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Chapter 22
Copyright Š 2010 Pearson Education, Inc.
Organic Chemistry, 7th
Edition
L. G. Wade, Jr.
Condensations and Alpha Substitutions of
Carbonyl Compounds
Chapter 22 2
Alpha Substitution
ī‚§ Alpha substitution is the substitution of one of the
hydrogens attached to the alpha-carbon for an
electrophile.
ī‚§ The reaction occurs through an enolate ion
intermediate.
Chapter 22 3
Condensation with an Aldehyde or
Ketone
ī‚§ The enolate ion attacks the carbonyl group to form an
alkoxide.
ī‚§ Protonation of the alkoxide gives the addition product:
a β-hydroxy carbonyl compound.
Chapter 22 4
Condensation with Esters
ī‚§ The enolate adds to the ester to form a tetrahedral
intermediate.
ī‚§ Elimination of the leaving group (alkoxide) gives the
substitution product (a β-carbonyl compound).
Chapter 22 5
Keto–Enol Tautomers
O
H
H
OH
H
keto form
(99.99%)
enol form
(0.01%)
ī‚§ Tautomerization is an interconversion of
isomers that occur through the migration of a
proton and the movement of a double bond.
ī‚§ Tautomers are not resonance form.
Chapter 22 6
Base–Catalyzed Tautomerism
ī‚§ In the presence of strong bases, ketones and aldehydes act as
weak proton acids.
ī‚§ A proton on the Îą carbon is abstracted to form a resonance-
stabilized enolate ion with the negative charge spread over a
carbon atom and an oxygen atom.
ī‚§ The equilibrium favors the keto form over the enolate ion.
Chapter 22 7
Acid-Catalyzed Tautomerism
ī‚§ In acid, a proton is moved from the Îą-carbon
to oxygen by first protonating oxygen and
then removing a proton from the carbon.
Chapter 22 8
Racemization
ī‚§ For aldehydes and ketones, the keto form is greatly
favored at equilibrium.
ī‚§ If a chiral carbon has an enolizable hydrogen atom, a
trace of acid or base allows that carbon to invert its
configuration, with the enol serving as the
intermediate. This is called racemization.
Chapter 22 9
Acidity of Îą Hydrogens
ī‚§ pKa for Îą H of aldehyde or ketone ~20.
ī‚§ Much more acidic than alkane or alkene
(pKa > 40) or alkyne (pKa = 25).
ī‚§ Less acidic than water (pKa = 15.7) or
alcohol (pKa = 16–19).
ī‚§ Only a small amount of enolate ion is
present at equilibrium.
Chapter 22 10
Formation and Stability of
Enolate Ions
ī‚§ The equilibrium mixture contains only a small
fraction of the deprotonated, enolate form.
Chapter 22 11
Energy Diagram of Enolate
Reaction
ī‚§ Even though the keto–enol tautomerism equilibrium
favors the keto form, addition of an electrophile shifts
the equilibrium toward the formation of more enol.
Chapter 22 12
Synthesis of Lithium
Diisopropylamine (LDA)
ī‚§ LDA is made by using an alkyllithium reagent
to deprotonate diisopropylamine.
Chapter 22 13
Enolate of Cyclohexanone
ī‚§ When LDA reacts with a ketone, it abstracts
the Îą-proton to form the lithium salt of the
enolate.
Chapter 22 14
The Îą Halogenation of Ketones
ī‚§ When a ketone is treated with a halogen and a base,
an Îą halogenation reaction occurs.
ī‚§ The reaction is called base-promoted, rather than
base-catalyzed, because a full equivalent of the base
is consumed in the reaction.
Chapter 22 15
Base-Promoted Halogenation
Mechanism
ī‚§ The base-promoted halogenation takes place by a
nucleophilic attack of an enolate ion on the
electrophilic halogen molecule.
ī‚§ The products are the halogenated ketone and a
halide ion.
Chapter 22 16
Multiple Halogenations
ī‚§ The Îą-haloketone produced is more reactive than
ketone because the enolate ion is stabilized by the
electron-withdrawing halogen.
ī‚§ The second halogenation occurs faster than the first.
