Esters, Amides, and Nitriles

17

Esters, Amides, and Nitriles

An Overview of Carboxylic Acid Derivatives and Their Chemistry

Study Hints The reactions of derivatives of carboxylic acids may seem overwhelming in number, until the pattern of the common mechanism that unites them is recognized. Then, the importance of some of the members of these families in chemical synthesis, medicine, and commerce can be appreciated. Problem-solving Skills After completing this chapter you should be able to do the following: 1.

Learn to formulate IUPAC and common names of esters, amides, nitriles, and salts of carboxylic acids (review root names in the common name system from Chapter 16 if needed)

2.

(Problems 1, 2)

Use ir spectra characteristics and mass spectra fragmentation patterns to solve structure problems (3, 13, 27, 47)

3.

Remember the relative reactivities of the classes of carboxylic acid derivatives and apply them to plan syntheses that utilize substitution at acyl groups (4, 5, 7, 9, 14, 16, 22, 23, 24, 42)

4.

For derivatives of carboxylic acids, use resonance to interpret the trend of reactivity; for amides, know the implications of TT-bonding character on conformation and on acidic and basic properties (6, 8, 20, 26, 27)

Chapter Seventeen

5.

316

Know means of shifting the equilibrium in Fischer esterification to improve conversion to products, understand the role of base in driving saponification to completion, and the effect of acid in amide hydrolysis (31, 32)

6.

Plan syntheses utilizing esters and nitriles with organometallic reagents and reducing agents (17, 18, 19, 25, 44)

7.

Understand the importance of thiolesters, phosphate esters, and phosphoric anhydrides in biochemical synthesis and energy storage (21, 33, 46)

8.

Apply the mechanism of acyl substitution to the full variety of functional group interconversions that it makes possible (IMPORTANT!) (8, 10, 15, 29, 30, 34, 35, 39, 40, 41, 42, 45)

Reversibility of substitution in carboxyl derivatives When you read that "less reactive carboxylic acid derivatives (esters, amides) may be readily synthesized from the more reactive ones (acid chlorides, anhydrides)", you might conclude that interconversions between these classes of compounds are like one-way streets, and we can only do syntheses that go the right way.

However, it turns out later that Fischer esterification may be driven either

to products or to reactants by adjustment of reaction conditions.

Furthermore,

amides, the most stable class, can be hydrolyzed completely back to a carboxylic acid. There is no inconsistency in these statements.

There is a thermodynamic

order within these classes of compounds that is based on their relative bond strengths.

That establishes the direction in which reactions can be made to occur

with relative ease.

But, the art of the chemist is to make things happen that

appear to be unfavorable, by adjusting the conditions in a local region (inside a flask) to temporarily make things go the "wrong" way if necessary.

Besides

adjusting concentrations, adjusting the acidity of the media can also provide a lot of leverage to move things the way you want, utilizing the acid-base characteristics of members of these classes to your advantage.

317

Student Solutions Manual for Organic Chemistry

Resonance in carboxylic acid derivatives Stabilization by electron delocalization through resonance is maximum in amides, quite significant in esters and free acids, but quite weak in acid halides.

Q:

:0:"

/
NR£

<

>

/%■ R

NR£

You should be prepared to draw structures analogous to those above and justify the order of stability of these classes.

You may be asked to include

resonance forms in a mechanism of substitution, which lengthens the mechanism but is not difficult. The tetrahedral intermediate 2 Due to its SJD hybridization, the carbonyl group of the acid derivatives is planar, thus relatively susceptible to nucleophilic addition on either side. The change to the SJD3-hybridized tetrahedral intermediate increases steric repulsions, but to a much lesser extent than occurs in SXT2 substitution, where the N configuration changes from four substituents to five in the transition state. Consequently, these reactions often occur more easily and with fewer complications of side reactions. In one sense, these reactions are like S I reactions, except for the sequence of steps.

In S i reactions, it is elimination of the leaving group to achieve the 2 unhindered, SJD -hybridized cation intermediate, followed by addition of the 2 nucleophile. These carboxyl derivatives, already SJD , permit addition-elimination. Important Terms symmetrical vibration

bactericide

asymmetrical vibration

cephalosporin

acyl group

penicillin

acylating a nucleophile

saponification

primary amide

glycerine

Chapter Seventeen

secondary amide

glyceride

tertiary amide

transesterification

tetrahedral intermediate

hydrophobic

Fischer esterification

hydrophilic

hydrolysis

biodegradable

intramolecular esterification

tautomers

lactone

thiolester

a, ß ,y

, or ä lactone

acetyl-CoA

polyester

sulfate ester

neat mixture

phosphate ester

ami no acid

phosphoric anhydride

lactam

adenosine triphosphate, ATP

antibiotic

phosphorylating agent

New Mechanisms Acyl group substitution :0: II C

:0:; Cv

R T Y

Nu

:Nu~

R

+ :Y Nu

(basic form)

The form above is the most direct.

