79720762 chapter 11 solutions

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JWCL234-11 JWCL234-Solomons-v1 December 11, 2009 17:31

CONFIRMING PAGES

11ALCOHOLS AND ETHERS

SOLUTIONS TO PROBLEMS

Note: A mixture of bond-line and condensed structural formulas is used for solutions in

this chapter so as to aid your facility in using both types.

11.1 These names mix two systems of nomenclature (functional class and substitutive; see

Section 4.3F). The proper names are: isopropyl alcohol (functional class) or 2-propanol

(substitutive), and tert-butyl alcohol (functional class) or 2-methyl-2-propanol (substitu-

tive). Names with mixed systems of nomenclature should not be used.

11.2

1-Propanol

or propan-1-ol

(Propyl alcohol)

1-Methoxypropane

(Methyl propyl ether)

2-Methoxypropane

(Isopropyl methyl ether)

Ethoxyethane

(Diethyl ether)

1-Butanol

or

butan-1-ol

(Butyl alcohol)

2-Methyl-1-propanol

or

2-methylpropan-1-ol

(Isobutyl alcohol)

2-Methyl-2-propanol

or

2-methylpropan-2-ol

(tert-Butyl alcohol)

2-Butanol

or

butan-2-ol

(sec-Butyl alcohol)

2-Propanol

or propan-2-ol

(Isopropyl alcohol)

Methoxyethane

(Ethyl methyl ether)

(a)

(b)

OH

OH

OH

O

OH

O

O O

OH

OH

11.3 (b) OH(a)

OH

(c)

OH

204



CONFIRMING PAGES

ALCOHOLS AND ETHERS 205

11.4 A rearrangement takes place.

(a)+ H H

H

O+

+

1,2-methanide

shift

HO

H

+

2,3-Dimethyl-2-butanol

(major product)

OH2 OH2+

OH

(b) (1) Hg(OAc)2/THF-H2O; (2) NaBH4, OH−

(oxymercuration-demercuration)

11.5

Stronger

acid

NaNH2

Stronger

base

(a) +

Weaker

base

NH3

Weaker

acid

+

Stronger

acid

Stronger

base

(b) +

Weaker

base

Weaker

acid

+

Weaker

acid

Weaker

base

(c) +

Stronger

base

Stronger

acid

+

OH

OH

OH

O− Na+

O− Na+

O− Na+O−

Na+

H Na+− H H

Ο

OH

Ο

Weaker

acid

NaOH

Weaker

base

(d) +

Stronger

base

H2O

Stronger

acid

+OH O− Na+

11.6OH OH

2

Br

H Br

+

+−H2O

+

Br−



CONFIRMING PAGES

206 ALCOHOLS AND ETHERS

11.7 (a) Tertiary alcohols react faster than secondary alcohols because they form more stable

carbocations; that is, 3◦ rather than 2◦:

H

O H

X

+

+

−

H2O

X

+

(b) CH3OH reacts faster than 1◦ alcohols because it offers less hindrance to SN2 attack.

(Recall that CH3OH and 1◦ alcohols must react through an SN2 mechanism.)

11.8PCl5

(−POCl3, −HCl)

PCl5

(−POCl3, −HCl)

H3C SO2OH SO2Cl

CH3OH

base (−HCl)

CH3SO2Cl

OSO2Me

base (−HCl)

base (−HCl)

SO2OCH3

CH3SO2OH

(a)

(b)

(c)

CH3SO2Cl

OSO2Me

H3C

H3C

OH

OH

11.9

OH(a)

HO

X

OTs

Y

OHOTs

TsCl

(pyr.)

retention

(−HCl)+

OTs−+

−

inversion

SN2

(b)

cis-2-Methyl-

cyclohexanol

OTs−+

TsCl

retention

(pyr.)