ī‚§ Because of the tendency for multiple halogenations
this base-promoted halogenation is not widely used
to prepare monohalogenated ketones.
O
H
Cl
Cl2
OH , H2O
_
O
Cl
Cl
O
Cl
ClCl
O
Cl
ClCl
Cl
Chapter 22 17
Bromoform Reaction
ī‚§ A methyl ketone reacts with a halogen under
strongly basic conditions to give a carboxylate
ion and a molecule of haloform.
ī‚§ The trihalomethyl intermediate is not isolated.
Chapter 22 18
Mechanism of Haloform
Formation
ī‚§ The trihalomethyl ketone reacts with hydroxide ion to
give a carboxylic acid.
ī‚§ A fast proton exchange gives a carboxylate ion and a
haloform.
ī‚§ When Cl2 is used, chloroform is formed; Br2 forms
bromoform ; and I2 forms iodoform.
Chapter 22 19
Positive Iodoform Test
for Alcohols
ī‚§ The iodine oxidizes the alcohol to a methyl
ketone and it will give a positive iodoform
test.
ī‚§ Iodoform (CHI3) is a yellow solid that will
precipitate out of solution.
Chapter 22 20
Propose a mechanism for the reaction of 3-pentanone with sodium hydroxide and bromine to give 2-
bromo-3-pentanone.
In the presence of sodium hydroxide, a small amount of 3-pentanone is present as its enolate.
The enolate reacts with bromine to give the observed product.
Solved Problem 1
Solution
Chapter 22 21
Acid-Catalyzed Îą Halogenation
ī‚§ Ketones also undergo acid-catalyzed Îą halogenation.
ī‚§ Acidic halogenation may replace one or more alpha
hydrogens depending on how much halogen is used.
ī‚§ Acetic acid serves as both the solvent and the acid
catalyst.
Chapter 22 22
Mechanism of Acid-Catalyzed
Îą Halogenation
ī‚§ The mechanism of acid-catalyzed halogenation
involves attack of the enol form of the ketone on the
electrophile halogen molecule.
ī‚§ Loss of a proton gives the haloketone and the
hydrogen halide.
Chapter 22 23
Propose a mechanism for the acid-catalyzed conversion of cyclohexanone to 2-chlorocyclohexanone.
Under acid catalysis, the ketone is in equilibrium with its enol form.
The enol acts as a weak nucleophile, attacking chlorine to give a resonance-stabilized intermediate.
Loss of a proton gives the product.
Solved Problem 2
Solution
Chapter 22 24
Hell–Volhard–Zelinsky (HVZ)
Reaction
ī‚§ The HVZ reaction replaces a hydrogen atom with a
bromine atom on the alpha-carbon of a carboxylic
acid (Îą-bromoacid).
ī‚§ The acid is treated with bromine and phosphorus
tribromide, followed by hydrolysis.
Chapter 22 25
Hell–Volhard–Zelinski Reaction:
Step 1
ī‚§ The enol form of the acyl bromide serves as a
nucleophilic intermediate.
ī‚§ The first step is the formation of acyl bromide, which
enolizes more easily than does the acid.
Chapter 22 26
Hell–Volhard–Zelinski Reaction:
Step 2
ī‚§ The enol is nucleophilic, so it attacks bromine
to give the alpha-brominated acyl bromide.
ī‚§ In the last step of the reaction, the acyl
bromide is hydrolyzed by water to the
carboxylic acid.
Chapter 22 27
Alkylation of Enolate Ions
ī‚§ Because the enolate has two nucleophilic sites (the
oxygen and the Îą carbon), it can react at either of
these sites.
ī‚§ The reaction usually takes place primarily at the
α−carbon, forming a new C—C bond.
Chapter 22 28
Îą Alkylation of Enolate Ions
ī‚§ LDA forms the enolate.
ī‚§ The enolate acts as the nucleophile and attacks the
partially positive carbon of the alkyl halide, displacing
the halide and forming a C—C bond.
Chapter 22 29
Enamine Formation
ī‚§ Ketones or aldehydes react with a secondary
amine to form enamines.
ī‚§ The enamine has a nucleophilic Îą-carbon,
which can be used to attack electrophiles.