However, when the nucleophile is

uncharged because the medium is neither acidic nor basic, the following form is appropriate (:NuH is usually water, ammonia, or an alcohol, as is :YH): :0:

\||

I

R :NuH

ό Y

^NuH

An, Nu

(neutral form)

:0: II

Å'

R

+ Nu

YH

318

319

Student Solutions Manual for Organic Chemistry

Furthermore, when a reaction occurs in a strongly acidic medium, which is often, the form follows the model below.

„/ o

H

o./x

0/ IK C

I C

*

x

—r—*

7

Nu (acidic form)

II

C +

I C

0

\*

R

*

+ NuH

NuH / 0)

o'

I C

^

C

Nu

R

+ H

+

Nu

Steps A and E, in which addition of the nucleophile and then elimination of the leaving group occurs, are always present.

They are reversible in principle,

but are indicated here with single arrows because they (or at least one of them) is rate determining.

H

transfers occur very fast, and are indicated as reversible

steps to emphasize that they are always at equilibrium. 2.

Reaction of Grignard reagents with esters :0:Γ

K OEt OH

•4

R

The points to remember about this mechanism are: 1)

The first step is the expected nucleophilic addition at the carbonyl

2)

Loss of ethoxide ion occurs in the second step.

group.

ethoxide being a weaker base than the carbanion R: .

This is a consequence of

A really poor leaving group

ordinarily, it is displaced in this case by an even worse one. irreversible.

The step is

Chapter Seventeen

3)

320

Addition of the second Grignard carbanion has a larger rate constant k«

than k1 associated with the first addition.

This means that ketone molecules are

reactive intermediates in the strict sense and cannot be isolated. 3.

Nucleophilic additions to nitrile Hydrolysis and Grignard reaction of nitriles are similar in principle, both

involving nucleophilic attack on carbon of the ΟÎÎÍ group.

However, the presence

of water promotes subsequent steps in the hydrolysis. A.

Hydrolysis Q-j-H«— :0H"

OH

:0H" R—C = N:

Ν

>R

>R

Ν:-*Η-Κ)Η

ΛΝΗ ^H—OH

0

:0:

:6T

c

c

c

II

/ \ B.

I

II

_._ 9f_^ + / \jH-

h

Grignard addition.

<

,/

\H

The anhydrous medium prevents the type of tautomeric

rearrangement that occurs in the hydrolysis. :N:" II s \

„ n* K

MgX+ ii

eine

R:

MgX

The last step above is an imine hydrolysis (Chapter 1 1 ) . Reactions For Synthesis 1.

Esterification O

II

0

,

II

RCOH + HOR' Examples: Variations:

N

RCOR 1 + H 2 0

Eqs. 17.1, 17.23, 17.24, 17.25, 17.31, 17.32, 17.33 The reaction may be called Fischer esterification.

It is

catalyzed by acids, especially sulfuric acid or p-toluenesulfonic acid. 2.

Pyrolysis of ammonium carboxylate salts

321

Student Solutions Manual for Organic Chemistry

0

II .

R C0

~

0

NH

> "hut (no solvent)

4

Examples:

II

RCNH

2

+ H

2°

Eqs. 17.2, 17.35

Variations:

Since other methods of amide formation occur in good yield under

milder conditions, this rather inelegant method is not popular. 3.

Nucleophilic substitution at acyl groups (or acylation of a nucleophile)

R

CY + NuH

Examples:

>

RCNu + HY

Eqs. 17.1, 17.5, 17.6, 17.7f 17.15, 17.16, 17.17, plus most other

substitution reactions in the chapter. 4.

See the discussion under New Mechanisms.

Hydrolysis

R

0

0

II

II

CX + H 2 0

Examples:

>

RCOH + HX

Eqs. 17.30, 17.37, 17.38, 17.43

Variations:

All of the carboxylic acid derivatives may be hydrolyzed back to

the carboxylic acid from which they are derived, with variable ease.

Note that

this is hydrolysis in the literal sense of the word—decomposition of the derivatives into two parts by water, with incorporation of water atoms into the products. 5.

Saponification 0

II

Q

_

II _

RCOR1 + OH Examples:

>

RCO

+ HOR1

Eqs. 17.7, 17.41

Variations:

This is a special case of hydrolysis, limited to esters as

reactant with sufficient base present to convert all the carboxylic acid to its carboxylate salt (soap). 6.

