. H2C C O

N H

SO3

O

23. HO

O

O

O 22.

O S N H O

-HCl

O H2C C

NH2Ph

~H

OH H2C

NHPh

+

taut

HOAc

O NHPh

17.1 Reactions


25. dimethyl phthalate + HO

H

OH

O transesterification O

O

O

O

O

O

O

O

O

O

a polyester

O O

O 26. HO

OH

+

Cl

Cl - n HCl

O

O

O

O

O

O

O O

OH 27.

H

O

O

OH

O

O C N Me

O

O

N Me

O

N Me taut

~H

O O 1. Li

LiCu

O

O 3. Ph

Cl

2. CuI

2

O

Ph

O

O

28.

N Me H

4. H3O

Gilman reagent

Cl

O O 29. F3C

NHMe

Me N

2O

1.

OH

-

2. LiAlH4 3. H

O

O O F3C

17.1 Reactions

Me N

Solutions • 391

HO

O O

O 30.

H2N

O :NH3

O

O

+

NH2

OH OH OH

O O

OH C CH

31.

1.

C CH

OH, H2O

2. H3O N OH

+ HOAc + NH3OH

O

O MeO 32.

MeO

1. LiAlH4

N H

2. H3O

HO

N H

HO

O H2N

O

H2N

1. H2O, OH 2. SOCl2

S

O 33.

N N

CF3

HO

N N

HO O

O

F + HSCH2F +

O

OH

F

F

CO2Na MeO

O OH

H3O

F O

CF3

NH2

HO

34.

35.

H O

F O O

S

O

3. H2N

S

O

N

O 1. PhLi

O Ph

O

2. H3O -H2O

MeO

Ph

17.1 Reactions


O

O O S O

N

36. N

Me

N

HN

N

H N H

H3O

N

Me N H

OEt

O 37.

1. base

O

N

HN

N N

OEt

O

O NH2

2.

-Br

CO2H

Br

O HO S + O

N H

O

SN2

OH

Br O

O NH

38.

O

OH

NH O2C

~H

NH2

O2C

~H

NH2

N H O

O O2C

O

H N

N H

O

O NH

39.

O

1. Cl

N

OEt

N

OEt

2. Et2NH

OEt

N

OH

H N

N H3O

N

Cl

-CO2

N

N

Cl

Cl

O NH

NEt2

N

-HOEt

O

N 40.

O

N

-HCl O

NH

etc.

N H

N

H N

Ph

H Ph

O 1. Ph

41.

OEt

2. LiAlH4

-HOEt

3. H

Cl Cl

17.1 Reactions

Cl

Cl Cl

Cl

Solutions • 393

H2N OMe

OMe O CO2H

1. SOCl2

42.

OMe O

N

O Cl

OMe

S H N

CO2H

2. -HCl

OMe

OMe

S N

O

CO2H

CF3

O

O

N H

OH H3O

43.

+

CF3 O

CO2H

N H

CF3

H3N

O

CO2H 44.

CF3 H3N

O

H2N

H3N

Ph

O N

hydrolysis

N O

phenylalanine

N

CO2H

N Cl

N + H N O H a carbinolamine

N

O N

-CO2

H N

N

MeOH

Ph

CO2H N

exhaustive

N

OH +

CO2H

aspartic acid

Cl

N

H3N +

OMe

O

45.

CO2H

H3O

N H

N

CO2H

+ N

Cl

+

H

N

N H2N

O

O

OH

1. NaBH4 46.

CO2H

OH

OH OH a tetrose

O

OH H

47.

OH

- 2 H2O

2. H3O 3. SOCl2 OH

O

O

O

OH

OH Cl 4. LiAlH(t-BuO)3

CN OH OH a cyanohydrin

O

H

HO

OH

OH

1. HCN CN

O

+

2. H

O

HO

OH

O

5. H

HO OH HO O H a pentose

17.1 Reactions


R

O 48. a.

O

O b.

O

Cl

:NH2

R N H

H

CO2H

O N H O

O

O

+H OH

O H

R' R

R

O

-HCl

R N H

CO2H

N H

(H

O N H O

-H OH

R'

+

O

H2N O

O

-CO2

N H

H)

OH

O

R N H H

R'

O N H O

OH R'

OH

O 49.

+

SCoA

-HSCoA transesterification

CO2 NMe3 O

CO2 O NMe3

CH3(CH2)14CO2H, H

50.

O

-H2O

CH3(CH2)14

HO

O

O 51.

H3O

HN O

NH4

+

CO2

+

HO

N H

O 52.

O

H3N

1. ATP OH

O

O P O OR' O

NAS

2. HSCoA -HPO4R'

17.1 Reactions

OH

2. H2N

-HPO4R' (AMP) O S CoA

O N H

OH

Solutions • 395

+H

O

53.

O O H glucose

O H)

O

N

H

transesterification

O glucose

N

1. SOCl2 2. NH3

OH 55. S O

O

N N H

N O

F

OH

O

N

H3O

S

H3N

OH

O OH taut

+ S

-CO2

NHMe

taut SO3H

O

O NHMe

H SO3H

N

OH NHMe SO3H

NHMe SO3H

H OH

NHMe

C O

NH2

O

O

N

N

O

F

CO2H

F

O

N 3. SOCl2

N

HO-glucose

O +

O

O N

N

N

54.

-H

O

SO3H

OH O

H3O

MeNH3

+

H SO3H

(aldehyde gives a positive Tollens' test)

17.2 Syntheses O 1.

1. H3O -HOAc

O

3. Li

2. PCl3

4. CO2

Cl

O 2.

O

1. H2O, OH R

NH2 2. SOCl2 O

1. H3O R

5. MeOH, H

R

CO2

Cl

R

4. H O

3. LiMe2Cu

R

3. H, -H2O

O 3.

H O

or 2. MeLi (2 equiv) Me

O

O

3. LiAlH(O-t-Bu)3

2. SOCl2 OH

MeO

R

OH

O NH

1. LiAlH4 2. H

NH

3. Ac2O

N

-HOAc

17.2 Syntheses


.

3. CrO3, H [O]

1. BH3 THF 4.

O H

2. H2O2, OH

CO2H

5.

Cl

4. PCl3

1. D-A

C

5. AlCl3, -HCl O

2. Ac2O, or

CO2H

'

C

F-C acylation

O

O

H2SO4, -H2O

HO2C

O

CO2H O

6.

HO

1. BH3

H O

or 1.

O

7. Ph

O

1. H3O NH2

Ph

2. SOCl2

H

b. H3O O

3. LiAlH(O-t-Bu)3 Cl

O 8.

2. a. LiAlH4

CO2H O

O

Ph

4. H O

1. AgNO3, EtOH

OH

Cl

O

Cl

O

O

3. H

Me

-H2O

O

Ph

NCH3

Cl , AlCl3

Cl

O

Cl

H NCH3

O Cl

Ph O

H

F-C acylation

Cl CH3 O N

O

OH

O

NCH3

Ac2O

H

(2 equiv)

O NHCH3

5. HO

O

2. MeLi

(Tollens' reagent)

9.