Chapter 22 30
Mechanism of Enolate Formation
ī‚§ An enamine results from the reaction of a
ketone or aldehyde with a secondary amine.
Chapter 22 31
Electrostatic Potential Map of an
Enamine
ī‚§ The electrostatic potential map (EPM) of a simple
enamine shows a high negative electrostatic potential
(red) near the Îą-carbon atom of the double bond.
ī‚§ This is the nucleophilic carbon atom of the enamine.
Chapter 22 32
Alkylation of an Enamine
ī‚§ Enamines displace halides from reactive alkyl
halides, giving alkylated iminium salts.
ī‚§ The alkylated iminium salt can be hydrolyzed
to the ketone under acidic conditions.
Chapter 22 33
Acylation of Enamines
ī‚§ The enamine attacks the acyl halide, forming an acyl
iminium salt.
ī‚§ Hydrolysis of the iminium salt produces the β-
diketone as the final product.
Chapter 22 34
Aldol Condensation
ī‚§ Under basic conditions, the aldol condensation involves the
nucleophilic addition of an enolate ion to another carbonyl
group.
ī‚§ When the reaction is carried out at low temperatures, the β-
hydroxy carbonyl compound can be isolated.
ī‚§ Heating will dehydrate the aldol product to the α−β unsaturated
compound.
Chapter 22 35
Base-Catalyzed Aldol
Condensation: Step 1
ī‚§ During Step 1, the base removes the Îą-
proton, forming the enolate ion.
ī‚§ The enolate ion has a nucleophilic Îą-carbon.
Chapter 22 36
Base-Catalyzed Aldol
Condensation: Step 2
ī‚§ The enolate attacks the carbonyl carbon of a
second molecule of carbonyl compound.
Chapter 22 37
Base-Catalyzed Aldol
Condensation: Step 3
ī‚§ Protonation of the alkoxide gives the aldol
product.
Chapter 22 38
Dehydration of Aldol Products
ī‚§ Heating a basic or acidic aldol dehydration of the
alcohol functional group.
ī‚§ The product is a Îą,β-unsaturated conjugated
aldehyde or ketone.
ī‚§ An Aldol condensation, followed by dehydration,
forms a new carbon–carbon double bond.
Chapter 22 39
Crossed Aldol Condensations
Chapter 22 40
Successful Crossed Aldol
Condensations
Chapter 22 41
Propose a mechanism for the base-catalyzed aldol condensation of acetone (Figure 22-2).
The first step is formation of the enolate to serve as a nucleophile.
The second step is a nucleophilic attack by the enolate on another molecule of acetone. Protonation
gives the aldol product.
Solved Problem 3
Solution
Chapter 22 42
Aldol Cyclization
ī‚§ Intramolecular aldol reactions of diketones are often
used for making five- and six-membered rings.
ī‚§ Rings smaller or larger than five or six members are
not favored due to ring strain or entropy.
Chapter 22 43
Retrosynthesis of Aldol
Condensation
Chapter 22 44
Claisen Condensation
ī‚§ The Claisen condensation results when an
ester molecule undergoes nucleophilic acyl
substitution by an enolate.
Chapter 22 45
Dieckman Condensation
Chapter 22 46
Crossed Claisen
ī‚§ Two different esters can be used, but
one ester should have no Îą hydrogens.
ī‚§ Useful esters are benzoates, formates,
carbonates, and oxalates.
ī‚§ Ketones (pKa = 20) may also react with
an ester to form a β-diketone.
Chapter 22 47
Crossed Claisen Condensation
ī‚§ In a crossed Claisen condensation, an ester
without Îą hydrogens serves as the
electrophilic component.
Chapter 22 48
Crossed Claisen Condensation
with Ketones and Esters
ī‚§ Crossed Claisen condensation between ketones and
esters are also possible.
ī‚§ Ketones are more acidic than esters, and the ketone
component is more likely to deprotonate and serve as
the enolate component in the condensation.
Chapter 22 49
Crossed Claisen Mechanism
ī‚§ The ketone enolate attacks the ester, which
undergoes nucleophilic acyl substitution, and
thereby, acylates the ketone.
