Jones M., Fleming S.A. Organic Chemistry

19.8 Addition of Acid Derivatives to the Position: The Claisen Condensation 989

␣

the condensation step is removal of one of the doubly α hydrogens to give the highly

resonance-stabilized salt of the β-keto ester (Fig. 19.102).

Note resonance

stabilization

..

O

RO

2

..

C

CH

3

–

..

OR

..

–

..

OR

..

HOR

..

HOR+

..

O

RO

..

C

CH

H

CH

3

O

..

C

endothermic

exothermic

..

O

RO

..

C

CH CH

3

O

..

C

–

(–) (–)

pK

a

~ 11

Doubly



hydrogen

pK

a

~17

..

–

Catalyst (RO ) is

destroyed!

FIGURE 19.102 Alkoxide is destroyed

in the last step of the Claisen

condensation as a doubly α proton is

removed to give a stable anion.

If the doubly α proton is removed, all is well thermodynamically; now the product

is more stable than the starting material, but the catalyst for the reaction has been destroyed!

A Claisen condensation using a catalytic amount of base must fail, because it can only

generate an amount of product equal to the original amount of base.The simple reme-

dy is to use a full equivalent of alkoxide, not a catalytic amount. A thermodynamically

favorable Claisen condensation is not run with catalytic base, but requires a full equiv-

alent of base.Be certain you are clear as to the reason why the aldol condensation requires

only a catalytic amount of base but the Claisen needs a full equivalent (Fig. 19.103).

..

OR

..

–

..

O

RO

2

..

C

CH

3

–

..

OR

..

HOR

..

HOR+

O

RO

..

C

CH

2

CH

3

O

..

C

..

O

RO

..

C

H

CH

3

O

..

C

–

(–) (–)

Doubly 

Removes doubly  proton;

the catalyst is destroyed

Claisen

O

H

2

..

C

CH

3

Aldol

–

..

OH

..

H

2

O

..

OH+

O

H

..

C

CH

2

CH

3

..

CH

–

..

OH

Can carry out

another aldol

FIGURE 19.103 The aldol

condensation can be carried out using

a catalytic amount of base because

the final protonation step regenerates

a molecule of hydroxide. In the last

step of the Claisen condensation,

the alkoxide ion is consumed by

removing the doubly α proton.

Because we are left with a carbanion after the Claisen condensation, acidification

of the reaction mixture is required in order to form the neutral β-keto ester. What

prevents this product,thermodynamically unstable with respect to the starting ester mol-

ecules, from re-forming the starting esters? Certainly the Claisen condensation is

reversible.But the β-keto ester that is formed in the basic conditions is much more like-

ly to form the resonance-stabilized enolate than go back to starting material.And once

the stable enolate is formed it is not likely to react with alkoxide,a nucleophile  nucleo-

phile reaction. And after acidification there is no base present! Without the base, there

can be no reverse Claisen condensation, and the product ester is obtained (Fig. 19.104).

OH

2

H

..

H

3

O

..

+

H

2

O

..

H

2

O

+

..

O

RO

..

C

H

CH

3

O

..

C

protonation

–

..

O

RO

..

C

CH

3

CH

2

O

..

C

No base present!

-Keto ester

FIGURE 19.104 Acidification of the

Claisen condensation produces a

β-keto ester in the absence of base.

PROBLEM 19.32 Draw the mechanism for the Claisen condensation of ethyl phenyl-

acetate using the appropriate base. Include the acidification process at the end.

Figure 19.105 shows Energy versus Reaction progress diagrams for the aldol and

Claisen condensations.

Energy

..

H

2

O

..

H

2

O

H

..

C2

CH

3

O

H

..

C

CH

2

–

..

–

..

OH

..

+

O

H

..

C

Reaction progress

The aldol condensation for acetaldehyde is approximately thermoneutral

CH

3

+

..

O

H

..

C

CH

3

O

..

CH

+

–

..

HO

..

Energy

O

RO

..

C2

CH

3

O

..

C

CH

2

–

..

+

O

..

C

Reaction progress

CH

3

+

RO

..

RO

..

C

O

..

O

..

C

CH

3

+

–

..

RO

..

–

..

RO

..

