Jones M., Fleming S.A. Organic Chemistry

17.6 Syntheses of Carboxylic Acids 839

17.6 Syntheses of Carboxylic Acids

We already know a number of routes to carboxylic acids,and this section will add

an important new one, the reaction of organometallic reagents with carbon

dioxide.

17.6a Oxidative Routes As you know, there are several oxidative reactions

that produce carboxylic acids (Fig. 17.16). Primary alcohols and aldehydes can

be oxidized to acids by a variety of oxidants including HNO

3

, KMnO

4

, CrO

3

,

K

2

Cr

2

O

7

, or RuO

4

(p. 802). In Figure 17.16, these reagents are generalized by

“[O]”, which stands for oxidant. Alkenes with a hydrogen attached to one or

both of the doubly bonded carbons can undergo ozonolysis using an oxidative

workup to give acids (p. 436). In addition, the side chains of alkyl aromatic com-

pounds can be oxidized to acid groups with KMnO

4

(p. 613).

CH

3

Cl

COOH

Cl

KMnO

4

(77%)

H

2

O

100 ⬚C

R

O

OH

C

KMnO

4

H

2

O

..

HOOC

(63%)

1. O

3

2. HOOH/H

2

O

1. O

3

2. HOOH/H

2

O

CC

R

H

R

2 R C

OH

O

..

or

(100%)

RC

O

..

OH

..

OH

..

O

..

OH

..

H

2

CrO

4

CH

3

COOH

(77%)

KMnO

4

H

2

SO

4

[O]

RCH

2

OH

..

RC

H

O

..

O

H

..

O

OH

..

SPECIFIC EXAMPLES

THE GENERAL CASES

FIGURE 17.16 Oxidative routes to acids.

840 CHAPTER 17 Carboxylic Acids

Alkenes can also be oxidatively cleaved by potassium permanganate to give car-

boxylic acids (Fig. 17.17). We already know that alkenes form vicinal diols when

treated with basic permanganate or OsO

4

(p. 443).The reaction with permanganate

can go further to form two acids. The trick is to use the polyether 18-crown-6

(p. 254) to make KMnO

4

soluble in benzene.The crown ether has a great ability to

bind potassium ions, and this property allows the negatively charged permanganate

ion (MnO

4



) to follow the bound cation K



(crown) into solution, where it is more

available for reaction with the organic substrate.

KMnO

4

2

18-crown-6

25 ⬚C

..

OH

..

O

C

H

(100%)

FIGURE 17.17 Another oxidative route to carboxylic

acids.

17.6b Reaction of Organometallic Reagents with Carbon Dioxide

Carboxylic acids can be obtained from the Grignard reaction. Like any other

compound containing a carbonyl group, carbon dioxide reacts with organometal-

lic reagents through an addition reaction (Fig. 17.18). The initial product is a

carboxylate salt that is converted into the acid when the reaction mixture is

acidified. This reaction is very general and a source of many acids. The typical

procedure for these reactions is simply addition of dry ice to the Grignard

reagent.

Mg

ether

Cl

..

MgCl

1. CO

2

, –12 ⬚C

2. 25% H

2

SO

4

..

COOH

(81%)

A SPECIFIC EXAMPLE

H

+

RMgX

..

O

..

O

(or RLi)

..

OMgX

R

C

–

..

O

..

O

R

C

+

..

H

2

O

H

3

O

..

+

THE GENERAL CASE

FIGURE 17.18 A general route to carboxylic acids uses the carboxylation of Grignard

reagents with carbon dioxide.

17.7 Reactions of Carboxylic Acids 841

OH

?

(a)

OH

O

(b)

OH

O

(c)

OH

(d)

OH

O

Protonation of the carbonyl oxygen of the carboxylic acid

..

+

H

ROH

..

+

H

2

OR

R

C

..

+

O

R

C

..

+

O

..

R

H

Protonation of the carbonyl oxygen of a ketone (from Chapter 16)

..

+

OH

..

+

OH

O

C

..

R

C

R

C

R

+

O

..

R

H

O

C

O

..

R

H

O

C

O

..

ROH

..

+

H

2

OR

..

R

H

(a)

(b)

FIGURE 17.20 (a) The first step in

the Fischer esterification mechanism

is protonation of the carbonyl oxygen

to give a resonance-stabilized

intermediate. (b) The analogous

process with an aldehyde or ketone.