Chapter 22 50
Propose a mechanism for the self-condensation of ethyl acetate to give ethyl acetoacetate.
The first step is formation of the ester enolate. The equilibrium for this step lies far to the
left; ethoxide deprotonates only a small fraction of the ester.
The enolate ion attacks another molecule of the ester; expulsion of ethoxide ion gives ethyl
acetoacetate.
Solved Problem 4
Solution
Chapter 22 51
In the presence of ethoxide ion, ethyl acetoacetate is deprotonated to give its enolate. This exothermic
deprotonation helps to drive the reaction to completion.
When the reaction is complete, the enolate ion is reprotonated to give ethyl acetoacetate.
Solved Problem 4 (Continued)
Solution (Continued)
Chapter 22 52
Show what ester would undergo Claisen condensation to give the following β-keto ester.
First, break the structure apart at the ι, β bond (ι, β to the ester carbonyl). This is the bond formed in
the Claisen condensation.
Solved Problem 5
Solution
Chapter 22 53
Next, replace the Îą proton that was lost, and replace the alkoxy group that was lost from the carbonyl.
Two molecules of methyl 3-phenylpropionate result.
Now draw out the reaction. Sodium methoxide is used as the base because the reactants are methyl
esters.
Solved Problem 5 (Continued)
Solution (Continued)
Chapter 22 54
Chapter 22 55
Malonic Ester Synthesis
ī‚§ The malonic ester synthesis makes substituted
derivatives of acetic acids.
ī‚§ Malonic ester is alkylated or acylated on the carbon
that is alpha to both carbonyl groups, and the
resulting derivative is hydrolyzed and allowed to
decarboxylate.
Chapter 22 56
Decarboxylation of the
Alkylmalonic Acid
ī‚§ Decarboxylation takes place through a cyclic
transition state, initially giving an enol form
that quickly tautomerizes to the product.
Chapter 22 57
Example of the Malonic
Synthesis
Chapter 22 58
Dialkylation of Malonic Ester
Chapter 22 59
Show how the malonic ester synthesis is used to prepare 2-benzylbutanoic acid.
2-Benzylbutanoic acid is a substituted acetic acid having the substituents Ph–CH2– and CH3CH2–.
Adding these substituents to the enolate of malonic ester eventually gives the correct
product.
Solved Problem 6
Solution
Chapter 22 60
Acetoacetic Ester Synthesis
ī‚§ The acetoacetic ester synthesis is similar to
the malonic ester synthesis, but the final
products are ketones.
Chapter 22 61
Alkylation of Acetoacetic Ester
ī‚§ Ethoxide ion completely deprotonates acetoacetic
ester.
ī‚§ The resulting enolate is alkylated by an unhindered
alkyl halide or tosylate to give an alkylacetoacetic
ester.
Chapter 22 62
Hydrolysis of Alkylacetoacetic
Ester
ī‚§ Acidic hydrolysis of the alkylacetoacetic ester initially
gives an alkylacetoacetic acid, which is a β-keto acid.
ī‚§ The keto group in the β-position promotes
decarboxylation to form a substituted version of
acetone.
Chapter 22 63
Show how the acetoacetic ester synthesis is used to make 3-propylhex-5-en-2-one.
The target compound is acetone with an n-propyl group and an allyl group as substituents:
Solved Problem 7
Solution
Chapter 22 64
Hydrolysis proceeds with decarboxylation to give the disubstituted acetone product.
With an n-propyl halide and an allyl halide as the alkylating agents, the acetoacetic ester synthesis
should produce 3-propyl-5-hexen-2-one. Two alkylation steps give the required substitution:
Solved Problem 7 (Continued)
Solution (Continued)
Chapter 22 65
Conjugate Additions: The
Michael Reaction
ī‚§ Îą,β-unsaturated carbonyl compounds have unusually
electrophilic double bonds.
ī‚§ The β-carbon is electrophilic because it shares the
partial positive charge of the carbonyl carbon through
resonance.
Chapter 22 66
1,2-Addition and 1,4-Addition
ī‚§ When attack occurs at the carbonyl group,
protonation of the oxygen leads to a 1,2-addition.