RO

..

C

Endothermic

to here

Exothermic

to here

O

..

O

..

C

CH

3

CH

+

–

..

HOR

RO

..

–

..

O

..

C

CH

3

O

..

C

RO

..

OR

..

CH

2

CH

2

–

..

O

H

..

C

CH

3

O

..

CH

2

CH

2

FIGURE 19.105 Energy versus

Reaction progress diagrams for the

aldol and Claisen condensations.

990 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

PROBLEM SOLVING

Summary

A full equivalent of base makes the Claisen condensation a practical source of

acetoacetic esters (β-keto esters).The five steps of the process are summarized in

Figure 19.106:

1. Enolate formation.

2. The condensation itself (addition).

3. Loss of alkoxide from the tetrahedral intermediate (elimination).

4. Removal of the doubly α hydrogen.

5. Acidification to produce the β-keto ester.

19.8 Addition of Acid Derivatives to the Position: The Claisen Condensation 991

␣

THE GENERAL CASE

A SPECIFIC EXAMPLE

Enolate

formation

1

Loss of

alkoxide

3

Condensation

2

Removal of

“doubly ”

hydrogen

4

Acidification

5

2. Acidification

1.

8 h, 78 ⬚C

..

H

3

O

..

+

H

2

O

..

O

RO

..

C

CH

2

O

RO

..

C

CH

R

+

..

O

RO

..

OR

..

C

CH

2

R

O

..

C

..

–

..

–

(–)

(–)(–)

..

O

RO

..

C

CH CH

2

R

O

..

C

R

RR

R

..

O

RO

..

C

CCH

2

R

O

..

C

..

O

RO

..

C

H

CH

2

R

O

..

C

+

–

O

..

C

CH

3

CH

2

(75%)

O

..

C

..

O

CH

3

CH

2

OCH

3

CH

2

O

..

CH

3

CH

2

ONa

..

C

CH

3

–

..

RO

..

–

..

RO

..

FIGURE 19.106 The Claisen

condensation.

“All” β-keto esters are Claisen products or products of closely related reactions.

Every time you see a β-keto ester, think—“What Claisen condensation would

have formed it?”

GO

992 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

+

–

..

O

CH

3

CH

2

O

..

OCH

2

CH

3

..

HOCH

2

CH

3

Na

..

C

Ethyl 2-methylpropanoate

(full equiv.) No useful amount of

Claisen condensation

product

CH(CH

3

)

2

However, the product A contains no doubly α hydrogens! There can be no for-

mation of the stable, highly resonance-stabilized anion that accounts for the suc-

cess of the Claisen condensation. Thermodynamics dictates that the starting

materials be more stable than the product, and the reaction settles out at a point

at which there is little product. The mechanism of the reverse reaction is written

for you: Just read the scheme backward.

19.9 Variations on the Claisen Condensation

So, that’s the Claisen condensation: Two esters combine with an equivalent of base

to make a β-keto ester. Now it’s time to look at how changes in structure of the start-

ing materials affect the structure of the product and the mechanism of its formation.

19.9a Intramolecular Claisen Condensations The Dieckmann condensa-

tion, the cyclic version of the Claisen condensation, is named after Walter Dieckmann

(1869–1925). The reaction transforms an acyclic diester into a cyclic β-keto ester.

Formation of five- and six-membered rings is favored, and there is little beside the

intramolecular nature of the reaction to distinguish the Dieckmann from the standard

–

O

..

C

..

+

..

elimination

addition

CH

3

CH

2

O CH(CH

3

)

2

..

O

..

C

..

....

CH

3

CH

2

OC(CH

3

)

2

..

O

..

C

..

CH

3

CH

2

O CH(CH

3

)

2

..

O

..

C

..

CH

3

CH

2

O

OCH

2

CH

3

CH

3

H

3

C

CH(CH

3

)

2

..

O

..

No hydrogens here

O

..

C

A

..

CH

3

CH

2

O

OCH

2

CH

3

CH

3

H

3

C

CH(CH

3

)

2

..

O

..

–

..