Cl−

inversionOH

A

OTs

B

Cl



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11.10 Use an alcohol containing labeled oxygen. If all of the label appears in the sulfonate ester,

then one can conclude that the alcohol C O bond does not break during the reaction:

R +O H18

R O SO218

R′ R′SO2Clbase

(−HCl)

11.11

HA −H2O

(1° only)

A−

−HA

HO+

ORR

HO+

H2O

+ OHR

This reaction succeeds because a 3◦ carbocation is much more stable than a 1◦ car-

bocation. Consequently, mixing the 1◦ alcohol and H2SO4 does not lead to formation of

appreciable amounts of a 1◦ carbocation. However, when the 3◦ alcohol is added, it is rapidly

converted to a 3◦ carbocation, which then reacts with the 1◦ alcohol that is present in the

mixture.

11.12 (a)

(1) L−

CH3

+CH3 L+

(L = X, OSO2R, or OSO2OR)

(2) + L−

(L = X, OSO2R, or OSO2OR)

CH3O

O O

CH3OL

−

−

(b) Both methods involve SN2 reactions. Therefore, method (1) is better because substitution

takes place at an unhindered methyl carbon atom. In method (2) where substitution

must take place at a relatively hindered secondary carbon atom, the reaction would be

accompanied by considerable elimination.

+++ L−CH3OH

L

HCH3O −

11.13 (a) HO− H2O Cl–O

++ +HOCl

− OCl



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(b) The O − group must displace the Cl− from the backside,

OH −

trans-2-Chlorocyclohexanol

Cl

HH

O −

Cl

HH

OH

O

SN2

Cl

OH

O

=

(c) Backside attack is not possible with the cis isomer (below); therefore, it does not form

an epoxide.

cis-2-Chlorocyclohexanol

H

OHH

Cl

Cl

OH

=

11.14 K∞

(-H2)

K2CO3

A

C

B

OH

OTs

D

O− K+ O

TsCl

pyr.

Br

OH

O



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11.15 (a)

OHBr

OBr

O

H

BrO S

OH

HO

O

H2OHSO4−

H3O+

+

+

+

+

(b)

OH

H2O

H

O

H

O S

OH

HO

O HSO4

−+

+

+

+

O

11.16 (a)

HI I+ +

+

−

+

Me

MeI

O

Me

O

H

OH

SN2 attack by I− occurs at the methyl carbon atom because it is less hindered; therefore,

the bond between the sec-butyl group and the oxygen is not broken.



CONFIRMING PAGES


(b)

HI I−

I−

+ +

MeOH +

+

OMe

I

OMe

H

+

In this reaction the much more stable tert-butyl cation is produced. It then combines with

I− to form tert-butyl iodide.

11.17 O

Br

Br

H Br

OH

H Br

O

H

Br

O

H

H

+

+

+

+

−

Br−

H2O



CONFIRMING PAGES


11.18

OCH3

H

H Cl

Cl

Cl

Cl−

H

HCl

CH3O

+

CH3OH+

−

+

+

+

11.19 (a) HO

H

HAO O+

HO

Methyl Cellosolve

O+

Me

H

Me

A−

HO

OMe

(b) An analogous reaction yields ethyl cellosolve,HO

O.

(c)

O −O

I

HO

I

I H2OOH−

+

−

(d)

O −O

+ NH3

HO

NH2

NH3

(e)

O −O

OMe

OMe HO

OMe

MeOHCH3O−

+

−

11.20 The reaction is an SN2 reaction, and thus nucleophilic attack takes place much more rapidly

at the primary carbon atom than at the more hindered tertiary carbon atom.

MeOH

fast Major

productMeO

O

+ MeO OH−


JWCL234-11 JWCL234-Solomons-v1 January 9, 2010 13:56

CONFIRMING PAGES


MeOMeOH

slow Minor

productO

+ HO OCH3−

11.21 Ethoxide ion attacks the epoxide ring at the primary carbon because it is less hindered, and

the following reactions take place.