ī‚§ When attack occurs at the β-position, the oxygen
atom is the fourth atom counting from the
nucleophile, and the addition is called a 1,4-addition.
Donors and Acceptors
Chapter 22 67
Chapter 22 68
1,4-Addition of an Enolate to
Methyl Vinyl Ketone (MVK)
ī‚§ An enolate will do a 1,4-attack on the Îą,β-
unsaturated ketone (MVK).
Chapter 22 69
Show how the following diketone might be synthesized using a Michael addition.
A Michael addition would have formed a new bond at the β carbon of the acceptor. Therefore,
we break this molecule apart at the β,Îŗ bond.
Solved Problem 8
Solution
Chapter 22 70
Robinson Annulation
ī‚§ With enough base, the product of the Michael
reaction undergoes a spontaneous intramolecular
aldol condensation, usually with dehydration, to give
a six-membered ring—a conjugated cyclohexenone.
Chapter 22 71
Robinson Mechanism

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Organic Chemistry Chapter 22 Summary

  • 1. Chapter 22 Copyright Š 2010 Pearson Education, Inc. Organic Chemistry, 7th Edition L. G. Wade, Jr. Condensations and Alpha Substitutions of Carbonyl Compounds
  • 2. Chapter 22 2 Alpha Substitution ī‚§ Alpha substitution is the substitution of one of the hydrogens attached to the alpha-carbon for an electrophile. ī‚§ The reaction occurs through an enolate ion intermediate.
  • 3. Chapter 22 3 Condensation with an Aldehyde or Ketone ī‚§ The enolate ion attacks the carbonyl group to form an alkoxide. ī‚§ Protonation of the alkoxide gives the addition product: a β-hydroxy carbonyl compound.
  • 4. Chapter 22 4 Condensation with Esters ī‚§ The enolate adds to the ester to form a tetrahedral intermediate. ī‚§ Elimination of the leaving group (alkoxide) gives the substitution product (a β-carbonyl compound).
  • 5. Chapter 22 5 Keto–Enol Tautomers O H H OH H keto form (99.99%) enol form (0.01%) ī‚§ Tautomerization is an interconversion of isomers that occur through the migration of a proton and the movement of a double bond. ī‚§ Tautomers are not resonance form.
  • 6. Chapter 22 6 Base–Catalyzed Tautomerism ī‚§ In the presence of strong bases, ketones and aldehydes act as weak proton acids. ī‚§ A proton on the Îą carbon is abstracted to form a resonance- stabilized enolate ion with the negative charge spread over a carbon atom and an oxygen atom. ī‚§ The equilibrium favors the keto form over the enolate ion.
  • 7. Chapter 22 7 Acid-Catalyzed Tautomerism ī‚§ In acid, a proton is moved from the Îą-carbon to oxygen by first protonating oxygen and then removing a proton from the carbon.
  • 8. Chapter 22 8 Racemization ī‚§ For aldehydes and ketones, the keto form is greatly favored at equilibrium. ī‚§ If a chiral carbon has an enolizable hydrogen atom, a trace of acid or base allows that carbon to invert its configuration, with the enol serving as the intermediate. This is called racemization.
  • 9. Chapter 22 9 Acidity of Îą Hydrogens ī‚§ pKa for Îą H of aldehyde or ketone ~20. ī‚§ Much more acidic than alkane or alkene (pKa > 40) or alkyne (pKa = 25). ī‚§ Less acidic than water (pKa = 15.7) or alcohol (pKa = 16–19). ī‚§ Only a small amount of enolate ion is present at equilibrium.
  • 10. Chapter 22 10 Formation and Stability of Enolate Ions ī‚§ The equilibrium mixture contains only a small fraction of the deprotonated, enolate form.
  • 11. Chapter 22 11 Energy Diagram of Enolate Reaction ī‚§ Even though the keto–enol tautomerism equilibrium favors the keto form, addition of an electrophile shifts the equilibrium toward the formation of more enol.
  • 12. Chapter 22 12 Synthesis of Lithium Diisopropylamine (LDA) ī‚§ LDA is made by using an alkyllithium reagent to deprotonate diisopropylamine.