17.7 Reactions of Carboxylic Acids

17.7a Formation of Esters: Fischer Esterification An ester is a derivative of

a carboxylic acid in which the hydrogen of the hydroxyl group has been replaced with

an R group. We will examine the nomenclature, properties, and chem-

istry of esters in detail in Chapter 18, but one of the most important

syntheses of these species involves the acid-catalyzed reaction of car-

boxylic acids with excess alcohol, and must be given here. The reaction

is called Fischer esterification after the great German chemist Emil

Fischer (1852–1919) (Fig. 17.19).

We can write the mechanism for this reaction rather easily,

because its important steps are quite analogous to reactions of other

carbonyl-containing compounds (Fig. 17.20). As in the reaction of

..

ROH

acid

R

C

O

..

HOH

..

O

..

R

C

O

..

O

H

R

+

An ester

(RO replaces HO)

FIGURE 17.19 Fischer esterification, the acid-

catalyzed reaction of a carboxylic acid with an

alcohol, yields an ester.

PROBLEM 17.9 Starting from inorganic reagents, 1-butanol,CO

2

,and tosyl chloride

as your only sources of carbon, devise syntheses of the following molecules:

Fischer esterification

842 CHAPTER 17 Carboxylic Acids

PROBLEM 17.10 Explain why treatment of acetic acid with leads to

exchange of both oxygen atoms. Red highlight indicates

18

O.

18

O

+

H

3

>

18

OH

2

..

18

OH

3

18

OH

2

H

3

C

OH

..

O

C

Labeled O appears

in both positions

+

H

3

C

OH

..

O

C

..

H

3

C+

OH

..

O

C

The second and third steps in Fischer esterification are also completely analo-

gous to parts of aldehyde or ketone chemistry. A molecule of alcohol acts as a nucleo-

phile and adds to the carbonyl carbon of the protonated carbonyl group (Fig. 17.21).

For a ketone, this addition is followed by deprotonation, and gives the hemiacetal;

for the acid, a somewhat more complicated intermediate with one more hydroxyl

group is formed. In both reactions, however, a planar, sp

2

hybridized carbon has been

converted into an intermediate with a tetrahedral carbon, called the tetrahedral

intermediate.

H

sp

2

O

C

O

..

C

OH

..

+

C

..

O

+

O

..

H

..

+

H

2

OR

R

+

O

..

H

R

OH

Protonation

of the

carbonyl

Addition of

alcohol to the

protonated

carbonyl Deprotonation

Tetrahedral

intermediates

OH

C

..

R

OH

ROH

..

ROH

..

OR

sp

2

sp

3

O

C

..

C

OH

..

+

C

..

O

H

R

R = Group on original carbonyl compound

R = Group on alcohol

..

+

H

2

OR

R

OH

ROH

..

ROH

C

..

OR

R

OH

sp

3

(b)

(a)

2

1

Protonation

of the

carbonyl

1

Addition of

alcohol to the

protonated

carbonyl

2

3

Deprotonation

Hemiacetal

3

FIGURE 17.21 (a) The second step in Fischer esterification involves addition of the alcohol to the protonated acid,

which is followed by deprotonation to give a tetrahedral intermediate. (b) The parallel process occurs with aldehydes or

ketones, in which case the hemiacetal is formed.

an aldehyde or ketone in acid, the first step is surely protonation of the carbonyl

group by the acid catalyst, to form a resonance-stabilized intermediate. Sulfuric

acid is often used as the acid catalyst.

17.7 Reactions of Carboxylic Acids 843

Now what can these tetrahedral intermediates do in acid? The hemiacetal

has only two options. It can reprotonate the OR group and then revert to the

starting ketone, or it can protonate the OH group and proceed to the full acetal

(Fig. 17.22).

+

..

H

2

O

protonate

OH

Hemiacetal

(tetrahedral

intermediate)

deprotonate

add

alcohol

Acetal

Ketone

C

..

OR

R

O

H

+

C

..

O

R

..

O

R

C

R R

C

..

O

H

protonate

OR

R

OH

C

..

O

R

C

..

R

..

R

O

H

+

C

..

O

R

OH

+

C

..

OH

R

+

C

..

OH

R

C

..

