Solutions Manual 3rd c1 7

7/28/2019 Solutions Manual 3rd c1 7

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Solutions Manual to the Third Edition of Organic Synthesis

This book contains answers to most of the end-of-the-chapter problems found in Organic Synthesis, 3rd edition.

The vast majority of problems and answers are taken from the literature, and the appropriate citation is given for

each answer. For reactions, only the product is shown, along with the literature reference. For discussion type

problems and for mechanistic questions, a brief explanation is usually offered, again with the literature citation.

The only exception is chapter 10, where answers are not provided. In a few cases, literature citations are given

where a synthesis can be found. In most cases, however, I believe that solutions to synthesis problems are best

discussed with your instructor because there is no one correct answer.

If there are errors, corrections, and suggestions, please let me know by EMail or normal post. Any errors will

be posted.

Thank you again. I hope this new solutions manual is useful to you in your studies using Organic Synthesis.

Michael B. Smith, Storrs, ConnecticutApril, 2010

University of Connecticut, Department of Chemistry, 55 N. Eagleville Road, Storrs, Connecticut 06269-3060

EMail: [email protected]: 860-486-2981homepage: http://orgchem.chem.uconn.edu/home/mbs-home.html

Table of Contents

Answers to Chapter 1 Problems 3Answers to Chapter 2 Problems 13Answers to Chapter 3 Problems 26Answers to Chapter 4 Problems 37Answers to Chapter 5 Problems 52Answers to Chapter 6 Problems 60Answers to Chapter 7 Problems 66Answers to Chapter 8 Problems 77

Answers to Chapter 9 Problems 96Answers to Chapter 10 Problems 117Answers to Chapter 11 Problems 119Answers to Chapter 12 Problems 133Answers to Chapter 13 Problems 143


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2 Organic Synthesis Solutions Manual

CHAPTER 1

1. The calculations are shown for each molecule using values from Table 1.5 in Chapter 1.

NH2Me2HC

CC-CH3NH2

CCMe

CHMe2

A

NH2

CCMeMe2HC

B

(a) HA = ACHR2 + AC = 2.1 + 0.2 = 2.3 kcal mol-1 If A + B = 1

HB = ANHR+ GC + GCHR2 = 1.3 + 0 + 0.8 = 2.1 kcal mol-1 then, A = 1-B

H = HB HA = 2.1 2.3 = 0.2 kcal mol-1 Keq= BA= B

1-B

At 150C, 2.303RT = 2.303(1.987)(423)* = 1.936 kcal mol-1 Keq(1-B) = B

[* T is in Kelvin = C + 273] and B =Keq

1+Keq, via

therefore G = 0.2 = 1.936 log Keq 1.27(1-B) = B

log Keq=-0.2

-1.936= +0.103 1.27 - 1.27B = B

Keq= 100.103 = 1.27 1.27 = B+1.27B

1.27 = B(2.27)

1.272.27

= B = 0.56

Therefore, 56% of B and 100-56 = 44% of A.

Since A has two axial groups and B has only one, an initial glance suggests that B will be lower in energy and

be the greatest contributor to the chair population. Conformation A has one axial group (CHMe2) on the top and

one axial group (CCMe) on the bottom so ACHR2 and AC are used from Table 1.5. In B there is only one axial

group (NH2) so AOR is used. The two equatorial groups in B (CHMe2 and CCMe) are on adjacent carbons, so

there are two G terms, GC and GCHR2.


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Chapter 4 3

MeO

H 3C O

Cl

Cl

O

MeO

CH3

A

O

Cl

H3C

OMe

B

(b) HA =34 (ACH2R+ ACl) = 0.1.8 + 0.4 = 1.65 kcal mol

-1 If A + B = 1

HB =34 (AOR+ GCH2R+ GCl) =

34 (1.8 + 0.4 + 0.5) = 2.03 kcal mol

-1 then, A = 1-B

H = HB HA = 2.03 1.65 = 0.38 kcal mol-1 Keq=

BA

=B

1-B

at 25C, G = 0.38 = -1.364 log Keq Keq(1-B) = B

log Keq=0.38

-1.364 = -0.279 and B =Keq

1+Keq

Keq= 10-0.279 = 0.526

0.5261.526

= B = 0.345

Therefore, 34.5% of B and 100-34.5 = 65.5% of A.

Although B has two axial groups, it is actually lower in energy because the axial chlorine has a lower interaction

that the combined G value interactions in B. It accounts for only 35% of the population of chair conformers.

Conformation B has two adjacent and diequatorial groups, so GCH2R and GCl are used from Table 1.5. Since B

has one axial methoxy group, AORis used.

(c)

OMe

OMe

MeO

MeCl

Cl OMeMeO

OMe

Me

A B

Me

OMe

OMe

MeO

Cl

(c) HA = AOR+ ACl + GCH2R+ GOR= 0.8 + 1.8 + 0.4 + 0.2 = 3.2 kcal mol-1 If A + B = 1

HB = UOR+ UOR+ ACH2R+ GCl + GOR= 0.8 + 0.8 + 1.8 + 0.5 + 0.2 = 4.1 kcal mol-1 then, A = 1-B

H = HB HA = 4.1 3.2 = 0.9 kcal mol-1 Keq=

BA

=B

1-B

at 25C, G = 0.9 = 1.364 log Keq Keq(1-B) = B

log Keq=0.9

-1.364= 0.66 and B =

Keq1+Keq

Keq= 10-0.66 = 0.22

0.221.22

= B = 0.18

Therefore, 18% of B and 100-18 = 82% of A.


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The three axial groups in B, along with the two G-interactions make it much more sterically demanding than the

two axial groups and the two G-interactions in A. Therefore, A accounts for the greater percentage of chair

conformations.

PhMe3C

Ph

Me3C

Ph

CMe3A B

(e) HA = no interactions = 0 kcal mol-1 If A + B = 1

HB = Aaryl + ACR3 = 3.0 + 6.0 = 9.0 kcal mol-1 then, A = 1-B

H = HB HA = 9.0 0 = 9.0 kcal mol-1 Keq=

BA

=B

1-B

at 25C, G = 9.0 = 1.364 log Keq Keq(1-B) = B

log Keq=9.0

-1.364= -6.598 and B =

Keq1+Keq

Keq= 10-6.598 = 3x10-7

3x10-7

1.00= B =3x10-7

Therefore, 0.00003% of B and 100-0.00003 = 99.99997% of A.

Since A has two large equatorial groups and B have two large axial group, the equilibrium is pushed in the

direction of A, in essentially 100%.

2. The absolute configuration for each chiral center in the following molecules is shown beside the appropriate

chiral center.

O

O

H H

OH

HO H

H

HH

HH

O

O

O

O

O

O

H

H

O

Br

HH

H

OAc

(a)

(b)(c)

(+)-absinthinseeJ. Am. Chem. Soc.,2005, 127, 18

biepiasterolideseeJ. Org. Chem.,2004, 69, 9100

(+)-laurencinsee Org. Lett., 2005, 7, 75

3.


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Chapter 4 5

OO

O

O

O

O

N

N

MeO2C

OO

O

O

OH OH

O OH OH

O

(a)(b) (c)

amphidinolide XseeJ. Am. Chem. Soc.,2004, 126, 15970

(+)-lapidilectine BseeJ. Org. Chem.,2004, 69, 9109

mycolactone Csee Org. Lett. 2004, 6, 4901

4. After donor and acceptor sites for disconnect fragments 22 and23, fragment B represents an umpolung reagent

(acyl anion equivalent - see Chap. 8, Sec. 8.6.B), F. Fragment A is simply the alkyl halide,

O

Br

A

F

S

S

OB

AND

C

CuLi

H

Cl

O

DO

2

dd aa

2223

E G E. The other d/a combination is C (which is the equivalent of organocuprate, G) andD (which is the equivalent of

acid chloride H). Both are viable processes but we not will choose the reaction ofE andF since alkylation of

dithiane reagents can be sluggish (see Chap. 8, Sec. 8.6.B.ii) and because an extra step is required to convert the

dithiane back to the carbonyl (see Chap. 7, Sec. 7.3.B.ii). Acid chloride H contains

J

O

Cl HO

O OH

H I

the four carbons of the starting material, and it is obviously derived from isobutyric acid (I), available from Aldrich

(2000-2001), $25.50/L. Assume it must be made from 2-methylpropene, however. Accord-ing to Figure 1.1, one

route to an acid is by oxidation of an alcohol (see Chap. 3, Sec. 3.2.A). The alcohol (J) can be prepared from the

alkene by hydroboration (see Chap. 5, Sec. 5.4.A). Once acid chloride H is available, it is reacted with


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organocuprate G, derived from bromide E. In this analysis, E is not formally prepared from 2-methylpropene

(although a synthesis could be designed if desired), andG is considered to be a reagentin this synthesis (an 'off the

shelf' compound that is reacted with the molecules derived from the retrosynthesis, just like B2H6 or CrO3 in the

sequence shown). If the reagents provided in this synthesis are not familiar, careful reading of the remainder of this

book and the literature will explain these choices.

5. There is more than one "correct" answer. One possible solution is shown for each transformation. The letters

(a), (f), etc., beside each reagent refer to the lettered transformations from Figure 1.1 in Chapter 1.

(a)OH1. B2H6

2. NaOH/H2O2(w)

(b)

ORCO3H

OH

CN

1. NaCN , THF

2. hydrolysis(v) (c)

(c)

Br OH CHO

dilute NaOH CrO3 , H+

The first reaction is a poor one since elimination will be a major process. Limitingthe choices to those in Figure 1.1 is a problem since there are other ways to do this.

(c) (a)

(d)

OHPBr3

BrCO2Me

CO2MeKOH , EtOH 1. O3 2. H2O2

(d) (f)

(x) & no reactions for conversion to acid derivatives

2. SOCl2 ; MeOH

(e) O MeMgBrMe

OH1. O3 2. Me2S

(x) (3b)see Table 1.1

(f) CN CONH2 No reagents are provided in Figure 1.1 -they are part of the conversion to acid derivatives

1. H3O+

2. SOCl23. NH3

6. There are potentially many examples for each part (but not always; see the last sentence for this answer). This is

a literature searching question, and there are no correct or incorrect answers. The appropriate sections and pages

from Vol. 11of the Compendium are indicated for each category.


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Chapter 4 7

(a) Acids from Nitriles. Section 28. Volume 11: p. 15 (0 examples).

(b) Aldehydes from Nitriles. Section 58. Volume 11: p. 103 (0 examples).

(c) Amines from Halides. Section 100. Volume 11: pp. 289-294 (28 examples).

(d) Amides from Nitriles. Section 88. Volume 11: p. 257 (5 examples).

(e) Ethers from Halides. Section 130. Volume 11: pp. 347-348 (6 examples).

(f) Halides from Amines. Section 142. Volume 11: p. 369 (0 examples)

(g) Ketones from Olefins. Section 179. Volume 11: p. 409 (3 examples).

(h) Olefins from Aldehydes. Section 199. Volume 11: pp. 440-443 (17 examples).

The real lesson from this exercise, apart from learning to use this literature resource, is that some functional

group exchange reactions are well studied and there are many variations. Others yield only a handful of

possibilities. Despite the request for three examples in the question, noexamples were found for (a), (b) or (f). In

this case, further literature searching in the older literature is a necessity.

7. The bicyclo[2.2.1]heptane unit (A) is very rigid and greatly influences the stereochemistry for the molecule.

The appended six-membered ring B assumes a conformation that is close to a twist-boat, and it cannot 'ring flip'

easily due to the conformationally immobile bicycloheptane unit. This constraint forces the methyl group (Me2)

into a pseudo-axial position. The other methyl group (Me1) is at a bridgehead position of the bicycloheptane and is

effectively perpendicular to that ring, as shown. The hydroxyl group is formally in a pseudo-axial position relative

to ring B.

Me

2

H

HO

Me1

A

B

A

A

Me2

Me2

8. For each molecule, a representative model from the indicatied perspective using Spartan is shown.


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(plus 'top' [above page] and 'bottom [below page] views)OO

AcO

H

1

2

3

4

(+)-mycoepoxydieneseeJ. Org. Chem., 2004, 69, 8789

top view

1

bottom view

2

3 4

(plus 'top' [above page] and 'bottom [below page] views)

N

OO

Cl

OH

H(b)

1

2 3

halochlorineseeJ. Org. Chem., 2004, 69, 7928

top

bottom

3


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Chapter 4 9

1

2

The purpose of this drawing exercise is to cast yourself in the role of a reagent approaching a molecule.

Different angles of approach means the reagent "sees" a different set of groups and atoms for each portion of the

molecule, which will have an important effect on the reactivity of the incoming reagent. By drawing the molecule

from several perspectives, the goal is to see how angle of approach and topography may influence reactivity in a

given molecule. Also, it emphasizes that the two-dimensional drawings we usually use often have little meaning in

terms of the real structure of a molecule and determining its reactivity.

9. The R or S configuration for the chiral axis of each molecule that has a chiral axis rather, as well as that of

stereogenic centers chiral center, is shown for (a) - (e).

C

Cl H

Br CH 2CH2Cl

OH

OMe

NH2

CH2Cl

Me2HC CMe3

O

CHMe2H

OHHOH2C

NH

MeH

Cl Et

(d)(c)(b)(a) (e)

For (a), Cl is 1 and H is 2; Br is 3 and CH2CH2Cl is 4. For (b), OH is 1 and Caryl is 2; NH2 is 3 and C

aryl is 4.

For (c), Cl is 1 and Et is 2; Me is 3 and H is 4. For (d), CH2Cl is 1 and Et is 2; CMe3 is 3 and CHMe2 is 4. For (e)

the epoxy O is 1 and MHMe2 is 2; OH is 3 and CH2OH is 4.

10.

O

Mg

O

Me

Me

Me

MeBr

Me H

HP

PMe

Me

Me

Me

Ph

Et

Ar

OC

FeMe

Ph3P

Cp

O

O

OEtO

EtO2CPh

OiPr

R

R

(a) (b) (c) (d)

11. When the hydroxy-acid cyclizes to form the lactone (see A) the two methyl groups are 1,3-diaxial, with the tert-

butyl group equatorial. In the presence of acid, the hydroxyl group can be protonated and eliminate water and form

a cation.

In the presence of water, the alcohol is re-formed, but both epimers are generated since reaction with water leads


10/79


to the methyl group being axial or equatorial in the lactone. Under equilibration conditions, the lactone with one

axial methyl and one equatorial methyl will be lower in energy due to diminished A-strain, and the equilibrium will

shift to favor that lactone (net epimerization of the alcohol-bearing carbon in the hydroxy-acid). Acid catalyzed

opening of the lactone gives, of course, the hydroxy-acid.

O O

t-Bu

Me

H

H

MeH

O

O

MeMe

t-Bu

A

OHO

H

B

12. The diaxial conformation allows hydrogen bonding (see B ). Such hydrogen bonding counterbalances the

energy difference of the diaxial (which has U-strain) vs. diequatorial conformations. The net result is that although

the diaxial conformation has U-strain, the hydrogen bonding makes it the dominant species.

13.

O H

OH

Ph

NN-H

Ph

(c)(b)

(a)

a

bc

rea

b

c

a

c b

a

bc

si

re

si

14. The correlation is .

Taken fromEur. J. Org. Chem. 2004, 2398

15.

OH

Br Br

ABr Cl

B

1

2 43

6

71

2

3

4

5 5

6

7

OH

Br Br

A

1

2 43

6

5

7

ery thro and syn

Br ClB

71

2

3

4

5

6

threo and anti

2S, 3 R, 5R

2S, 3S, 5S

16. (a) diastereoselective (b) enantiospecific (c) regiospecific and diastereoselective.


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Chapter 4 11

17. Cyclodecane has a relatively small internal cavity where transannular interactions are maximized. This

transannular interaction disappears in cyclooctadecane where the additional carbons make the internal cavity large

enough for the "internal" hydrogens to have little interaction. Compare 182B and184 on page ??? in the text.

Inspection of Figure 1.13 of the text can answer the second part of this question. Cyclohexadecane is an even-

membered ring and, as such, will fit better on the diamond lattice. Since cyclopentadecane has an odd-membered

ring, there is a "twist" in the ring when attempting to "lay it on the diamond lattice". This "twist" makes the energy

of the ring slightly higher.

18.

H

H

H

H

SeeJ. Am. Chem. Soc., 1994, 116, 10306.


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CHAPTER 2

1. This solvolysis reaction proceeds with loss of the leaving group to generate a cation. The aromatic ring will

stabilize that cation either by -donation from the ring or by formal resonance participation via a non-classical ion.

In either case, an electron-donating substituent on the aromatic ring will stabilize the charge and, thereby, stabilize

the intermediate cation. This stability will be reflected in the relative rate of the solvolysis reaction. Both OMe and

Me are electron donating and therefore enhance the rate. Since OMe is a more powerful electron donating group,

the effect is large. The Cl substituent is mildly electron withdrawing and, therefore, slows the relative rate by

slightly destabilizing the intermediate cation.

This problem is taken fromJ. Org. Chem., 1985, 50, 821.

2. Since the OH group in 3-pentanol is an extremely poor leaving group and iodide is a nucleophile, an SN2

reaction is not facile. Despite the presence of the water, solvolysis of the alcohol moiety to generate a cation is

somewhat slow. If an acid catalyst is added, however, protonation of the OH occurs and loss of water generates the

cation. The cation is then trapped under SN1 conditions by iodide to give 3-iodopentane.

OHO

IHH

- H2O + I

H+

3. This bromide is in fact a neopentyl-like molecule [R3C-CH(R')X] and, therefore, very sterically hindered to

nucleophilic displacement (see Sec. 2.6.A.i). Neopentyl halides react much slower with nucleophiles under SN2

conditions than do tertiary halides (see Table 2.11 on page 105 in the text). The use of KOH in DMF promotes

substitution, and the DMF does not allow facile ionization to give an SN1 displacement. Elimination with the basic

KOH dominates if the substitution reaction is too slow to compete.

4. (a) Since the phenyl ring is somewhat electron-withdrawing, the three phenoxides are less basic than the

cyclohexanol anion. The OMe group is electron releasing, and this makes the availability of electron density on O-

greater than on phenoxide. Similarly, the nitro group is electron withdrawing, making the electron density on O

less than in phenoxide.

(b) When comparing formic acid and acetic acid, the electron releasing methyl group in acetic acid diminishes

the positive character of the acidic hydrogen. It is therefore less acidic. In addition, once ionized to the

carboxylate, the presence of the electron releasing methyl group slightly diminishes the adjacent positive character


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Chapter 4 13

of the carboxyl carbon, destabilizing the anion relative to the formate anion. The smaller formate anion is probably

better solvated than the acetate anion, which also contributes to enhanced acidity.

ortho-Methoxyphenol is more acidic thanpara-methoxyphenol due to the "ortho effect." The proximity of the

OMe in the ortho derivative allows internal hydrogen bonding with the O-H moiety, making that bond more

polarized and more acidic. Such internal hydrogen bonding is not possible in the para derivative, although

intermolecular hydrogen bonding may occur.

Acetic acid is more acidic in water, although THF is a stronger base. Ionization is much easier in water than in

THF, and water can better stabilize the ionic products. This makes the ionization (loss of H+) easier, enhancing

acidity relative to THF where ionization is less efficient.

5.

OMe

N(CH2Ph)2

Et

O

N

Et

H

O O

SiMe3

O

Cl

MeO OMe

O

O

Cl NHi-Pr

ONEt2

OMe

OMe

MeO

O

O

N

HPh

Me

Me(CH2)9

ON

O

HN

O

O

O

O

O MeO

seeJ. Org. Chem., 1999, 64, 4980 seeJ. Org. Chem., 1999, 64, 7586see Tetrahedron Lett., 2000, 41, 733

see Tetrahedron Lett., 2000, 41, 1975

seeJ. Am. Chem. Soc., 1994, 116, 9921

seeJ. Org. Chem., 1999, 64, 2450

seeJ. Org. Chem., 1999, 64, 3736 seeJ. Am. Chem. Soc., 1994, 116, 9921

(a)

(b)

(c)

(d) (e) (f)

(g) (h)

6. Taken from Stock, L.M., Aromatic Substitution Reactions, Prentice-Hall, Englewood Cliffs, N.J., 1968, p. 91.

This reaction proceeds via a benzyne intermediate (X). Addition of NH2 to either carbon of the triple bond leads to

the two products shown and the 14C label on both carbons relative to the ipso carbon bearing the amine unit.


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NH2Cl

+

NH2

*

*

*

*

H2N

* =14

C X

H2O

7. This cyclohexane derivative exists in two chair forms, A andB. An E2 reaction cannot occur from A because

the bromine is equatorial. A trans-diaxial relationship between the bromine and any -hydrogen(s) is a

requirement. Only when the bromine is axial in conformation B is one -hydrogen is axial. This means that the E2

reaction can give only one alkene with the double bond toward the methyl group rather than the ethyl group. The

stereochemistry in the halide makes this a regiospecific elimination.

Me

Br

Me

Et

Br

Me

Me

H

Et

H

H

Br

Me

Me

Et

H

A B

Me

Me

Et

8. See Tetrahedron Lett., 2000, 41, 3411. The tertiary allylic alcohol reacts with the acidic resin (H+) to give the

oxonium ion. There are two pathways. An SN2' displacement of water by the primary alcohol unit leads to the

product directly. This is the mechanism presented by the authors of this paper. An alternative mechanism would

be ionization to an allylic tertiary cation followed by cyclization and loss of a proton, as shown in the diagram.

HO

OH+ H

+H2O

OH

OH

O

H

H2O

H+

H2O

O

SN2'

vinylogousSN1-like

Tetrahedron Lett., 2000, 41, 3411

9. This reaction proceeds via a mercury-stabilized secondary cation. In this paper, a detergent (sodium dodecyl

sulfate) was added and it had an important effect. The cation formed in the reaction can react with water (from the

aqueous solvent) to give an alcohol after reductive cleavage of the C-Hg bond with sodium borohydride. The ether

is formed by attack of octanol on the cation. There is a large excess of water, however, so the alcohol is the major

product. Increasing the proportion of octanol leads to increased amounts of ether. Even when 10 equivalents of


15/79

Chapter 4 15

octanol are used, water is present in a large excess. In fact, water and octanol are close in nucleophilic strength.

The detergent leads to enhanced local concentrations of octanol that can overcome the bulk concentration effects of

the excess water, leading to more ether product.

Taken fromJ. Org. Chem., 1987, 52, 5039.

10. This problem was taken fromJ. Org. Chem., 1987, 52, 260; 1984, 47, 4855.

(a) Product A is the usual product of reaction of an alkene with chlorine. Formation of chloronium ion X was

followed by addition of chloride ion to give A, as shown. Product B arises by a cationic mechanism that involves

participation of the aromatic ring. If chloronium ion X opens to form cation Y, the benzene ring attacks the

positive charge to form the bridged cation Z. If chloride ion attacks the three-membered ring, as shown, B results.

Cl

ClMeO

Cl

MeO

Cl

Cl

MeO

Y

Cl

MeO

Cl

X

ClCl

B

MeO

Cl

Cl

MeO

A

Z

(b) If the OMe group, which is electron-releasing, is replaced with Cl, which is electron-withdrawing, product Bshould be more difficult to form. Cation Z is stabilized by the electron releasing OMe group, but an electron

withdrawing Cl would destabilize this intermediate, making formation ofZ more difficult and, thereby, formation

ofB.

11. See Tetrahedron Lett., 1997, 38, 3469. When drawn in a form that approximates the three-dimensional shape it

is clear that the cyclobutanone unit flattens the tricyclic system. Protonation of the glycol unit is followed by acid-

catalyzed addition of methanol to the ketone. A Grob-like fragmentation leads to formation of the methyl ester unit

and a cation. This cation traps water and with proton transfers loses ethylene glycol to form the ketone unit found

in the final product.


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O

O

O

H

H

H

O

OO

H

H

O

OOH

RT

H

H

OMe

OOH

O

H

H

CO2Me

O

H

H

OMe

OOH

O

OOH

O

OH

H

H

H

OMe

H

H

OMe

O

O

H

H

OMe

OOH

O

OH2

+ H+

+ MeOH;

H+ transfer

H+

+ H2O

H+;+H

+

ethylene glycol

H+

MeOH , 1N HCl

12. See Tetrahedron Lett., 2000, 41, 403. This reaction proceeds by protonation of the alcohol and loss of water

from A to form a carbocation 1. There are two potential sites for rearrangement to a tertiary cation, to 2 or to 3.

To form cation 3 requires that the sp2 carbon be flattened to a trigonal planar shape. This would subject the

tricyclic system to a great deal of strain, and the result is that cation 2 is formed preferentially. Loss of a proton via

an E1 mechanism leads to the alkene product, B.

H OH

H

H

H

p-TsOH , benzene

reflux

70%

A B

H

H

H

H

H

high energy dueto flattening thetricyclic r ing system-less favorable

1

2

3

lower in energy

H+

+ H+ ; H2O

13. This reaction is taken from Org. Lett., 2003, 4, 59. Initial deprotonation of the alcohol leads to the alkoxide


17/79

Chapter 4 17

shown. As the carbonyl of the aldehyde is formed, the four-membered ring is formed by transannular displacement

of the bride, which is properly positioned for reaction.

O

Br

O

H

O

CHO

O

Br

OH

H

KH

14. Initial attack of hydroxide leads to opening of the oxazolone ring, and the nitrogen attacks the epoxide, forming

the indolizidine ring and generating an alkoxide. Protonation to give the alcohol is followed by a second attack of

hydroxide on the carbonate ester. Carbonic acid is formed as the alkoxide is released in an acyl substitution

reaction. The carbonic acid decomposes to water and carbon dioxide and protonation by ethanol gives the final

product.

N

H

OBn

OH

O

O

HO

NO

O

OBn

O

OH

HO

N

O

O

OBn

O OH

N

H

OBn

OH

O

HO O

OH

HO

O

OH

N

H

OBn

OH

OH

EtOH

N

H

OBn

O

O

O

HO

N

H

OBn

OH

O

EtOH

seeJ. Org. Chem., 2000, 65, 9129

15. (a) This is the Hofmann elimination reaction (Sec. 2.9.C.i) and demands a syn transition state (an eclipsed

conformation for reaction). Because of the requirement for an eclipsed conformation (where the leaving group on

the -carbon and the "base" can be in close enough contact) the lowest energy eclipsed conformation will lead to

the major product. In this case, the lowest energy transition state is A rather than B, and ethene is the major alkene

product, not 4-methyl-2-pentene. The i-PrMe interaction in B destabilizes that conformation relative to the


18/79


HH interactions in A. In both A andB, the NR3H interactions are about the same.

NEt2C4H9

HH

H

HH

NEt3

HMe

H

Me

H

A B

(b) Taken from J. Org. Chem., 2004, 69, 7616. The strongly electronegative fluorine atoms withdraw electron

density, making the C=C unit very electrophilic. This inductive effective make the alkene carbon more susceptible

to attack than the epoxide carbon, despite the fact that there is modest steric hindrance. Such attack drives the SN2'

reaction to the allylic alcohol shown, as a mixture of E and Z isomers.

PhOF

F

Li

Ph

OH

F

FBu E + Z2nd step of

workup included

(c) This is an SN2 reaction, and inspection of the Walden inversion transition state shows that these two neutral

reactants produce a charged transition state where a positive charge builds on the nitrogen (not on carbon) and a

negative charge builds on iodine (see A). Separation of these two charges (water promotes separation

C

C3H 9

HH

Et3N I C

R

RR

X I

A B

+ +

of charges) favors formation of the final products since the carbon is transferred to the positive nitrogen, with

iodide as the counter-ion. This contrasts with transition state B, the "normal" Walden inversion that arises when a

charged nucleophile attacks a neutral substrate. Water as a solvent will separate charges and promote ionization.

Separation of charges in B will slow the reaction since the incoming charged nucleophile is separated from the

developing charge on the central carbon. In A, however, separation of positive and negative charges favors product

formation. For this reason, the reaction of an amine and a halide to give an ammonium halide is faster in aqueous

media than in non-aqueous media (which cannot separate charges).

(d) The proximity of the ammonium group to the carboxyl carbon is the key to this answer. The aminopropanoic

acid has the amino group two carbons away, whereas the amino group is four carbons away in aminopentanoicacid. Both inductive effects and field effects are strongest when the ammonium groups are close. For this reason,

amino propanoic acid is more acidic than amine pentanoic acid.

(e) The nitro group in 4-nitrophenol is electron withdrawing and, therefore, stabilizes the charge in the phenoxide

product. No such stabilization is possible with phenol . The electronic effects also weaken the

OH bond, enhancing acidity.


19/79

Chapter 4 19

16. 3-Bromo-4-methylhexane reacts with hydroxide (a base for an E2 reaction) by removing Ha. The orientation

of the molecule does not matter because the important feature is the anti-relationship of the Br and Ha. When Ha is

removed, the transition state for the E2 reaction will retain the stereochemical relationship of the groups. Since the

two ethyl groups are on the same side in the anti-orientation, they will be on the same side in the transition state,

and this will lead to cis-3-hexene as the major product, with the two ethyl groups sill on the same side. If this

reaction were carried out under E1 conditions, ionization of the bromine would lead to planar carbocation C, and

removal of Ha could occur from either face since the bond to the carbon bearing Ha is free to rotate. This leads to

scrambling of the stereochemistry and a mixture of cis- and trans-alkenes.

Ha

Br Me

H

A

B

Ha

Br Me

H

OH

MeH

Ha

Me

H65

4

3

21

65

43

21

rotation about this bond

leads to a mixture of

cis- and trans-alkenes

17. SeeJ. Org. Chem., 1997, 62, 641.

18. See Synthesis, 1996, 219. Since NBS is a source of bromine, reaction with the alkene unit generates a

bromonium ion. The oxygen of the alcohol is fixed on the bottom of the molecule relative to the Br, and properly

positioned to open the bromonium ion to form the ether unit. This places the bromine on the top of the molecule.

HO

OOH

BrNBS , CH2Cl2

25C RT

Br

O

Br

H

Br2 from NBS H+

H

H

H

H

19. This sequence is taken from Org. Lett., 2003, 5, 3361. The alkene unit reacts with iodine to give the diiodide

in situ, and the proximal iodide is displaced by the amine in an internal SN2 reaction to give the bicyclic amine. A


20/79


second internal SN2 reaction displaces the remaining iodide to form an aziridinium iodide. the nucleophilic iodide

ion attacks the methyl group of the ester, displacing the carboxylate group, and the electron flow is such that the

oxygen opens the aziridinium ion to form the lactone unit in the product.