  • 13. Chapter 22 13 Enolate of Cyclohexanone ī‚§ When LDA reacts with a ketone, it abstracts the Îą-proton to form the lithium salt of the enolate.
  • 14. Chapter 22 14 The Îą Halogenation of Ketones ī‚§ When a ketone is treated with a halogen and a base, an Îą halogenation reaction occurs. ī‚§ The reaction is called base-promoted, rather than base-catalyzed, because a full equivalent of the base is consumed in the reaction.
  • 15. Chapter 22 15 Base-Promoted Halogenation Mechanism ī‚§ The base-promoted halogenation takes place by a nucleophilic attack of an enolate ion on the electrophilic halogen molecule. ī‚§ The products are the halogenated ketone and a halide ion.
  • 16. Chapter 22 16 Multiple Halogenations ī‚§ The Îą-haloketone produced is more reactive than ketone because the enolate ion is stabilized by the electron-withdrawing halogen. ī‚§ The second halogenation occurs faster than the first. ī‚§ Because of the tendency for multiple halogenations this base-promoted halogenation is not widely used to prepare monohalogenated ketones. O H Cl Cl2 OH , H2O _ O Cl Cl O Cl ClCl O Cl ClCl Cl
  • 17. Chapter 22 17 Bromoform Reaction ī‚§ A methyl ketone reacts with a halogen under strongly basic conditions to give a carboxylate ion and a molecule of haloform. ī‚§ The trihalomethyl intermediate is not isolated.
  • 18. Chapter 22 18 Mechanism of Haloform Formation ī‚§ The trihalomethyl ketone reacts with hydroxide ion to give a carboxylic acid. ī‚§ A fast proton exchange gives a carboxylate ion and a haloform. ī‚§ When Cl2 is used, chloroform is formed; Br2 forms bromoform ; and I2 forms iodoform.
  • 19. Chapter 22 19 Positive Iodoform Test for Alcohols ī‚§ The iodine oxidizes the alcohol to a methyl ketone and it will give a positive iodoform test. ī‚§ Iodoform (CHI3) is a yellow solid that will precipitate out of solution.
  • 20. Chapter 22 20 Propose a mechanism for the reaction of 3-pentanone with sodium hydroxide and bromine to give 2- bromo-3-pentanone. In the presence of sodium hydroxide, a small amount of 3-pentanone is present as its enolate. The enolate reacts with bromine to give the observed product. Solved Problem 1 Solution
  • 21. Chapter 22 21 Acid-Catalyzed Îą Halogenation ī‚§ Ketones also undergo acid-catalyzed Îą halogenation. ī‚§ Acidic halogenation may replace one or more alpha hydrogens depending on how much halogen is used. ī‚§ Acetic acid serves as both the solvent and the acid catalyst.
  • 22. Chapter 22 22 Mechanism of Acid-Catalyzed Îą Halogenation ī‚§ The mechanism of acid-catalyzed halogenation involves attack of the enol form of the ketone on the electrophile halogen molecule. ī‚§ Loss of a proton gives the haloketone and the hydrogen halide.
  • 23. Chapter 22 23 Propose a mechanism for the acid-catalyzed conversion of cyclohexanone to 2-chlorocyclohexanone. Under acid catalysis, the ketone is in equilibrium with its enol form. The enol acts as a weak nucleophile, attacking chlorine to give a resonance-stabilized intermediate. Loss of a proton gives the product. Solved Problem 2 Solution
  • 24. Chapter 22 24 Hell–Volhard–Zelinsky (HVZ) Reaction ī‚§ The HVZ reaction replaces a hydrogen atom with a bromine atom on the alpha-carbon of a carboxylic acid (Îą-bromoacid). ī‚§ The acid is treated with bromine and phosphorus tribromide, followed by hydrolysis.
  • 25. Chapter 22 25 Hell–Volhard–Zelinski Reaction: Step 1 ī‚§ The enol form of the acyl bromide serves as a nucleophilic intermediate. ī‚§ The first step is the formation of acyl bromide, which enolizes more easily than does the acid.
