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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
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PROBLEMS
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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: ________
1.1 Hybridization, formulas, physical properties
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
1.1 Hybridization, formulas, physical properties
4 • Chapter 1 The Basics
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
6 • Chapter 1 The Basics
b. Identify the weak and strong acids and bases. c. Is Keq about equal to, less than, or greater than 1? 7. Which compound has the lowest pKa? a. EtOH
b. HOAc
c. H2O
d. PhOH
e. H2
f. NH3
8. Which species has the ability to quantitatively (completely) remove the proton Ha (pKa 22) from R 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
8 • Chapter 1 The Basics
5. Draw a resonance structure that is more stable than the one given. Use curved arrows to derive. H N a.
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
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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
14 • Chapter 2 Alkanes
4. Draw the conformer of isopentane that corresponds to the highest minimum in a plot of the potential energy vs. rotation about the C2-C3 bond (use a Newman projection).
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.
2.3 Conformational analysis, acyclic
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
18 • Chapter 3 Cycloalkanes
3.3 Conformational analysis, cyclic 1. Draw the most stable conformer of Me
Me =
=
OH
OH
i-Pr
i-Pr
menthol
neomenthol
2. In each of the following predict whether Keq is about equal to, greater than, or less than one: a. trans-1,3-diphenylcyclohexane
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.
3.3 Conformational analysis, cyclic
20 • Chapter 3 Cycloalkanes
7. Glucose, like cyclohexane, exists in a chair conformation. Two configurations of glucose are possible; they are drawn below: 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?
3.3 Conformational analysis, cyclic
CHAPTER 4 REACTION BASICS 1. Which type of reaction – addition, elimination, rearrangement, substitution, reduction [H], or oxidation [O] – best describes each of the following? OLi a. MeLi
+
O Me 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
24 • Chapter 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
26 • Chapter 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
30 • Chapter 5 Alkenes and Carbocations
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
32 • Chapter 5 Alkenes and Carbocations
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
34 • Chapter 5 Alkenes and Carbocations
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
36 • Chapter 5 Alkenes and Carbocations
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
38 • Chapter 5 Alkenes and Carbocations
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
40 • Chapter 5 Alkenes and Carbocations
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
42 • Chapter 5 Alkenes and Carbocations
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
44 • Chapter 5 Alkenes and Carbocations
CH2N2, hQ 21.
H
22.
H
23.
isocomene (from goldenrod)
24. Hydride shifts and alkyl migrations occur in many enzyme-catalyzed reactions in all living species – including you as you are working these problems! Below is one such biochemical reaction (see 14.3, 6 for perhaps the very best example). Account for the formation of all intermediates leading to the product. (Hint: positive sulfur, like positive oxygen, is a good leaving group, i.e., it easily leaves a carbon to which it is attached, taking with it both bonding electrons.) R
R' S 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
46 • Chapter 5 Alkenes and Carbocations
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
48 • Chapter 5 Alkenes and Carbocations
36. Carbonyl groups greatly affect the acidity of nearby (D-) protons. For example, the pKa of cyclohexane is about 60, but the pKa of Ha in cyclohexanone is about 20. This dramatic increase in acidity is largely a consequence of resonance stabilization of the conjugate base of the latter (for an example of the additive effect on pKas of 1,3-dicarbonyls, see problem 1.3, 7), and allows an easy exchange of 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
52 • Chapter 6 Alkynes
10. styrene
(E)-1-phenyl-1-butene
11. diphenylacetylene
cis-1,2-diphenylcyclopropane
PhCHO
Et
Cl
12. 3-hexyne Cl Et
n-Bu 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
54 • Chapter 6 Alkynes
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
58 • Chapter 7 Stereochemistry
7. How many a. pairs of enantiomers exist for bromochlorocyclopentane?
b. geometric diastereomers exist for 1,3-dichloro-2,4-dimethylcyclobutane?
c. pairs of enantiomers are possible for chlorofluorocyclobutane?
OH
OH ?
d. meso stereoisomers and how many enantiomeric pairs exist for Cl
e. meso stereoisomers exist for 2,3,4,5-tetrachlorohexane?
8. a. D-Xylose is a common sugar found in maple trees. Because it is much less likely to cause tooth decay than sucrose, D-xylose is often used in the manufacture of candy and gum. D-Xylose is the C4-epimer of the enantiomer of A. Draw its structure. CHO HO
H
H
H
OH
MeHN
Me
H
OH
H
OH
CH2OH
Ph (-)-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
60 • Chapter 7 Stereochemistry
13. Another example of different enantiomers having remarkably different biochemical activities is penicillamine. The (S)-form has anti-arthritic properties, whereas the (R)-form is toxic. Which form is the following configuration? 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
62 • Chapter 7 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
7.2 Reactions and stereochemistry
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
7.2 Reactions and stereochemistry
64 • Chapter 7 Stereochemistry
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”?
7.2 Reactions and stereochemistry
CHAPTER 8 ALKYL HALIDES AND RADICALS 8.1 Reactions Draw the structural formula of the major organic product(s). Show stereochemistry where appropriate. 1. How many different dichlorides could be isolated by ordinary physical methods (e.g., fractional distillation) from the following reaction? Would each, as collected, be optically active or inactive? H
Cl2, hv
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
68 • Chapter 8 Alkyl Halides and Radicals
5. a. The vinylcyclopropane – cyclopentene rearrangement proceeds by a free radical mechanism. Explain. Hint: the cyclopropyl C-C bond is easily homolyzed.
'
b. Predict the product: '
6. Aspirin, as well as other non-steroidal anti-inflammatory drugs (NSAIDS), blocks the synthesis of certain inflammation-mediating prostaglandins by inhibiting the enzyme cyclooxygenase (COX – see 5.1, 6d). COX converts arachidonic acid to the prostaglandin PGG2, which subsequently undergoes reduction to give PGH2. Other prostaglandins derive from the latter. Outline a mechanism for the synthesis of PGG2. Hint: begin by a free radical removal of one of the doubly allylic hydrogen atoms. 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
72 • Chapter 9 SN1, SN2, E1, and E2 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
74 • Chapter 9 SN1, SN2, E1, and E2 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
76 • Chapter 9 SN1, SN2, E1, and E2 Reactions
9.3 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. Ph
Ph 1. Ph
Ph 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
78 • Chapter 9 SN1, SN2, E1, and E2 Reactions
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
80 • Chapter 9 SN1, SN2, E1, and E2 Reactions
8. When treated with hydroxide, trans-A yields B. However, when cis-A is treated with hydroxide, no B is observed. Explain. O HO
Cl
A
B
_______________________________________________________ Problems 4 and 5 above illustrate the concept of “neighboring group participation” (NGP), wherein an internal nucleophilic atom (e.g., N and O, respectively, in those examples) facilitates the ejection of the leaving group by an intramolecular SN2 attack to form an unstable intermediate. This type of mechanism is often evidenced by (1) rearrangement (problem 4), (2) stereochemistry (problem 5), or (3) kinetic data (problem 9 below). Problems 9 – 16 are additional examples. Account for the observations mechanistically. 9. Unlike most primary alkyl halides the molecules below, types of sulfur and nitrogen mustard gases, do NOT undergo second order hydrolysis, but rather first order: -d[RX]/dt = k[RX]. Yet their rates of hydrolysis are enormously faster than those of most primary alkyl halides. Cl
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
82 • Chapter 9 SN1, SN2, E1, and E2 Reactions
16. DNA is stable in dilute aqueous hydroxide solution, but RNA rapidly hydrolyzes. A mechanistic clue is provided in the observation that hydrolysis of the latter yields not only 3’-phosphates but also 2’phosphates. Explain.
O 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
84 • Chapter 9 SN1, SN2, E1, and E2 Reactions
d. F-PP can isomerizes to nerolidol pyrophosphate (N-PP). F-PP and N-PP undergo a “head-to-head” reductive coupling by an E1 reaction to form the C30-triterpene squalene. Outline the mechanisms for each of these events. Hint: reductive coupling is initiated by hydride attack on N-PP as shown below. :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
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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
90 • Chapter 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
92 • Chapter 10 NMR
28. Pettit (UT) observed that the protonation of cyclooctatetraene (COT) yields a carbocation (homotropylium ion) that possesses homoaromatic stabilization. (Homoaromatics refers to S systems that are interrupted by a saturated center but in which the geometry still permits significant overlap of the p orbitals across a gap.) 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
96 • Chapter 11 Conjugated Systems
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
98 • Chapter 11 Conjugated Systems
11.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates. 1. Phenolphthalein in solutions below pH 8.5 is colorless, but in solutions above pH 8.5 is a deep 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
100 • Chapter 11 Conjugated Systems
7. An early stage reaction in Paquette’s (OSU) total synthesis of dodecahedrane employed the following “domino” Diels-Alder: R
R 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
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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
108 • Chapter 12 Aromatics
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
110 • Chapter 12 Aromatics
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
112 • Chapter 12 Aromatics
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
114 • Chapter 12 Aromatics
12. Dyes such as indigo blue (see 19.1, 34) do not bond well to cotton and tend to wash off after repeated laundering; they are known as surface dyes. On the other hand, reactive dyes bind covalently to cotton, resulting in greater color retention (‘fastness’). The following process illustrates the latter. An 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
116 • Chapter 12 Aromatics
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
120 • Chapter 13 Alcohols
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
122 • Chapter 13 Alcohols
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
124 • Chapter 13 Alcohols
21. Berson (Yale) discovered that the bicyclic carbocation A undergoes a clever rearrangement in which the cyclopropyl ring circumambulates around the cyclopentenium ring. Beginning with B, synthesize a deuterium-labeled species that would support this observation.
etc. 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
126 • Chapter 13 Alcohols
9. Aflatoxin B1 is one of the most potent carcinogens known. In the presence of water and acid, compound A is formed. O
O
O 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
128 • Chapter 13 Alcohols
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
132 • Chapter 14 Ethers
16. Going back to problem 14.1, 13, if the ortho positions are blocked the initial Claisen rearrangement product may be followed by a Cope rearrangement. Fill in the brackets.
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
134 • Chapter 14 Ethers
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
136 • Chapter 14 Ethers
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
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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
142 • Chapter 15 Aldehydes and Ketones
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
144 • Chapter 15 Aldehydes and Ketones
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
146 • Chapter 15 Aldehydes and Ketones
44. Ammonia is produced in the mitochondria primarily by the oxidation of glutamate to produce an imine, which is subsequently hydrolyzed: NH2 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
148 • Chapter 15 Aldehydes and Ketones
Catabolic reactions that produce formaldehyde, as above, generally occur in the presence of another vitamin derivative, tetrahydrofolic acid (FH4). The later detoxifies formaldehyde by reacting with it to produce A. On the other hand, many anabolic reactions require formaldehyde as a building block (e.g., biosyntheses of the nucleoside bases). In those instances A undergoes hydrolysis to yield FH4 and formaldehyde in situ. Draw the structure of FH4.
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
150 • Chapter 15 Aldehydes and Ketones
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
152 • Chapter 15 Aldehydes and Ketones
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
154 • Chapter 15 Aldehydes and Ketones
15.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates. 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
156 • Chapter 15 Aldehydes and Ketones
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
158 • Chapter 15 Aldehydes and Ketones
20. Another protecting group for alcohols (in addition to TMS or vinyl ether derivatives) is MOM (methoxymethyl). MOM is stable to base, but can be cleaved upon treatment with mild acid. The following sequence illustrates its use: Cl
Cl
1. Li 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
160 • Chapter 15 Aldehydes and Ketones
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
162 • Chapter 15 Aldehydes and Ketones
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
164 • Chapter 15 Aldehydes and Ketones
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
166 • Chapter 15 Aldehydes and Ketones
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
170 • Chapter 16 Carboxylic Acids
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
172 • Chapter 16 Carboxylic Acids
15. acetylene
hexanoic acid
OH
OH OH
O
16.
16.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates. 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
174 • Chapter 16 Carboxylic Acids
9. The biosynthesis of the amino acid phenylalanine involves an acid-catalyzed decarboxylation of prephenic acid: CO2H
HO
H
CO2H
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
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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
180 • Chapter 17 Carboxylic Acid Derivatives
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
182 • Chapter 17 Carboxylic Acid Derivatives
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
184 • Chapter 17 Carboxylic Acid Derivatives
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
186 • Chapter 17 Carboxylic Acid Derivatives
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
190 • Chapter 17 Carboxylic Acid Derivatives
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
192 • Chapter 17 Carboxylic Acid Derivatives
b. morphine
heroin
c. codeine
hydrocodone
d. In aqueous solution codeinone exists in dynamic equilibrium with its EJ-unsaturated isomer, neopinone, hydration of which yields oxycodone. Write a mechanism for the equilibration. CH3O
CH3O H
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
194 • Chapter 17 Carboxylic Acid Derivatives
6. Phenylisothiocyanate (A, PITC, Edman reagent) can be used to sequence proteins, i.e., to determine the order of amino acids (primary structure). For example, treatment of dipeptide B with A in the presence of acid yields C (a phenylthiohydantoin, or PTH, derivative of the amino acid). Characterization of C identifies the first (from the N-terminal end) amino acid, in this case alanine. valine residue
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
196 • Chapter 17 Carboxylic Acid Derivatives
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
198 • Chapter 17 Carboxylic Acid Derivatives
16. The final step in Sheehan’s (MIT) total synthesis of penicillin V involved the formation of a strained 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
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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
204 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates
18.2 Syntheses Supply a reagent or sequence of reagents that will effect the following conversions. 1. butyric acid
ethylpropanedioic acid
2. dimethyl malonate
G-valerolactone (valeric acid is a common name for pentanoic acid)
3. dimethyl malonate
butanedial
O NH
4. dimethyl malonate O
O
N H
seconal (a sedative)
5. dimethyl malonate
2-benzylbutanoic acid
O 6. dimethyl malonate
7. dimethyl malonate
+
+
styrene
Ph
methyl acrylate
Br
(acrylic acid is propenoic acid)
8. ethyl acetoacetate
18.2 Syntheses
OH
s-Bu-CONH2
Br
Problems • 205
O
O OH
9. ethyl acetoacetate O
O
O O
10. ethyl acetoacetate
O
O
Cl
Cl
11. ethyl acetoacetate
12. ethyl acetoacetate
s-butyl methyl ketone
2-methylbutanoate
O 13. cyclopentanone OAc
O 14. cyclopentanol
CO2H
O
O
15. 2,4-pentanedione
18.2 Syntheses
206 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates
16. methyl acetoacetate
2,4-pentanedione
via an organoselenium cmpd
17. 3-pentanone
ethyl vinyl ketone
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
208 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates
O H
6.
O
O
7. The vitamin biotin is necessary for many metabolic carboxylation reactions. It reacts initially with CO2 to form unstable A, which then “donates” CO2 to a substrate. Outline the mechanism for carboxylation of pyruvic acid to oxaloacetic acid (OAA). O 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
212 • Chapter 19 Carbonyl Condensation 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
214 • Chapter 19 Carbonyl Condensation Reactions
31. Excessive accumulation of acetyl CoA can lead to metabolic ketosis by the following pathway: H3O Claisen A O 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
216 • Chapter 19 Carbonyl Condensation 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
218 • Chapter 19 Carbonyl Condensation Reactions
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
220 • Chapter 19 Carbonyl Condensation Reactions
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
222 • Chapter 19 Carbonyl Condensation Reactions
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
224 • Chapter 19 Carbonyl Condensation Reactions
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
226 • Chapter 19 Carbonyl Condensation Reactions
26. The CD – CE bond in E-hydroxyketones is easily cleaved via a retro-aldol reaction; the carbonyl – CD bond is unreactive. In D-hydroxyketones, however, the CD – CE bond is unreactive; but, in the presence of thiamine, the carbonyl – CD bond can be cleaved (2). OH (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
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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
232 • Chapter 20 Amines
1. Br2 2. NaNO2, HCl
19. p-toluidine (p-aminotoluene)
3. KI
1. Cl2, FeCl3 2. NaNH2 / NH3
20. benzonitrile
3. KNO2, H 4. CuCN 5. H3O
1. KMnO4, H 2. fuming nitric acid 3. Fe, HCl
21.
4. NaNO2, HCl 5. HBF4
1. Br2, Fe 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
234 • Chapter 20 Amines
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
236 • Chapter 20 Amines
20.3 Mechanisms Outline a detailed mechanism for each of the following. No other reagents than those given are necessary. Use arrows to explain the flow of electrons and show all intermediates. H 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
238 • Chapter 20 Amines
9. The Fischer indole synthesis involves an isomerization known as a [3,3] sigmatropic rearrangement, shown by the arrows below: O H2SO4 PhNH-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
240 • Chapter 20 Amines
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
1.1 Hybridization, formulas, physical properties
244 • Chapter 1 The Basics
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.
1.1 Hybridization, formulas, physical properties
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
248 • Chapter 1 The Basics
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
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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
252 • Chapter 2 Alkanes
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
2.3 Conformational analysis, acyclic
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
2.3 Conformational analysis, acyclic
254 • Chapter 2 Alkanes
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
2.3 Conformational analysis, acyclic
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
256 • Chapter 3 Cycloalkanes
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
258 • Chapter 3 Cycloalkanes
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
3.3 Conformational analysis, cyclic
(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
3.3 Conformational analysis, cyclic
260 • Chapter 3 Cycloalkanes
b. for conformational chair-chair flipping, Keq = 1 for configurations 1, 3, and 5: 3e/3a
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
3.3 Conformational analysis, cyclic
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
262 • Chapter 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
264 • Chapter 5 Alkenes and Carbocations
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
266 • Chapter 5 Alkenes and Carbocations
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
268 • Chapter 5 Alkenes and Carbocations
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
270 • Chapter 5 Alkenes and Carbocations
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
272 • Chapter 5 Alkenes and Carbocations
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
274 • Chapter 5 Alkenes and Carbocations
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
276 • Chapter 5 Alkenes and Carbocations
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
278 • Chapter 5 Alkenes and Carbocations
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
280 • Chapter 5 Alkenes and Carbocations
34. a. no. deg. unsat: C16H34 - C16H30 = H4 => H4/2 = 2 deg. hydrogenation: C16H34O - C16H30O = H4 => H4/2 = 2 DB c. E => a cis- DB at C12; F => a trans- 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
282 • Chapter 6 Alkynes
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
284 • Chapter 6 Alkynes
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
This page intentionally left blank
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
288 • Chapter 7 Stereochemistry
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
290 • Chapter 7 Stereochemistry
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
7.2 Reactions and stereochemistry
292 • Chapter 7 Stereochemistry
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
7.2 Reactions and stereochemistry
H
OH
H
OH
enantiomers => 1 fraction: R
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
1. H2 / Lindlar catalyst
O
H
H
2. Br2 H H
Ph
+
Br
Ph
Br Br
Ph
Ph cis-
enantiomers => 1 fraction: R
H
Br Ph
anti-add'n
H
OMe
+
-H
4. a. Ph
enantiomers => 1 fraction: R
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)
7.2 Reactions and stereochemistry
294 • Chapter 7 Stereochemistry
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
+ _) (+
7.2 Reactions and stereochemistry
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
296 • Chapter 8 Alkyl Halides and Radicals
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
298 • Chapter 8 Alkyl Halides and Radicals
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
300 • Chapter 9 SN1, SN2, E1, and E2 Reactions
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
302 • Chapter 9 SN1, SN2, E1, and E2 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
304 • Chapter 9 SN1, SN2, E1, and E2 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
306 • Chapter 9 SN1, SN2, E1, and E2 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
308 • Chapter 9 SN1, SN2, E1, and E2 Reactions
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
310 • Chapter 9 SN1, SN2, E1, and E2 Reactions
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
312 • Chapter 9 SN1, SN2, E1, and E2 Reactions
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
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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
316 • Chapter 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
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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
320 • Chapter 11 Conjugated Systems
'
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
322 • Chapter 11 Conjugated Systems
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
324 • Chapter 11 Conjugated Systems
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
326 • Chapter 11 Conjugated Systems
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
328 • Chapter 12 Aromatics
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
330 • Chapter 12 Aromatics
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
332 • Chapter 12 Aromatics
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
334 • Chapter 12 Aromatics
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
336 • Chapter 12 Aromatics
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
338 • Chapter 12 Aromatics
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
340 • Chapter 12 Aromatics
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
342 • Chapter 13 Alcohols
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
344 • Chapter 13 Alcohols
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
346 • Chapter 13 Alcohols
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
348 • Chapter 13 Alcohols
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
350 • Chapter 13 Alcohols
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
352 • Chapter 14 Ethers
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
354 • Chapter 14 Ethers
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
356 • Chapter 14 Ethers
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
358 • Chapter 15 Aldehydes and Ketones
[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
360 • Chapter 15 Aldehydes and Ketones
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
362 • Chapter 15 Aldehydes and Ketones
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
2. H2 / Lindlar catalyst
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
364 • Chapter 15 Aldehydes and Ketones
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
366 • Chapter 15 Aldehydes and Ketones
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
368 • Chapter 15 Aldehydes and Ketones
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
370 • Chapter 15 Aldehydes and Ketones
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
372 • Chapter 15 Aldehydes and Ketones
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
374 • Chapter 15 Aldehydes and Ketones
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
376 • Chapter 15 Aldehydes and Ketones
+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
378 • Chapter 15 Aldehydes and Ketones
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
380 • Chapter 16 Carboxylic Acids
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
382 • Chapter 16 Carboxylic Acids
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
384 • Chapter 16 Carboxylic Acids
+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
386 • Chapter 16 Carboxylic Acids
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
388 • Chapter 17 Carboxylic Acid Derivatives
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
390 • Chapter 17 Carboxylic Acid Derivatives
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
392 • Chapter 17 Carboxylic Acid Derivatives
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
394 • Chapter 17 Carboxylic Acid Derivatives
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
396 • Chapter 17 Carboxylic Acid Derivatives
.
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
398 • Chapter 17 Carboxylic Acid Derivatives
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
400 • Chapter 17 Carboxylic Acid Derivatives
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
402 • Chapter 17 Carboxylic Acid Derivatives
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
404 • Chapter 17 Carboxylic Acid Derivatives
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
406 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates
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
408 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates
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
410 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates
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
412 • Chapter 18 Carbonyl Į-Substitution Reactions and Enolates
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
414 • Chapter 19 Carbonyl Condensation 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
416 • Chapter 19 Carbonyl Condensation 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
418 • Chapter 19 Carbonyl Condensation 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
420 • Chapter 19 Carbonyl Condensation Reactions
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
422 • Chapter 19 Carbonyl Condensation Reactions
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
424 • Chapter 19 Carbonyl Condensation Reactions
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
426 • Chapter 19 Carbonyl Condensation Reactions
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
428 • Chapter 20 Amines
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
430 • Chapter 20 Amines
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
432 • Chapter 20 Amines
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
434 • Chapter 20 Amines
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
436 • Chapter 20 Amines
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