O

ClOEt

Cl+

O

*

OEt

**

OEt

O−

−

11.22+ H3O+

H2O

−H3O+

H

H

H2O

H

H

OHOH+

H2O+

H(plus enantiomer,

by attack at the other

epoxide carbon)H

OH

OH

H

H

O

11.23 A: 2-Butyne

B: H2, Ni2B (P − 2)

C: MCPBA

D: O

E: MeOH, cat. acid

11.24

O OO O

O OO

O

O

12-Crown-415-Crown-5

(a) (b)

Problems

Nomenclature

11.25 (a) 3,3-Dimethyl-1-butanol or 3,3-dimethylbutan-1-ol

(b) 4-Penten-2-ol or pent-4-en-2-ol

(c) 2-Methyl-1,4-butanediol or 2-methylbutan-1,4-diol

(d) 2-Phenylethanol

(e) 1-Methylcyclopentanol

(f) cis-3-Methylcyclohexanol



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11.26

(a)

(c)H

OH

HO

H

(e)

(g)

(i)

OH

OH

OH

Cl

O

O

(b)

(f )

(d)

(h)

( j)

OHHO

HO H

OH

O

O

O

Reactions and Synthesis

11.27 (a)

(b)

(c)

(d)

or

11.28 (a)

(b)

(c)

(d)

11.29

H2O2/OH−

(oxidation)

(b)OH−

(c)

HBr

ROOR

OH−

OH

OH

Br OH

Cl

BH3 : THF

(hydroboration)(a)

ClOK OH

3 B3



CONFIRMING PAGES


H2

Ni2B (P-2)

[as in (a)]

(d)

OH

11.30

(c) See (b) above.

PBr3(a) + H3PO3+3

OH

3

Br

PBr3(b) OH Br

OK

OH

HBr

(no peroxides)

Br

(d)H2

Ni2B (P-2)

HBr

(no peroxides)

Br

11.31

(1) BH3 : THF

(2) H2O2, OHOH

T

BD3 : THF

D

TB

R

D

O

OT

(1) BH3 : THF

(2)

NaOH ONa

OBr

(a)

(b)

(c)

(d)

OR

OT

[from (a)]

11.32

OH

SOCl2+

Cl

SO2+ HCl+(a)

HCl+

Cl

(b)



CONFIRMING PAGES


enantiomer+

Br

(c)HBr

(no peroxides)

(d)(1) BH3 : THF

(e)

(2) H2O2, OH− H

H

OH

Br

(1) BH3 : THF

(2) H2O2, OH−

OK

OH

OH

11.33 CH3Br +(a) (c)

(d)+(b)

Br

Br

BrBr

Br (2 molar equivalents)Br

Br

11.34 A: O− Na+

B: O

C:

OSO2CH3

D:O

CH3

+ CH3SO3− Na+

E:

OSO2

F:I SO3

− Na++

G: O

H: SiO

I: SiFOH +

J: Br

K: Cl

L: Br

11.35 A: O− Na+

B: O

C: OSO2CH3

D: OCH3 CH3SO3

−Na++

E: OSO2

F: ISO3

−Na++



CONFIRMING PAGES


G: O+

H: OSi

I:

OH SiF+

J: Br

K: Cl

L: Br

11.36

Br2

heat, hv(a)

(b) Br

Br

ONa

OH

(g)CH3ONa

CH3OH

(h)

(i) NaCN

( j) CH3SNa

CH3

Br

Br

Br

Br

O

O

CN

SCH3

(f )HBr

(no peroxides)Br

(1) (BH3)2

(2) H2O2, OH−

(e)

or

OH−

H2O

OH

OH

(d)KI

acetone

IBr

Br

(c)HBr

peroxides

Br

O

O

ONa

O

OH



CONFIRMING PAGES


Br2

H2O

Br

HO

OOH−

(k)

OsO4, pyr. NaHSO3

H2O

OH

HO

(l)

NH3

excessBr

HO

NH2

HO

(m)

–OEt

EtOH

O

HO

O(n)

11.37

NaNH2

liq. NH3

A (C9H16)

NaNH2

liq. NH3

B (C9H15Na)

C (C19H36)

D (C19H38)

MCPBAH2

Ni2B (P-2)

H H H Na

Br

H

Na Br4

4

H H

4

H H

4

O Disparlure (C19H38O)

11.38

OH

1. PBr3

2. NaOC(CH3)3

3. H2O, H2SO4 cat.

OH

(a)

1. NaOC(CH3)3

2. CH3C(O)OOH

Br O

(b)

1. NaOC(CH3)3

2. H2O, H2SO4 cat.I

OH(c)



CONFIRMING PAGES


or

H2O, H2SO4 cat.