O

O

OH H

OH

H

H

2. H

O HO

HO

Cl

O

Cl

H2O,

OH

-HCl, NAS H3C

CH3 O N NH3 (SN2) -HCl

17.2 Syntheses

Cl

O

O

N NH2

H -H2O

Cl

N

Solutions • 397

O 10.

HO

2. NaBH4 CO2H

1. LiAlH4 2. H

O 11. Ph

O

1. ' (-CO2)

CO2H CO2H

H CO2H

3. H

O

4. KCN (SN2) Ph

OMe

OTs

3. TsCl

6. CH2N2

O

Ph 5. H3O

O

Ph

-N2

OH

O NaO

O

-H2O

OCH3

O F3C

N

F

O

N

12. O

nucleophilic aromatic subst'n

O

F3C

OH, H2O

O O

N H

OEt

Cl

O

OEt

NH2

O

F3C

F3C

1. LiAlH4 H H N

:H

H N

-EtO

H

H H N

+H:

OEt O O

+AlH3 H

O H N

H N OAlH3

O

CH3

H N

+H: 2. H3O

F3C

:H - OAlH3 2CH2

O 13. Ph

N

1. NBS

Br

peroxide

Ph

2. Li

N

O

O

N

4. EtOH, H

EtO

NMe2

N Ph

Ph

3. CO2

HO

Cl

NMe2

2. a. NaBH4 b. H

1. Me2NH

14.

O2C

NMe2

3. PCl3

conj. add'n O F3C

Me N

O H

5. Cl 6.

O

OH, H2O -CO2

NMe2 4.

OEt F3C

F3C

O SN2

(continued on next page)

17.2 Syntheses


14. (cont.) Mechanism for step 5: O

NMe2 Cl

F3C

-Cl

OEt

Cl

CH3

NAS EtO

O

+Cl

CH3

N

Me N

O

-CH3Cl

O

F3C

O

OEt

O Et

1.

O

Cl

15.

Br

2. NBS, R2O2

Et

AlCl3

3. H2 / Ni (or W-K)

HO2C

BrMg

5. CO2

4. Mg

6. H

O

O 1. KCN, HCN

3. BH3

2. H3O

16. CN O

OH

4. H

CO2H

O

7. MeLi OH

O

O

5. HO

O

6. PCC

8. H3O

O

O

O

OH , H

OH

H O

O 1. KMnO4, H

17.

HO

O OH

O

2. ' OH

-CO2

O

O 3. OH

O OH

O

H O

Cl 18.

4. LiAlH4

OH

O

OH

O

5. H3O

O

O

1. Li 2. CO2

CO2H

4. SOCl2

Cl

5. Et2NH

NEt2

-HCl

3. H

H N Ac

O NH2

HO

1. KOH

Me

NH2

O

3.

19. N H

17.2 Syntheses

OH, H

2. MeI

N H

more nucleophilic than

2O

-HOAc NAS

Me

O

N H

Solutions • 399

O 20.

Br

O

H2 / Pt

CN

(XS) HBr

(XS) NaCN / DMF

high pressure/temp

Br Cl

1. H3O 2 . SOCl2

CN

H2 / Ni

CN

Cl B

O CN

SN2

O

H2N

NH2

or 1. LiAlH4 2. H

C

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

HOAc, H, Hg2+

2. H3O

HC CH OH

OH

or saponification

OH

OAc OAc OAc

poly(vinyl alcohol)

poly(vinyl acetate) CH3O

O

HO

22. a.

1. NaOH

O

2. CH3-I

O

N HO

HO

HO O

CH3O

O

O O

O

O

O

NAS

N

N

O

O

O heroin

O N (H

H

hydrocodone

CH3O

CH3O

d.

O codeinone

1. H2 / Pd 2. PCC

codeine

morphine

(XS)

c.

O

SN2

N

N

b.

etc.

~H taut

~H

O N H)

O

CH3O

CH3O

H

taut

+H2O

O N O neopinone

H

O

-H

N OH O oxycodone

17.2 Syntheses


17.3 Mechanisms O

O

HO

OH2

+H

O

1.

H ~H

O

O

H) O

O (H

HO

OH2

O H

OH O H

=

-H O

-label appears in both carboxyl oxygens - but NOT in alcohol oxygen

O

-H O OH

HO

+

O

O

Cl

O

R

:R

O

O

1. (XS) RMgX 2.

OH

HO

O

+

R

R

R

:R

R R

:R 2. H

2. H

t-BuOH

R3COH

OH O

-H2O

O

3.

D-lactone does NOT form because of ring strain

O

OH

O

O -H2O OH HO2C

intermolecular condensation

O

-H2O

O

O

intramolecular condensation

=

O

O

O much less ring strain

O

O

4.

O

+H

H

O

H

O

Et

O

Et

-H

+ (H

O

Et

O

30 carbon O

OEt O

5.

O PhMgCl

OEt

Ph

OEt

O Ph

O

- OEt

O

O

O

Ph

(1 equiv) Ph:

ketone more reactive than ester

S O 6.

Ph N C S H2N

R=

PhN N H

NH2

R R N H

H PhN

~H

Ph + H2N

17.3 Mechanisms

CO2H

N O

+H

NH

R N H

O S

CO2H

S H N HO

-H

Ph H)

NH

HN

O H

R

S NH

S

Ph

N

O HN H R

NH

~H

Solutions • 401

Me

O

O

NH2

NH2

7.

Me O 1. Me3O BF4 Me -Me2O

OMe H2O

NH2 OMe

OMe

-NH4

O

S O

Cl

Cl

-Cl

S

Me R b. R

O

:NR3

O (H

(H

Cl

+Cl

Cl

S Cl

+ CO2

O

Cl H

O

O

Me 8. a.

~H

NH3

O O

OMe NH2 OH2

2. H3O

:NR3 -NHR3

R

(H

-NHR3

-Cl

R

O S

-Me2S

:C O:

+

O R

R

S Cl

O 9.