Grignard reactions of esters 0

R'äoEt

2

™*X>

ether

R(L |

è"

M3 g X + - ^ -+ * RCjR

OH

R1

R1

Chapter Seventeen

Examples:

322

Eqs. 17.45, 17.46

Variations:

Although this reaction is generally restricted to the preparation

of tertiary alcohols, use of formate ester permits synthesis of symmetrical secondary alcohols (see Prob. 17d). 7.

Acid chloride reaction with lithium dialkylcuprate

ii

R2CuLi

2 R'CCl —

Examples: Comment:

ii

> 2 R—C—R'

Eqs. 17.47 This provides the ketone synthesis that cannot be accomplished by a

Grignard reaction with esters. 8.

Reductions with active hydride reagents § LiAlH H-0+ > RCH + R RCOR' ether 2°" '°" — — > RCH20H + R'OH (as aluminum salts) Examples:

Eqs. 17.48, 17.49, 17.50, 17.51, 17.54, 17.55, 17.56, 17.57

Variations:

Incorporation of deuterium atoms by using active deuteride

reagents (Eqs. 17.50, 17.51), reduction of carboxylic acids (17.54), use of B.H^ (17.55), reduction of amides or nitriles to corresponding amines (17.56, 17.59), reduction of acid chlorides to aldehydes by selective, sterically hindered hydride (17.57). 9.

Grignard reaction with nitriles R'—C = N ^her* R'—C=N"

Example:

H.0+ ° MgX+ — — * RCR'

Eq. 17.60

Special purpose 1.

Acetyl transfer by acetyl-CoA 0 II

CH3CSCoA Example:

0 M M · " m

'

II

> CH3CNu + Eq.

17.63

SCoA

323

Student Solutions Manual for Organic Chemistry

This process is ubiquitous in the body.

Its importance to a student at this

point depends upon how much emphasis is given to biochemistry. 2.

Hydrolysis of ATP 0

0

II

II

0

i

u

Ad—0—P—0—P—0—P—OH I.

I.

0

0

Example:

n

H

?

L

I.

0

u

II

0

II

> Ad—0—P—O—P—OH

0

I.

0

I.

+

.

0 II

0—P—OH

0

l-

0

Eq. 17.68

The same comment about biochemical emphasis applies.

The reaction itself,

occurring with the evolution of heat, is a thing to be avoided within the body. is the potential for energy release that is the vital characteristic of ATP.

It

uses this potential instead to phosphorylate other molecules, such as sugars, to activate them for essential metabolic processes. Answers to Problems 17-1

Name

Alcohol [Amine]

Carboxylic Acid 0

a.

propyl benzoate

CH 3 CH 2 CH 2 OH

C^H.COH

b.

2-methyl-l-butyl 5-chloro-

CKCH^.COH 2 4

pentanoate

CHoCHoCHCH^0H 3 2i 2 CH,

3-methylbenzamide

[NH3]

c.

b b

CH3

λ~νΙ!θΗ 0

d.

2-bromo-N,N-diethylethanamide [(CH3CH2) NH]

BrCH2COH

(Í,Í-diethylbromoacetamide)

e.

2-bromoethyl 3-bromobenzoate

BrCH^CH^OH

f.

propyl 3-cyclopentene-

CH 3CH2 CH2 OH

carboxylate

*°" OH

It

Chapter Seventeen

17-2

0

CH~ 0 I 3 || CH 3 N CCH 3

II CH 3 OCCH 2 CH 3

cm

Cl C H 3 C H 2 C H 2 C H — 0 — C C H 2 C HIC H 2 C H 2 C H 3

s

O

ocSc

Br

17-3

324

CH.

HO-

r^

Q

sy

-NHC(CH 2 ) 6 CH 3

0 Cl II I \^C—UCH 2 CH 2 CH 3

Compound I would give a mass spectrum with fragments CH-CH^CED , m/e = 57,

and CH NHCHD , m/e = 5 8 .

On the other hand, the fragments from II are different:

CH 3 (S0 + , m/e = 43, and CH3CH2NHC3D+, m/e = 72.

C.H_CSCH0CH. and HCl

17-4

d.

CH3CO

and CH 3 NH 2

CH CH CN(CH ) and HCl I 0 11 c. 17-5

„

y—C0CH2CH

and CH 0

0

ii

CH3CH2CNHCH3 and CH3CH COH

Resonance causes the reactant to be more stable than the product.

reaction is thermodynamically unfavorable.

a

-

:0:

, CoH

Thus, the

It goes rapidly in the opposite

direction. 17-6

CH

:0H^

R — C — N — H -2—>· R — C — N — H <—> R—C-^^N—H < I I + 1 H H H

:0H > R—C=N^ I H

325

Student Solutions Manual for Organic Chemistry

The ion resulting from protonation on oxygen is well stabilized by resonance. below.

Comparable resonance is impossible if protonation occurs on N, as shown

In fact, this may even reduce the usual resonance 0 0 H JL R—C—N—H - -^ R—C—^ì-Ç I

I

H

H

X

X

within the carbonyl group (^C=0 <

-

> + C-0

), which would cause positive charges

to occur on adjacent atoms. b.

0

Qo:

:0V

+

CH3C—UU—H + H20 ; = = f H30 + CH3C—NH" <

> CH3C=NH

Acetamide anion has two resonance forms comparable to the two resonance contributors in acetate ion.

However, the first one shown is less stable than the

other, with the negative charge on 0. match that of acetate.

Thus, the stability of the ion does not

Consequently, the equilibrium lies more to the left; the

amide is a weaker acid. 17-7

Several combinations will work, provided that the reactivity of the acyl

reactant is greater than that of the product acyl group. 0 a.

II

C,Hc0H and CH,(CH_),CC1

V , V \\ ^- 00 H H,

c.

and and

002^N-< -/'

0 b.

(CH ) CHNH

and CH (CH2> COEt 0

II

(CH 3 ) NH and CH (CH2> CCl

NH0 and C1(CH 0 ) „COEt 3 24

> 9

H 0 N(CH 0 ) „COEt 2 2 4

>

III

\-Q >—CC1

x

Chapter Seventeen

17-8

0 CH

3

J

H2NCH3

\ 0:

CH,

CH

ll

CH3

0 ^ ^CH~ +NHCH3

CHf

X

J

07 V

^CHNHCH^

NHCH3

c

pci

II

H2Ö:~^H \ J

0 x

I

x

CH^

^CH3

°

r?

o

:0:

II

:0:

CHf t Cl

CH3^

N(CH 3 ) 3

N(CH 3 ) 3

N(CH 3 ) 3

:C1:

Reaction can occur via the same mechanism, but the product i s u n s t a b l e . cannot lose a methyl group as most amides lose a hydrogen atom, so i t remains p o s i t i v e l y charged on N adjacent to the p a r t i a l p o s i t i v e charge of the carbonyl carbon atom.

There i s no resonance s t a b i l i z a t i o n e i t h e r , so reverse r e a c t i o n

dominates the e q u i l i b r i u m . 17-9

N-acylation with (CH CO) 0

H(W/

V ^ H 0 —»-> HO-/

\ - i lH,

0C0CH-

^ CH,

-^ HO-/

V-N K—CCH3

+

HO-//

V^H-C7

0C0CH,

CH*

CH3C00"

O-acylation 0C0CH-

H tH3

r-> H 2 N--/

N

V<>-^CH3

+

OLC00

CH3

3

327

Student Solutions Manual for Organic Chemistry

The rate-controlling steps are designated A for N-acylation, A1 for O-acylation. 1

than A

If this is treated as a competitive reaction, then A should be faster

because A leads to development of a more stable cation and should therefore

occur with a lower E

. The same result should be obtained if the product is

thermodynamically controlled, since resonance in the amide makes it more stable than the ester. 17-10

a.

0

M

II

18n

18

0

0

H? 0 || no || CH 3 CH 2 C0Et—~—>CH 3 CH 2 C , 0 0H and CH3CH2C0H

b.

18

+ HOEt

0 i s r e p r e s e n t e d by a s o l i d c i r c l e H+

( · ) for

H

H+

C:Q

:0:^ CH^CH9C0Et 3 Ü

convenience.

OH /

CH.CHoCOEt 3 c£

OH H > CH-CH-C-MMt 3 2.··

> CH~CH 0 C—0—Et 3 ^ i -

•

C*·*

LLUn

>

:·Η.

t

H20 > CH^CHoC—%W

3 2

X~\ N

*

17-11

+

II

CH3CH2C—OH

0H

CH 3 C—0--C(CH 3 ) 3 - IuL+^ C H 3 C -I^ - 0 — C ( C H 3 ) 3 Jj

18

+

=w r u r

Λ

r f ru

\

+ Λ

°

> CH3C0H || + (CH 3 ) 3 C+ H 9 0 -> v r u rr\u

_L

f ru

\ r

C-

18

( C H 3 ) 3 C — 0 H 2 - ± U > (CH 3 ) 3 C— 0H

The r e a c t i o n i s r a p i d and i n c o n s i s t e n t w i t h t h e t r e n d of t h e p r i m a r y and s e c o n d a r y a l k y l a c e t a t e s b e c a u s e i t can f o l l o w a d i f f e r e n t

mechanism.

Since the

t - b u t y l c a t i o n i s s t a b l e , t h e p r o t o n a t e d a c e t a t e p a r t a c t s a s a l e a v i n g group i n an S N 1 mechanism ( j u s t a s p r o t o n a t e d OH forms w a t e r a s a l e a v i n g g r o u p ) . T h i s f i n a l l y 18 leads to 0 i n c o r p o r a t i o n i n t o t h e a l c o h o l p a r t i n s t e a d of t h e a c i d p o r t i o n .

C h a p t e r Seventeen

12

CH

The ß - l a c t o n e i s s t r a i n e d

r-

HOCH2CH2C ^

328

because

of s m a l l r i n g s i z e and t h u s does n o t form c o m p e t i t i v e l y w i t h t h e more s t a b l e δ-lactone.

13

-1620 cm"

Cl^Q>-C H CNH 2

1780 cm 1700 cm

OH 14

C H . ( C H 0 ) . .CONa 3 2 14

HOCH2CHCH2OH

sodium

glycerine

CH3(CH2)7

CH 3 (CH 2 ) 1 6 CONa sodium

c<

15

C H C

sodium o l e a t e 0

^1 Me

-

C

~:0Et

a.

(CH 2 > 7 CONa

c=:c

stearate

6 5 AT^

16

palmitate

EtOOCCH2COO

cc

6

H

Me 5 ^

II

->C 6 H 5 C—OEt

+

:0Me

OEt

+ EtOH

c.

CH 0 CH o C00 + CH 0 NHC,H. 3 2 3 6 5 CH- CH 0 3 3

d.

H2NCH2CH—CHCH COO

I

I

329

Student Solutions Manual for Organic Chemistry

17-17

a.

Ester

Grignard reagent

CH3CH2COEt

CH3CH2CH2MgBr

>. a - ^ V_<:oEt

CH3CH2MgBr

C6H5MgBr

CA

OEt

0

/

I!

V-MgBr

CH3COEt 0

r—,

EtOC-<^

17-18

0

CH MgBr

y-COEt

iMethylmagnesium bromide a c t s as a base toward propanoic a c i d or propanamide.

The gas i s methane.

0 CH3CH-2C—
CH3CH2C—NH^-H*-:CH3"

17-19

a.

0

0 MgBr*

MgBr*

> CH 4 (g) + CH3CH2C0:"

> CH 4 (g) + CH3CH2CNH"

+

MgBr+ -^->

MgBr+ - ^ - > CH3CH2C0NH2

0

II II CH3CH2CC1 and (CH 3 CH 2 CH 2 ) 2 CuLi, or CH 3 CH 2 CH 2 CC1 and (CH3CH2)2CuLi

b.

CH3CH2C00H

C 6 H 5 CC1 and [CH 3 (CH 2 ) 4 ] 2 CuLi, or CH 3 (CH 2 ) 4 CC1 and (C ί H 5 ) 2 CuLi

Chapter Seventeen

? § Et0CCH 2 CH 2 CH 2 CCl

(CH3)2CuLi ft ft — » EtOCCH 2 CH 2 CH 2 CCH 3

H20

330

H+ —L^-E->

0

0

II

II

H0CCH 2 CH 2 CH 2 CCH 3

17-20

:NH

NH

:0:

II

II

R—C—OH <

R—C—NH 2 <- -* R — C = N H 0

> R—C=0H

Resonance in the amide form is shown at left, while that in the imidic acid is at right.

The charged resonance form in the amide is a better contributor,

since the charges are placed more appropriately, considering the electronegativities of N and 0. 17-21

0

0

II

enzyme —-OLOPOEt 2 I

+

OEt

17-22

II

HOPOEt I

OEt

The r e a c t i o n i s p r a c t i c a l l y i r r e v e r s i b l e .

The amide i s more s t a b l e than the

e s t e r because the nitrogen i s a b e t t e r e l e c t r o n - p a i r donor in resonance than i s oxygen (see s e c . 1 7 : 3 ) . 17-23

All the r e a c t i o n s produce CH^CH^OH by s u b s t i t u t i n g another nucleophile, plus

the following:

17-24

a.

CcH_COOH

f.

C 6 H 5 CONHNH 2

b.

CgH COO"

g.

C^H-COOCH-

c.

C6H

d.

C6H5CONH2

h.

C.H_C(CH0CH

e.

C6H5CON(CH3)2

i .

CcHcCHo0H 6 5 2

b

a.

b

a

c H C0Me

6 5

OH 1

COOCH 2 CH 2 CH 3

2 CH~CH„CHp MgBr

— h i t F

»

H20

1

¼

D

OH

CH,3CH2CH 2CCH2CH2CH3 ^6 H 5

Z

331

S t u d e n t S o l u t i o n s Manual f o r O r g a n i c C h e m i s t r y

fl CH3CH2CH2COH

w

S

S0C1 9

CH3CH2CH2COH

J

[CMCM-LCuLi ^ - ^

0 II CH3CH2CH2CC1

fH3

—

§

>CH 3 CH 2 CH 2 CCH 2 CH(CH 3 ) 2

f H 3 ft CH3CH2CHCH2CC1

S0C1 ? ^

°

> CH 3 CH 2 CHCH 2 C(CH 2 ) 3 CH 3

PBr fH3 ßÇ3 ^ CH3CH2CHCH2CH2Br J = ± * - M L * (CH3CH2CHCH2CH2)2CuLi

CHn | 3 0 > CH 3 CH 2 CHCH 2 CH 2 CCH 2 CH 2 CH 3

+

1»

/^^-CH^H

a.

C,H_CH o 0H a n d CH o CH o CH o 0H Z

3

0

f.

D

—

CrO, ?H3 \ 1-> CH3CH2CHCH2COH

fH3 CH3CH2CHCH2CH2OH

¼

[(CH.) ? CHCH ? ] 9 CuLi

^CH3CH2CH2CC1

fH3 CH3CH2CHCH2CH2OH

17-25

2 CH CH 2 MgBr H 0+ ?H — > - - - — > (CH 3 CH 2 ) 2 CCH 2 CH 2 CH 3

fl ; u n > CH3CH2CH2COMe

5

>

Z

^y

Z

C

+

QH

^6>

^^

3

γ

CD20H

c.

(CH 0 ).CHCD.OH a n d CH-OH

d.

DOCD2CH2CD OD a n d

e.

CH3(C.,2/4CD

ό

Z

Z

J

C^OD

,-CH20D and CH 3 CH 2 0D

17-26

The mechanism is S 2, with carboxylate ion serving as the nucleophile.

Loss

of resonance within the carboxylate ion occurs when the ester is formed, an unfavorable factor.

Favorable factors are the dipolar aprotic solvent to enhance

nucleophilicity and the fact that the halide ion leaving group is less nucleophilic than the carboxylate ion. halides.

Reaction would be limited to primary and secondary alkyl

Chapter Seventeen

17-27

Resonance within amides creates double bond character in the C-N bond.

332

This

restricts rotation, making the methyl groups diastereotopic and thus nonequivalent in nmr spectra, since one methyl group is eis to the carbonyl oxygen, the other trans. CH

CH3

yc_N\ (Ã

CH

*

/

*

CH3

CH3

\

:0-

CH 3

At elevated temperature, kinetic energy is sufficient to overcome the barrier to rotation.

When rotation occurs rapidly enough, the nmr spectrum would

simplify to one line for the two methyl groups bonded to N, another for methyl bonded to carbonyl. 17-28

a. C:0:R—C^-Y 4

:0: > R—C=Y

When Y is a good ÔÃ-electron donor, as the resonance structures illustrate, there is more single-bond character to the carbonyl group, so it has a stretching vibration at lower frequency, closer to that of a C-0 single bond. b.

Resonance involving the ÔÃ bond of the unsaturated group in conjugation

with the carbonyl group also increases the C-0 single-bond character, as illustrated by the following resonance forms. :0: I R—CH=CH—CY <* * R—CH—CH=CY ::0:~ : / Ë ° / — 1 \ 0' " / =CY <-^> /+ +

y, :Ö:" , >CY *—* /v \

17-29

0

:0p

II

vL / c.s%-> — > / c\

UC1

.

0

\4<

\

:Ö: >CY

v

R

ci"

N

Nucleophilic attack by pyridine produces the ion pair above, which precipitates.

333

Student Solutions Manual for Organic Chemistry

o

r:0:

:0: I

n

.A*

->

c.

R

R'

/ R'

H<-:B

?
c R^ Nh-R' :l

W

The catalytic effect of pyridine in this process is similar to the effect of I~ in catalyzing certain S 2 reactions.

It works because pyridine is both a good

nucleophile and, after becoming positively charged, a good leaving group as well.

17-30

Co

a.

/N)H

b.

H

V^H3

v_

cH

\_0^

3-r-\/H3 V_Q:

Qf^"3-

S...+

H0(CH2)3C0CH. vQU

17-31

^N

HOCH«

Several answers may be equally c o r r e c t for the s y n t h e s e s .

A v a r i e t y of

procedures have been i l l u s t r a t e d below t o t r y to show the complete scope of esterification

reactions.

/7-\ C

Λ^/™

I?

CH OH. H+ excess

J

?0

/Ã\ a

—COCH,

~\J^

Chapter Seventeen

ιί

CH 2 0H

ft

son.

^ CH 3 (CH 2 ) 4 CC1

CH 3 (CH 2 ) 4 C0H

-> C H 3 ( C H 2 ) 4 C 0 C H 2 -

OH I CH 3 CHCH 3 j — ^ (CH3)2CH0CCH2CH2C0H

d

·

H+

CHQ

I BrCH2COH

p

B^3 II0 *-> BrCH 2 CBr

HHQH»3 \=^

COOH

00"

2 OH COOH

3°,

OH CH 3 CHCH 3

Cl 0 cnr1 Cl 0 Cl 0 OH 0 I // SOCI9 | a Ftnu j // I II ηìLLUn CH~CH9CHC ^CH~CH 9 CHC > CH9CH9CHC - ^ L · * CH~CH 9 CH—COEt 3 2 \ 3 2 v _ DMF 3 2 3 2 \ OH Cl OEt CH 3

e.

2H

334

x£

COO

>

BrCH 2 C0-/ /

V7 2V

\/ô\

/-CH3

C00CH2CH20" 00CH 2 CH 2 0

C00CH2CH20H C00CH2CH20H

17-32

There is a steric effect that makes it difficult for the nucleophile to add

at carbon.

The c a r b o x y l i c acid group is twisted out of the plane of the benzene

ring by the adjacent methyl g r o u p s .

To accomplish t h e transformation of a c a r b o x y l i c acid into an acylium ion requires both a strong proton donor and a dehydrating agent.

Concentrated

sulfuric

335

Student Solutions Manual for Organic Chemistry

acid performs both t a s k s .

Methanol can r e a c t with the acylium ion, which i s l e s s

hindered. 17-33

17-34

c

0

:0:

Ski

ii

..C CH3 / Λ · · / CH3 i] UN HO

/ HO

C \

y

CH 3

CK, 3

+ :N I CH 3

CH 0 '3

0

/

0 C.

\

+ HNC1L CH 3

I

3

CH 3

From the viewpoint of the t e t r a h e d r a l i n t e r m e d i a t e , t h e r e may be loss of e i t h e r hydroxide or dimethylamide ion.

Since dimethylamide i s a b e t t e r nucleophile

than hydroxide, i t i s a worse leaving group.

Therefore equilibrium i s to the l e f t

in t h i s r e a c t i o n . 17-35 18 0 i s represented by · . iU

Vt H

]

CH3 +N)Et

5

CH3

CH3/+X0Et

CH3

Steps in which H followed by loss of

2

CH3 \)Et

3

Zwf \ E t

^~

OEt

is transferred are relatively fast.

Therefore, step 3,

O-water, is competitive with protonation of the OEt group,

leading to the product.

This evidence of reversibility of the formation of the 18 tetrahedral intermediate is imminently reasonable. Loss of O-water is structurally equivalent to loss of HOEt.

Chapter Seventeen

17-36

336

In VIII the nonbonding electron pair on N is at right angles to the plane of

the carbonyl group ð bond.

Consequently, there is no resonance stabilization and the bond is weaker than the usual amide bond stabilized by resonance.

The

high stretching frequency of the carbonyl group means that it is not being given single-bond character as it would if resonance was effective.

17-37

CH20C(CH2)16CH3

There are several possible

I° I °

structures, such as the one at right.

CHOC(CH2)7CH=^CH(CH2)7CH3

The unsaturation might be found in one

CH2OC(CH2)7CH^=CH(CH2)7CH3

linoleyl group, as an alternative to two oleyl groups.

The center carbon of the glycerine part is the chiral center.

will be chiral in glycerides having different fatty acid parts on each end.

It The

molecule above becomes achiral and yields only stearic acid after hydrogenation. 17-38

a.

0

c.

d.

ccr

-ocH?^/\rj:H2c a polyester

a polyami de

The eis isomer can also form a lactone. 17-39 1)

b.

CH3CNHNH2

0

0

II

II

2) CH3CNHNHCCH3 The nitrogen of hydroxyl amine is more nucleophilic than the oxygen.

Thus, the rate of forming hydroxamic acids such as IX is greater. once formed, they are more stable due to amide group resonance.

Furthermore,

337

Student Solutions Manual for Organic Chemistry

0

17-40

S0 2 NH 2

02NH2

0

v

NH

CH,

17-41

:N:

^

^^"COOH

o 0

a. ^

A

NHCH^

^

OH

0

/

Na

:

^

0

0 il CH30CC1

17-42

:H 3 OCN(CH 2 CH 3 ) 2

Since XI is a carbonate, there are two potential leaving groups, but

ethanethiolate is the better leaving group.

There is less resonance of S

nonbonding electrons in large and diffuse third period orbitals that do not overlap effectively with second period 7T-bonding atoms, so the C-S bond is weaker.

(TO:

4v

MeO'

I SEt

NH,

:0:> MeO

Os+SEt NH 3

0

• A II

MeO X

NH 2

+ 0 II C

MeOH + Η 2 Ν ^

17-43

a.

HSEt

SEt

Since A has one degree of unsaturation, two oxygens, and is cleaved by

aqueous base, an ester is suggested. propyl acetate.

Isolation of 1-propanol proves it to be

Chapter Seventeen

338

CH3CH2CH2OH 0 CH^COCH CH CH. ι υ υ υ π 0^ π 0υ . . 3

2

2

^ether

H20

3

NaQH

+ 0 u n H 'I i, + 3 ° CH3CO" Na — - — > CH3C0H

watery

b.

The combination of one degree of u n s a t u r a t i o n , one N in the formula, and

gas evolution from b a s i c h y d r o l y s i s suggests an amide. ammonia or a v o l a t i l e s h o r t - c h a i n amine.

The gas would be e i t h e r

I s o l a t i o n of 4 - m e t h y l - l - p e n t a n o l means

t h a t the sequence of s t e p s i s as f o l l o w s : NH„

3(9)

Ð (CH3 CHCH 0 0^ Ã Ì ,0CH Ã 0ÃCNH "2

H

?0 NaOH II . (CH3)2CHCH2CH2CO

+

H

^°

II

Na — — > (CH3)2CHCH2CH2COH D

ii LiAlH, +> (CH3)2CHCH2CH2COMe ethel>

M ðì

D H c.

(CH3)2CHCH2CH2CH2OH

+ MeOH

E

Since the last step (I —> J )

F has five degrees of unsaturation.

indicates nitration of an aromatic ring, and the apparent esterification with the loss of one Cl (F — > G ) suggests an acid chloride for F, the following sequence is suggested:

0

n

„., /,>,, N ....

/Γ^ H (CH^CHCH.OH *~\ Ú ClUCH^NHg 3 ?—> C1-^V-C0CH2CH(CH3)2 - ^ — ^ Ci-^A—CCl — - ^ F /Γ~λ

S ü

M+

/r~\

n

C1-^^H(CH2)3CH3-A_C1-^J^C0H

HNO3 º

^

JT\

°

Cl-fJ-COH 09N

H

I

2

J

The position of Cl on the ring is proven by the last step. would lead to more than one nitration product.

Any other isomer

339

17-44

Student Solutions Manual for Organic Chemistry

a.

CH2CH3 ,K Ç'â OH CH3

CH^ (inversion)'

Br

(S)-(+)-2-butanol

Cl·^^

A , \^H CH3

DMF (inversion)

C H ^

,.'X H [} CsN CH3

A

H2S04 ^ 4

B,(S)

A \ H ί C0OH CH3 (+)-C,(S)

MsCl

pyridine (retention)

X»

CH,CH, I 2 3 NaCN

,a\

CH ? CH ? „ n 1 2 3 H 2° ,

TMSÖ

H'ό OTs (inversion) CH~

A·..

N = C |} H CH~

P

f H 2 CH 3 '/I \ Hj COOH

u

f2 3

+

*

M

CH«

?\ u n L) M U

2 '

CrL·

M

+>

H

··// \ H'J XH20H

HOOC

Λ·,

βH CrL

(-)-C

PBr3 (retention)'

CrL·

F

? H 2 CH 3

(-)-G,(S)

? H 2 CH 3

(CH,),CuL1

H'2f CH?Br CH 3

H'$ CH-CH, CH 3

H

I 3-methylpentane (optically inactive)

I2

KCN (retention)

3

H /J CH9C=N c CH3 J

17-45

^H2CH3

1) LiAlH4

.·# \ H
(+)-C

„

E

( H CH

CH30H

HL2S0. 4

CH?CH. '2 3

H

2°

v

*

H

I2

3

H'£/ CH9C00H CH3 * (+)-K,(S)

The accelerated rate suggests a catalytic effect for those structures with a

carboxylate group geometrically capable of intramolecular displacement.

nucleophilic

The release of phenoxide before the free dicarboxylic acid is formed

suggests this hypothesis.

Chapter Seventeen


340

Mhh

^0C6H5 0:"

0

0 XIV

H20

XV is 100% in the eis geometry needed for intramolecular participation in phenoxide displacement.

However, XIV will be predominantly in the more stable

anti-conformation, which cannot react via the intramolecular mechanism. 17-46

The amine reacts with phenyl acetate by adding at carbonyl carbon, followed

by the elimination of phenoxide ion, the usual mechanism of ammonolysis of carboxylate esters.

However, phosphate esters generally react by an S 2

displacement on the alkyl group (eq. 17.66).

Since S 2 displacement on a phenyl

ring does not occur at all, triphenyl phosphate is inert.

17-47 o6.8-8.1,m,4H H,

,H

«*t%3

0CH2CH2CH3 ol.0,t,3H 5l.8,m.2H

59.5,s,H

<54.2,t,2H