O

R

lose

water

lose

alcohol

ROH

O

R

FIGURE 17.22 For the hemiacetal intermediate, there are two further possibilities: Protonation of the OH

leads to water loss and formation of the full acetal, whereas protonation of the OR leads back to the original

starting ketone.

protonate

OH

Tetrahedral

intermediate

deprotonate

Carboxylic acid

C

..

OR

R

O

H

+

H

C

..

O

H

protonate

OR

R

OH

C

..

O

R

OH

+

C

..

O

R

C

..

O

R

OH

lose

water

lose

alcohol

..

ROH

R

OH

..

OH

..

+

deprotonate

Ester

H

+

C

..

O

R

C

..

O

R

OR

..

ROH

..

ROH

2

+

..

ROH

2

..

OR

..

OH

..

FIGURE 17.23 For the tetrahedral intermediate from Figure 17.21a, there are also two further

possibilities: Protonation of the OR leads back to the starting acid, whereas protonation of either

of the OH groups leads to water loss and formation of the ester.

The tetrahedral intermediate formed from the acid has more options, because

it has one OR and two OH groups. Protonation of the OR simply takes us back

along the path to the starting acid. Protonation of one of the OH groups leads to

an intermediate that can lose water, deprotonate, and give the ester (Fig. 17.23). In

excess alcohol, the reaction is driven in this direction.

844 CHAPTER 17 Carboxylic Acids

PROBLEM 17.11 Write an arrow formalism mechanism, showing the electron-

pushing, for the acid-catalyzed formation of an organic acid from the reaction of

an ester with excess water.

..

+

C

..

OH

..

OH

R

+

C

..

R

C

R

C

..

O

OH

R

ROH

..

ROH

R

..

+

H

2

OR

..

+ H

2

O

Acid

..

OH

..

OH

..

OH

+

..

C

..

OH

R

OH

O

BD

A

C

E

H

C

..

R

OH

OR

C

..

OH

R

O

H

+

..

OR

C

..

O

R

C

..

OH

R

..

OR

..

OR

C

..

OH

R

..

OR

C

..

OH

R

..

OR

..

+

+ H

3

O

Ester

+

..

+

H

2

OR

Centerpoint of this mechanism

WEB 3D

FIGURE 17.24 The

full mechanism for

Fischer esterification

as well as for the

reverse reaction,

hydrolysis of an ester

to a carboxylic acid.

..

C

..

O

OH

R

ROH

..

H

2

O

C

..

O

R

..

OR

..

+

H

3

O

The complete mechanism for Fischer esterification is shown in Figure 17.24.

Note the symmetry of the process about the tetrahedral intermediate A. Note also

the similarity of B and C, the intermediates on either side of A. In intermediate B,

the OR is protonated. This pathway leads back to the starting carboxylic acid. In

intermediate C, it is an OH that is protonated. This path leads to the ester. To the

left of B and C are D and E, the two resonance-stabilized intermediates resulting

from protonation of the acid or ester carbonyl.

..

?

..

O

ROH

..

ROH

+

CH

3

S

..

O

..

O

Cr

..

O

OH

Cl

..

PROBLEM 17.12 Ester formation is not new to you.Give products and mechanisms

for the following reactions, which appear in earlier chapters.

We have spent a lot of time and space on Fischer esterification and its reverse,

acid hydrolysis of the ester.We paid this close attention because this reaction is pro-

totypal and an anchor for further discussion.As for other anchor reactions (for exam-

ple, the S

N

2 reaction and addition of Br

2

to alkenes), knowing this reaction, quite

literally backward and forward, allows you to generalize and to organize the many

following reactions that show a similar pattern, although the details may be differ-

ent. Fischer esterification is definitely a reaction to know well.

17.7 Reactions of Carboxylic Acids 845

Note here what makes the chemistry of aldehydes and ketones different from that

of acids (or esters). In the aldehydes and ketones there is no leaving group present,

whereas in an acid (or ester) there is (Fig. 17.25). The chemistry of aldehydes and

ketones on the one hand, and the chemistry of acids and esters on the other, diverge

only after the nucleophilic addition of the alcohol to the carbonyl (see Fig. 17.21). In

the acids and esters there is a group that can be lost, providing a route to other mol-

ecules; in the aldehydes and ketones there isn’t, which means that the acids and esters

can undergo an addition–elimination reaction that the aldehydes and ketones can’t.

..