  • 26. Chapter 22 26 Hell–Volhard–Zelinski Reaction: Step 2 ī‚§ The enol is nucleophilic, so it attacks bromine to give the alpha-brominated acyl bromide. ī‚§ In the last step of the reaction, the acyl bromide is hydrolyzed by water to the carboxylic acid.
  • 27. Chapter 22 27 Alkylation of Enolate Ions ī‚§ Because the enolate has two nucleophilic sites (the oxygen and the Îą carbon), it can react at either of these sites. ī‚§ The reaction usually takes place primarily at the α−carbon, forming a new C—C bond.
  • 28. Chapter 22 28 Îą Alkylation of Enolate Ions ī‚§ LDA forms the enolate. ī‚§ The enolate acts as the nucleophile and attacks the partially positive carbon of the alkyl halide, displacing the halide and forming a C—C bond.
  • 29. Chapter 22 29 Enamine Formation ī‚§ Ketones or aldehydes react with a secondary amine to form enamines. ī‚§ The enamine has a nucleophilic Îą-carbon, which can be used to attack electrophiles.
  • 30. Chapter 22 30 Mechanism of Enolate Formation ī‚§ An enamine results from the reaction of a ketone or aldehyde with a secondary amine.
  • 31. Chapter 22 31 Electrostatic Potential Map of an Enamine ī‚§ The electrostatic potential map (EPM) of a simple enamine shows a high negative electrostatic potential (red) near the Îą-carbon atom of the double bond. ī‚§ This is the nucleophilic carbon atom of the enamine.
  • 32. Chapter 22 32 Alkylation of an Enamine ī‚§ Enamines displace halides from reactive alkyl halides, giving alkylated iminium salts. ī‚§ The alkylated iminium salt can be hydrolyzed to the ketone under acidic conditions.
  • 33. Chapter 22 33 Acylation of Enamines ī‚§ The enamine attacks the acyl halide, forming an acyl iminium salt. ī‚§ Hydrolysis of the iminium salt produces the β- diketone as the final product.
  • 34. Chapter 22 34 Aldol Condensation ī‚§ Under basic conditions, the aldol condensation involves the nucleophilic addition of an enolate ion to another carbonyl group. ī‚§ When the reaction is carried out at low temperatures, the β- hydroxy carbonyl compound can be isolated. ī‚§ Heating will dehydrate the aldol product to the α−β unsaturated compound.
  • 35. Chapter 22 35 Base-Catalyzed Aldol Condensation: Step 1 ī‚§ During Step 1, the base removes the Îą- proton, forming the enolate ion. ī‚§ The enolate ion has a nucleophilic Îą-carbon.
  • 36. Chapter 22 36 Base-Catalyzed Aldol Condensation: Step 2 ī‚§ The enolate attacks the carbonyl carbon of a second molecule of carbonyl compound.
  • 37. Chapter 22 37 Base-Catalyzed Aldol Condensation: Step 3 ī‚§ Protonation of the alkoxide gives the aldol product.
  • 38. Chapter 22 38 Dehydration of Aldol Products ī‚§ Heating a basic or acidic aldol dehydration of the alcohol functional group. ī‚§ The product is a Îą,β-unsaturated conjugated aldehyde or ketone. ī‚§ An Aldol condensation, followed by dehydration, forms a new carbon–carbon double bond.
  • 39. Chapter 22 39 Crossed Aldol Condensations
  • 40. Chapter 22 40 Successful Crossed Aldol Condensations
  • 41. Chapter 22 41 Propose a mechanism for the base-catalyzed aldol condensation of acetone (Figure 22-2). The first step is formation of the enolate to serve as a nucleophile. The second step is a nucleophilic attack by the enolate on another molecule of acetone. Protonation gives the aldol product. Solved Problem 3 Solution
  • 42. Chapter 22 42 Aldol Cyclization ī‚§ Intramolecular aldol reactions of diketones are often used for making five- and six-membered rings. ī‚§ Rings smaller or larger than five or six members are not favored due to ring strain or entropy.
  • 43. Chapter 22 43 Retrosynthesis of Aldol Condensation
  • 44. Chapter 22 44 Claisen Condensation ī‚§ The Claisen condensation results when an ester molecule undergoes nucleophilic acyl substitution by an enolate.
