Chapter 2 1 CHAPTER 2 · Chapter 2 1 CHAPTER 2 1. This solvolysis reaction proceeds with loss of...

Chapter 2 1

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 from J. 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.

OH O 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

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.

Copyright © 2011 Elsevier Inc. All rights reserved.

2 Organic Synthesis Solutions Manual

ortho -Methoxyphenol is more acidic than para-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 OSiMe3

O

Cl

MeOOMe

O

O

Cl NHi-Pr

ONEt2

OMe

OMe

MeO

O

O

N

HPh

Me

Me(CH2)9

ON

O

HN

O

O

O

OO MeO

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

see Tetrahedron Lett., 2000, 41, 1975see J. Am. Chem. Soc., 1994, 116, 9921 see J. Org. Chem., 1999, 64, 2450

see J. Org. Chem., 1999, 64, 3736 see J. 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.

NH2Cl

+

NH2

**

**

H2N

* = 14C X

H2O

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


Chapter 2 3

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

MeEt

Br

Me

Me

H

Et

H

H

Br

MeMe

EtH

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.

HOOH + H+

H2OOH

OHO

H

– H2O

– H+

– H2O

O

SN2'

vinylogous SN1-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

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 from J. Org. Chem., 1987, 52, 5039.

10. This problem was taken from J. 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

ClX

Cl—Cl

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 B

should 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 of Z more difficult and, thereby, formation

of B.

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.


Chapter 2 5

O

O

O

H

HH

O

OO

H

H

OO

OH

RT

HH

OMe

OOH

OH

H

CO2Me

O

H

H

OMe

OOH

O

OOH

O

OH

H

HH

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

Hhigh energy dueto flattening thetricyclic ring system-less favorable1

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

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

OO

HO

NO

O

OBn

O

OH

HO

NO

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

OO

HO

N

H

OBn

OH

O

EtOH

see J. 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-Pr Me interaction in B destabilizes that conformation relative to the

H H interactions in A. In both A and B, the NR3 H interactions are about the same.


Chapter 2 7

NEt2C4H9

HH

H

H H

NEt3

HMe

H

MeH

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 effect makes 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

C3H9

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 aminopentanoic

acid. 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

O—H bond, enhancing acidity.

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 still 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

MeH

65

43

21

65

43

21

rotation about this bondleads to a mixture ofcis- and trans-alkenes

17. See J. 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

–25°C RT

Br

O

Br

H

Br2 from NBS – H+

H

HH

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

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.


Chapter 2 9

NHCO2Me

Me

t-BuMe2SiO

t-BuMe2SiO Me

Nt-BuMe2SiO

t-BuMe2SiO Me

OO

MeH

H

N CO2Me

Met-BuMe2SiO

t-BuMe2SiO Me

I

HN Me

t-BuMe2SiO

t-BuMe2SiO Me

H

O

OH3C

NHCO2Me

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

see Eur. J. Org. Chem., 2000, 1889

(d)

N

NH

N

N

PhH2CO

Me

H

H

O

OCH2Ph

OCH2OMe

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)

C12H25 OCH2OCH2CH2OMe

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

OO

Me

OMe HO

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

(p)

MeO

NMe

HEt

J. Chem. Soc., Perkin Trans. 1, 2001, 2398 (q)

ON3AcO

AcO OAc

HOOMe

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

O

Me3SiO

O

Org. Lett., 2004 6, 2961

(s)

N

H

HN

H

O

Me2N

HN

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

OH

O

CO2Me

I

HO

H

MeO2C

OH

H

I

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

Br

HO

Tetrahedron Lett., 2000, 41, 2573


Chapter 2 11

(v)

OMe

MeO

Me

Me

Me

OH

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

Me

see J. 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

J. Am. Chem. Soc., 2005,

127, 5596 (aa)

MeO2C CO2Me

HH

HNNH

O OChem. Commun., 2004, 2404

(bb)

Ph CH3

OOHNHS

O

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

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

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

Ph Ph

Me

BrH

H

EtBr

H

Ph

EtMe

H

Br

H

Ph

EtMe

H–OH

–HBr

Ph

HEt

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

PhMeOAc

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

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 and B. 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 mixture

of 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.


Chapter 2 13

PhS

O

S

Ha

Hb

O Ph

HCHb(Me)C5H11H

•

Ha

Me

Ha

PhS

O

CHHaMeH

•

Hb

Me

C5H11A B

HbC

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 C CMeBr Br

a b

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

(b) This cyclohexane derivative exists as an equilibrium mixture of A and B. 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

regioisomeric products, C and D.

BrMeEt

Br

MeEt

Br

Me

Et

HH

A B

MeEt

C

MeEt

D

(c) In this example, the cyclohexane bromide also exists as A and B, and A 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.

BrEtMe

BrEt

MeBr

Et

Me

HH

A B

EtMe

C

25.

(a) CHOBr Br

C CMea b

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



(b)

NO2NO2

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)

(d)

OH Br CN OOH

ONEt2

a b c d

(a) PBr3 (b) NaCN , DMF (c) 6N HCl , heat (d) 1. SOCl2 2. Et2NH

(e)

NO2NO2

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. 200°C


Chapter 2 15

(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


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