NH

CO2Me

Me

t-BuMe2SiO

t-BuMe2SiO Me

N

t-BuMe2SiO

t-BuMe2SiO Me

OO

Me

H

H

N CO2Me

Met-BuMe2SiO

t-BuMe2SiO Me

I

HN Me

t-BuMe2SiO

t-BuMe2SiO Me

H

O

OH3C

NH

CO2Me

Me

t-BuMe2SiO

t-BuMe2SiO Me

I I

I

I2 , CH2Cl2/ether

rt , 2 d

20.

(a)

N

EtEt

C3H7

via OH Br followedby internal SN2 by nitrogenJ. Org. Chem., 2003,

68, 4371 (b)

OPMB

Me

MeCH2O OSiMe2t-Bu

J. Org. Chem., 2003, 68, 8129 (c)

O OHO OH

note inversion of configurationat the site of ether formation

seeEur. J. Org. Chem., 2000, 1889


21/79

Chapter 4 21

(d)

NNH

N

N

PhH2CO

Me

H

H

O

OCH2Ph

O

CH2OMe

O

CH2OMe

HO

J. Am. Chem. Soc.,2002, 124, 3939 (e)

NH

N

Et

HO

J. Am.Chem. Soc., 2002,

124, 4628 (f)

OCH2Ph

NHCH2Ph

J. Am. Chem. Soc., 2002,

124, 8584

(g)

THPOBnO

J. Org. Chem., 2003, 68, 6905 (h)

N

CO2t-Bu

Br

J. Org. Chem., 2003, 68, 6279 (i)

C12H25OCH2OCH2CH2OMe

OH

J. Am. Chem. Soc., 2004, 126, 36

(j)

NC

OCH2OMe

Org. Lett. 2002, 4, 937

(k)

O

O

O

O

BuOH

J. Org. Chem., 2003, 68, 4039

(l)

OO

J. Nat. Prod., 2002, 65, 909

(m)

O

OH

H

HBr

J. Org. Chem., 2004, 69, 1744 (n)

O

CN

CO2MePhH2CO

HOH

J. Org. Chem.,2003, 68, 7422 (o)

N

O

O

Me

OMe HO

J. Org. Chem., 2004, 69, 2191


22/79


(p)

MeO

NMe

H

Et

J. Chem. Soc., Perkin Trans. 1,

2001, 2398 (q)

O

N3AcO

AcO OAc

HOOMe

Org. Lett. 2001, 3, 3353 (r)

O

Me3SiO

O

Org. Lett., 20046, 2961

(s)

N

H

HN

H

O

Me2N

HN

J. Org. Chem., 2002, 67, 7147 (t)

OH

O

CO2Me

I

H

OH

MeO2C

OH

H

I

Org. Lett. 2003, 5 , 4385 (u)

Br

HO

Tetrahedron Lett.,

2000, 41, 2573

(v)

OMe

MeO

Me

Me

Me

OH

see Tetrahedron Lett., 2000, 41, 1151 (w)

Me

seeJ. Org. Chem., 1993, 58, 2186 (x)

Nt-BuO2C

I

J. Org. Chem., 2002, 67, 6181

(y)

N

OMeOMeO

J. Am. Chem. Soc.,2001, 123, 3214 (z)

BnO

TBSO SiMe3


127, 5596 (aa)

MeO2C CO2Me

HH

HNNH

O O

Chem. Commun., 2004, 2404

(bb)

Ph CH3

OOHNHSO

p-Tol

Org. Lett., 2003, 5, 3855

21. (a) The transition state is shown in brackets. Hydroxide removes the -hydrogen only when that hydrogen is in


23/79

Chapter 4 23

a conformation that places it anti to the leaving group (here, bromine). In this transition state, the relative position

of the groups are fixed and this is translated to the alkene, where a single product is formed. The reaction is,

therefore, stereospecific.

Me

HEt

BrH

PhPh

Me

Br

H

H

EtBr

H

Ph

EtMe

H

Br

H

Ph

EtMe

HOH

HBr

Ph

H

Et

Merotate

(b) The major product is trans-1,2-diphenyl-1-propene. (c) This is an E2 reaction.

22.

O

HO

MeO CO2H

Ph

HO

Ph

MeOAc

NC

Me(a) (b) (c) (d) (e)

(a) The most acidic hydrogen is the phenolic hydrogen (pKa 10) vs. pKa 17 for the primary alcohol.

Deprotonation gives the phenoxide, which reacts with iodomethane to give the anisole derivative shown.

(b) Two equivalents of base deprotonate both the carboxyl first (most acidic) and the alcohol second (least acidic).

Of the two anions, the carboxyl anion is resonance stabilized and less nucleophilic than the alkoxide, where the

charge resides almost entirely on oxygen. Since the alkoxide oxygen is more nucleophilic, it will react

preferentially with one equivalent of allyl bromide to give the ether shown.

(c) There are two leaving groups in this molecule that can be displaced by the nucleophilic cyanide in an SN2

reaction. The mesylate group is a much better leaving group than acetate. For this reason, one expects the

cyanoacetate product shown to predominate rather than the alternative cyanomesylate.

(d) The dianion shown has a carbon nucleophile and an oxygen nucleophile. The carbon nucleophile is more

nucleophilic, despite the fact that it is resonance stabilized by the adjacent phenyls. The product will therefore be

the methylated derivative shown. Since the nucleophilic strength of the carbanion is diminished by resonance,

some alkoxide may be formed.

(e) In this case, the primary iodide is treated with base. Normally, primary iodides undergo elimination slowly,

and substitution predominates when a nucleophilic base is used. In this case, however, DBU is a non-nucleophilic

base, and reaction will give the alkene shown as the major product.

(f) The product is that shown. Initial formation of the enolate anion and quenching with PhSecl generated the

phenylselenide. Oxidation with hydrogen peroxide gave the selenoxide in situ, which eliminated spontaneously to


24/79


give the alkene unit in the conjugated lactam product.

N

O

CO2t-Bu

OSiPh2t-Bu

J. Am. Chem. Soc. 2002, 124, 14826

23. The product is the less substituted alkene (4,6-dimethylhept-2-ene, C) via syn elimination of an intermediate

sulfoxide. The syn elimination demands an eclipsed transition state, and the two pertinent eclipsed conformations

where a -hydrogen can be removed are A andB. From these Newman projections, A (for removal of Ha) is less

sterically hindered due to decreased torsion strain than is B (for removal of Hb). For this reason, A will have a

higher population at a lower energy and will account for the major product, C. Note that C is shown as a mixtureof cis and trans isomers, despite the fact that the starting iodide contained a stereogenic center. Removal of the two

Ha's in A will lead to the isomeric mixture of alkenes, C.

PhS

O

S

Ha

Hb

O Ph

H

CHb(Me)C5H11H

Ha

Me

Ha

PhS

O

CHHaMeH

Hb

Me

C5H11A B

Hb

C

24. (a) The Fischer projection for the starting material represents a chiral, non-racemic bromide. Once the

bromine and -hydrogen are in the proper anti conformation, the base removes the hydrogen and expels the

bromine. The other groups are fixed in this transition state, leading to trans-2-phenyl-3-methyl-pent-2-ene.

CHO CCMe

Br Br

a b

(a) CBr4 , PPh3 (b) BuLi (c) H2O

(b) This cyclohexane derivative exists as an equilibrium mixture ofA andB. For an E2 reaction, the bromine

and a -hydrogen must be trans and diaxial. In A, the bromine is equatorial, so there is no chance for elimination.

In B, the axial bromine is axial to two axial hydrogens. Elimination will therefore lead to a mixture of two


25/79

Chapter 4 25

regioisomeric products, C andD.

Br

MeEt

Br

MeEt

Br

Me

Et

HH

A B

MeEt

C

MeEt

D

(c) In this example, the cyclohexane bromide also exists as A andB, andA cannot react via an E2 reaction

because the bromine is equatorial. In B, however, only the hydrogen attached to the methyl-bearing carbon is trans

and diaxial with the bromine, so there is but one product, C.

Br

EtMe

BrEt

MeBr

Et

Me

HH

A B

EtMe

C

25.

(a) CHOBr Br

CCMea b

(a) CBr4 , PPh3 (b) BuLi (c) H2O

(b)

NO2 NO2

Cl

NHAc

Cl

NHAc

Cl

SO3H

NH2

Cl

SO3H

NH2

Cl

a b c d

e f(a) HNO3/H2SO4 (b) Cl2/AlCl3 , heat(c) H2 , Pd/C (see chap. 4, sec. 4.8.D) (d) Ac2O(e) SO3/H2SO4 (f) aq acid

(c)

ClCl

NO2

Cl

NH2

Cl

OH

a b c d

(a) Cl2 , AlCl3 (b) HNO3 , H2SO4 (c) H2 , Pt (d) NaNO2 , HCl ; H2O (reflux)

+ ortho(separate)


26/79


(d)

OH Br CN OOH

ONEt2

a b c d

(a) PBr3

(b) NaCN , DMF (c) 6N HCl , heat (d) 1. SOCl2

2. Et2

NH

(e)

NO2 NO2

Br

NH2

Br

OH

Br

a b c d

(a) HNO3 , H2SO4 (b) Br2 , AlCl3 (c) H2 , Ni (d) NaNO2 , HCl ; H2O (reflux)

(f)

O

OH

CN

OH

CN

a b

(a) 1. NaCN , THF 2. aq acid (b) 1. PhCO2Na , PPh3 , DEAD 2. saponify

(g)

OH OEta b

(a) aq H2SO4 (b) 1. NaH , THF 2. EtI

(h)OH Et

a(a) 1. NaH 2. CS2 3. MeI 4. 200C

(i)

OHBr CN

CO2H

O

NMe2

a

b c

d e

(a) POCl3 , pyridine (b) HBr , h (c) NaCN , DMF (d) H3O+ , heat (e) i. SOCl2 i.. HNMe2


27/79

Chapter 4 27

CHAPTER 3

1. The product is the aldehyde, and the mechanism is analogous to the DMSO-based oxidations discussed in

Section 3.2.C. A reasonable mechanism is shown. Pyridine N-oxide attacks the bromomethyl moiety via an SN2

mechanism. Upon heating, pyridine N-oxide (or eventually the pyridine by-product) removes the hydrogen, as

shown, with displacement of pyridine (the leaving group) to generate the aldehyde. This is related to DMSO

oxidations of alcohols in that a leaving group is attached to the oxygen in A, making the -hydrogen susceptible to

removal by a base. SeeJ. Org. Chem., 1999, 64, 3778.

R

Br

R = C5H11OTHPR

O NH

R

OH

base

NO N

Br

2. These reagents are used for the Sharpless asymmetric epoxidation. Using the Sharpless model shown,

()-DET will deliver the epoxide oxygen from the front of the (R)-enantiomer of the racemic alcohol to give the

epoxide shown. Since the (S)-enantiomer is mismatched for this chiral additive, it will react much slower so it is

possible to convert the (R)-enantiomer to the epoxide while the (S)-enantiomer does not react. Therefore, the

authors in the cited paper isolated the unreacted enantiopure alcohol for use in their synthesis. This process is

calledkinetic resolution.

OMOM

OBnOH

MOMO

OBnOH

A

MOMO

OBn

OH

H

MOMO

OBnOHO

+

Ti(OiPr)4 , D-()-DETt-BuOOH , MS 4

20C , 4 d

see Synthesis, 1993, 615

"O"D-()-DET

via

3. (a) This reaction is taken fromJ. Am. Chem. Soc., 2002, 124, 9199. The epoxidation must take place from the

top face, as the molecule is drawn, to give the proper stereochemistry of the alcohol unit. The alcohol is formed by

removal of the ketone a-hydrogen with the base (DBU - sec. 2.9.A), formation of the C=C unit and opening the

epoxide ring. The stereochemistry of epoxidation is discerned from the model (C=C alkene carbons A and B are

marked. It is not completely obvious from the model that the top face is less hindered because of the methyl group,


28/79


but the fused five-membered rings are somewhat puckered, and this blocks approach of the bulky meta-

chloroperoxybenzoic acid from the bottom. remember that the transition state for this epoxidation is rather bulky

(sec. 3.4.C).

A

O

OH

H

H

B

O

OH

H

H

OH PhH

O

OH

H

H

OH

A

B

1. mcpba , CH2Cl2

2. DBU

(b) This is a Baeyer-Villiger rearrangement, and the carbon best able to bear a positive charge is the one that

migrates. The tertiary bridgehead carbon therefore migrates in preference to the primary carbon.

(c) Oxidation of phenol with Fremy's salt shows a preference for the para quinone. The reason is formation of

the intermediate Ar-ON(SO3K)2. This rather bulky substituent shows less steric hindrance with the oxygen in the

para position than it does in the ortho position. Relief of steric hindrance therefore drives this reaction to give the

para intermediate and, thereby, thepara quinone.

(d) In general, alkenes bearing electron withdrawing groups react slower than simple alkenes. There is also a

steric effect that may lay a role, since dihydroxylation usually occurs at the less sterically hindered site. SeeJ. Am.

Chem. Soc., 1999, 121, 7582

4. (a) In this reaction, the active reagent is the hydroperoxide anion (HOO). Conjugate addition to the ,-

unsaturated carbonyl occurs from the face of the molecule opposite the methyl groups in order to minimize steric

hindrance. The resulting enolate anion attacks the electrophilic oxygen to generate an epoxide, with loss of

hydroxyl. Steric hindrance with the methyl groups dictates delivery of HOO from the bottom face of the

molecule, and the reaction proceeds with high diastereoselectivity for the product shown.


29/79

Chapter 4 29

Me

MeO

O

H

H

OHOOH

too hindered

(b) The reagents will induce cis hydroxylation of the alkene. As drawn, the reagent will be delivered from the

less sterically hindered exo face to give the major product. The primary source of this steric hindrance is the

hydrogen bridging ether unit on the bottom side of the ring, which interacts with any reagent approaching from that

face. In a simple bicyclo[2.2.1]heptene, about 20-30% delivery of regent from the endoface is common, but here

the bridging ether effectively prevents this.

O

Br cat OsO4 , NMO , aq THF

10C RT

see Synthesis, 1996, 219 O

Br

OHOH

H H

(c) The major product described in J. Am. Chem. Soc., 2002, 124, 9726 is the diol shown. There may be a

neighboring group effect involving the allylic alcohol unit to direct the dihydroxylation via path 1. Inspection of

the model suggests that the top face is less hindered, and that approach to carbons A/B (path 1) may be somewhat

less hindered than approach to carbons C/D (path 2). It is likely that the regioselectivity arises from a combination

path 1 being less hindered and the neighboring group assistance provided by the allylic OH.

O

OH

O

OHHO

HO

OsO4

69%

A

C

B

D

A

B

C

D

12

(d) In this reaction, the presence of the hydroxyl group might be expected to provide a neighboring group

effect, placing the epoxy-oxygen syn to the OH. A quick look at the 3D model, however, shows that the

conformation of the 8-membered ring places the OH more or less at right angles to the -bond so one face is not


30/79


favored over the other via coordination. This reaction is dominated by a steric effect, and the top face (A) is less

hindered, leading to the stereochemistry shown.

N

O

OH

O

NO

OH

O

O

MCPBA , CH2Cl

2

74%

seeJ. Org. Chem., 2000, 65, 9129

A

1

2

1

2

(e) In the first reaction, the mild Dess-Martin procedure converts the allylic alcohol unit to a conjugated ketone. In

the second step, the AD-mix-

delivers dihydroxylation from the top face to give the diol shown, with highdiastereoselectivity and enantioselectivity. Using the Sharpless model, AD-mix- should deliver the hydroxyls

from the bottom but in that model, bottom is relative to the methyl groups at the allylic position. Therefore,

delivery opposite the methyl groups leads to the stereochemistry shown. This sequence is take from Lee's synthesis

of amphidinolide B1 (see reference).

OSiiPr3

OSiMe2t-BuOPMB

OSiMe2t-Bu

OH

OSiiPr3

OSiMe2t-BuOPMB

OSiMe2t-Bu

O

OH

HO

OSiiPr3

OSiMe2t-BuOPMB

OSiMe2t-Bu

O

?a

?b

a

b

(a) Dess-Martin periodinane , pyridine, CH2Cl2(b) AD-mix- , MeSO2NH2 , aq t-BuOH


5. These three reactions involve Sharpless asymmetric epoxidation. The model in Figure 3.2 is used to predict

delivery of the reagent from the re or si face.

OHt-BuOOH , ()-DET

Ti(Oi-Pr)4 , CH2Cl2

OHO

seeJ. Org. Chem., 2000, 65, 1738


31/79

Chapter 4 31

(a) When oriented according to Figure 3.2, () tartrate delivers O from the bottom face to give the epoxide with

the stereochemistry shown.

(b) Using the same model from Figure 3.2, the allylic alcohol is aligned as shown, and ()-tartrate should

approach from the back for best selectivity. This would lead to the stereochemistry shown with the epoxy unit to

the rear and the methyl projected to the front. Notice that the allylic acetate unit was not epoxidized under these

conditions, only the allylic alcohol unit.

OAc

OH

OAc

OH

O

OAcOH

()-tartrate

t-BuOOH , ( )-DET

Ti(OiPr)4 , CH2Cl2

see Tetrahedron Lett.,2000

, 41, 2181

(c) Using the model from Figure 3.2, the orientation of the allylic alcohol using ()-DIPT delivers the oxygen

from the bottom, as shown. The smaller ethyl group is on that face, and the epoxide shown is generated with good

stereoselectivity.

OH

()-tartrate

OHO H

H

6. The major products of each reaction are shown in the following sequence.


32/79


(a)

O

O

OH

OMeOSiMe2t-Bu

O

OOMe

OSiMe2t-Bu

O

Ph

O

O

OOMe

OSiMe2t-Bu

O

Ph

OH

OH

OOMe

OSiMe2t-Bu

O

Ph

?a ?b

?c

ab

c(a) benzoyl chloride (b) MeOH, H+ (c) NaIO4

J. Am. Chem. Soc., 1999, 121, 5589

(b)

OH OMe

O

OMeH

O

OMeOH

?a ?b ?c

a b c

(a) BuLi , ether-DMSO ; MeI (b) O3 ; PPh3 (c) PDC , DMF

J. Org. Chem., 2000, 65, 3738

7. This sequence is taken from J. Am. Chem. Soc., 2002, 124, 9060. Swern oxidation (3.2.C.i) gives the ketone,

which eliminates the tosyl group in the presence of triethylamine (via removal of the acidic -hydrogen with

concomitant loss f the tosyl) to give the conjugated ketone. An internal conjugate addition of the pyrrole unit (also

see 9.7.A) leads to the observed product.


33/79

Chapter 4 33

NH

OH

TolO2S N

OHN

O

N

MeO2N

NO2NEt3

NH

O

N

O

HN

O

N

MeO2N

NO2

H+

NH

O

TolO2S N

O

HN

O

N

MeO2N

NO2

H

NH

OO

N

MeO2N

N N

O

NO2

DMSO , (COCl)2

67%

Ts

8. The initial reaction is the expected oxidation of the benzylic alcohol to the aldehyde. This is susceptible to

attack by the pendant OH unit, to form a protonated hemiacetal, and loss of the proton gives the hemi-acetal. If the

OH unit is oxidized further with MnO2 that is still present, the observed lactone is obtained.

MnO2

OH

OH

OH

O

H

O

OH

H

OH

CHO

H+

O

OH

MnO2

O

O

seeHeterocycles, 1996, 42, 589

+

MnO2 , CH2Cl2

2% 98%


34/79


9.

(a)

HO

OPMB

OSiMe2t-Bu

O

H

O

HH

A B

OPMB

OSiMe2t-Bu

O

H

O

HH

OH

(b)

N

MeO

NAc

OMe H

Org. Lett. 2002, 4, 443

(c)

O O

(i-Pr)3SiO

J. Org. Chem., 2003,68, 4215

(d)

O

OH

H

O

HO

J. Org. Chem., 2002, 67, 2566

(e)

O

OMeMeO

O

Org. Lett., 2002, 4, 19

(f)

O

J. Org. Chem.,2003, 68, 1030

(g)

H OSiMe2t-Bu

OPMB

PMB -p-methoxybenzoylJ. Am. Chem. Soc., 2002, 124, 5654

O

(h)

NH

NO

H

SiMe3

O

NHCO2t-Bu

seeJ. Am. Chem. Soc.,1999, 121, 9574

(i)

CHOCHO

J. Org. Chem., 2003,68, 1242

(j)

Cl

seeJ. Chem. Soc., Perkin Trans 1,

1993, 1095

OH

OH

(k)

O

O

CHO

Me

Me Me

H

HO

seeJ. Am. Chem. Soc.,1979, 101, 4400

(l)

OHC

O

O

OHO

J. Org. Chem., 2003, 68, 7428

(m)

HO

O H

BrO2N

Tetrahedron,2003,

59, 9239


35/79

Chapter 4 35

(n)

NO

PMBO

O

H

OCH2PhPMB =p-methoxybenzyl J. Org. Chem., 2003, 68, 7818

OH

OH

(o)

OHO

O

seeJ. Org. Chem., 2000, 65, 9129

O

(p)

O

O

Me

Me

Me

OBn

Et

O

O

seeJ. Am. Chem. Soc.,

1987, 109, 5878

(q)

O

O

OSiMe2t-Bu

CHO

seeJ. Org. Chem., 2000, 65, 3432

(r)

O

C12

H25

OAc

O

J. Org. Chem., 2003, 68, 7548

(s)

see p 137(Cope elimination)

(t)

O

Me

HO

Me

Me


1999, 121, 5087

(u)

O

O HO

AcO

t-BuMe2SiO

O

J. Am. Chem. Soc., 2004, 126, 2194

(v)

seeJ. Org. Chem.,2002, 67, 7774

N

CO2t-Bu

HO

O2C(4-NO2-C6H4)

HO OH

(w)

MeO OAc

H O

Org. Lett. 2003, 5, 3931

(x)

CO2Me

OAc

Me

Me

t-BuPh2SiO

OHC

Org. Lett. 2002, 4, 1543

(y)

N O

Si(i-Pr)3

OHCN

HN

O

CO2t-Bu

Angew. Chem. Int. Ed.,2003, 42, 694

(z)

O

see Chem. Commun., 2000, 837

O

(aa)

N

CHO

CO2t-Bu

Me


(ab)

OMe

OH

OMe

Org. Lett. 2002, 4, 909


36/79


(ac)

OSiMe2t-Bu

CHO

t-BuMe2SiO

J. Am. Chem. Soc.,2002

, 124, 11102 (ad)

Me CO2Et

Me

O

O

see Org. Lett.,2000

, 2, 3177 (ae)

NMe

OMeMeO2C

OHHO

see Org. Lett.,2000

, 2, 3039

10. In each case one example of a suitable synthesis is shown. In many, perhaps most, cases there are other

synthetic approaches that are reasonable.

(a) The shortest approach is to use the appropriate Grignard reagent with the aldehyde derived from oxidative

cleavage of a diol, derived from hydrolysis of the starting epoxide. The Grignard reaction is discussed in Section

8.4.C.

OOH

OH

O

Ph

CHO OH

Ph

a c

(a) aq. H+ (b) OsO4 , NaIO4 (c) PhCH2CH2MgBr ; H2O (d) PCCd

b

(b) See the actual synthesis in Chem. Lett., 1979, 1245. This pertinent reactions are outlined below.

Me Me

Me

H H

CHO

Me MeH H

MeO2C

Me Me

H H

HO2C

O

Me Me

H H

MeO2C

O

Me Me

H H

MeO2C

HO

Me Me

H H

MeO2C

a b c d

e(a) O3 ; H2O2 (b) SOCl2 ; MeOH (c) NaBH4 ; H3O

+(d) POCl3, pyridine (e) O3 ; Me2S

(c) It is very possible that the hydroxy acid will spontaneously cyclize to the lactone. The acid catalysis in step dis

added as a formalism since six-membered ring lactones are somewhat harder to form than five-membered ring

lactones, which spontaneously form from hydroxy acids in virtually all cases. Step c is a reduction and the

functional group reaction wheel in Chapter 1 (Figure 1.1) provides several possible reagents, including sodium


37/79

Chapter 4 37

borohydride, which will be discussed in Section 4..4.A.

Me

Br

Me

HO2C

O

Me

HO2C

OH

Me

O

Me

O

a b cd

(a) KOH , EtOH (b) O3 ; H2O2 (c) NaBH4 ; H3O+ (d) H+

(d) This reaction uses a Baeyer-Villiger rearrangement (Sec. 3.6.A) to set the oxygen on the cyclohexane ring.

Eventual oxidation leads to the ketone that can be converted to its dioxolane ketal.

OO

O

Oa b d

(a) MCPBA (b) i. aq KOH ii. aq H+ (c) PCC (d) 1,2-ethanediol, cat H+

O

OOHc

(e) The conversion of the alcohol to the alkene involves a Chugaev elimination (see Sec. 2.9.C.iv). Other syn

elimination methods could be used if the alcohol were converted to another functional group.

Ph Ph

O

Ph

OHPh Ph Oa b c d

(a) O3 ; Me2S (b) NaBH4 ; H3O+ (c) i. CS2 ii. MeI iii. 200C (d) MCPBA

(f) An E2 reaction gives the alkene, allowing a selenium dioxide oxidation to the allylic alcohol. Oxidation to theacid with PDC in DMF is followed by conversion to the acid chloride and quenching with ammonia to give the

amide.

BrOH

OH

O

NH2

Oa b

c d

(a) KOH , EtOH (b) SeO2 (c) PDC , DMF (d) i. oxalyl chloride ii. NH3

(g) An E2 reaction gives cyclohexene and epoxidation followed by an acid-catalyzed ring opening in the presence

of methanol gives 2-methoxy cyclohexanol. Oxidation gives the ketone and Swern oxidation was used here,

although most of the milder conditions in this chapter would suffice.


38/79


Br

O

OH

OMe

O

OMea

b c d

(a) KOH , EtOH (b) MCPBA (c) MeOH , cat H+ (d) Swern oxidation

(h)

O

CN

OH

CN

OMe

CO2H

OMe(a)

(a) MCPBA (b) NaCN , DMF ; hydrolysis (c) i. NaH ii. MeI (d) i. aq NaOH ii. H3O+

(b) (c) (d)

(i)

OHCO2H

NMe2

Oa b c

(a) POCl3 , pyridine (b) O3 ; H2O2 (c) i. SOCl2 ii. HNMe2

(j) Oxidation of the secondary alcohol in the presence of the tertiary alcohol requires a mild oxidizing agent.

Several reagents are available, including tetrapropylperruthenate and the Dess-Martin reagent shown.

HO

OHO

OH

(a) OsO4 , NMO (b) Dess-Martin periodinanea b

(k) Elimination of the alcohol with POCl3 (Sec. 2.8.A) and pyridine gives the alkene, and ozonolysis leads to the

methyl ketone. The final step is a Baeyer-Villiger rearrangement.

OH

O

MeOAc

b c

(a) POCl3 , pyridine (b) O3 , Me2S (c) MCPBA

a

(l)

O OAc OH O OH

N3a b c e

(a) MCPBA (b) i. aq KOH ii. aq H+

(c) POCl3 , pyridine (d) MCPBA (e) NaN3 , THF

d

(m) This diol to ketone rearrangement is the pinacol rearrangement (see Sec. 12.3.A).


39/79

Chapter 4 39

OH

HO Oa b

(a) OsO4 ; NMO (b) H+


40/79


CHAPTER 4

1. In each case, the Cram model is shown first and then the Felkin-Ahn model, both in Newman projection. The

diastereomer that is predicted to be the major product is also shown.

(a)

OHO H

n-C 3H7Hn-C3H7n-C 3H7

H

H H

CRAM

n- C3

H7

FELKIN-AHN

H

H O

HHOO

O

(b)

Naphth

M e

OHNaphth

Me

OH

M eH

Naphth

NaphthM eH

M e

H H

Naphth

CRAM

MeNaphth

FELKIN-AHN

Me

H

HO

M e

HHO

O

Me

O

M e

(c) In this case, reduction does not generate a chiral center, so the model used is irrelevant. Nonetheless, the

models are shown.

H

O H

M e O

P hH

H

C H2OM e

H

H

HO

P h

H

C H2OM e

O

H

Ph

H

O H

C H2OM e

PhH

HHO

HPh

MeOH2C

O

H

H

CRAM FELKIN-AHN

(d)


41/79

Chapter 4 41

H

Et

O

Me

H

N

O

N

O

CRAM

O

Me

Et

Et

H

Me

H

OH

H Me

HHO

N

O

N

O

Et

N

O

Me

OH

H

N

O

Me

OH

H

FELKIN-AHN

2. (a) The conformation of this molecule is shown in the usual half-chair of a cyclohexenone derivative. Since

cerium borohydride will give selective 1,2-reduction, the main issue is stereochemistry. The bulky siloxymethyl

group on the bottom blocks approach from that face. Therefore, approach is from the top to give the

stereochemistry shown for the alcohol product.

N

O

EtO

Ts OSiPh2t-Bu

NaBH4CeCl3

N

OH

EtO

Ts OSiPh2t-BuB

N

O

H

OSiPh2t-Bu

Ts

H

EtO

A


via

(b) To predict the stereochemistry of this reduction, we can examine a 3D model of the ketone. The gem-

dimethyl unit as well as the other bridgehead methyl sterically block path A, but path B is relatively open and

predicts the major product. Alternatively, a LUMO map of the ketone shows a more intense blue color exposed to

face B, so delivery of hydride will be from that face.

Me

Me

H

MeMe

OBnO

B

A

Me

Me

H

MeMe

OBnOH

LiAlH4 , THF

0C

seeJ. Org. Chem., 2000, 65, 7865

HH


42/79


(c) Coordination of the alcohol moiety with zinc borohydride modifies the conformation, as shown, to deliver

hydride from behind. The result is the diastereomeric trans diol shown.

Me

H

OH

H

O

Me

H

O

H

OZn

H4B BH4

Me

H

OH

OH

(d) The ethyl group effectively blocks one face of the azabicyclooctane ring. Delivery of hydride from the

direction of the arrow leads to the diastereomeric alcohol shown.

N

Et

O

Naphth

H N

Et

Naphth

HO H

H

3.

Me

CO2H

O Me O ONaBH4 , CeCl3

see J. Org. Chem., 1998, 63, 1259

Cerium borohydride gives selective 1,2-reduction of the conjugated ketone, delivering hydride from the bottom

face to give the alcohol. This alcohol unit then reacts with the free carboxyl to form the lactone. Examination of

the 3D figure clearly suggests that attack from the top face (A) is blocked by both the methyl group and the -

CH2COOH group. The bottom face (B) is unencumbered, leading to the stereochemistry indicated for reduction of

the ketone to the alcohol and shown in the lactone. The LUMO map shows that attack from the bottom face, to

give the alcohol precursor required to give the lactone product, is slightly preferred.


43/79

Chapter 4 43

A

B

topbottom

4. The first step is to look at the actual conformation of the steroid. Using the usual model for conjugated ketones

with R being the steroid ring, predictions can be made for reduction of the carbonyl. Diisobutylaluminum hydride

will coordinate to the carbonyl oxygen, restricting the angle from which hydride will be delivered. This

coordination leads to delivery of hydride via a complex that gives anti-Cram selectivity and generates the

diastereomer labeled. Selectride, however, does not coordinate, and delivery will be from the less sterically

hindered pathway (Cram selectivity), as shown, to give the -diastereomer.

see Tetrahedron Lett., 1985, 26, 69.

H

Me

Me

THPO

H

H

H

HO

Me

Me H

Me

R

MeH

O

Al

H i-Bu

i-Bu

Selectride

H

MeOH

R

MeH

H

H

MeH

R

MeH

HO

H

Me

R

MeH

O

Al

HiBu

iBu

Selectride

5. The first step of this reaction is reduction of the ketone unit to give an alcohol. Treatment with the basic reagent

tert-butoxide leads to an alkoxide product, as shown. With the tosylate unit properly positioned, a Grob-like

fragmentation is possible, leads to the aldehyde and the alkene units in the final product.


44/79


N

O

CO2t-BuPh

H H

CO2Et

OTs

LiBEt3H

N

OH

CO2t-Bu

Ph

H H

CO2Et

TsO

KOt-Bu

N

O

CO2t-BuPh

H H

CO2Et

TsO

N

CO2t-Bu

Ph

H H

CO2Et

CHO

see J. Org. Chem., 1999, 64, 4304

6. The reaction products shown are taken from the cited reference. The first step is epoxidation of the C=C unit

with meta-chloroperoxybenzoic acid. The Spartan model shown can be interpreted in several way: the methyl

group on the adjacent allylic carbon can provide steric hindrance, or the bridgehead methyl on the lower face could

block the reaction. In fact, the lower face is more hindered, and epoxidation occurs from the top face, in 81% yield.

Lithium borohydride opens the epoxide, this time from the lower face, and regioselectively at the less sterically

hindered carbon (see model) to give the alcohol shown.

O

O

OSiMe2Thex

H

A

Na2CO3

B

mCPBA

LiBEt3H , THF

O

O

OSiMe2Thex

H

O

O

O

OSiMe2Thex

H

OH

seeJ. Org. Chem., 2003, 68, 3831

7. Inspection of the 3D model shows that path A is blocked by the methyl group, so reduction with LiAlH4 occurs

by attack via path B, which gives the stereochemistry shown for the allylic alcohol. Both 1,2- and 1,4-addition are

possible, so LiAlH4 is used in ether at low temperature to maximize 1,2-addition. The second reaction is simply a

Mitsunobu reaction with the alcohol, first forming the benzoate ester with inversion of configuration, and the


45/79

Chapter 4 45

treatment with methanolic KOH to give the requisite alcohol.


O

O H

LiAlH4 , ether20C

PPh3 , DEAD , PhCOOHthen KOH/MeOH

OH

A

B

8. (a) Aluminum hydride coordinates effectively to the oxygen of the carbonyl. 1,4-Reduction demands

delivery of hydride to carbon via path B, whereas 1,2-reduction demands delivery via path A. Once bound to

oxygen, the distance between H and the terminal C of the C=C unit via path B is rather long, making delivery

difficult. Delivery via path A is facile due to the relatively short distance between the carbonyl C and H.

Coordination to oxygen therefore inhibits 1,4-addition and promotes 1,2-addition.

H

H

H

H

O

Al

H H

H

A

B

(b) L-Selectride is much more bulky than sodium borohydride. On approach to the carbonyl carbon, this steric

bulk will maximize steric interactions with the indane ring system and lead to greater selectivity relative to NaBH4.

Using the Cram model, the diastereomer shown is predicted to be the major product.

H

H

O

O

H

HH

H

OHH

HH

H

(c) In this reaction, sodium transfers an electron to benzene to generate radical anion A. The proximity of the

single electron and the negative charge (2 electrons localized on that carbon) destabilize A due to electrostatic


46/79


repulsion. The resonance form (B) diminishes electrostatic interactions and is energetically favored. Transfer of

hydrogens to B leads to the final product, 1,4-cyclohexadiene.

A B

(d) Transfer of an electron from sodium to anisole can generate two different regioisomeric radical anions, A

and B. The proximity of the negative charge to the electron donating OMe group destabilizes A, making B

energetically more favorable. This leads to the product shown. Similar reaction with benzoic acid leads to C andD

as the possible radical anions. In this case, the electron withdrawing carboxylate group stabilizes the adjacent

negative change in C, making it more stable than D where the charge is distal to the carboxylate group. For this

reason, C leads to the major product shown.

OMe OMe OMe

CO2

H CO2

CO

2

A B

C D

OMe

CO2H

(e) Lithium aluminum hydride is a powerful reducing agent due to the charge distribution and polarity of the

AlH moiety. It will reduce ester groups, acid groups, and carboxylate anion groups. Since LiAlH4 will reduce

both the ester and the acid, the product is a diol. Borane coordinates with the acid but not with the ester. It will

therefore transfer hydride only to the acid and not to the ester, and the acid is reduced and the ester is not. The

product is a hydroxy-ester.

(f) This reduction occurs via a six-centered transition state, as shown.

R

R

HN

N

H

When the alkene is symmetrical and not polarized, electron density is readily transferred to hydrogen, leading

eventually to expulsion of N2, which drives the reaction. When the alkene is conjugated to a carbonyl, the electron

donating ability of the alkene is diminished, slowing the reaction by destabilizing the requisite six-centered

transition state. SeeJ. Am. Chem. Soc., 1961, 83, 4302.


47/79

Chapter 4 47

(g) In A there is a neighboring group effect where the OH group coordinates with zinc borohydride and delivers

hydride from the same face as the hydroxyl group. When the OH is blocked as the silyl derivative, zinc

borohydride cannot coordinate, and the usual steric effects lead to delivery of hydride from the face opposite the

OSiR3 group.

(h) This transformation is taken fromJ. Am. Chem. Soc., 2002, 124, 12416. Birch reduction of the naphthalene

unit containing the electron releasing OMe groups is expected to give the reduced product shown. Aqueous

hydrolysis converts the vinyl ether to the ketone, but the C=C unit in the ketone-bearing ring will move into the

other ring to form the aromatic ring, rather than into conjugated with the carbonyl. aromatization is the driving

force leading to the major product.

OMe

MeO

OMe

A

OMe

MeO

O

B

OMe

OMe

MeO

O

OMe

MeO

Na , EtOH , reflux

aq HClreflux

9. The first transformation can be effected with Sn/HCl (Stephen reduction) or with LiAlH(Ot-Bu)3. The second

transformation can be effected with LiAlH4 or with LiBHEt3.

10.

(a)

H

OH

OO

Org. Lett., 2003, 5, 4741

(b)

CHO

(c)

O

O

(CH 2)4OPiv

Me

HO

OPMB

OPMB

OPMBMe

MeOOTIPS

PMB =p-methoxybenzylPiv = pivaloyl

J. Am. Chem. Soc.,2003, 125, 12844

(d)

MeO

MeO CO2H

NH2

J. Org. Chem., 2002. 67, 8284

(e)

N

CO2t-Bu

CO2MeMeO2C

Org. Lett., 2003,5, 999

(f)

Br

OMe

MeO

MeONHPMB

PMB -p-methoxybenzyl

Eur. J. Org. Chem.,2003, 1231


48/79


(g)

O

t-BuMe2SiO

O

OH

Org. Lett., 2002, 4, 1451

(h)

N Ph

HN

CO2t-Bu

MeO

Org. Lett. 2003, 5, 1927

(i)

N

N

HO

H

MeBr

J. Org. Chem.,2004, 69, 1283

(j)

NH

NHCO2t-Bu

CO2Me


125, 5628

(k)

t-BuMe2SiO

Me

OH

Me

seeJ. Org. Chem., 1999, 64, 4477

(l)

NHO

CO2t-Bu

CO2Me

H

H

Org. Lett., 2003,5, 447

(m)

NH

N H

J. Org. Chem. 2003,68, 4523

(n)

O

O

PhMe2SiCH2OH

OH

J. Org. Chem., 200368, 2572

(o)

N

H

Me

CH3

OH

seeJ. Org. Chem., 1999, 64, 2184

(p)

O

C15H31 CN

see Tetrahedron Lett.,

2000, 41, 3467

(q)

N

O

H

OSiMe2t-Bu

Org. Lett. 2002, 4, 177

(r)

CHO

OSiPh2t-Bu

J. Am. Chem. Soc., 2002, 124, 4257

(s)

OO

OHO

HO

Org. Lett.2003

, 5, 3357

(t)

O

H

I

OH

O

O


1999, 121, 5176

(u)

OO

Me

Me

OH

seeJ. Chem. Soc., Perkin Trans. 1

2000, 2645


49/79

Chapter 4 49

(v)

OH OH

Ph

Me3Si


126, 2194 (w)

OMe

MeO

NO2

OMe

Br

J. Org. Chem. 2003, 68, 8162 (x)

NO

OHC

CHO

J. Org. Chem.,2004, 69, 1598

(y)

see Can. J. Chem.,

1997, 75, 621

O

(z)

C13H27

OSiMe2t-Bu

NHAc

OH

see Tetrahedron Lett., 2000, 41, 2765 (aa) seeJ. Org. Chem., 2002, 67, 4127

TBSO

O

O

O

(ab)

Ph Ph

see Synlett, 1999, 182 (ac)

N N

CH2OH

CH3

Cl CO2Et

J. Am. Chem. Soc.,2002,124, 9476 (ad)

t-BuMe2SiO

t-BuMe2SiO OH OH

J. Am. Chem. Soc., 2002, 124, 12806

11. In each case a synthesis is shown. These are not necessarily the only approaches. It is very likely there are

many approaches for several of these questions.

(a) The acid hydrolysis of the vinyl ether to the ketone may also convert the methyl ester to the acid. If this

occurs, an esterification step (thionyl chloride; methanol) would be required. Mild acid hydrolysis of the vinyl

ether should be possible, however.

Me

CO2Me

MeO

Me

CO2Me

MeO

Me

CO2Me

O

Me

CO2Me

O

O

O

ab c (a) Na , NH3 , EtOH

(b) aq H+

(c) O3 ; H2O2

(b) The first step requires a chain extension and converting the alcohol to its tosylate introduces the requisite

leaving group, allowing a subsequent reaction with NaCN to give the corresponding nitrile. DIBAL-H reduction of

the nitrile gives the aldehyde.


50/79


HO

OPMB

NC

OPMB

OHC

OPMB


a b

(a) 1. TsCl , pyridine 2. NaCN , DMSO(b) DIBAL-H , THF

(c) All steps in this synthesis are taken from Org. Lett., 2002, 4, 1379. Bromination of the allylic alcohol (2.8)

and SN2 displacement with cyanide gives the nitrile (2.6.A.i). Basic hydrolysis to the carboxylic acid, and

esterification with diazomethane (2.5.C; 13.9.C), was followed by selenium dioxide oxidation to the aldehyde

(3.8.A). Reduction of both the ester and aldehyde with LiAlH4 gave the diol (4.2.B), and selective chlorination of

the allylic alcohol with N-chlorosuccinimide (2.8) gave the final target.

OH Br CN

CO2Me

OHC

CO2Me

OH

OH

CO2H

Cl

OH

a b c d

e f g

(a) PBr3 t-BuOMe (b) KCN , DMSO (c) 25% KOH , MeOH (d) CH2N2 , ether

(e) SeO2 , t-BuOOH , CH2Cl2 (f) LiAlH4 , THF (g) NCS , Me2S , CHCl2

(d) The Wittig olefination step is discussed in chapter 8.

OH

OMe

O

OMe

OH

OMe OMe

OHC

OMe

Oa b c d

e f (a) NaH ; MeI (b) Na , NH3 , EtOH (c) H2 , Pd(d) O3 ; Me2S (e) Ph3P=CH2 (f) LiAlH4

(e) All reagents are taken from the cited reference. The first reaction reduces the ester unit to an alcohol. This

is followed by conversion to a tosylate that allows Super-Hydride reduction to give the methyl group. The O-SiR3


51/79

Chapter 4 51

unit is cleaved with aqueous acid to give the corresponding alcohol (see chap. 7, sec. 7.3.A.i). The primary alcohol

is then converted to the aldehyde by a Swern oxidation (see chap. 3, sec. 3.2.C.i).

CO2Me

OSiMe2t-Bu

CH2OTS

OSiMe2t-Bu

Me

O

H

CH2OH

OSiMe2t-Bu

CH3

OSiMe2t-BuCH3

OH


ab

c d

(a) DIBAL-H , toluene , 78C (b) TsCl , DMAP , NEt3 (c) LiEt3BH , THF(d) AcOH , aq THF (e) DMSO , (COCl)2 , NEt3

e

(f) All reagents are taken from the cited reference. Initial reduction of the acid to the alcohol was followed by

treatment with tosic acid. This led to an internal transesterification to the lactone, and conversion of the dioxolone

to the alcohol. Treatment with the dimethyl acetal of formaldehyde led to the ether shown (a MOM group - see.

chap. 7, sec. 7.3.A.i), and DIBAL-H reduction converts the lactone to a lactol.

CO2H

O

O

O

O

O

O

OH

OO

HO

OO

OO

OHO

OO

seeJ. Org. Chem, 2000, 65, 9129

a

b c d

(a) BH3 , THF ; H3O+ (b)p-TsOH , CHCl3 (c) CH2(OMe)2 , P2O5 (d) DIBAL-H , CH2Cl2

(g) All reagents are taken from the cited reference. Initial reduction of the ester unit (using DIBAL-H) gives the

allylic alcohol. To insert the proper stereochemistry for the new alcohol unit, Sharpless asymmetric epoxidation

using (-)-DIPT (see Sec. 3.4.D.i) gives the epoxide and selective reduction of the less sterically hindered carbon of

the epoxide with Red-Al gives the final product.


52/79


CO2Et

OO SiMe

3

O

O SiMe3

OH

OH

OO SiMe3

OH

OO SiMe3

OH

O


a b

c(a) DIBAL-H , CH2Cl2 , 78C

(b) ()-DIPT , t-BuOOH , Ti(OiPr)4 (c) Red-Al , THF

(h) A Friedel-Crafts acylation inserts the ketone moiety, and the Wolff-Kishner reduction removes the carbonyl.

Catalytic hydrogenation reduces not only the benzene ring but also the nitro group in a single step.

O

n-C3H7n-C3H7 n-C3H7

NO2

n-C3H7

NH2

a

b

cd

(a) butanoyl chloride , AlCl3 (b) N2H4 , KOH (c) HNO3 , H2SO4 ;separate ortho product(d) excess H2 , Ni(R)

Section12.4.Dp 1090

(i) All reagents were taken fromEur. J. Org. Chem., 2003, 4445. Reduction of the acid with borane (4.6.A) was

followed by bromination with carbon tetrabromide and triphenylphosphine (2.8.A).

NCl OMe

CO2H

NCl OMe

HO

NCl OMe

Br

a b

(a) BH3SMe2 , B(Me)3 , THF (b) CBr4 , PPh3 , benzene

(j) All reagents are taken from the cited reference. The first problem is how to remove the OH group. If the

OH is first converted to a tosylate, Finkelstein exchange (chap. 2) generates an iodide which can be reduced with

tin hydride. The ester is reduced to an alcohol with LiBH4, and aqueous acid converts the dioxolane unit (a ketal)

to the carbonyl (see chap. 7, sec. 7.3.B.i). The final step is an oxidation. Several methods can be used from chap.

3, sec. 3.2, but the Dess-Martin periodinane reagent was used in this citation.


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Chapter 4 53

N

OO

O

OMe

EtO2C

OH

N

OO

O

OMe

EtO2C

OTs

N

OO

O

OMe

EtO2C

N

O

OMe

OHC

O

N

O

OMe

HOH2C

O

N

OO

O

OMe

HOH2C

seeJ. Am. Chem. Soc., 1999,121 , 7778

a b

c

de

(a) TsCl , DMAP (b) 1. NaI , acetone 2. Bu3SnH , AIBN(c) LiBH4 , THF (d) HCl , aq. AcOH (e) Dess-Martin

(k) Hydrogenation with a Lindlar catalyst sets the cis alkene geometry.

H H Bu H Bu BuBu Bua b

(a) NaNH2 ; n-C4H9-I (b) NaNH2 ; n-C4H9-I (c) H2 , PdCO3 , quinoline

c

12. In each case a synthetic solution is shown. There are other approaches based on other disconnections. The

Aldrich catalog (2000-2001) was used as a source for the starting material for convenience. There are, obviously,

other sources of chemicals.

(a) An attractive starting material is the commercially available benzylacetone at $95.40/Kg, but it has more than

the required five carbons. One more disconnection leads to acetone, which can be converted to benzylacetone via

enolate alkylation (see Sec. 9.3.A). Acetone is available from Aldrich ($18.30/L) and has less than five carbons.

The Wittig olefination step(a) is discussed in Section 8.8.A.


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O Me

Me

MePh

Br

Br Me

O Ph

Me

MePh

Me

MePh

Me

O Ph

Me

MePh

Br

Br Me

O Me

Mebenzylacetone

b c

(a) LiN(iPr)2 ; BnBr (b) Ph3P=CHMe (b) Br2 , CCl4

a

acetone

(b) The starting material is 3-methyl-2-butanol at $43.10/100 mL. Initial conversion to a mesylate allows an SN2

reaction with the anion of 1-propyne. The alcohol could also be converted to a bromide using PBr3 or a similar

reagent. Lindlar reduction of the alkyne leads to the cis-alkene.

OMsOH

OH

a b c

(a) MeSO2Cl , NEt3 (b) MeCC

Na+

, DMF (c) H2 , Pd-BaCO3-quinoline (Lindlar)

(c) The requirement that the starting material be five carbons or less makes this a very cumbersome synthesis. The

point of this exercise is first to give practice for various reactions but also to show that setting arbitrary starting

material requirements can have profound consequences. A shorter approach, for example, would use phenetole

(ethoxy benzene) and Birch reduction. In the synthesis shown, 2-ethoxy-propanoic acid was not found to be

commercially available (someone probably sells it if sufficient time was taken to find a source), but the expensive

-propiolactone at $76.20/5 mL can be opened with ethanol (transesterification) to give the requisite alcohol-ester.

Wittig olefination (see Sec. 8.8.A) can be done with a carboxyl substituent.

EtO2C NHBu

O

O

EtO2CCO

2H

O

EtO2

C CHO EtO2COH

O

O

EtO2C

CO2H


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Chapter 4 55

O

O

EtO2COH

EtO2C CHO

EtO2CCO2H

O EtO2C NHBuO

OEtO2C

CO2Ha b c

d e

(a) EtOH , H+ (b) PDC , CH2Cl2 (c) Ph3P=CHCH2CO2 Na+ (d) MCPBA (e) BuNH2 , DCC

(d) This synthesis begins with benzene at $31.00/L and bromination followed by reaction with cuprous cyanide

gives benzonitrile. Rosenmund reduction gives benzaldehyde (remember that we had to begin with material

containing only 6 carbons). A Wittig olefination (see Sec. 8.8.A) gives styrene and a radical HBr addition gives the

primary bromide. An SN2 displacement with cyanide allows selective reduction to the aldehyde.

PhCHO

PhBr PhCHO

PhCHO Ph PhBr

PhCN

PhH PhBr PhCN PhCHO

PhH

PhCHO

d e f g

d) Ph3P=CH2 (e) HBr , ROOR (f) NaCN , DMF (g) LiAlH(Ot-Bu)3

a b c

(a) Br2 , FeBr3 (b) CuCN , heat (c) Sn , HCl

13. (a) NaBH4 (b) Na , NH3 , EtOH (c) LiAlH(Ot-Bu)3 - some diol will also be formed with virtually any

reagent (d) NaBH3CN , pH 7

(e) The first step requires reduction of the ketone to an alcohol, and the mild reagent NaBH 4 was used. The second

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