  • 45. Chapter 22 45 Dieckman Condensation
  • 46. Chapter 22 46 Crossed Claisen ī‚§ Two different esters can be used, but one ester should have no Îą hydrogens. ī‚§ Useful esters are benzoates, formates, carbonates, and oxalates. ī‚§ Ketones (pKa = 20) may also react with an ester to form a β-diketone.
  • 47. Chapter 22 47 Crossed Claisen Condensation ī‚§ In a crossed Claisen condensation, an ester without Îą hydrogens serves as the electrophilic component.
  • 48. Chapter 22 48 Crossed Claisen Condensation with Ketones and Esters ī‚§ Crossed Claisen condensation between ketones and esters are also possible. ī‚§ Ketones are more acidic than esters, and the ketone component is more likely to deprotonate and serve as the enolate component in the condensation.
  • 49. Chapter 22 49 Crossed Claisen Mechanism ī‚§ The ketone enolate attacks the ester, which undergoes nucleophilic acyl substitution, and thereby, acylates the ketone.
  • 50. Chapter 22 50 Propose a mechanism for the self-condensation of ethyl acetate to give ethyl acetoacetate. The first step is formation of the ester enolate. The equilibrium for this step lies far to the left; ethoxide deprotonates only a small fraction of the ester. The enolate ion attacks another molecule of the ester; expulsion of ethoxide ion gives ethyl acetoacetate. Solved Problem 4 Solution
  • 51. Chapter 22 51 In the presence of ethoxide ion, ethyl acetoacetate is deprotonated to give its enolate. This exothermic deprotonation helps to drive the reaction to completion. When the reaction is complete, the enolate ion is reprotonated to give ethyl acetoacetate. Solved Problem 4 (Continued) Solution (Continued)
  • 52. Chapter 22 52 Show what ester would undergo Claisen condensation to give the following β-keto ester. First, break the structure apart at the Îą, β bond (Îą, β to the ester carbonyl). This is the bond formed in the Claisen condensation. Solved Problem 5 Solution
  • 53. Chapter 22 53 Next, replace the Îą proton that was lost, and replace the alkoxy group that was lost from the carbonyl. Two molecules of methyl 3-phenylpropionate result. Now draw out the reaction. Sodium methoxide is used as the base because the reactants are methyl esters. Solved Problem 5 (Continued) Solution (Continued)
  • 55. Chapter 22 55 Malonic Ester Synthesis ī‚§ The malonic ester synthesis makes substituted derivatives of acetic acids. ī‚§ Malonic ester is alkylated or acylated on the carbon that is alpha to both carbonyl groups, and the resulting derivative is hydrolyzed and allowed to decarboxylate.
  • 56. Chapter 22 56 Decarboxylation of the Alkylmalonic Acid ī‚§ Decarboxylation takes place through a cyclic transition state, initially giving an enol form that quickly tautomerizes to the product.
  • 57. Chapter 22 57 Example of the Malonic Synthesis
  • 58. Chapter 22 58 Dialkylation of Malonic Ester
  • 59. Chapter 22 59 Show how the malonic ester synthesis is used to prepare 2-benzylbutanoic acid. 2-Benzylbutanoic acid is a substituted acetic acid having the substituents Ph–CH2– and CH3CH2–. Adding these substituents to the enolate of malonic ester eventually gives the correct product. Solved Problem 6 Solution
  • 60. Chapter 22 60 Acetoacetic Ester Synthesis ī‚§ The acetoacetic ester synthesis is similar to the malonic ester synthesis, but the final products are ketones.
  • 61. Chapter 22 61 Alkylation of Acetoacetic Ester ī‚§ Ethoxide ion completely deprotonates acetoacetic ester. ī‚§ The resulting enolate is alkylated by an unhindered alkyl halide or tosylate to give an alkylacetoacetic ester.
  • 62. Chapter 22 62 Hydrolysis of Alkylacetoacetic Ester ī‚§ Acidic hydrolysis of the alkylacetoacetic ester initially gives an alkylacetoacetic acid, which is a β-keto acid. ī‚§ The keto group in the β-position promotes decarboxylation to form a substituted version of acetone.
  • 63. