864 CHAPTER 20 ENOLATE AND OTHER CARBON NUCLEOPHILESaskthenerd.com/NOW/CH20/20_3-11.pdf · enol,...

which precipitates as a yellow solid, provides a positive test for the presence of a methylketone. The reaction can also be used in synthesis to convert a methyl ketone to a car-boxylic acid with one less carbon. An example is provided in the following equation:

PROBLEM 20.3Show the products of these reactions:

20.3 Alkylation of Enolate AnionsEnolate anions generated from ketones, esters, and nitriles can be used as nucleophilesin SN2 reactions. This results in the attachment of an alkyl group to the �-carbon in aprocess termed alkylation. Aldehydes are too reactive and cannot usually be alkylatedin this manner. Alkylation of cyclohexanone is illustrated in the following equation:

(62%)

CH2CHœCH2H

Br–CH2CHœCH2..

. .

Na NH2+ –

H

H

O ..

. .

H. .

–O

.. ..

. .

H

–O

O

A strong base must be used to ensure complete deprotonation in this step. The solvent must not have any acidic hydrogens. An ether (diethyl ether, DME, THF, dioxane) or DMF is commonly used.

Because this is an SN2 reaction, it works only when the leaving group is attached to an unhindered carbon (primary or second-ary). When the leaving group is attached to a tertiary carbon, E2 elimination occurs rather than substitution.

1 2

1 2

b)excess Br2

NaOH

H2SO4W

W

CH3

CH3

OXC–CH3H3C–C

OXC–

CH3Br2

Br

a) �CH3CO2H

OXC–

CH3

OXC–

OH3 Cl2�

NaOH

A methyl ketone

(88%)HCl

A carboxylic acid

864 CHAPTER 20 ■ ENOLATE AND OTHER CARBON NUCLEOPHILES

Hornback_Ch20_858-917 12/16/04 12:05 PM Page 864

The base that is used must be strong enough to convert all of the starting ketone (orester or nitrile) to the enolate anion. If it is not strong enough to do so, unwanted reac-tions of the enolate nucleophile with the electrophilic carbon of the remaining ketonemay occur (see Sections 20.5 and 20.6). Lithium diisopropylamide (LDA) is often thebase of choice. It is a very strong base but is not prone to give side reactions in whichit acts as a nucleophile because of the steric hindrance provided by the bulky isopropylgroups. The alkyl halide (or tosylate or mesylate ester) is subject to the usual restric-tions of the SN2 mechanism. The leaving group may be bonded to a primary or sec-ondary carbon but not to a tertiary carbon.

PROBLEM 20.4What reaction would occur if one attempted to use butyllithium to form the enolate an-ion of cyclohexanone?

To avoid the formation of two products, deprotonation of the ketone must produce a single enolate ion. Therefore, the ketone must be symmetrical, like cyclo-hexanone in the preceding example, or have a structure that favors the formation of the enolate ion at only one of the �-carbons, as is the case in the following example:

Esters and nitriles can also be alkylated by this procedure:

This last example shows how two alkyl groups can be added in sequence:

(86%)

1) LDA, DME

2)

3) LDA4) CH3–I

Br

O

O

O

O

(85%)

Br

1) LDA

N2)N

1) LDA, THF

2) CH3(CH2)6CH2–I(90%)

OX

COCH3

OX

COCH3CH3(CH2)6CH2

(88%)

Br

1) NaH

2)

O

Ph

O

Ph

20.3 ■ ALKYLATION OF ENOLATE ANIONS 865


PRACTICE PROBLEM 20.1

Show the product of this reaction:

Strategy

The key to working problems of this type is the same as it has been in previous chap-ters. First identify the nucleophile and the electrophile. In these particular reactions thenucleophile is generated by removal of the most acidic hydrogen (one on the carbon �to the carbonyl or cyano group) to generate an enolate or related anion. This is the nu-cleophile in an SN2 reaction.

Solution

The diisopropylamide anion acts as a base and removes an acidic hydrogen from thecarbon adjacent to the carbonyl group. The resulting enolate anion reacts as a nucleo-phile in an SN2 reaction, displacing the bromine at the primary carbon.


e)1) LDA

2) BrO

O

ClCH2CH2CH2CNNaNH2d)PhCCH2CH2CH3

OX 1) LDA

2) CH3CH2Brc)

1) LDA

2) b)

O

O

Br

1) NaNH2

PhCH2CN

2)

a)Br

H

H

– H CH2CH2CH3

H

. . Br–CH2CH2CH3

Enolate ionnucleophile

O OO

N(i-Pr)2

. .

..–

1) LDA

2) CH3CH2CH2Br

O



20.4 Alkylation of More Stabilized AnionsThe nucleophiles described in the preceding section are strong bases and therefore arequite reactive. This high reactivity sometimes causes problems. We have seen beforethat one way to solve problems caused by a nucleophile that is too reactive is to attacha group to the nucleophilic site that decreases its reactivity. After this new, less reactivereagent, which causes fewer side reactions, is used in the substitution, the extra groupis removed. Although it involves more steps, this overall process provides the sameproduct, often in higher yield, as would be obtained by using the original nucleophile.

One example of this strategy is the preparation of alcohols using acetate anion asthe synthetic equivalent of hydroxide ion in SN2 reactions (see Section 10.2 and Figure 10.1).

In a similar manner the conjugate base of phthalimide is used as the synthetic equiva-lent of amide ion for the preparation of primary amines in the Gabriel synthesis (seeSection 10.6 and Figure 10.5).

In both of these cases a carbonyl group(s) is attached to the nucleophilic atom. Resonancedelocalization of the electron pair makes the anion more stable. It is easier to generate,and its reactions are easier to control. After the substitution reaction has been accom-plished, the carbonyl group(s) is removed, unmasking the desired substitution product.

Now suppose that we want to use the enolate anion derived from acetone (pKa � 20)as a nucleophile in a substitution reaction. This anion requires the use of a very strongbase to generate it, and its high reactivity often causes low yields of the desired prod-uct. Instead, we may choose to use its synthetic equivalent, the enolate anion derivedfrom ethyl acetoacetate (pKa � 11):

Alkylation of the enolate anion derived from ethyl acetoacetate followed by removalof the ester group is known as the acetoacetic ester synthesis and is an excellentmethod for the preparation of methyl ketones. The product of an acetoacetic ester syn-thesis is the same as the product that would be produced by the addition of the same

Enolate anion ofethyl acetoacetate

is the syntheticequivalent of

–

Enolate anion ofacetone

–

OX

. .

± ±C

CH3 CH

OX

± ±C

OCH2CH3

OX

. .

± ±C

CH3 CH2


. .

..NH2

O

O

. ...N– –

CH3–C–O

OX is the synthetic

equivalent of

. .. . ..

. .. . ..H–O

– –

20.4 ■ ALKYLATION OF MORE STABILIZED ANIONS 867


alkyl group to the �-carbon of acetone. First, the enolate ion is generated from the �-ketoester by the use of a moderate base such as sodium ethoxide. Then the enolate ionis alkylated by reaction with an alkyl halide (or alkyl sulfonate ester) in an SN2 reac-tion. This step is subject to the usual SN2 restriction that the leaving group be on a pri-mary or secondary carbon.

The ester group is removed by treating the alkylated �-ketoester with aqueous base, fol-lowed by treatment with acid and heat:

Let’s examine the mechanism of the last part of the acetoacetic ester synthesis,which results in the loss of the ester group. Treatment of the �-ketoester with aqueousbase results in saponification of the ester to form, after acidification, a �-ketoacid. Themechanism for this step was described in Chapter 19.

When the �-ketoacid is heated, carbon dioxide is lost. This step, a decarboxylation,occurs by a mechanism that is quite different from any other that we have encounteredso far. Three bonds are broken and three bonds are formed in a concerted reaction thatproceeds through a cyclic, six-membered transition state. The product of this step is anenol, which tautomerizes to the final product, a ketone:

Note that simple carboxylic acids are quite stable and do not lose carbon dioxide whenheated. For carbon dioxide to be eliminated, the acid must have a carbonyl group at the�-position so that the cyclic mechanism can occur.

CH

H

Oœ

±–

–O

C

OWC

CH3–

OXC

CH3 CHCH2CH2CH2CH3±–C

H

CH3 CH WCH2CH2CH2CH3

–

–

–– �

WCH2CH2CH2CH3 H

W

O

OC

œ

œWW

OW

�

An enol A methyl ketone

CH OCH2CH3±

±–

OXC

OXC

CH3–

OXC

CH3 CH WCH2CH2CH2CH3

±–

OXC

OH CH3CH2OH�±–

WCH2CH2CH2CH3

A �-ketoester A �-ketoacid

1) NaOH, H2O

2) H3O+

CH OCH2CH3±

±–

OXC

OXC

CH3–

OXC

CH3 CH2±CH2CH2CH2CH3±–

WCH2CH2CH2CH3

(61%)1) NaOH, H2O

2) H3O+, �

CH2 OCH2CH3±

±–

OXC

Ethyl acetoacetate

OXC

CH3–

OXC

CH3 CH±–

OXC

OCH2CH3 WCH2CH2CH2CH3

±–1) NaOEt, EtOH

2) CH3CH2CH2CH2Br(72%)



Another example of the acetoacetic ester synthesis is shown in the following equation:

Note, again, that the final product, 2-hexanone, is the same product that would resultfrom the reaction of the enolate anion of acetone with 1-bromopropane.

The malonic ester synthesis is similar to the acetoacetic ester synthesis. It beginswith deprotonation of diethyl malonate (pKa � 11) to produce an enolate anion that isthe synthetic equivalent of the enolate anion derived from acetic acid:

In the malonic ester synthesis this enolate ion is alkylated in the same manner as in theacetoacetic ester synthesis. Saponification of the alkylated diester produces a diacid.The carbonyl group of either of the acid groups is at the �-position relative to the otheracid group. Therefore, when the diacid is heated, carbon dioxide is lost in the same man-ner as in the acetoacetic ester synthesis. The difference is that the product is a car-boxylic acid in the malonic ester synthesis rather than the methyl ketone that isproduced in the acetoacetic ester synthesis. The loss of carbon dioxide from a substi-tuted malonic acid to produce a monoacid is illustrated in the following equation:

As was the case in the decarboxylation that occurs in the acetoacetic ester synthesis, itis the presence of a carbonyl group at the �-position of the carboxylic acid that allowscarbon dioxide to be lost when the compound is heated.

Examples of the malonic ester synthesis are provided in the following equations:

1) NaOEt

Br W

2) CH3CHCH2CH3

EtOCCH2COEt EtOCCHCOEt WCH3CHCH2CH3

(84%)

(65%)

OX

CH3CH2CHCH2COH

CH3W

OX

OX

OX

OX

EtOCCH2COEt

(74%)

OX

CH3(CH2)5CH2–CH2COH

OX

OX

1) KOH, H2O

2) H3O+, �

1) NaOEt

2) CH3(CH2)5CH2Br3) KOH, H2O4) H3O

+, �

H

Oœ

±–

–O

C

OWC

HO–

OXC

HO CH2±R±–

OWC

O

OC

H

HO CH WR

CH WR

–

–

–– �

œ

œWW

�


Enolate ion ofdiethyl malonate

Enolate ion ofacetic acid

. .

CH–± ±–

OXC

EtO–

OXC

HO

. .

CH2–––

OEt

OXC

CH2 OEt±

±–

OXC

OXC

CH3–

OXC

CH3 CH±–

OXC

CH3 CH2±CH2CH2CH3±–

OXC

OEt WCH2CH2CH3

±–

2-Hexanone

(47%)

1) NaOEt

2) CH3CH2CH2Br

1) NaOH, H2O

2) H3O+, �



In both the acetoacetic ester synthesis and the malonic ester synthesis, it is possibleto add two different alkyl groups to the �-carbon in sequential steps. First the enolateion is generated by reaction with sodium ethoxide and alkylated. Then the enolate ionof the alkylated product is generated by reaction with a second equivalent of sodiumethoxide, and that anion is alkylated with another alkyl halide. An example is providedby the following equation:

Although the acetoacetic ester synthesis and the malonic ester synthesis are used toprepare ketones and carboxylic acids, the same alkylation, without the hydrolysis and decarboxylation steps, can be employed to prepare substituted �-ketoesters and �-diesters. In fact, any compound with two anion stabilizing groups on the same car-bon can be deprotonated and then alkylated by the same general procedure. Several ex-amples are shown in the following equations. The first example shows the alkylation ofa �-ketoester. Close examination shows the similarity of the starting material to ethylacetoacetate. Although sodium hydride is used as a base in this example, sodium ethox-ide could also be employed.

This next example shows the alkylation of a �-diketone (pKa � 9). Because this compoundis more acidic than a �-ketoester or a �-diester, the weaker base potassium carbonate wasused. However, sodium ethoxide would also be satisfactory as the base for this reaction:

This last example shows the addition of two alkyl groups to a dinitrile (pKa � 11). Be-cause the alkyl groups to be added are identical, they do not have to be added in se-quence. Instead, the reaction is conducted by adding two equivalents of base and twoequivalents of the alkylating agent, benzyl chloride, simultaneously:

CH2

±–NPC CPN

C±

–NPC CPN

CH2Ph(75%)±

PhCH2–

2 NaH, DMSO

2 PhCH2ClA dinitrile

CH2 CH3±

±–

OXC

OXC

CH3–

CH CH3 (77%)±

±–

OXC

OXC

CH3–

WCH3

K2CO3CH3I

A �-diketone

1) NaH, DMF

2)

HOXCOEt

Br OEt(85%)

O

O

OXCOEt

O

OEt

OA �-ketoester

EtOCCHCOEt

(74%)

OX

W CH3

(83%)

CH3(CH2)8CH2CHCOH

OX

H3C W

OX

EtOCCH2COEt

OX

OX 1) NaOEt

2) CH3(CH2)8CH2Br3) KOH, H2O4) H3O

+, �

1) NaOEt

2) CH3Br





Show a synthesis of 2-heptanone using the acetoacetic or malonic ester synthesis:

Strategy

Decide which synthesis to use. The acetoacetic ester synthesis is used to prepare methylketones, and the malonic ester synthesis is used to prepare carboxylic acids. Both syn-theses provide a method to add alkyl groups to the �-carbon. Therefore, next identifythe group or groups that must be added to the �-carbon. Remember that the �-carbonis the nucleophile, so the groups to be attached must be the electrophile in the SN2 re-action; they must have a leaving group bonded to the carbon to which the new bond isto be formed.

CH3CH2CH2CH2CH2CCH3

OX

2-Heptanone

e) NPCCH2CCH3

OX 1) NaOEt, EtOH

2) CH3CH2CH2Cl

d) CH3CCH2COEt

OX

OX 1) NaOEt, EtOH

2) CH3CH2CH2CH2Br3) NaOEt, EtOH4) CH3I

1) NaOH, H2O

2) H3O+, �

a) CH3CCH2COEt

OX

OX

b) EtOCCH2COEt

OX

OX

c)OEt

1) NaOEt, EtOH

2) PhCH2Br

1) NaOEt, EtOH

2) CH3CH2Br

1) NaOH, H2O

2) H3O+, �

1) NaOEt, EtOH

2) CH3CH2CH2Br

1) NaOH, H2O

2) H3O+, �

O O


Click Coached Tutorial Problemsto practice Alkylations ofEnolate Anions.


Solution

The acetoacetic ester synthesis is used to prepare methyl ketones such as this. In thisexample, a butyl group must be attached to the enolate nucleophile.

Use a base to generate the enolate anion nucleophile. Then add the alkyl group with aleaving group on the electrophilic carbon. Finally, decarboxylate the alkylated product.

PROBLEM 20.7Show syntheses of these compounds using the acetoacetic or malonic ester syntheses:

PROBLEM 20.8Show how these compounds could be synthesized using alkylation reactions:

d)N

N

c)

N

N

b)

O

O

O

Ph

a)

OO

e)

O

d) OH

O

c) OH

O

b) Ph

O

a) CH3CHCH2CO2H

CH3W

CH3CCH2COEt

OX

OX

CH3CCHCOEt

OX

CH3CCH2CH2CH2CH2CH3

OX

W CH2CH2CH2CH3

OX1) NaOEt, EtOH

2) CH3CH2CH2CH2Br

1) NaOH, H2O

2) H3O+, �

This is the bond to be formedin this acetoacetic ester synthesis.

CH3CH2CH2CH2–CH2CCH3

OX



20.5 The Aldol CondensationYou may have noticed that aldehydes were conspicuously absent from the examples ofalkylation reactions presented in Sections 20.3 and 20.4. This is due to the high reac-tivity of the carbonyl carbon of an aldehyde as an electrophile. When an enolate anionnucleophile is generated from an aldehyde, under most circumstances it rapidly reactswith the electrophilic carbonyl carbon of an un-ionized aldehyde molecule. Althoughthis reaction, known as the aldol condensation, interferes with the alkylation of alde-hydes, it is a very useful synthetic reaction in its own right. The aldol condensation ofethanal is shown in the following equation:

The product, 3-hydroxybutanal, is also known as aldol and gives rise to the name forthe whole class of reactions.

The mechanism for this reaction is shown in Figure 20.3. For this mechanism to oc-cur, both the enolate anion derived from the aldehyde and the un-ionized aldehyde mustbe present. To ensure that this is the case, hydroxide ion is most commonly used as thebase. Because hydroxide ion is a weaker base than the aldehyde enolate anion, only a small amount of the enolate anion is produced. Most of the aldehyde remains

(75%)2 CH3CH CH3CH–CH2CH

Aldol(3-hydroxybutanal)

Ethanal

OH W

OX

OXNaOH

20.5 ■ THE ALDOL CONDENSATION 873

CH2–C–H

OX

H–

H–

O–H

. .. ...

O–HWH

. .

..

. .

..

CH3–CH–CH2–C–H

OW

. .

.. OX

. .

....

CH3–CH–CH2–C–H

OW

. .

OX

. .

....

O

. .

..WW

O

CH3–C–H

. ...WW

–

–

CH2–C–H. .

–CH2–C–H

OW

. .

.. ..

–

–

This part of the mechanism is just like the mechanism for the addition reactions of Chapter 18. The enolate nucleophile adds to the carbonyl carbon of a second aldehyde molecule, and the negative oxygen removes a proton from water. This step regenerates hydroxide ion, so the reaction is base catalyzed.

The base, hydroxide ion, removes an acidic hydrogen from the �-carbon of the aldehyde. The conjugate base of the aldehyde is a stronger base than hydroxide, so the equilibrium for this first step favors the reactants.

However, enough enolate ion nucleophile is present to react with the electrophilic carbonyl carbon of a second aldehyde molecule.

1

1

2

2

3

3

Figure 20.3

MECHANISM OF THE ALDOL CONDENSATION.

Click Mechanisms inMotion to view theMechanism of theAldol Condensation.


un-ionized and is available for reaction as the electrophile. Note that the strong basesdescribed in Section 20.3, which would tend to convert most of the aldehyde moleculesto enolate ions, are not used in aldol condensations. The addition of the enolate nu-cleophile to the aldehyde follows the same mechanism as the addition of other nucleo-philes that were described in Chapter 18. Remember that the �-carbon of one aldehydemolecule bonds to the carbonyl carbon of a second aldehyde molecule, as illustrated inthe following example:

If the aldol condensation is conducted under more vigorous conditions (higher tem-perature, longer reaction time, and/or stronger base), elimination of water to form an�,�-unsaturated aldehyde usually occurs. This elimination is illustrated in the followingexample. Note that the �-carbon of one molecule is now doubly bonded to the carbonylcarbon of the other. (This text uses the symbol for heat, �, to indicate the vigorous con-ditions that cause eliminations to occur in these aldol condensations, even though otherconditions might have been used.)

This elimination occurs by a somewhat different mechanism than those described inChapter 9. Because the hydrogen on the �-carbon is relatively acidic, it is removed bythe base in the first step to produce an enolate anion. Then hydroxide ion is lost from theenolate ion in the second step. Because this step is intramolecular and the product is sta-bilized by conjugation of its CC double bond with the CO double bond of the carbonylgroup, even a poor leaving group such as hydroxide ion can leave. (This is an exampleof the E1cb mechanism described in the Focus On box on page 333 in Chapter 9.) Mostaldol condensations are run under conditions that favor dehydration because the stabil-ity of the product helps drive the equilibrium in the desired direction, resulting in ahigher yield. For example, the reaction of butanal shown previously results in a 75%yield of the aldol product. If the reaction is conducted so that dehydration occurs, theyield of the conjugated product is 97%.

(97%)2 CH3CH2CH2CH CH3CH2CH2CHœCCH H2O� WCH3CH2

OX

OXNaOH

�

CH2CH3

±±

– OHW

H

OX

CHCH3CH2CH2CH±C

OHW

WCH3CH2

OX

CH3CH2CH2CH±C±CH

OX

CH3CH2CH2CH–CCH WCH3CH2

H2O�

An �,�-unsaturated aldehyde

–OH

�

. .. . ..H–O

. .

–

–

–

(75%)2 CH3CH2CH2CH CH3CH2CH2CH–CHCH

OH W

WCH3CH2

OX

OXKOH



Another example is shown in the following equation:



Strategy

The key to determining the products of an aldol condensation is to remember that thenucleophile is an enolate anion, which is formed at the �-carbon of the aldehyde, andthe electrophile is the carbonyl carbon of another aldehyde molecule. Therefore theproduct has the �-carbon of one aldehyde molecule bonded to the carbonyl carbon ofanother aldehyde molecule. Under milder conditions an OH group remains on the car-bonyl carbon of the electrophile, whereas under vigorous conditions the �-carbon andthe carbonyl carbon are connected by a double bond.

Solution


PROBLEM 20.10Show all of the steps in the mechanism for this reaction:

2 CH3CH2CH2CH

OX

CH3CH2CH2CHCHCH

OX

OHW

WCH3CH2

NaOH

c) 2 PhCH2CH

OX NaOH

�b)

NaOH

�H

O

a) CH2CH2

OX KOH

�-Carbonnucleophile

Carbonyl carbonelectrophile

CH3CH2CH CH3CHCH–

O OX

. . CH3CH2CHCHCH

OHW

OX

W CH3

WW

2 CH3CH2CH

OX NaOH

(79%)HNaOEt

�

O

2 H

O



Ketones are less reactive electrophiles than aldehydes. Therefore, the aldol conden-sation of ketones is not often used because the equilibrium is unfavorable. However, theintramolecular condensation of diketones is useful if the size of the resulting ring is fa-vorable (formation of five- and six-membered rings).

Often, it is desirable to conduct an aldol condensation in which the nucleophile andthe electrophile are derived from different compounds. In general, such mixed aldolcondensations, involving two different aldehydes, result in the formation of severalproducts and for this reason are not useful. For example, the reaction of ethanal andpropanal results in the formation of four products because there are two possible eno-late nucleophiles and two carbonyl electrophiles:

Mixed aldol condensations can be employed if one of the aldehydes has no hydro-gens on the �-carbon, so it cannot form an enolate ion and can only act as the elec-trophilic partner in the reaction. Aromatic aldehydes are especially useful in this rolebecause the dehydration product has additional stabilization from the conjugation of thenewly formed CC double bond with the aromatic ring. This stabilization makes theequilibrium for the formation of this product more favorable.

(80%)� CH3CH2CH

OX

C–H

CH3CHCH3

OX

CH3–C–CH X

C–H

OX

NaOH

�

CH3CHCH3

�

CH3CH

CH3CHœCHCH CH3CH2CHœCHCH�

OX

Ethanal

Propanal

CH3CH2CH

OX

OX

OX

CH3CHœCCH WCH3

WCH3

CH3CH2CHœCCH��

OX

OX

NaOH

�

(42%)

A diketone

–

±

OXC

CH2 CH3

CH2–CCH3

O

–

––

œ

CH3

NaOH

H2O�

O



With an aromatic aldehyde as the electrophilic partner, the nucleophilic enolateion can also be derived from a ketone or a nitrile. As illustrated in the following ex-amples, this enables the aldol condensation to be used to form a wide variety ofcompounds:



Strategy

Again, the key is to identify the nucleophile (the enolate anion) and the electrophile (thecarbonyl carbon).

� CH3CCH3

OX NaOH

�

±H

OXC

(91%)� H

O

Ph

PhPh

N

Ph

NaOEt

�

N

�

O X

PhCH (75%)

CHPhNaOH

�

O O

(94%)2

�NaOH

�

O

Ph Ph

OO

HPh

(85%)�

H

NaOH

�

O

Ph

Ph

O

Ph

O

HPh



Solution

In this example the enolate anion can be derived only from acetone. The electrophile isthe more reactive carbonyl carbon, that of benzaldehyde.



� � H2ONaOH

�

OX

PhCH

CH Ph

OO

e) �KOH

H2O, �

±H

OXCCH2CN

CH3

d) NCCH2COEt �

Br

±œHO

C

OX KOH

EtOH, �c)

NaOH

H2O, �

O O

b) � ±CH3CH2

OXC

NaOH

�H3C

±H

OXC

a) CH3CPh�

OX NaOH

�

±H

OXCO

±

±

±

H

H CCH3CXC

OX

CH2CCH3–

OX. .

WW

±H

O

C




Show how the aldol condensation could be used to synthesize this compound:

Strategy

First identify the carbon–carbon bond that could be formed in an aldol condensation.This is the bond between the �-carbon of the carbonyl group of the product and the car-bon that is either doubly bonded to it or has a hydroxy substituent. Disconnection of thisbond gives the fragments needed for the aldol condensation. Remember, the �-carbonof the product is the nucleophilic carbon of the enolate anion and the carbon to whichit is bonded is the electrophilic carbonyl carbon.

Solution

So the synthesis is

PROBLEM 20.13Show how the aldol condensation could be used to synthesize these compounds.

d)

O

c)CHPh

O

b)

O

OHW

a) CH3CHCH2CHCHCH

OX

CH3W

WCH3CH W CH3

2 CH3CH2CH2CH2CH2CH

OX

OHW

WCH3CH2CH2CH2

CH3CH2CH2CH2CH2CH–CHCH

OXNaOH

�-Carbon

Disconnect here

CH3CH2CH2CH2CH2CH–CHCH

OX

CH3CH2CH2CH2CH2CH CHCH–

O OX

OHW

WCH3CH2CH2CH2

WCH3CH2CH2CH2

. .WW

CH3CH2CH2CH2CH2CHCHCH

OX

OHW

WCH3CH2CH2CH2


Click Coached Tutorial Problemsto practice more AldolCondensations.


20.6 Ester CondensationsSo far, we have seen that an enolate anion is able to act as a nucleophile in an SN2 re-action (Sections 20.3 and 20.4) and also in an addition reaction to the carbonyl groupof an aldehyde in the aldol condensation (Section 20.5). It also can act as a nucleophilein a substitution reaction with the carbonyl group of an ester as the electrophile. Whenan ester is treated with a base such as sodium ethoxide, the enolate ion that is producedcan react with another molecule of the same ester. The product has the �-carbon of oneester molecule bonded to the carbonyl carbon of a second ester molecule, replacing thealkoxy group. Examples of this reaction, called the Claisen ester condensation, areprovided by the following equations:


Focus On Biological Chemistry

The Reverse Aldol Reaction in MetabolismThe initial product of an aldol condensation has a hydroxy group on the �-carbon to acarbonyl group. Sugars also have hydroxy groups on the �-carbon to their carbonylgroups, so they can be viewed as products of aldol condensations. In fact, a reverse al-dol condensation is used in the metabolism of glucose (glycolysis) to cleave this six-carbon sugar into two three-carbon sugars.

To cleave a six-carbon sugar into two three-carbon fragments by a reverse aldolcondensation, there must be a carbonyl group at C-2 and a hydroxy group at C-4.Therefore, glucose is first isomerized to fructose during its metabolism. The substratefor the cleavage reaction is the diphosphate ester, fructose-1,6-bisphosphate. In step

, a proton is removed from the hydroxy group on C-4. The bond between C-3 and C-4 is then broken in step , which is the reverse of the aldol condensation, produc-ing glyceraldehyde-3-phosphate (GAP) and the enolate ion of dihydroxyacetone phos-phate (DHAP). This enolate ion is protonated in step . GAP and DHAP can beinterconverted by the same process that interconverts glucose and fructose and thusprovide a common intermediate for further metabolism.

..B–

Fructose-1,6-bisphosphate

CH2OPO32�

WCœO WCHOH WCH–O–H WCHOH WCH2OPO3

2�

CH2OPO32�

WC O WCHOH WCH–O WCHOH WCH2OPO3

2�

1

2

3

4

5

6

. .

... .

CH2OPO32�

WCœO WCHOH

CH O WCHOH WCH2OPO3

2�

CH2OPO32�

WCœO W

H–CHOH. .

H–B

GAP

�

––

––

–

–

DHAP

1 2

3

3

2

1


The mechanism for this reaction, shown in Figure 20.4, has similarities to those ofboth an aldol condensation (see Figure 20.3) and an ester saponification (see Figure 19.4).As was the case with the aldol condensation, the presence of both the enolate ion and the

EtOH�CH3–C–CH2–C–OEt (80%)

OX

OX

2 CH3–C–OEt

OX 1) NaOEt, EtOH

2) H3O+

EtOH�CH3CH2CH2CCHCOEt (76%)

OX

OX

WCH3CH2

2 CH3CH2CH2COEt

OX 1) NaOEt, EtOH

2) H3O+

20.6 ■ ESTER CONDENSATIONS 881

The reverse aldol reaction is catalyzed by an enzyme called aldolase. One of theroles of the enzyme is to stabilize the enolate anion intermediate because such ions aretoo basic to be produced under physiological conditions. In animals, aldolase accom-plishes this task by forming an imine bond between the carbonyl group of fructose-1,6-bisphosphate and the amino group of a lysine amino acid of the enzyme. As a re-sult, the product of the reverse aldol step is an enamine derived from DHAP rather thanits enolate anion. (Section 20.8 shows that enamines are the synthetic equivalents ofenolate anions.) The formation of the strongly basic enolate anion is avoided. Thisprocess is outlined here:

CH2OPO32�

WCœO WCHOH WCH–O–H WCHOH WCH2OPO3

2�

NH2(CH2)4– Enzyme

CH2OPO32�

WC WCHOH WCH–O–H WCHOH WCH2OPO3

2�

+Enzyme EnzymeNH(CH2)4–

CH2OPO32�

WC–

CHOH

CH O WCHOH WCH2OPO3

2�

NH(CH2)4–––

––

– –

Fructose-1,6-bisphosphate

..B

An amino group of a lysine amino acid of the enzyme reacts with the carbonyl group of fructose-1,6-bisphosphate to form a protonated imine. See Figure 18.3 for the mechanism of this reaction.

When the reverse aldol reaction occurs, an enamine, rather than a strongly basic enolate anion, is produced.

This enamine is hydrolyzed to DHAP, freeing the enzyme to catalyze another reverse aldol cleavage.

1 2

1 2


neutral ester is necessary for the reaction to occur. Therefore, ethoxide ion is used as thebase because it is a weaker base than the enolate ion. Step 4 of the mechanism, in whichethoxide ion removes the acidic hydrogen, is of critical importance. The equilibria for thefirst three steps of the reaction are all unfavorable. But the equilibrium for step 4 is veryfavorable because the product of this step, the conjugate base of a �-ketoester, is the weak-est base in the reaction. The formation of this weak base drives the equilibria to the prod-uct. If step 4 cannot occur, no significant amount of the �-ketoester is present in thereaction mixture and the condensation fails.

When sodium ethoxide is used as the base, the ester condensation fails with esters thathave only one hydrogen on the �-carbon. The equilibrium favors the reactants because theequilibrium driving step (step 4 of the mechanism in Figure 20.4) cannot occur.

EtOH�2 CH3CHCOEt

OX

H3C W

CH3CHC–C––C–OEt

No H, so step 4 cannot occur

OX

H3C W

OX

CH3W

W CH3

NaOEt NaOEt

step 4


CH2–C–OCH2CH3

OX

H–

OCH2CH3–

. .. ...

CH3–C–CH2–C–OCH2CH3

OW

W OCH2CH3

OX

. ... ..

CH3–C–CH2–C–OCH2CH3

OX

OX

CH3–C–CH–C–OCH2CH3

OX

OXH3O

+

equilibriumdriving step

. .

CH3–C–CH–C–OCH2CH3

OWW

OX

WH

�

OCH2CH3

. .. ...

CH2–C–OCH2CH3–

CH3–C–OCH2CH3

OX

O

. .

WW

workupstep

–

–

–

The base, ethoxide ion, removes an acidic hydrogen from the �-carbon of the ester.

The enolate anion reacts as a nucleophile, attacking the electrophilic carbonyl carbon of another ester molecule. This is another example of the substitution mechanism presented in Chapter 19.

Ethoxide ion leaves as the electrons on the oxygen reform the carbonyl double bond.

Workup with acid protonates the anion, producing the �-ketoester.

The equilibria for , , and are all un-favorable. However, the formation of the weakest base in this reaction , drives the overall equilibrium to completion. If this step cannot occur, the equilibrium will be unfav-orable and the reaction will not occur.

The hydrogen on the carbon between the two carbonyl groups of the �-ketoester is quite acidic and is removed by ethoxide ion.

1

2

1 2

3

4

4

5

4

5

1 2 3

3

Active Figure 20.4

MECHANISM OF THE CLAISEN ESTER CONDENSATION. Test yourself on the concepts in this figure at OrganicChemistryNow.


However, good yields of the condensation product can be obtained if a very strong baseis used. In this case the equilibrium is driven toward the products because the ethoxideion that is formed on the product side of the equation is weaker than the base on the re-actant side of the equation. Note that only enough base is used to deprotonate one-halfof the ester.

In fact, even with an ester that gives an acceptable yield of the condensation productwith sodium ethoxide as the base, a better yield is often obtained when a stronger baseis employed.

Intramolecular ester condensation reactions are called Dieckmann condensationsand are very useful ring-forming reactions. Examples are shown in the following equa-tions. In the second equation the yield is only 54% if sodium ethoxide is used as thebase.


d) EtOC(CH2)5COEt

OX

OX 1) NaOEt, EtOH

2) H3O+

a) 2 CH3CH2COEt

OX

b) 2 CH3CH2CHCOEt

OX

WCH3

c) 2

OX

PhCH2COEt

1) NaOEt, EtOH

2) H3O+

1) NaOEt, EtOH

2) H3O+

1) LDA

2) H3O+

(80%)OEt

OEt

OEt

OEt

(90%)

Ethyl 3-methyl-2-oxo-cyclohexanecarboxylate

O OO

O

1) NaNH2

2) H3O+

1) NaH

2) H3O+

OO

OEt

OEt

O

O

CH3CHCOEt2 KH�

OX

H3C W

CH3CHC–C––C–OEt �

OX

H3C W

OX

CH3W

H2 �–OEtK

+

W CH3



PROBLEM 20.15The second example of a Dieckmann condensation shown earlier produces ethyl 3-methyl-2-oxocyclohexanecarboxylate in 90% yield. What other cyclic product mighthave been formed in this reaction? Explain why the actual product is favored rather thanthis other product.

As was the case with the aldol condensation, mixed ester condensations can be use-ful if one of the components can only act as the electrophile—that is, if it cannot forman enolate anion (no hydrogens on the �-carbon). The following esters are most com-monly employed in this role:

The nucleophile, an enolate or related anion, can be obtained by deprotonation of an es-ter, ketone, or nitrile. Examples are provided by the following equations:

Finally, note that the products of most of these reactions are �-dicarbonyl com-pounds. They can be alkylated in the same manner as ethyl acetoacetate and diethyl

(75%)� EtOC–COEt

OX

OX

PhCH2–CPN

NPC W

OX

OX

PhCH–C–C–OEt1) NaOEt

2) H3O+

1) NaOEt

2) H3O+ (71%)�

O O

PhEtOPh

O O

PhPh

OX

OX

OX

PhCH2CH2CH2COEt EtOC–COEt�

OX

OX

OX

EtOC±C

PhCH2CH2CHCOEt (81%)

(74%)H–C–OEt�

OX H

(94%)EtO–C–OEt�

OX

W

OEt

OXC–

1) NaOEt

2) H3O+

1) NaH

2) H3O+

1) NaH

2) H3O+

O O O

O O

H–C–OEt

Ethylformate

OX

EtO–C–OEt

Diethylcarbonate

OX

EtO–C–C–OEt

Diethyloxalate

OX

OX

Ph–C–OEt

Ethylbenzoate

OX



malonate, and they can also be decarboxylated if one of the two carbonyl groups is anester group. This makes them quite useful in synthesis.



Strategy

Again the best approach is to identify the site where the nucleophilic enolate anionforms, the �-carbon with the most acidic hydrogen. This carbon becomes bonded to thecarbonyl carbon of the ester electrophile in the final product.

Solution

PROBLEM 20.16Show the products of these reactions.


HCOEt�

OX H–1) NaOEt, EtOH

2) H3O+

O O OXC

d) EtOCOEt�

OX

�c)

OX

PhCOEt1) NaOEt, EtOH

2) H3O+

1) NaOEt, EtOH

2) H3O+

1) NaOEt, EtOH

2) CH3 I

O

O

b)

OX

� HCOEtPhCCH3

OX 1) NaOEt, EtOH

2) H3O+a) PhCH2CN EtOCOEt�

OX 1) NaOEt, EtOH

2) H3O+

EtOC–COEt CH3CHCOEt

O OX

OX

. . EtOC–C–CHCOEt

OX

OX

OX

W CH3

WW –

CH3CH2COEt �

OX

EtOC±COEt

OX

OX 1) NaOEt, EtOH

2) H3O+



PROBLEM 20.18Show how ester condensation reactions could be used to synthesize these compounds:

e)

OXCOEtH3C

O

d)

OXCOEt

O

c)H

O OCH3

a)

Cl

b)

O O

Ph

O O


Focus On

An Industrial Aldol ReactionThe development of the perfumery ingredient Flosal, which has a strong jasminelikeodor, provides an interesting example of how discoveries are sometimes made in an in-dustrial setting. A chemical company was producing substantial amounts of heptanalas a by-product of one of its processes and needed to find some use for this aldehyde.The company’s chemists ran a number of reactions using this compound to determinewhether any of the products might be useful. They found that the product of a mixedaldol condensation between heptanal and benzaldehyde has a very powerful jasmineodor. This compound, which was given the trade name Flosal, became an important in-gredient in soaps, perfumes, and cosmetics and was at one time prepared in amountsin excess of 100,000 pounds per year.

The synthesis must be carefully controlled to minimize the formation of the aldolproduct that results from the reaction of two molecules of heptanal (2-pentyl-2-nonenal)because this compound has an unpleasant, rancid odor.

PROBLEM 20.19Show the structure of the product of the aldol condensation of heptanal under vigorous con-ditions. How would you minimize the formation of this product in the synthesis of Flosal?

�CH3(CH2)4CH2CH PhCH Ph � H2O

FlosalBenzaldehydeHeptanal

OX

OX

X CH

CH3(CH2)4CCH

OX

NaOH

�


20.7 Carbon and Hydrogen Leaving GroupsPreviously we learned that the carbonyl group of an aldehyde or a ketone does not undergo the substitution reactions of Chapter 19 because hydride ion and carban-ions are strong bases and poor leaving groups. There are, however, some special sit-uations in which these species do leave. Some of these exceptions are described inthis section.

A carbanion can act as a leaving group if it is stabilized somehow. We have alreadyseen several examples of this. For example, CX3�, a methyl carbanion stabilized by theinductive effect of three electronegative halogen atoms, leaves in the haloform reaction(see Section 20.2).

Another example is provided by the equilibrium in the aldol condensation. Exami-nation of the mechanism for this reaction (see Figure 20.3) shows that an enolate an-ion leaves in the reverse of the second step of this reaction. Again, it is thestabilization of the carbanion, this time by resonance, that enables the enolate anionto leave.

A similar process is described in the Focus On box titled “The Reverse Aldol Reaction in Metabolism” on page 880. The Claisen ester condensation also has anequilibrium step in which an enolate anion leaves in the reverse of the step (see Fig-ure 20.4).

Reactions in which hydride leaves are less common but can occur if other reac-tions are precluded and the hydride is transferred directly to an electrophile. One example occurs when an aldehyde without any hydrogens on its �-carbon is treatedwith NaOH or KOH. (If the aldehyde has hydrogens on its �-carbon, the aldol con-densation is faster and occurs instead.) In this reaction, called the Cannizzaro reac-tion, two molecules of aldehyde react. One is oxidized to a carboxylate anion and theother is reduced to a primary alcohol. The mechanism for this reaction is shown in Figure 20.5. The reaction begins in the same manner as the reactions described inChapter 18; a hydroxide ion nucleophile attacks the carbonyl carbon of the aldehydeto form an anion. The reaction now begins to resemble the reactions in Chapter 19.

CH3COCH2CH3 CH2COCH2CH3�WW

O OX

OX

. ...

CH3C–CH2COCH2CH3W

W

O

CH3CH2O

. .

.. ..–

–

. .CH3CH CH2CH�

WWO O

XOX

. .

..

CH3CH–CH2CHWO

. .

.. ..–

–. .

CH3CH2CH2C–OH CI3–

–

�WW

O

. .

..

..CH3CH2CH2C–CI3W

W

O

. .

.. ..

OH

20.7 ■ CARBON AND HYDROGEN LEAVING GROUPS 887


As the electrons on the negative oxygen reform the double bond, hydride ion beginsto leave. But hydride ion is too poor a leaving group to leave without help, so it istransferred directly to the carbonyl carbon of a second aldehyde molecule, as shownin path A of the mechanism. An acid–base reaction completes the process. Undermore strongly basic conditions, the mechanism may change slightly and follow pathB. In this case the initially formed anion loses a proton to the base to form a dianion.Although the concentration of the dianion is less than that of the monoanion, it do-nates hydride more rapidly than the monoanion because of its two negative charges.After protonation of the alkoxide anion by the solvent, the same products are pro-


. .. .

O

. .

..WW

Ph–C–H

O

. .

..WW

Ph–C–O–H �

O

. ...WW

Ph–C–H

Path A

Path B

Ph–C–H

OW

W

. .

.. ..–

O–H

. .. ...

–

O–H. ...Ph–C–H

OW

W

. .

.. ..–

H

. .

O

. .

..WW

Ph–C–O �

O

. .

..WW

Ph–C–H

Ph–C–H

OW

W

. .

.. ..–

–O. ... ..

Ph–C–HW

H–O

W

. .

..

. .

..–

H

O–H

. .. ...

–

In path B, the initial anion loses a proton to the base, forming a dianion.

Transfer of a hydride from the dianion to the second aldehyde molecule leads to the carboxylate anion and the conjugate base of the alcohol, which then obtains a proton from the solvent. Although the dianion is less stable than the initial anion, it is a stronger hydride donor because of its two negative charges.

The Cannizzaro reaction begins with the attack of the hydroxide ion nucleophile at the electrophilic carbon of the aldehyde, as described in Chapter 18.

In path A, the electrons on the negative oxygen reform the double bond as hydride leaves. Hydride is too basic to leave by itself, so it is transferred tothe electrophilic carbonyl carbon of another aldehyde molecule in a concerted step. This step is relatively slow and occurs only when no other reaction pathways, such as an aldol condensation, are available.

An acid–base step completes the reaction.

The reaction mayproceed by eitherpath A or path B,depending on thereaction conditions.

1

1

2

2

3

3

5

5

6

6

4

4

Figure 20.5

MECHANISM OF THE CANNIZZARO REACTION.


duced from path B as from path A. An example of the Cannizzaro reaction is providedin the following equation:

PROBLEM 20.20Show the product of this reaction:

PROBLEM 20.21�-Ketoesters that have two substituents on the �-carbon undergo fragmentation whentreated with ethoxide anion as shown in the following equation. Suggest a mechanismfor this reaction.

20.8 EnaminesBecause of the importance of carbon nucleophiles in synthesis, organic chemists havespent considerable effort developing others in addition to the enolate anions that havealready been described. Several of these other carbon nucleophiles are presented in thisand the following section. This section describes the use of enamines.

As discussed in Section 18.8, enamines are prepared by the reaction of a secondaryamine with a ketone in the presence of an acid catalyst. The equilibrium is usually dri-ven toward product formation by removal of water.

TsOH

toluene�

An enamine

(93%)

O

NWH

. .N

Nucleophiliccarbon

–. .

N+

H2O�

NaOEtCH3C–C–COEt EtOH�

OX

OX W

W

CH3

CH3

CH3CH–COEt

OXW

CH3

CH3COEt �

OX

H3C

OXC

H– 1) NaOH

2) H2SO4

�

XO

C–H

63%

2 O

XO

C–OHO

C–HO W

W

H

OH

63%

1) NaOH

2) H2SO4

20.8 ■ ENAMINES 889


Because of the contribution of structures such as the one on the right to the resonancehybrid, the �-carbon of an enamine is nucleophilic. However, an enamine is a muchweaker nucleophile than an enolate anion. For it to react in the SN2 reaction, the alkylhalide electrophile must be very reactive (see Table 8.1). An enamine can also be usedas a nucleophile in substitution reactions with acyl chlorides. The reactive electrophilescommonly used in reactions with enamines are:

After the enamine has been used as a nucleophile, it can easily be hydrolyzed backto the ketone and the secondary amine by treatment with aqueous acid. This is simplythe reverse of the process used to prepare it. Overall, enamines serve as the syntheticequivalent of ketone enolate anions. Examples are provided in the following equations:


c)

OX

N 1) CH3CH2CCl

2) H3O+a)

1) BrCH2CO2Et

2) H3O+

N

Brb)

O

TsOH1) NWH

2)

3) H3O+

2) H3O+

(67%)

N CH3OO

1)CH3

–

OXC

Cl

(81%)

(66%)

. .N O

CH3

1) CH3I

2) H3O+

TsOH1) NWH

2) CH2œCHCH2–Br3) H3O

+

O O

CH2CHœCH2

CH3–I

Methyliodide

CH2œCHCH2

Allylichalides

XW

Benzylichalides

X

CH2CR

Halides on�-carbons of

ketones and esters

OX

RCCl

Acylchlorides

OX

XWCH2

–



20.9 Other Carbon NucleophilesMany other carbon nucleophiles have been developed. Only two additional types are in-troduced here, but both provide interesting variations on the themes that have been pre-sented so far.

The first of these nucleophiles is derived from a dithiane. A dithiane can be preparedby the reaction of an aldehyde with 1,3-propanedithiol. This reaction, described in Sec-tion 18.9, is the sulfur analog of acetal formation and requires a proton or Lewis acidcatalyst:

The hydrogen on the carbon attached to the two sulfur atoms is weakly acidic (pKa �31) and can be removed by reaction with a strong base, such as butyllithium. (Butyl-lithium is also a nucleophile, and therefore it is not used to generate enolate anions fromcarbonyl compounds. However, the dithiane is not electrophilic, so butyllithium can beused as the base in this reaction.)

The acidity of the dithiane can be attributed to the stabilization of the conjugate base bythe inductive effect of the sulfurs.

The dithiane anion is a good nucleophile in SN2 reactions. After it has been al-kylated, the thioacetal group can be removed by hydrolysis using Hg2� as a Lewis acidcatalyst.

The following equation provides an example of the overall process:

H3C CH2Ph (67%)

OXC– –H3C H

OXC– –

1) , BF3

2) BuLi3) PhCH2–Br4) Hg2�, H2O

SH SH

(98%) (80%)

H3C CH2Ph

OXC– –

1) BuLi

2) PhCH2–Br

Hg2�

H2OH3C H

––S

CS

H3C CH2Ph––

SC

S

WH3C

CS... .

BuLi

H3C H. . –––

SC

S S ... .

OXC

HH3C

BF3� (86%)

A dithiane1,3-PropanedithiolAcetaldehyde

––

H3C H––

SC

SSH SH

20.9 ■ OTHER CARBON NUCLEOPHILES 891


The nucleophile obtained by deprotonation of a dithiane serves as the synthetic equiv-alent of an acyl anion, a species that is too unstable to be prepared directly.

In all of the reactions that have been presented until this one, a carbonyl carbon has al-ways reacted as an electrophile. An acyl anion, however, has a nucleophilic carbonylcarbon. Thus, the use of a nucleophile obtained by deprotonation of a dithiane providesan example of the formal reversal of the normal polarity of a functional group. Such polarity reversal is termed umpolung, using the German word for reversed polarity.

Additional examples of the use of a dithiane to generate an acyl anion syntheticequivalent are provided by the following equations:

The final type of carbon nucleophile that is discussed in this chapter is a dianion. Insome cases, treatment of an anion with a very strong base can remove a second protonto form a dianion. As an example, the reaction of 2,4-pentanedione with one equivalentof base removes a proton from the carbon between the two carbonyl groups. If this an-ion is treated with a second equivalent of a strong base, such as potassium amide, a sec-ond proton can be removed to form a dienolate anion:

OXC

CH3 CH2

(82%)

2,4-Nonanedione

2,4-Pentanedione Dienolateanion

±–

OXC

CH2±CH2CH2CH2CH3±–

1) 1 CH3CH2CH2CH2Br

2) H3O+

CH2 CH3±

±–

OXC

OXC

CH3–

OXC

CH3 CH3CH– – –

±–

OXC±–

More basic and more nucleophilic site

. .

OXC

CH3 CH2CH±–

OXC±–

. . . .

KNH2 KNH2

O

Hg2+

H2OBr

(66%) (80%)

Ph

(74%)

1) BuLi2)

3) BuLi4)5) Hg2+, H2O

Br

Ph H––

SC

S

H H––

SC

S

S S

Ph

1) BuLi

2)

O

Br

OXC

H3C

An acyl anion


–

WH3C

C–

–

. ...S S



When this dianion is reacted with one equivalent of an alkyl halide, the more basic siteacts as a nucleophile. The addition of acid neutralizes the remaining anion. Overall, thisprocess allows the more basic site to be alkylated preferentially. The following equationshows another example:

The same strategy can be extended to the alkylation of carboxylic acids at the �-carbon,as illustrated in the following example:


PROBLEM 20.24Suggest methods to accomplish these transformations. More than one step may be necessary.

c) PhOH

O

PhOH

O

a)

Ph

Ph

b) H

OEt

O O O

Ph

OO

d) HPh

O

2) BuLi3)4) Hg2+, H2O

SH SH1) , BF3

Br

c) CH3CCH2COCH3

OX

OX 1) NaH

2) BuLi3) CH3CH2Br4) H3O

+

1) BuLi

2)

3) Hg2+, H2O

b)H

S S Ia) 2

1) 2 LDA

2)OH

O

Br

1) 2 LDA

2) PhCH2CH2Br

3) H3O+

CH–C–OH PhCH2CH2–C––COH

CH3W

W CH3

OXH3C

H3C(90%)

OX–

–

1) NaH2) BuLi

3)

4) H3O+

(85%)OCH3

O O

Br

OCH3

O O

20.9 ■ OTHER CARBON NUCLEOPHILES 893


20.10 Conjugate AdditionsThe conjugate addition of nucleophiles to �,�-unsaturated carbonyl compounds at the �-position was described in Section 18.10. Enolate and related carbanion nucleo-philes also add in a conjugate manner to �,�-unsaturated carbonyl compounds in aprocess known as the Michael reaction or Michael addition. In many of the exam-ples the enolate ion is one that is stabilized by two carbonyl (or similar) groups. The�,�-unsaturated compound is called the Michael acceptor.

The mechanism for the Michael reaction is shown in Figure 20.6. Only a catalyticamount of base is needed because the initial adduct is itself an enolate anion and is ba-sic enough to deprotonate the dicarbonyl compound, allowing additional reaction to oc-cur. Other examples of the Michael reaction are provided in the following equations:

With stronger bases, less stable enolate anions can be generated and used in the Michaelreaction:

� (70%)NaNH2

Ph

O

Ph

2-Propenenitrile 2-Phenylcyclohexanone

O

N

N

�

CO2Et WCH2 WCO2Et

(90%)

–

–CO2Et

CO2Et

CH

�

CPN WCH2 WCO2Et

NPCW C W

EtO2C

CH2 W

CH XCH2

OEt

OXC

(73%)––

H–CH2

OEt

OXC––

NaOEt

NaOEt

OO

�

C–OCH2CH3

C–CH3

–CH2

CH2W

CH3

OXC

Michaelacceptor

(92%)

OX

W CH2W

XO

CH3CH2O–C

H3C–C

OX

W CHW

XO

– –

CH XCH2

CH3

OXC–– NaOEt





Solution

Identify the electrophile and the nucleophile. The base (NaOEt) removes the mostacidic hydrogen, the one on the carbon between the carbonyl groups of ethyl acetoac-etate, to generate the enolate anion. This nucleophile then attacks at the �-carbon of theMichael acceptor:

CH2 CHCOEtCH3C–CH–COEt �

OX

OX

OX

. . CH3CCH–CH2CH2COEt

OX

OX

W CO2Et

–––

CH2œCHCOEt �

OX

CH3CCH2COEt

OX

OX NaOEt

EtOH

20.10 ■ CONJUGATE ADDITIONS 895

CH3–C

HC–H

EtOC

OX

W

W

XO

CH3–C

EtOC

HC

OX

W

W

XO

O–CH2CH3

. .. ...

..

H3C–C–CH–CH2–CH

OX . .

OX

CCH3W

W COEt

XO

H3C–C–CH–CH2–CH

OX

HW

OX

CCH3W

W COEt

XO

OX

H3C–C–CH CH2CH3–C

HC–H

EtOC

OX

W

W

XO

––

–

–

–

The enolate anion of the �-dicarbonyl compound is generated in the usual manner.

The enolate nucleophile adds to the �-carbon of an �,�-unsaturated ketone (or ester or nitrile), which is called the Michael acceptor.

The product, formed in 92% yield in this case, has a bond from the �-carbon of the original enolate ion to the �-carbon of the Michael acceptor.

The product of this addition is an enolate ion.

This ion reacts as a base with ethyl aceto-acetate to regenerate another enolate ion, so only a catalytic amount of base is needed for the reaction.

Enolate anion

Michael acceptor

1

1

2

2

3

3

Figure 20.6

MECHANISM OF THEMICHAEL REACTION.



PROBLEM 20.26Explain why the Michael reaction of 2-phenylcyclohexanone with 2-propenenitrilegives the product shown in the equation on page 894 rather than this product:

The Michael reaction in combination with an aldol condensation provides a usefulmethod for the construction of six-membered rings in a process termed the Robinsonannulation. In the following example a tertiary amine is used as the base to catalyzethe conjugate addition. Then, treatment with sodium hydroxide causes an intramolecu-lar aldol condensation to occur.

Often, the Michael addition product is not isolated. Instead, the intramolecular aldolcondensation occurs immediately, and the new six-membered ring is formed, as shownin the following equation. (However, when you are attempting to write the product ofsuch a reaction, it is best to first write the product of Michael addition and then writethe final product that results from the aldol condensation.)

Michaeladdition

Aldolcondensation

(85%)(70%)

O

OO

O

O

NaOH

�R3N

Ph

O

N

a) CH2œCHCCH3 EtOCCH2COEt�

OX

OX

OX

b) PhCHœCHCPh EtOCCH2CN�

OX

OX

c) CH2œCHCN�NaNH2

NaOEt

EtOH

NaOEt

EtOH

O


Click Coached Tutorial Problemsfor more practice with theMichael Reaction.


PROBLEM 20.27Show the intermediate aldol product in the Robinson annulation reaction of ethyl 2-oxocyclohexanecarboxylate with 1-penten-3-one.


Show the product of this Robinson annulation:

Solution

The base removes the most acidic hydrogen, and the resulting enolate anion undergoesa conjugate addition with the Michael acceptor:

This product then undergoes an aldol condensation to give the final product:

H2C–. .

O

O

O O

O

H–. .

O

O

OH3CO

O

O

�

OO

O

KOH

�

(59%)�

CO2EtO

NaOEt

�

Ethyl2-oxocyclohexanecarboxylate

1-Penten-3-one

CO2Et

O O

20.10 ■ CONJUGATE ADDITIONS 897


PROBLEM 20.28Show the product of this reaction:

20.11 SynthesisThe reactions presented in this chapter are very important in synthesis because they all re-sult in the formation of carbon–carbon bonds. As we have seen, the best way to approacha synthesis problem is to employ retrosynthetic analysis, that is, to work backward fromthe target molecule to simpler compounds until a readily available starting material isreached. (Elias J. Corey, winner of the 1990 Nobel Prize in chemistry for “his develop-ment of the theory and methodology of organic synthesis,” coined the term retrosyntheticanalysis and formalized much of its logic. He also developed numerous new reagents, in-cluding the dithiane anion nucleophile discussed in Section 20.9, and synthesized a largenumber of natural products, including many prostaglandins [Section 28.9].) Recall that itis helpful in retrosynthetic analysis to recognize that certain structural features in the tar-get suggest certain reactions. For example, we learned in Chapter 18 that an alcohol tar-get compound suggests that a Grignard reaction might be used in its synthesis.

Let’s look at the reactions presented in this chapter in terms of their products so thatwe might more easily recognize the synthetic reactions that are suggested by the pres-ence of certain features in the target compound.

Alkylations of ketones, esters, and nitriles add an alkyl group to the �-carbon ofcompounds containing these functional groups. Therefore, a target molecule that is aketone, ester, or nitrile with an alkyl group(s) attached to its �-carbon suggests the useof one of these alkylation reactions, as illustrated in the following equations using ret-rosynthetic arrows:

The acetoacetic ester synthesis produces a methyl ketone with an alkyl group(s)substituted on the �-carbon, whereas the malonic ester synthesis produces a

ketonealkylation

R'–CH2–C–R"

OX

R'–CH–C–R"

Starting MaterialTarget Compound

OX

RW

esteralkylation

R'–CH2–C–OEt

OX

R'–CH–C–OEt

OX

RW

nitrilealkylation

R'–CH2–CPNR'–CH–CPN

RW

�

OO

CH3

O

KOH

�



carboxylic acid with an alkyl group(s) substituted on the �-carbon. Note that these targets can sometimes also be synthesized by the direct alkylation of a ketoneor ester:

The presence of a carbon–carbon double bond conjugated to a carbonyl group (an�,�-unsaturated aldehyde, ketone, and so on) in the target compound suggests that analdol condensation be employed:

If the target compound has two carbonyl groups attached to the same carbon, an es-ter condensation is suggested. Note that such a target could also be prepared by alkyla-tion of a �-dicarbonyl compound:

The presence of two carbonyl groups (or other functional groups that are capable ofstabilizing a carbanion) in a 1,5-relationship suggests the use of a Michael addition toprepare that target compound:

Michaeladdition

12

3

5 4

OX

OX

�

R CH2W

CH2W

R'–C–CH–C–R"

OXC

OX

OX

R'–C–CH2–C–R"

–– R CHX

CH2

OXC––

estercondensation

R'–CH2–C–R"

OX

R–C–CH–C–R"

OX

alkylationR–C–CH2–C–R"

OX

OX

OX

W R'

OX

�

R–C–OEt

aldolcondensation

R'–CH2–C–R

OX

OX

�RR'

CXC

OXC

R"–C–H

–– –

HR"––

acetoacetic estersynthesis

EtO–C–CH2–C–CH3

OX

OX

R'–CH–C–CH3

OX

RW

malonic estersynthesis

EtO–C–CH2–C–OEt

OX

OX

R'–CH–C–OH

OX

RW

20.11 ■ SYNTHESIS 899


Finally, the presence of a cyclohexenone ring in the target suggests that a Robinsonannulation might be employed in its synthesis:

Let’s try a synthesis. Suppose the target is ethyl 2-methyl-3-oxo-2-propylpentanoate.The presence of the �-ketoester functionality suggests employing an alkylation reactionand/or an ester condensation. In one potential pathway, the propyl group can be attachedby alkylation of a simpler �-ketoester. Further retrosynthetic analysis suggests that thenew target (ethyl 2-methyl-3-oxopentanoate) can be prepared from ethyl propanoate bya Claisen ester condensation.

Written in the forward direction, the synthesis is

As another example, consider the following target compound. A Michael reaction issuggested by the observation that the carbonyl group of the ketone and the ester car-bonyl group (or the carbon of the cyano group) are in a 1,5-relationship. The compound

1) NaOEt

2) CH3CH2CH2Cl

2 CH3CH2COEt CH3CH2CC

OX

COEt

OX

H3C H

OX

––––

CH3CH2CC

OX

COEt

OX

H3C CH2CH2CH3

––––

1) NaOEt

2) H3O+

alkylation

Ethyl 2-methyl-3-oxo-2-propylpentanoate(target)

Ethyl 2-methyl-3-oxopentanoate

Ethyl propanoate

estercondensation

COEtCH3CH2C

CH2CH2CH3H3CC

OX

OX

–––

– COEtCH3CH2C

CH3CH2COEt

HH3CC

OX

OX

OX

–––

–

Robinsonannulation

�

R'R'

R

RR

OO

R

O



required for the Michael addition is an �,�-unsaturated ketone, suggesting that a mixedaldol condensation be used to prepare it.

Written in the forward direction, the synthesis is

Like any other endeavor, the only way to become proficient in designing synthesesis practice. Work the problems and, when working them, examine each target compoundfor structural features that suggest certain reactions. Remember that most targets can bereached by numerous pathways, so if your route does not match the one presented in theanswers, do not despair. Check all of the reactions that you use to ensure that they areappropriate. If your pathway seems a reasonable route to the target, then it may be asgood as, or even better than, the one shown in the answer.

PROBLEM 20.29Show syntheses of these compounds from the indicated starting materials:

f) from cyclohexanonePh Ph

O

e) from cyclopentanone

OXC

OCH3

O

–

d) from butanal

O

c) from ethyl butanoate

Ph

OH

O

b) from butanalH

O

a) from ethyl acetoacetate

O

1)

NaOEt

2) H3O+

Ph

NaOH

� CPN

CO2EtPh

OCPN

CO2Et

Ph

O

Ph H

�

O

O

Michaeladdition

Target

aldolcondensation

CPN

CO2Et Ph H

�

PhPh

O

O

O O

20.11 ■ SYNTHESIS 901


Review of Mastery GoalsAfter completing this chapter, you should be able to:

Show the products of any of the reactions discussed in this chapter. (Problems20.30, 20.31, 20.32, 20.41, 20.47, 20.54, and 20.55)

Show the mechanism for any of these reactions. (Problems 20.33, 20.34, 20.35, 20.36,20.40, 20.44, 20.45, 20.46, 20.51, 20.52, 20.53, 20.56, 20.58, 20.62, and 20.63)

Use these reactions in combination with reactions from previous chapters to syn-thesize compounds. (Problems 20.37, 20.38, 20.39, 20.42, 20.43, 20.48, 20.49,20.50, and 20.57)

Visual Summary of Key Reactions

A large number of reactions have been presented in this chapter. However, all of these re-actions involve an enolate ion (or a related species) acting as a nucleophile (see Table 20.2).This nucleophile reacts with one of the electrophiles discussed in Chapters 8, 18, and 19(see Table 20.3). The nucleophile can bond to the electrophilic carbon of an alkyl halide (orsulfonate ester) in an SN2 reaction, to the electrophilic carbonyl carbon of an aldehyde orketone in an addition reaction (an aldol condensation), to the electrophilic carbonyl carbonof an ester in an addition reaction (an ester condensation) or to the electrophilic �-carbonof an �,�-unsaturated compound in a conjugate addition (Michael reaction). These possi-bilities are summarized in the following equations:

SN2�

OX

–C–C– W

R–L

. .

OX W

–C–C–R W

aldolcondensation

�

OX

–C–C– W H

. .

O

R–C–H

OX

OHWW

–C–C–CHR W H

OX W

–C–CœCHRWW

�

OX

–C–C– W

. .

O

R–C–OEt

OX

OXW

–C–C–C–R W

WW

�

OX

–C–C– W

. .

OX

–C C–C– W W

OX

OXW W

–C–C–C–C–C– W W W

––

HW

–

–

–

–

estercondensation

Michaelreaction

h) from cyclopentanone

OXC

OCH3

O

–g) from diethyl malonate

Ph

NH2

O


Click Mastery Goal Quiz to testhow well you have met thesegoals.


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