R

C

..

O

R

..

C

..

O

OH

R

ROH

C

..

O

R

C

..

O

R

B

HB

..

OR

C

..

R

B

..

OH

C

R

B

C

..

R

B

OH

R

Potential leaving groups

Not a leaving group

Additions Eliminations

..

HOH+

C

..

O

R

B

+

..

OH

Protonated

alcohol

deprotonation

+

..

ROH

H

S

N

2

O

C

O

..

+

O

..

R

H

O

C

O

..

R

H

O

C

O

..

R

O

C

O

..

R

H

O

C

O

..

R

..

+

H

2

OR

H

FIGURE 17.26 A potential mechanism for Fischer esterification. Here the acid is the nucleophile and the protonated

alcohol is the electrophile. Watch out for conventions! The figure shows proton removal from one resonance form,

which is purely for convenience sake—the resonance form has no individual existence.

But we should wait a moment. Mechanisms other than the one shown in Figure

17.24 for Fischer esterification can be imagined, and it is not fair to dismiss them

without evidence. For example, why not use the protonated alcohol as the electrophile

and the carbonyl oxygen of the carboxylic acid as the nucleophile rather than the

protonated carboxylic acid as the electrophile and the alcohol as the nucleophile?

Displacement of water using the acid as nucleophile could give the ester (Fig. 17.26).

FIGURE 17.25 Unlike aldehydes and ketones, acids and esters bear OH and OR groups that are

potential leaving groups.

846 CHAPTER 17 Carboxylic Acids

The key difference between the mechanism for Fischer esterification shown

in Figure 17.24 (which is correct) and this hypothetical one in Figure 17.26 is the

site of carbon–oxygen bond breaking. In the first (correct) mechanism, it is a

carbon–oxygen bond in the carboxylic acid that is broken; in the new (incorrect)

mechanism, it is the carbon–oxygen bond in the alcohol that breaks.

WORKED PROBLEM 17.13 Design an experiment to tell these two mechanisms (Fig.

17.24 and Fig. 17.26) apart. Assume you have access to any isotopically labeled

compounds you may need.

ANSWER Using labeled alcohol will do the trick. If the mechanism involved

carbon–oxygen bond breaking in the alcohol (it does not), the use of

18

O-labeled

alcohol would not lead to

18

O incorporation in the product ester.

However, if the mechanism involves breaking of a carbon–oxygen bond in the

original acid (it does)

18

O will be incorporated, which it is.

Break this

C

O bond

..

O

OH

C

H

..

OH

C

18

O

..

H

R

+

protonations and

deprotonations

..

18

O

..

H

R

H

18

O

H

R

..

18

O

+

CO

R

18

O

++

R

18

OH

2

..

OH

C

..

+

H

R

18

O

OH

2

O

C

..

H

2

O

R

Why don’t esters react further under these conditions? Aldehydes and ketones

form acetals when treated with an acid catalyst and excess alcohol.Why don’t esters

go on to form ortho esters (Fig. 17.27)? Ortho ester is the rather misleading name

used for the functional group that has the general structure of RC(OR)

3

.Ortho esters

are not real esters.

C

..

O

R

C

..

R

..

R

OR

RO

C

..

O

R

..

HOR

C

..

R

An ortho

ester

An acetal An esterA ketone

..

OR

..

OR

..

OR

RO

..

+

H

2

OR

..

HOR

..

+

H

2

OR

FIGURE 17.27 The conversion of an

ester into an ortho ester is analogous

to the formation of acetals from

aldehydes and ketones.

Acid

..

O

OH

C

R

..

O

C

H

18

O

..

H

R

+

Ester

..

O

C

R

H

..

18

O

+

R

H

..

18

O

+

Break this C O bond

++

H

18

OH

..

R

17.7 Reactions of Carboxylic Acids 847

Ortho ester

C

..

O

R

protonation addition

addition

deprotonation

proton

transfers

lose

water

C

..

R

OR

..

OR

..

OR

C

..

R

..

OR

..

+

H

2

OR

..

+

C

..

OH

R

+

C

..

OH

R

C

..

OH

R

ROH

..

OR

..

OR

..

OR

C

..

O

R

OH

O

H

..

+

C

..

OR

R

OR

O

H

..

C

..

OR

R

O

H

+

(+)

..

OR

..

OR

..

HOR

..

HOR

..

HOR

..

Ester

FIGURE 17.28 A reasonable

mechanism for the conversion of an

ester into an ortho ester.

to those for acetal formation. However, stable ortho esters cannot be made this way.

The problem is one of thermodynamics.The sequence of Figure 17.28 is a series of equi-

libria, and at equilibrium it is the ester that is greatly favored, not the ortho ester. Note

that the starting material in this reaction, the ester, is stabilized by resonance but the

product, the ortho ester, is not. The ortho ester is analogous to the thermodynamically

unstable intermediate A in Figure 17.24.Aldehydes and ketones lack this ester resonance,

and are therefore less stable, and more prone to further addition reactions. Figure 17.29

A reasonable mechanism for the conversion of an ester into an ortho ester is out-

lined in Figure 17.28. Ortho esters are known compounds, and there is nothing funda-

mentally wrong with the steps outlined in Figure 17.28. The steps are quite analogous

Energy

Energy lowering

because of

ester resonance

C

..

R

..

R

OR

RO

HOR

+

C

O

R

An ortho ester

Ester resonance

Reaction progress

C

..

R

OR

..

OR

..

OR

..

Acetal

HOR

+

–

C

O

R

OR

C

O

R

OR

..

Several

intermediates

Several

intermediates

Ketone

FIGURE 17.29 A truncated

Energy versus Reaction

progress diagram comparing

acetal formation from a

ketone with ortho ester

formation from an ester.

848 CHAPTER 17 Carboxylic Acids

makes this point with a partial Energy versus Reaction progress diagram.The acetal

and the ortho ester are roughly equivalent in energy, but the ester lies well below

the ketone.The result is that acetals are thermodynamically accessible from ketones,

but ortho esters are disfavored relative to esters.

There are other mechanisms for ester formation, and some of them do not involve

breaking a carbon–oxygen bond in the starting carboxylic acid. Sometimes the car-

boxylate anion is nucleophilic enough to act as the displacing agent in an S

N

2 reac-

tion.The partner in the displacement reaction must be especially reactive.In practice,

this means that a primary halide or even more reactive species such as a methyl halide

must be used (Fig. 17.30).

..

O

..

O

R

I

..

O

..

O

R

S

N

2

..

O

–

R

..

I

..

I

dimethyl sulfoxide

24 ⬚C

(91%)

R

..

O

..

O

–

..

O

..

OCH

3

CH

3

O

..

O

H

3

C

+

THE GENERAL CASE

A SPECIFIC EXAMPLE

FIGURE 17.30 Carboxylate

anions can be used to

displace very reactive halides

in an ester-forming S

N

2

reaction.

Another example of a reaction in which the carboxylate acts as a

nucleophile is the ester-forming reaction of acids with diazomethane,

CH

2

N

2

. Diazomethane, though easily made, is quite toxic and a powerful

explosive. So, this method is generally used only when small amounts of

the methyl esters are needed (Fig. 17.31). Don’t try this reaction at home!

The first step in this reaction is deprotonation of the acid by the basic

carbon of the diazo compound (Fig. 17.32). The result is a carboxylate

anion and a diazonium ion, an extraordinarily reactive alkylating agent.

Next, an alkylation step similar to that in Figure 17.30 occurs between

the carboxylate and the diazonium ion. This reaction takes place

because nitrogen (N

2

) is a superb leaving group, perhaps the world’s best.

O

R

..

O

+

–

C CH

3

N

2

C

..

O

R

CH

3

..

O

N

S

N

2

..

Methyl ester

..

O

R

..

O

–

+

–

C

..

O

R

C

H

2

C

..

O

R

..

O

..

O

H

N

..

+

CH

2

N

A diazonium ionDiazomethane

Carboxylic

acid

N

..

+

++

H

FIGURE 17.32 In the first step of this methyl ester–forming reaction, diazomethane acts as a Brønsted

base and deprotonates the carboxylic acid. In the second step, the carboxylate does an S

N

2 substitution

on the very reactive methyl diazonium species.

(100%)

+

OH

CH

2

N

2

N

2

NO

2

..

OCH

3

C

NO

2

..

O

..

O

..

C

FIGURE 17.31 Diazomethane can be used to

make methyl esters from acids in yields as high

as 100%.

Jones M., Fleming S.A. Organic Chemistry

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