S

O S Cl

Cl N

N

O

-HPO42-

PO3H

NAS

H2N

H2N

H2N

HO

~H

HO

CO2H

N CO2H H H

O

O

O CO2H HN N H

CO2H

O Br2P Br 1. POBr3 O

11. CF3

N H

H) O HN

-H3O

CF3

~H

O

Br

N H

N H

N CF3

N

Br

CF3

N

(H

2. Li 3. CO2 4. H

5. C5H4NLi, -600 N CF3

CF3

CF3

CF3 CO2H

7. [H]

CO2H

-HO2PBr2

CF3

O

CO2H

OH

OPOBr2

+Br

-Br

N H

H2N

CO2H

O POBr2

CF3

mefloquine

HO

OH2

N H

O

Br

R

OH

O O

O

R

[O] (mechanism per 17.3, 8)

O

O 10. H2N

O

R2CHOH, NR3

6. H3O

N

CF3

CF3

17.3 Mechanisms


O 12. a. R

O

O Cl H :C N N: H

-Cl

-H

N N:

R H

N N:

R

(H

H :CH2N2

1. SOCl2

hQ

2. CH2N2

-N2

b. CO2H

C O H N N:

C

O

O

H

2. W-K

1. H2 / Pd hydrogenolysis of cyclopropyl bond O

13.

cleavage here H N

O

O

1. BrCN

N H

R

S

H N

R' -Br

O S

Br N

C

R

H

H N

N

-MeSCN

O O

R'

CN R

H N

H3N

+

R' N H

R

O O

O

O

R'

2. H3O iminium hydrolysis (see 17.3, 7 for mechanism)

a carbinolamine O

OH N

N H

H

14.

O

O

O

H2O2, H (see 15.3, 34)

N

N

O

O

O

N H

N O A

O

+H O

N

O HO

-H2O

17.3 Mechanisms

N O

O

N O

H O N

O H

-H

N O

N H

O

H O

Solutions • 403

Ac O

N

O O

O

15. N

O

-HOAc

CO2H

H) O

O

O

O

O

:py

Ac

O

O

-pyH

O

O

O

N

-CO2

O =

Me

N

O

OAc

O

O

S

S 16.

O

O

+H

N H

O

R

N

R = cyclohexyl

O

N H

O

CO2

RN C NR

+H

NHR

+ N H

PhO

S

O

N

taut

O H)

CO2

OH :P OH OH

~H

S

-H

N O

O H P OH OH

N N H H dicyclohexylurea

CO2

H

H

17.

O

- OAc

CO2

R = -CH2CH2NH2

nucleophilic form O R

(H

O

PCl2 Cl

H2N

R Cl

-HCl

OH O P OH OH O P OH OH

H

O

1. :PCl3

O

+Cl

PCl2

-H

(H O R P(OH)2 P(OH)2 O

2. +3H2O -3 HCl

O

H :PCl3

PCl3 P(OH)2 O

-HCl -H O

R

:PCl3 P(OH)2 O

O +

O OMe

H

R

O

OH P OH R Cl O (H

OH

+H O

:P(OH)3 Cl

OH

O 18.

R

-POCl2

H)

O

OMe

Me C O: an acylium ion

OMe

-H O

OH Me C O:

O H)

O

O

O Me

O

O Me

17.3 Mechanisms


19. a. Carboxylate as a nucleophile: O

O O

O

O ~H

O

O

O O

H

O

O

HOAc

+

OH H

18O-label

appears in salicylate

b. Carboxylate as a base: O H

O

O

O

O H

~H

OH

O

O

HOAc

OH

O O

+

O

no label!

HO c. Therefore, pathway b is preferred.

NH2

H

O

H

H 1. -H2O

20. CO2Me

N H

N H

O N N H

17.3 Mechanisms

:H

N C

H) N

-HOMe N H

OMe O

H OR

CO2Me

2. NaBH4, HOR 3. H

CHAPTER 18 CARBONYL Į-SUBSTITUTION REACTION AND ENOLATES 18.1 Reactions 1. LDA

1.

2. n-PrBr

O

O

O

H O 1. OMe

OMe

2. O

MeOH

O

O

O

1. LDA

3.

' -CO2

Ph

CO2Et

Ph 2. (PhCO)2O

3. EtI O

CO2Et

Cl 5. NC-CH2-CO2Et

H

1. OEt N

C

O

NO2

O

Ph

-CO2

NC

OH

HO2C

CN

CO2H

2. H3O, '

CO2Et

-Cl nucleophilic aromatic subst'n

-CO2 NO2

NO2

NaCH(CO2R)2 H

cis-

transO

O

O 1. H3O 2. CrO3, H

Me

HC (RO2C)2

OTs

7.

Ph

CO2Et

NC

SN2

O

O

H Me

H

Et

-CO2

H 6.

O

Et 4. H3O ' CO2Et

OEt

3. H3O, '

3. H3O

O

2. OEt

OEt

CO2Et

O

O

O

EtO

1. base

Ph CO2H

O 4.

CO2Me

O

NAS

CO2Et

Br

O

O O

2. Ph

O Ph

O

O

OMe

3. NaH 2 equiv O more reactive enolate

4. PhCH2Cl 1 equiv O

O

O

5. H O

Ph

Ph

18.1 Reactions


CO2Et

EtO2C

1. OEt

H

CO2Et

CO2Et

2.

8.

EtO2C

EtO2C

3. H3O, '

O

-CO2

O

OH

OH

O

O

O

O

1. (XS) OEt

9. EtO

O

O

2. Br(CH2)4Br

O

Ph Se

O

O

-PhSeOH

4. MeOH H

O O

3. LiMe2Cu

O

2. KO-t-Bu

O

O

H

1. Br2, H

11.

O O

O

OMe

O

O 4. H3O

O

O 3. H3O, '

2. (CN)2CH: SN2

Br

O

O

O 1. Br2, H

12.

-CO2

CO2Et Br

3. H2O2

O

O

3. H3O, '

-Br

OEt

1. LDA 2. PhSeBr

10.

O

CH(CN)2

OH

-CO2

O

O EtO2C

1. OEt

13.

EtO2C

H

2. n-Pr

EtO2C

EtO2C

O 14.

O Cl

NAS

EtO2C

-2 CO2

O O

OH 2. PCl3, Br2

haloform rx

3. H3O, '

EtO2C

O

1. a. Br2, OH (-CHBr3)

H

Cl Br

H-V-Z rx

O 3. MeOH

b. H

H 15. H

18.1 Reactions

OEt

H

+EtOH O -EtO

H

-H2O OH

OMe Br

Solutions • 407

O 16.

O

Cl

2. Cl2, H -Br

-H

Br

O

O

1. LDA

Br

OEt 1. HCl 17.

2.

O

3. H3O, '

O

Cl

conj. add'n

EtO2C

-Cl

-CO2 O

O O

O 18.

O 3. KMnO4

O OH

O

O

1. Cl2, H

1. Cl2, H

3. H3O

2. OH

2. E2

4. [O] O

19.

Cl Cl

C(CO2Me)2

-Cl

-Cl

LiH :CH(CO2Me)2

CH2(CO2Me)2

LiH

CO2Me

1. KOH

CO2Me

EtOH 2. '

Cl O NMe2

O 20.

O

O

base

O

NMe2

O Ph

1. SOCl2 2. Me2NH

O

3. mCPBA

O

CO2

O

O

taut

conj. add'n O

O

O

Ph

O

18.2 Syntheses O 1.

O

1. PCl3, Cl2

O OH

OH 2. H2O H-V-Z rx

Cl

3. KCN SN2

O OH

CN

4. H3O

OH CO2H

18.2 Syntheses


2.