NaOH, H2OO

OH

OH

(d)

MCPBAO

(e)

Cl Br

OH

1. NaOC(CH3)3

2. Br2, H2O( f)

11.39 (a) OH ClClSOCl2

(b) HBrOH BrBr

(c) OH O−Na+NaNH2

NH3+

(d) PBr3

BrOH

(e)1) TsCl, pyridine

2) NaSCH2CH3OH S

(f) OH INaI

H2SO4

11.40 (a)HI (excess) I

II

I

O

+

(b) HI (excess)I+

O OH

(c)H2SO4, H2O

+

OH

OHO



CONFIRMING PAGES


(d) NaOCH3

HO

OCH3

O

(e) HOCH3, H2SO4

OH

H3COO

(f)

+

1. EtSNa

2. H2OSEt

HO

O

SEtHO

OH

EtS

OHEtS

(g)HCl (1 equiv)O Cl HO+

(h)MeONaO

no rxn

(i)1. EtONa

2. MeIMeO

OEt

O

(j) HIHO

IO

11.41 (a)

1. EtSNa

2. MeIMeO

SEt

O+

SEt

MeO

(b)1. Na

+

3.

O O

2. H2O

I

−

(c)O

HBr (excess)

Br2



CONFIRMING PAGES


(d)+

1. HI

2. NaH

or

H2SO4 (cat)

O O O

11.42OH

Cl

Glycerol Epichlorohydrin

Cl Cl

OH−, l eq

(a)

(b)

Cl2

400°C

Cl2

H2O

OH−

xs

OH

HO OH ClO

11.43

CH3

OH

CH3

OTs

+ enantiomer

CH3

OH

I

+ enantiomer

+ enantiomerA =

C =

B =

=

(a)

CH3

+ enantiomer A and C are diastereomers.D =

H

H

OMs

OMs

CH3E =

H

CCH

H

CH3

HC

C

H

HCH3F =

(b)

H



CONFIRMING PAGES


(c)

O− Na+

G =

OMe

H =

OMs

I =

J =

OMe

H and J are enantiomers.

Mechanisms

11.44

OH

H

A−

HAOH2

+

H

+H

+

HA

+

+

H2O+ 3° Carbocation

is more stable

−H2O

11.45

H

O S

OH

HO

O

O S

O

HO

O

O O

OH OH

−

+

+



CONFIRMING PAGES


11.46+

+

OH OH

O Br

BrBr

Br

O

H

Br

Br−

11.47O

O

+

+

O

O

P

O

OHOHO

H

H

H3PO4 (cat), CH3CH2OH

OH

OH

O

OHOH

HOH

11.48 (a)O

HCl

Cl

OH

enantiomerH O OH

O

(b) The trans product because the Cl− attacks anti to the epoxide and an inversion of

configuration occurs.

HCl

H

H

enantiomer

Cl

Cl

H

H

OH

HH

O

H

O



CONFIRMING PAGES


11.49

Two SN2 reactions yield retention of configuration.

O

H

H Cl

1

2

O

H

H

SN2

SN2

OH

O

H

Cl

H

OH

HO

H

OH

O H

−

OH−

O−

−

11.50 Collapse of the α-haloalcohol by loss of a proton and expulsion of a halide ion leads to

the thermodynamically-favored carbonyl double bond. Practically speaking, the position

of the following equilibrium is completely to the right.

R

H O O

HXR′ R R′

X

+

11.51

A−

−HA

O HO+

HA −H2OOH OH2

+

HO OH

HO

+

OH

+

The reaction, known as the pinacol rearrangement, involves a 1,2-methanide shift to the

positive center produced from the loss of the protonated —OH group.



CONFIRMING PAGES


11.52 The angular methyl group impedes attack by the peroxy acid on the front of the molecule

(as drawn in the problem). II results from epoxidation from the back of the molecule—the

less hindered side.

11.53 For ethylene glycol, hydrogen bonding provides stabilization of the gauche conformer. This

cannot occur in the case of gauche butane.

H

O

H

H

H

H

H

δ−

δ+

O

only van der Waals

repulsive forces

H

H

H

H

Challenge Problems

11.54 The reactions proceed through the formation of bromonium ions identical to those formed

in the bromination of trans- and cis-2-butene (see Section 8.12A).

B

A

Me

MeH

OHH

Br

Me

HMe

OHH

Br−H2O

+ Br−

Me

MeH

H

Br

OH2

+

+ Br−

Me

HMe

H

Br

OH2

+

Me

HMe

H

Br

Br

−H2OHBr

HBr

HMe

MeH

Br

Br

HMe

MeH

Br

Br

meso-2,3-Dibromobutane

(Attack at the other carbon

atom of the bromonium ion

gives the same product.)

(a)

(b)

2,3-Dibromobutane

(racemic)

Me

Me

H

HBr+

(a) (b)

Me Me

HHBr+

Br−

Br−



CONFIRMING PAGES


11.55

H

SOCl2

−HCl

H OH

R H

H OS

O

Cl

R

H

H Cl

R−SO2

H

S

H

O

Cl

R

x x x

x

O− −

11.56

A B

HO

OH

OH

R R

HO

OH

OHachiral

S S

pseudoasymmetric

C

s OHHO

OH

R S

D

OH

OHHO

R S

r

A and B are enantiomers

A, C, and D are all diastereomers

B, C, and D are all diastereomers

C is meso

D is meso

11.57

O CH3

CH3O

H DMDO

(Z)-2-Butene

H

H3C

H3C

O CH3

CH3O

H

H

Concerted

transition state

H3C

H3C

O+

H

H

Epoxide

H3C

H3C

≠

CH3

CH3O

Acetone

11.58 The interaction of DMDO with (Z)-2-butene could take place with “syn” geometry, as

shown below. In this approach, the methyl groups of DMDO lie over the methyl groups of

(Z)-2-butene. This approach would be expected to have higher energy than that shown in



CONFIRMING PAGES


the solution to Problem 11.57, an “anti” approach geometry. Computations have been done

that indicate these relative energies. (Jenson, C.; Liu, J.; Houk, K.; Jorgenson, W. J. Am.

Chem. Soc. 1997, 199, 12982–12983.)

H3C

H3C

O

O

H

H

H3C

H3C

(Z )-2-Butene

with “syn” approach

of DMDO

DMDO

Concerted

transition state

H3CH

HH3C

H3C O

OH3C O+

H

H

Epoxide

H3C

H3C

H3C

H3CO

Acetone

Syn transition state Anti transition state

≠

QUIZ

11.1 Which set of reagents would effect the conversion,

?

OH

(a) BH3:THF, then H2O2/OH− (b) Hg(OAc)2, THF-H2O, then NaBH4/OH−

(c) H3O+, H2O, heat (d) More than one of these (e) None of these

11.2 Which of the reagents in item 11.1 would effect the conversion,

?

H

H

OHenantiomer+



CONFIRMING PAGES


11.3 The following compounds have identical molecular weights. Which would have the lowest

boiling point?

(a) 1-Butanol

(b) 2-Butanol

(c) 2-Methyl-1-propanol

(d) 1,1-Dimethylethanol

(e) 1-Methoxypropane

11.4 Complete the following synthesis:

(−CH3SO2ONa)

CH3SO2Cl

base (−HCl)

B C

D

NaH

(−H2)

A

(2) H3O+

(1)OOH

ONa

79720762 chapter 11 solutions

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