RO

19.8c Reverse Claisen Condensations As mentioned,formation of the ini-

tial condensation product in the Claisen condensation is usually endothermic. If no

doubly α hydrogens exist for anion formation, β-keto esters will revert to their

component ester starting materials when treated with alkoxide base.

ANSWER In this classic problem, Claisen condensation leads to product A through

the usual addition–elimination scheme:

WORKED PROBLEM 19.33 Even when a full equivalent of sodium ethoxide is used,

ethyl 2-methylpropanoate fails to give useful amounts of a condensation product.

Explain with a careful mechanistic analysis.

19.9 Variations on the Claisen Condensation 993

Claisen condensation (Fig. 19.107). Like the Claisen, the Dieckmann condensation

requires a full equivalent of base. Unless there is sufficient base available to remove

the doubly α hydrogen and drive the reaction to a thermodynamically favorable con-

clusion, the reaction is not successful. A final acidification liberates the β-keto ester

in the absence of base.

A MORE COMPLEX EXAMPLE

THE STANDARD DIECKMANN

..

O

C

–

(–)

..

OH

2

..

+

..

H

3

O

..

+

H

2

O

CH

2

C

CH

3

O

..

CH

3

O

..

CH

3

O

..

CH

3

O

H

..

CH

3

OH

..

HOCH

3

..

OCH

3

..

OCH

3

..

OCH

3

C

O

C

enolate formation

intramolecular

condensation

(addition phase)

acidification

elimination

phase

Removable

doubly 

hydrogen

C

O

..

O

..

–

..

OCH

3

O

..

CH

..

OCH

3

..

OCH

3

C

O

..

O

–

..

C

-Keto esterResonance-stabilized anion

Final product—

the -keto ester

deprotonation

C

H

..

CH

3

O

C

..

CH

3

O

C

H

..

O

..

O

..

O

..

CH

3

O

..

OCH

3

H

3

C

..

(97%)

..

CH

3

O

..

OCH

3

..

O

..

O

H

3

C

1. NaOCH

3

benzene

2. HCl/H

2

O

CH

C

WEB 3D

FIGURE 19.107 The Dieckmann

condensation—an intramolecular

Claisen condensation.

19.9b Crossed Claisen Condensations A naively designed crossed (or

mixed) Claisen condensation is doomed to the same kind of failure as is the

crossed aldol condensation. There are four possible products (Fig. 19.108), and

therefore no easy source of the selectivity we need for a practical synthetic reaction.

O

..

C

CH

3

+

O

..

C

CH

2

CH

3

–

..

/CH

3

OH

..

CH

3

O

..

CH

3

O

..

CH

3

O

..

CH

3

O

..

O

..

C

CH

3

CH

3

O

..

C

..

CH

3

O

..

C

CH

2

CH

3

CH

3

O

..

C

..

CH

3

O

..

C

CH

2

CH

3

O

..

C

..

CH

3

O

..

C

CH

3

O

..

C

+

Na

CH

2

CH

2

CH

FIGURE 19.108 There are four possible—and likely—products from a crossed Claisen condensation.

994 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

PROBLEM 19.34 Show in a general way how these products are produced and write

a mechanism for the formation of one of the four products shown in Figure 19.108.

CH

A SPECIFIC EXAMPLE

A RELATED EXAMPLE

THE “CROSSED CLAISEN”—THE GENERAL MECHANISM

..

H

3

O

..

+

–

(–)

H

2

O

..

H

3

O

..

H

2

O

acidification

..

O

H

2

C

..

C

enolate

elimination

remove doubly

 hydrogen

addition

OR

..

OR

..

OR

..

O

CH

3

CH

2

O

..

C

OCH

2

CH

3

..

O

X

X = Ph Benzoates

= H Formates

= RO Carbonates

..

C

OR

+

O

X

..

.. ..

C

CH

2

O

..

C

..

CH

3

O

..

CH

(56%)

(78%)

O

..

C

..

OR

O

X

..

C

CH

2

O

..

C

..

OR

O

X

..

C

CH

O

..

C

–

+

..

OR

O

X

..

C

CH

2

O

..

C

..

RO

..

OR

..

H

..

OCH

3

..

O

C

NaH/benzene

78 ⬚C

2.

1.

3. /

H

3

O

..

+

..

H

2

O

1.

2. /

..

CH

3

CH

2

C

..

OCH

3

OCH

3

..

O

..

OCH

2

CH

3

..

O

C

–

..

OCH

2

CH

3

..

+

Na

CN

C

N

C

(–)

CH

2

FIGURE 19.109 Successful crossed Claisen condensations.

As in the crossed aldol condensation, there are some simple steps we can take to

improve matters.If one of the starting esters has no α hydrogen,there can be no eno-

late formation from it, and it can act only as an electrophile, never as a nucleophile.

It is helpful to add the ester with an α hydrogen to a mixture of the partner without

an α hydrogen and the base. As the enolate is formed it is more likely to react with

the ester that has no α hydrogen since it is present in excess. Using this technique, a

successful crossed Claisen can be achieved. Benzoates, formates, and carbonates are

examples of esters without α hydrogens and are often used (Fig. 19.109).

19.9 Variations on the Claisen Condensation 995

(a)

H

3

O

+

/H

2

O

H

3

O

+

/H

2

O

..



1. NaOR/HOR

NaOR

HOR

2. H

2

O

..

O

..

COOR

COORROOC

CH

3

CH

3

(b)

(c)

(CH

3

OOC)

2

CH

2

NaOCH

3

/HOCH

3

1. NaOCH

3

/HOCH

3

2.

..

O

..

O

..

O

..

Ph

O

..

Ph

O

..

O

..

COOCH

3

COOCH

3

COO

–

CH

3

OOC

OH

We now have some new reactions to use in problems.Put another way, your lives

are now even more complicated than before,as even more sadistically difficult prob-

lems are within your abilities.There really aren’t too many new things to think about.

You only have to add the Claisen-like processes that involve the loss of a leaving

group from a carbonyl carbon that has been attacked by an enolate. In other words,

you always have to keep the addition–elimination reaction in mind.

But be careful—reactions are driven by thermodynamics,not our concepts of for-

ward and backward.The reverse Claisen is much harder to detect in problems than

the forward version, but no less reasonable mechanistically. As always in matters such

as these, experience counts a lot, and there is no way to get it except to work a lot

of problems.

PROBLEM 19.35 Provide mechanisms for the following reactions:

+

–

..

OCH

3

..

CH

3

O

Na

+

OCH

3

..

OCH

3

HOCH

3

H

CHO

O

..

O

..

O

..

CH

2

COO

–

Na

+

Mixture of stereoisomers

(35%)

PROBLEM 19.36 Provide a mechanism for the following transformation. Be care-

ful. This problem is much harder than it looks at first. Hint: The mechanism

involves a lactone intermediate.

+

(new bond in red)

–

– –

O

OO

–

O

S¬Enzyme

O SR

–

OO

O SR

S¬Enzyme

O

–

OO

O SR

OO

SR

OO

SR

CO

2

FIGURE 19.110 Loss of CO

2

drives this reaction toward product.

19.10 Special Topic: Forward and Reverse Claisen

Condensations in Biology

Both forward and reverse Claisen condensations feature prominently in the biochem-

ical synthesis and degradation of fatty acids (p. 862). A critical step in the construc-

tion of these long-chain carboxylic acids involves enzyme-mediated Claisen

condensations in which activated two-carbon fragments are sewn together. Of

course, Nature has a thermodynamic problem here—how is the endothermicity of

the Claisen condensation to be overcome? The trick is to use an activated malonate

in the condensation. Loss of carbon dioxide is used to drive the equilibrium toward

the product (Fig. 19.110).

996 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

NADPH

reduction

enzymatic

elimination

OO

O

OH

O

S R

FIGURE 19.111 A series of enzyme-

mediated reactions leads to an overall

reduction of one carbonyl group to a

methylene.

Nature’s work is not yet done; first,a reduction of the ketone occurs (Fig.19.111).

The reducing agent is NADPH, a molecule closely related to NADH, which we

saw in Chapter 16 (p. 814). Next, there is an enzyme-mediated elimination reac-

tion. Finally, NADPH participates again, reducing the newly formed double bond

to give the final product. Repetition of the steps in Figures 19.110 and 19.111 even-

tually leads to the fatty acid.

In Chapter 17 (p. 863), we discussed briefly the metabolism of fatty acids,

and raised a question: How is a fatty acid cleaved at the bond to give a car-

bon chain that is two carbons shorter? The problem is that the bond needs

to be activated in some way. If the β carbon were oxidized, we would have

α,β

19.10 Special Topic: Forward and Reverse Claisen Condensations in Biology 997

a β-keto ester all set up for a reverse Claisen condensation. That is exactly the

pathway Nature uses. Here, palmitate is first converted into the SCoA deriva-

tive. Then three enzymes act in sequence to dehydrogenate, hydrate, and dehy-

drogenate again to give the β-keto ester (Fig. 19.112). Now the reverse Claisen

does the first two-carbon cleavage reaction, and metabolism of this fatty acid

has begun.

CH

3

(CH

2

)

12

CH

2

CH

2

C

O

–

Palmitate

OH

α

β

CH

3

(CH

2

)

12

CH

2

CH

2

C

O

acyl-CoA

dehydrogenase

enoyl-CoA

hydrase

β-OH-acyl-CoA

dehydrogenase

ββ

A β-keto ester

SCoA

CH

3

(CH

2

)

12

CHCH

2

C

O

SCoA

CH

3

(CH

2

)

12

CH CH

C

O

SCoA

C

O

C

O

SCoA

+

Myristate

(ready for further degradation)

CH

3

(CH

2

)

12

C

O

–

S

Enzyme

CH

3

(CH

2

)

12

CH

2

C

O

SCoA

CH

2

C

–

O

S Enzyme

H

2

C

OH

S Enzyme

CH

3

(CH

2

)

12

C

SCo

A

Acetyl-CoA

H

3

C

O

C

Summary

There are several variations of the Claisen condensation.The intramolecular vari-

ation is called the Dieckmann condensation. A crossed Claisen condensation is

possible between two different esters, but this reaction can lead to multiple prod-

ucts. Claisen condensations can operate in the “forward” direction or in the

“reverse” direction, and these two processes lead to the construction and decon-

struction of fatty acids.

FIGURE 19.112 Metabolism of a fatty acid containing 16 carbons to one containing 14.

998 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

THE GENERAL CASE

A SPECIFIC EXAMPLE OF THE ROBINSON ANNULATION

..

O

+

–

..

H

2

O

H

2

Oprotonation

..

H

2

O

Na

+–

OCH

2

CH

3

HOCH

2

CH

3

–10 ⬚C

O

..

O

..

O

OH

–

..

OH

+

–

..

OH

–

..

OH

base

Michael

KOH

H

2

O

100 ⬚C

aldol

elimination

O

H

3

C

–

..

....

Note enolate formation

new

enolate

formation

enolate

formation

again

O

H

3

C

–

..

O

..

O

..

O

..

OH

..

O

..

HO

(52%) (85%)

..

O

..

O

..

HO

....

CH

3

CH

3

CH

3

..

O

CH

3

..

O

..

O

H

3

C

..

O

..

O

H

2

C

..

O

–

..

WEB 3D

FIGURE 19.113 The Robinson annulation creates a six-membered ring.

19.11 Condensation Reactions in Combination

Alas, neither Nature nor a typical chemistry teacher shows much mercy, and one is

more likely than not to find condensation reactions in combination. None of these

reactions is particularly difficult to understand when encountered by itself.

Similarly, the synthetic consequences of the reactions are relatively easy to grasp when

they are encountered one at a time, but are by no means simple when combinations

of these reactions are involved. Condensation problems, occasionally called “Fun in

Base” problems (a term that seems to encompass reactions run in acid as well as

base—there are no “Fun in Acid” problems), can be very hard. The best way to

become proficient is to work lots of them, but there are also some practical hints

that may help.

One common sequence is a combination of the Michael addition and aldol

condensations.Indeed, there is a synthetic procedure for constructing six-membered

rings, invented by Sir Robert Robinson (1886–1975) and called the Robinson

annulation, that uses exactly that combination. In this reaction a ketone is treated

with methyl vinyl ketone in base (Fig. 19.113). A Michael addition ensues and

Jones M., Fleming S.A. Organic Chemistry

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