CO2Me

1. OMe, MeOH

CO2Me

2. Br

(CO2Me)2 OH

CO2H

3. H3O -CO2

OH

-H2O

OH

O

O

O 3.

MeO2C

CO2Me 1. OMe CO2Me 2. Br

CO2Me

-CO2 4. PCl5

5. LiAlH(O-t-Bu)3

H H

6. H O

O

O 5. H2N

CO2Me

4. CO2Me

Cl Cl

O

1. OMe 2. 2-chloropentane

CO2Me

3. H3O

CO2Me

CO2Me

O

3. OMe 4. allyl chloride

NH2 NH

(urea) -2 MeOH

CO2Me

O

N H

O

O CO2Me

5.

CO2Me

1. OMe 2. EtI

Et

CO2Me

3. OMe

Et

CO2Me

5. H3O, '

H

CO2Me

4. PhCH2Br Bn

CO2Me

-CO2

1. BH3.THF

6.

2. H2O2, OH OH

O

1. OMe

CO2Me

CO2Me O

2.

(CO2Me)2

NAS

O

CH(CO2Me)2

-CO2

3. H3O (-CO2) 4. LiAlH4

OH

Br

6. HBr OH

5. H

Br

SN2

O 3. H2NNH2

1. NaOEt, HOEt 8. O

O 6. H3O, '

OMe

O

Et

5. (MeO2C)2CHNa

Cl

OH

8. H (-H2O)

7.

O

4. SOCl2

7. NaBH4

O

CO2Me

Ph

3. CrO3, H OH

OH

O

18.2 Syntheses

2. MeI

Et

O

O

OH, ROH W-K

4. NH3 Et

O

O

H2N

O

Solutions • 409

O

O

O

O

1. OEt

O

9. Et

O

O

O

2.

O

EtO

Et

O

OEt

O

O

1. KOEt / EtOH O

2. Cl

O

Et

O

O 2.

O

OH

O

3. OEt 4. MeI

4. NaBH4

O

-CO2

EtO2C O

O

O

1. OEt 2. EtI

12. O

O

3. H3O, ' O

Et

O

O

-HOEt transesterification

O 1. OEt

Et

O

3. H OH

O 11.

O

O

10. Et

OH

-CO2

OEt

O

O

3. H3O, '

Et

5. H 6. PCl3

Cl

O

O

Me

5. H3O, '

Et O

-CO2

O

Cl

6. I2, OH

O

haloform rx + HCI3

O

O

O 1. Cl2, H

O

3. H3O

13.

4. Ac2O

2. KO-t-Bu (E2) [or 3. HCl, 4. NaOAc (SN2)]

OH

OH

O 1. Jones reagent

O Br

14.

OAc

3. KCN

CO2H

4. H3O

2. Br2, H

[or 3. ethylene glycol, H, 4. Mg, 5. CO2, 6. H ]

O 15.

O

taut

O

O

O (H

O -H

+H

18.2 Syntheses


O

O

1. OMe

O

16. Me

O

O

2.

O

O

Me

O

O

O

1. LDA

17.

O

3. H3O ' -CO2

O

O

O

O

3. H2O2

O

-PhSeOH

2. PhSeBr Se

Ph

OH

6. CH2O, H O

OH

1.

Cl

O

O

Cl

OH

Cl

N -t-Bu H

1. KOH

Br

3. Br2, PCl3 4. H2O

CO2H

O OH

CO2H

O

19. 2.

O

-HCl

CO2H

O

Cl

NH2

3.

AlCl3

OH

4. H3O

O

2. Cl2, H

Cl

18.

O

5. H2 / Pt

O

O

Cl

Se

Ph

H-V-Z rx N H

O

O

7. LiAlH4 8. H

5. SOCl2 6. H2N-i-Pr

N H

Br

9. OH ((SN2)

20.

O

1. NaOMe / MeOH CO2Me 2. EtI

MeO

S

O OMe

5. H2N

CO2Me 3. NaOMe / MeOH 4. 2-iodopentane

21.

CN 1. NaNH2

NH2

thiourea

HN

NH

O

O

6. NaO-t-Bu (1 equiv)

HN

N

O

O

-2 MeOH

NC

H MeN(CH2CH2Cl)2

Ph

SN2 (-Cl , ) -H

NC intra-S N2

NMe

Ph Cl

NMe Ph

NMe

O 3. EtOH, H

EtO

NC Ph

-Cl

O

18.2 Syntheses

S Na

S

2.

O

NMe Ph

OH, ROH

Solutions • 411

18.3 Mechanisms O H

O

(H

+H

1. OH

O H

OH

OH

O

OH

OH

O 2.

2. O

-H

O

Et

O

O

3. Ph

OH

Ph

H O

Ph

OH

H

O OEt

O

-OEt O

I

O

I2

O

-H

O

I

O

I2

O

OEt O

O

I O

H

O (H

O

O

O

OH

H

OH

O H

O

1. OEt Et

-H

HO

H +H

O

Ph

OH

Ph

I3C:

H) O

Ph

Ph

-H

PhCO2

I

O

I

Ph

+

OH

I

O I3CH

PhCO2

+

I3C

O

I

Ph OH

(H O 4.

H

OH

+H

(H

-H

OH

-H +H

O

+H -H

O

-H +H

H achiral

H

H

O

O +MeO

5.

O

O MeO

H) OMe + -MeO

O O

MeO

O

H O

H

O taut

6. O

+H

O O H

O

~H

H O

OH

-H3O OH2 ~H

O

H) H)

18.3 Mechanisms


O

O O H

O taut

7. HO C 2

HO2C

HO

N

H) O

NH HO2C

H

OH

OH H

8.

H

taut

OH -HPO42-

taut

9.

18.3 Mechanisms

Me3N:

O

SN2 Si Cl

-Cl

O H

-H

O (H

H O

OH H

O O H PO3H

O O PO3H

OAA + biotin

-H

CO2H

S

H

CO2H

O

H)

O

taut

H O

Si -H

O TMS

CHAPTER 19 CARBONYL CONDENSATION REACTIONS 19.1 Reactions 1. 1. PhCHO + PhCOCH3

2. :CH(CO2R)2

OH Ph

-H2O aldol

Ph HO2C

O

H

H

taut

H

aldol

O

O

OEt

OH

O

t-Bu-CHO

O

O

O

O

O

O OMe

OMe

t-Bu

OH

O OMe

4.

or

t-Bu

Knoevenagel O

3. H3O

-CO2 Ph

H

H

O 3.

Ph

O

OH O

Ph

Michael

Ph

2.

CO2R O

RO2C

O

- OMe

-H O

CO2Me 5.

O

O

CO2Me

1. OMe

OMe

OMe

O

CO2Me

- OMe Dieckmann

O

O

O

O

2. H3O -CO2

O

O

OH, ROH

6.

conj. add'n

retro-aldol

OH O

O

O

HO

OH

7.

-H2O

-H O

Et

O

Et

O Et

O Et

19.1 Reactions


H

8.

O

1. BrCHCO2Me

O

O

Br

taut

CO2Me O H

-CO2

OH

H

O

-Br

CO2Me

H) O

2. H3O, '

O O

O 9. MeO

O

1. base

CHO

O

MeO

MeO

CO2 (H

O

O O

O

NAS

O

CO2H 2. acid

MeO

CO2

MeO

-CH3CH2CO2H

OCH3

OCH3

OCH3

base

10.

-H2O

Michael

O

aldol (Robinson annulation)

O

O O O

OH

OH

11. O2N CH3

PhCHO

O2N CH2

NO2

Ph

CO2Me

MeO

retro-Claisen

MeO

O

O

O 2. NH2NHPh conj. add'n

OH

O

O O

1.

O

O

O

-H2O

aldol

OMe Cl 2. O O

O

O

-Cl

CO2Me

O

15.

OH

O

Cl

1. :CH(CO2Me)2 Michael

mesityl oxide

19.1 Reactions

NHNHPh

O

O 3. H3O, '

H

O

aldol

-CO2

O O

OMe

O

1. NaOEt

14.

CO2Me

O

13.

NO2

Ph

O

OMe, MeOH

12.

or

H

O

O

O O

-MeO

OMe CO2Me

-H2O

2. H3O O CO2Me

-CO2

O dimedone

Solutions • 415

HO

CHO

CHO 16.

O

OH

O

-2 H2O

O

O

CHO HO

OH

OH

17. OH

O

OH

H

O

HO

H

retro-aldol

H

OH

(H

OH

HO

+

taut

O

HO

O

H

H

O O 1.

N

18.

2. H3O

3. NaOH

N

O

O

19.

O

O

O H

mixed

+

H

H

aldol

O

O again

H CH2OH

HOCH2

O again

H CH2OH

O

Ph

~H taut

H

C(CH2OH)4

O

OH

HO H OH

H

mixed Cannizzaro

OH

O

OH

20.

+

H OH

O HCO2

HO

O

H

-H

H mixed aldol

-H2O HO

Ph

Ph

OH CHO OH

OH

O

CHO O

+

O H

O OEt

-HOEt transesterification

O

OH

O

again

+ OHC O

O

OH

O

1. SCoA

O

O OHC

-2 H2O aldol

O

O CO2H

-H2O

KOH

+

23.

CO2Et

CO2Et

2. H

O

CHO 22.

CO2Et

CO2Et

1. :CH(CO2Et)2

21.

SCoA CO2H

2. H3O

OH CO2H

19.1 Reactions


OMe MeO

CHO

24.

+

OMe

CN

1. OEt

2. H3O

MeO HO

OMe

OMe

CN OMe MeO

O

(H

OMe

O

-CO2

MeO

-H2O HO

OMe

O

H

O

OH

O

1. H3O

25.

2. KO-t-Bu

conj. add'n

taut H

retro-aldol O (H

O O

H

retro-aldol

O

O

H

O

O

CO2

CO2 26.

taut

O2C

CO2

O H)

CO2

H

H

aldol O

O

O

CO2

CO2

CO2 HO O

C

OEt 27.

O

taut

C

O

O

O

O -H2O

O

H)

O taut OH

OH

19.1 Reactions

H OH

OH

H OH OH

OH

aldol O

O O

O2C

O

O

Michael EtO (H

O

HO

SCoA

O

O

O

CO2 hydrolysis

SCoA

28.

OMe

H

O

OH

~H OH aldolase

OH

OH

OH

Solutions • 417

O CO2

29. O C 2

+

CO2H

OH

O

CO2

O2C

SCoA

H3O

SCoA

HO

CO2H CO2H

O

O

OH

30.

O

O [O]

H3O

SCoA

O SCoA

OH SCoA

SCoA

O

HO

O

O

O

C

-CO2

O A

O 32. a.

O H

D-A

+

H EtO

OH

B

1. H2 / Ni (1 equiv)

EtO O

CoA

OH

H3O

SCoA CoAS

S

retro-Claisen

[H]

O Claisen

H)

taut

+

O 31.

SCoA

SCoA

O

HO H

O

OH

H

2. a. LiAlH4 b. H

EtO a vinyl ether

HO

HO H OH

H

H

aldol -H2O

O

H O

R O

taut C H CO2Et

2. MVK, base

1. H3O

HO H H

Michael

OH

O

R

R 3. +H C C OEt OH

2. a. EtO C C MgX b. H

R

OH

O

X , t-BuO

1.

b.

H

O

R

+H2O C C OEt OH

C C OEt

-H2O

-H

R C C OEt

19.1 Reactions


O

O

1. a. OsO4

c.

O

2. KIO4

OH OH

O

b. NaHSO3 OTs

OTs

O O

OTs

O

O

3. OMe -H

-OTs H

OTs O

O

O

O

1. RO2CCO2R

d.

O

I

2. I2, OH O

O

OR

I

O

CO2R

CO2R 3.

O O

I

O

I

O

C OH O

O OH

C

I

OH I

O

'

O

-CO2, -I

O

O

O ~H

taut

O

O

O

-CO2

O

O

I

I C O O

OAc

O

taut 4. KOAc SN2

OAc

OAc NC

OH

O O

O

1. HCN

e.

CN

2. -H2O

OAc

O

3. KMnO4,

OH

O O

OAc

4. H3O

cortisone

NC

OH -HCN

O

OAc O (H OH

O cortisone acetate

O 33.

CO2Me

1. - OMe Dieckmann

CO2Me

19.1 Reactions

O O OMe

2. H3O, ' -CO2

Solutions • 419

O OMe

34. O2N CH2 (H

O2N CH2

H OMe

OH

H

O O

OH O

H) NO2

O2N

35.

O N

Na O

NO2

O Et3N

-H

O N

O

O

+H

O

-H

O

Michael

O

taut

19.2 Syntheses O 1.

O

Ph

+

H

OH

O 1. H3O

Ph

2. H2 / Pt Ph

-H2O

Ph

Ph

Ph

O O 2.

N

1. H

+

-H2O

N H

N

2. O

O 3. H3O

O

O H

3. O

OH, ROH O

OH O

4.

O H

H

H3O

O

H

H

O

OH

O

-H2O

O O

O 5.

1. OH (aldol) -H2O

HO 2. H2 / Pd

3. H2SO4 (E1) 4. H2 / Pd

19.2 Syntheses


O 6.

1. LDA 2. MeI

O 4.

3. Cl2, H

HO

H (protect)

Cl

6. MeLi

PPh3 7. acetone Wittig

O

1. NaCH2CO2R

O 2. OR, HOR

OR

Claisen

CO2R

O

O

O

CO2R

Dieckmann

OR

O

O

O

O

O 1. Cl2, H

O

3. H3O

8. 2. KO-t-Bu (E2)

5. NaH (2 equiv)

4. CrO3, H

O

O

O

O 7. H

Ph

6. PhCH2Br

Ph

O

.

9.

4. H

N H

N

,H

O 5.

Cl

H

-H2O

3. PCC

(1 equiv)

O

O

1. BH3 THF 2. H2O2, OH

O

5. Ph3P:

O

O 7.

O

Cl

8. H3O

- via a Wittig, not a mixed aldol!

CO2R

O

OH

H

6. H3O

OH O 10.

1. CH2O,

H

O

OH

mixed aldol

H

O

OH 2. NaBH4

OH

3. H

O

O OR, ROH

+

11. O

O

1. O3

Michael

O

O

-H2O, aldol

O

3. KOH, EtOH

-H2O

aldol

2. Zn, H O

O

O

O

12.

19.2 Syntheses

OR, ROH

OH

Solutions • 421

1. Br2, H 2. ethylene glycol, H

O 13.

5.

O

O

O

PPh3

3. Ph3P: (SN2) 4. n-BuLi

O

6. H3O

- the reaction of cyclohexanone with acetone via an aldol would yield four possible products! O O

O

O 1. OR, HOR

14.

3.

O

OR

OR, -H2O

Michael

2. MeI O

H

2. Zn, H

H O

3. H

O

aldol

H

O

O

O

-H2O

H

O

O

O

OH, HOR

3

O

OH

O

15.

aldol

O

O

O

1. O3

16.

,

aldol

aldol, -H2O

-H2O

conj. add'n O

19.3 Mechanisms O

OH

OH +H

1.

O taut

-H

OH taut

retro-aldol

HO

H) O

O

O -H -H2O

O

O

O

OH

(H ~H

+H

OH2

OH

1,2-R: shift

O

aldol

O O

2. :CH2 N N:

N N:

+

N2

19.3 Mechanisms


O

O

O

H) OEt

retro-

3.

O O

O CO2Et

OEt

Claisen

OEt

O

OEt

H) OMe

4.

OMe

OMe

1. conj. add'n NH

O

O

Me

:NH2Me O

1. again

OMe O

Me

CHBr3

OH

O

Br3C:

O

N

-Br

Cl

H

OH

-HBr HO Br

CO2Et

CO2Et

1. OEt,

O

CO2Et -H2O

aldol O

O

Michael

Br

-Br

O

O

O

Br

HO

OH Cl

O

6.

O

Cl

CBr2 Br

Cl

CO2Et

2. OMe -H

OMe

Dieckmann

O Cl

O

OMe

- OMe N Me

N

CO2Me

O

O

MeO

5.

Me

O HO

O

(H O

H) OH

2. H2O, OH

O

3. ' O

O

O H

7.

O

OH

-CO2

O O

O

O H

1.

-H2O

2. H

aldol O

(H

O

-H

O

O

8.

OEt OEt

19.3 Mechanisms

retroClaisen

O

O OEt

+

EtO

O EtO O Ph

O

Claisen - OEt

EtO

Ph

Solutions • 423

O 9.

OH

O

OH, Michael

O

CHO

CHO

HO

H

OH

O

aldol

HO

O

O

(H

O

O

O

H O

~H

- OH -HCO2 RO (H

HO

HO

HO

HO

H O 10.

OH

OH

H

OH O

OH OH

(H

taut

O

H

OH

O

OH

(H

OH

taut

OH

OH

OH

H

O

OH

OH OH A

(H

O H

D-D-fructose

OH

retro-

H

O (H

OH A

R2N

R2N

CO2R (H Br

Br

OH

Ph

12.

O

base

Ph

O

-H

Br

R2N Br

O R2N

OR (H N

Ph

O

OR O

Ph

Ph - OR

Ph RO2C

O

1. OMe

O

(H

N

-H

O

OR

-Br

NH

CO2R

13.

OH

CO2R

NH

NH

O

~H taut

OH

CO2R

RO2C

OH O

H

HO

H

(H

+

aldol

O

RO2C

H

O

HO

11.

-H

HO

HO

R2N

OH

OH

OH

-H

Ph

CO2R

Ph

CO2

:B

O

O

O O

O

O

H) OMe

O

19.3 Mechanisms


OH

OH 2. H

13. (cont.)

-H

O

OR

O

retro-

O CO2R

H

O

OR

O ~H

CO2R Dieckmann

OR OR

OR

EtO

-H

O EtO

H)

O

EtO

OEt

EtO O

O EtO

O

O O

O

O

EtO (H O

O

O

O 1. a. O3

2. OH

16.

O

b. Zn, H

O

O

O

O

1.

H) NH, H

O

-H2O

OH H

-H N

3. H3O

A Et

A - a dienamine 2. EtI (SN2)

O

3. H3O

O Et

I

N

(H

2. EtI (SN2) N

aldol

OH

+H N

O

HO (H

-H2O

17.

O

O

O

O 15.

O

Claisen

OR

O

O

O

O

OR 14.

(H

O

taut

OH

N

Et

A I Et

O

O

O

H B

:B 18.

taut

Michael MeO

Michael OMe

O

19.3 Mechanisms

O

O

CO2Me

CO2Me

Solutions • 425

O

O

O

O

1. NaH

19.

2. H2O

MeO

MeO O

O

OH R'

R

R' =

O

HO HO

R'

taut, +H2O

HO

O

O

R'

O R

R'

21. a. Mannich

b.

O H

I

R' HO O

R

NH2 5. H R

[H]

N H H O

-H2PO4

HO

CO2H

O CO2H OH

HO

CO2H

O

CO2H

O H H3N

OH

H OH OH

3. -H2O, taut

HO

CO2H +H

aldol

-H2O NH2 H

HO2C

taut

O

O

H

HO

H, -H2O O

OH CO2H 4. H

aldol

H

CO2H

H3N

CO2Et

N H

H) O

HO

HO

OEt , H

OH (H

O H) -H2O

N R

-H2O

2. aldol

O

N H H

HO2C

OH

HO2C

R

SN2

O

OH

O

1. H3O

R O (H OH R' H

OH

H

(H

aldol- like

3. -NMe3

N H O

H

OH

CN

R

SN2

4. LiAlH4

R'

N

2. -I

CN

N H

H3N

-H2O

:N C:

NMe2 Me R

c.

23. H)

HO H) O

R'

N H

22. PEP

like

OH O R

HO O

OH

R O

(H O

R'

taut

HO O

O

R'

retro-aldol-

O (H

R'

O

MeO

OH

R OH

H O

R'

O

MeO

O

OH

20.

(H

HO

O

CO2H

CO2H

OH

CO2H

O

CO2H

-H (H H3N

N H

H3N

N H

19.3 Mechanisms


O

O

O

O

Cl

N

-Cl

N

N Cl O

HS

O

O

O

OMe

+MeOH

HS

- OMe taut

N

N

-H

O

H

O

O

Cl

O

O

H)

N

HS

O

-HOAc

~H O

b.

H)

H)

O

O

N

O O

-HOAc

O

NH

Cl

O AcO

O O(H

24. a.

O

O

OMe

HS N H

N MeO (H

O

O

O

O

O

O

-CO2

25.

O

SR

RS

SR

SR

R'

1.

S

R N

S

R N R N

S

OH

H

R

R N

OH HO

O

H

H

H) O

(H

S

S

- R N

2.

+ O

R'

R' O H

R'

R' 26.

O

- SR

O

-H

R

O (H OH

-H O R

H

H

OH O

O 27. a.

Claisen SCoA

OH aldol-

SCoA like O (H

-HSCoA O H

O

O

OH

SCoA SCoA

O

partial SCoA

hydrolysis CO2

O

SCoA

PO3H

b.

O O O

19.3 Mechanisms

P2O6H 2-

-CO2 -HPO4

O

2-

(H H

P2O6H 2-

~H

O

P2O6H 2-

CHAPTER 20 AMINES 20.1 Reactions O 1.

O

NH

O CO2Et H CO2Et

1. base N

2. ClCH(CO2Et)2 O

SN2

(CO2Et)2

3. base N

4.

Cl

O

O

O phthalic acid

CO2H

+

CO2H

5.

NH3

N

5. H3O

CO2H

-CO2 O

Cl NH2 2.

OH

1. (XS) CH3I

3. OsO4

2. Ag2O, H2O, '

4. NaHSO3

OH

O O

5. (XS) COCl2

O O

NH2 6. NH3

O

-2 HCl

NH2

Me 3.

1. (XS) MeI

OH

2. Ag2O, H2O

O

2. Ag2O, H2O

N

HO

'

O

-H2O

HO

H

Ph2N Me Et

HO OH

1. (XS) CH3I N

H

N

HO

O

Me

Me H)

H

Me 2. ' Cope

Et

Me

N HO

NPh2

O 1. H2O2

H2C CH2

+

3. ' -H2O

N

HO

5.

3. ' -H2O

1. (XS) MeI N H

6.

OH

2. Ag2O, H2O

4.

Me N Me

Me N

Cl O

-2 HCl O

N H

O

(cis-elimination)

H +

Et

Ph2NO

Me

20.1 Reactions


O CO2H

7.

O

1. SOCl2

3. ', -N2

N N N

2. NaN3

N N N

OH

O

-CO2

NH2

N H

H2O OH

OH OH

OH OH

1. (XS) MeI

N C O

Curtius

OH

taut

8. 2. Ag2O, H2O, ' NHMe

HO

HO

O D

D

H

H

9. H

H

anti-periplanar

Hofmann elimination

NMe2

NMe3 D Cope elimination

D

syn-

(H N O O 10.

Me2NH

+ H

O

H

+H

H

-H2O

H

O

HONO

Cl

1. Li 2. CO2

Ph

3. H

13.

Me

N H

4. SOCl2 CO2H

O

20.1 Reactions

5. NH3

O

O taut

HN

N H

Ph

O

NH2

6. Br2, OH

Me

NH2

a carbinolamine

N H

Ph

NH2

H2O

O

OH

Br2, OH H2O

N

-N2

OH NH2

O

OH H2O

N O

N H

12. Ph

OH

N N

N

N

OH

taut

NH2 11.

-H

N

N

O NH3

+

Me

H

Solutions • 429

1. NaNH2 14. Ph2CHOH

2.

Ph2CH

O

3. PBr3

O

O

Ph2CH

Br 4. Me2NH

O

Ph2CH

NMe2

O

O N 1. HBr

15.

2. Br

ROOR

O

O

3. H2NNH2

NH NH

+

NH2

O H

Me2N

O

(H O

OH

3. H

1. MeI

16.

2. Ag2O, H2O, '

MeO

pinacol-like rearrangement

MeO

Ph

Ph OH

17. OH

MeO

O

NH2

1. SOCl2 2. NH3

3. Br2, OH

O

OH

H2O

Ph

Ph 4.

N

I

2

NH2

OH

18.

CO2H

OH

CO2

1. a. Br2, PBr3 b. H2O (H-V-Z)

N 3. HCN, CN

4. H2 / Pt

Br 19.

Br NH2

1. Br2

Br

CN 20.

CN

1. Cl2 FeCl3 Cl

2. NaNO2 HCl

CN

2. NaNH2 (NAS via benzyne)

Br N2

3. KI

I

Br

-N2

Br

CN

3. KNO2, H

CN

CO2H

CO2H

CO2H 3. Fe, HCl

21. NO2

CO2H

5. H3O

4. CuCN NH2

CO2H 1. KMnO4, H 2. fuming nitric acid

H2N

conj. add'n

2. KO-t-Bu (E2)

NH2

CO2H

CO2H C

4. NaNO2, HCl

5. HBF4 N2

F

20.1 Reactions


NHAc 22.

Cl

1. Br2, Fe 2. Cl2, Fe 3. H2O, OH

N2

4. ICl, Fe 5. HONO

Br

Br

1. NaH O

N Me

-H2

N

O

N Me

2. H3O -OMe

O

N Br

OH

-HBr

N

H)N Me

NHMe

4. HBr

H

H

reductive amination

NaBH3CN HO

HO

-CO2 O

N

HO

N C NH

N Me

3. NaBH4

5. '

1.

I

O

O

O

Br

N N Me

HO

24.

5. H3PO2

I

CO2Me 23.

Cl

Cl NH2

OH

O

N HO

OH

O

2. HCl N H

Cl

HO

20.2 Syntheses 1. Br2, hv 2. Mg

1.

3. CO2 4. H

2.

3.

1. Cl2, '

1. Br2, hv 2. KOH (E2)

20.2 Syntheses

5. SOCl2 CO2H

Cl

6. NH3

NH2

7. Br2, OH

O

2. KCN

CN 3. LiAlH 4

SN2

4. H

4. potassium phthalimide

3. HBr R2O2

Br

5. H2O, OH

NH2

H2O

NH2

6. (XS) MeI NH2 7. Ag2O, H2O, '

Solutions • 431

1. KMnO4

4.

H

OMe MeO

2. SOCl2

OH

3. NH3

O

1. (XS) MeI OMe 2. Ag O, H O, ' 2 2

4. Br2, OH

NH2

OMe MeO

OMe OMe 5. H2NPh

5. 3. O3 NH2 4. Zn, H

NH2

H2O

O

MeO

OMe OMe

MeO

OMe

NaBH3CN H

O

H

NPh

NHPh

O N K 1. NBS, ROOR

6. O2N

2.

Br

O

NH2

3. H2O, OH

O2N

O2N

O CO2H

1. KMnO4, H

7. O2N

1. (XS) CH3I NH2

9. HO3S

NO2

H2O

3. O3

2. Ag2O, H2O

O

HO3S

N2

CO2Me

OH

2. CrO3, H O

3. PhNEt2

HO3S

3. '

N

O

N N

NEt2

4. MeI 5. Ag2O, H2O, '

-CO2

Ph

O

OH

O

N 1. H3O

10.

O2N

H

2. NaNO2, HCl

N

5. HO

4. Zn, H

1. SnCl2, H

NH2

NH2 4. Br2, OH

3. NH3 O N 2

O2N

8.

2. PCl3

O

6. repeat 4. and 5.

O

O

(double Hofmann)

O

11.

1. HONO2, H2SO4

Cl

Cl

3. SnCl2, HCl

Cl

Cl

5. H3PO2

Cl

Cl

4. NaNO2, HCl

2. Cl2, Fe NO2

N N

20.2 Syntheses


NH2 O2N

13.

NHAc

1. Ac2O

12.

2. SnCl2, HCl

NHEt2 O

4. H2O

H2N

1. CrO3, H

OH

Et2N

NHAc

3. KNO2, H HO

Cl

Et2N

2. SOCl2

NEt2 O

O

-HCl

NH

Fe, H NO2

NH2

1. HONO2, H2SO4

14.

NH2

2. Fe, HCl NO2

OMe

OH 3. NaNO2, H

5. KOH

4. H2O

6. MeI

N2 1. SnCl2, HCl

3. PhNH2

2. KNO2, HCl

-H

15.

N N

H2NMe (-H2O)

16. O

NH2

EAS

[H] N

NaBH3CN, H

NHMe

Me

(alternatively, 1. H2NMe, 2. H2 / Pd) H

H

N

N

N

O 1. mCPBA

17.

2. (XS) MeBr

S

S S

O O

O

1. HONO2, H2SO4 2. SnCl2, HCl 3. Ac2O

5. NO2

S

O

OH

O

OH, H2O

OH

NH2 O2N

OH 7.

N2

OH O2N

20.2 Syntheses

S

NHAc 4. HONO , H SO 2 2 4

18.

N

SN2 S

O

OH

N

Br

O

6. NaNO2, HCl

Solutions • 433

NO2

NO2

1. HONO2, 19. H2SO4

EAS

OH

D

5. H3PO2

D

D

F-C alkylation

OH 3. KNO2, HCl

2. H3O -HOAc

NHAc

D

OH

OH 1. propylene, H

20.

3. Fe, HCl 4. KNO2, HCl

2. DCl

4. H3PO2 NH3

NHAc

20.3 Mechanisms O 1.

N

O

OH

H

O

Br Br

-HBr

H

(H

O N C H OPh

-Br

~H N C

taut

O NH

O N C O

O H)

2.

Br N

N

O 1. Cl2

Ph

HO Ph

OH

CO2 -H

NCl OH, H2O

O

N O

CO2H NH2

CO2

-CO2 2. H

N CO2 H

Cl

-Cl

O CO2

OH taut

N C O

O CO2H 3. NH2

1. HONO

CO2H N N

2. pH 8

O

-H

N N

-CO2 -N2

3. D-A

20.3 Mechanisms


O

HO

HO

CN

1. HCN

2. H2

4.

3. NaNO2

Pt

CN H)

O

HCl

O

HO

HO

OH2

OH 5.

N

1. NH2OH

N

2. H

N O

OH taut

NH

OH2

1. PhMgX

O

N

N

MeO

XMg O

XMg O NMe

NMe

NMe

MeO

MeO

MeO

MeO

MeO

MeO Ph

Ph HO

NMe

+H2O

-H

MeO

MeO

6.

OH2 -H2O

-H2O

XMg

N2

-N2

-H

O

NH2

1.

NMe

XMg O

XMg O NMe

2. H MeO

MeO

MeO

MeO

MeO

Ph

Ph Hofmann :

HO

1. (XS) MeI NMe

HO

NMe2

2. Ag2O, H2O, ' MeO

Cl 7.

SN2, -Cl

N N N

20.3 Mechanisms

MeO

N N N

N N N

' -N2

N

Solutions • 435

H) N3 O

HO

N N N

N N N

8.

O

OH

H O

N N N

-N2

NH

~H

N

'

+H

9. N

H

Ph

N

Ph

(H

H

~H Ph

N

Ph

NH2

N

H

H

N NH2 H H

N

H

H

~H

H

H) Ph

-NH4

Ph N H

+H

10.

~H NH2 -H2O

NH2 H

N NH3 H -H

EAS NH

NH

H2C

H

H

NH

H

OH

O

O 11.

Ph

CH3

H taut

CH2O + NH3

(H

O Ph

Ph

O

CH2O, H

Ph

NH2

CH2 N H H)O

Ph

H2C NH2 H)O

O

O

-H Mannich

O Ph , -H

N 3

(again)

O

Ph

N CH2

CHO N (H H

12. CHO H2NMe O

N

Ph

~H

N

-2 CO2

N H

taut

Ph Ph

CO2 N

-H

CO2

O

OH

CO2

O2C

N CO2

-H

O(H

O2C O

-H

H)O

- OH

OH N CO2

O

O

Ph

(again)

acetone dicarboxylate OH

OH ~H

O CH2O, H

CO2

- OH taut

N OH

20.3 Mechanisms


OH2 +H

13. a.

N CMe

N

N CMe

O b.

OH2

-H

Me

taut

NHAc

O

+H

-HOAc

O

N C

O H O t-Bu

+H2O, -H

OMe

N C

taut

N H

OMe

OMe OH2

O 14. Ph S Cl O

O H Ph S N Ph O

Ph NH2 -HCl

PhSO3

O 15. a. R

(H

1. OH

+

O 2. CH3I Ph S N Ph SN2 O CH3 O 3. H2O, OH CH3 N Ph S N H Ph O

:B O

2. -HCl

CCl2 Cl

R

Cl

N3

Cl

N N N

b.

O

1.

Cl

:CCl3 -Cl

16. N H

O

NH

H

H N

H Ph H

C

O NH

N

H) N

O H

OH

R N

O

R

NAS

N3

C N H

-HCl

O

R N

-H +H

N H C

O

H N

O C H H C N

Ph Ph

20.3 Mechanisms

OMe

H N

Cl

O C

O

Me

2. -HCl

O

H C C

O

3. -HCl

H N

O

Cl

H N

Cl H)

NH H) R N

O

R

3. -Cl

O NH

Workbook for Organic Chemistry : Supplemental Problems and Solutions

Organic Chemistry Problems and Solutions

Solutions Manual for Organic Chemistry

Solutions Manual for Organic Chemistry

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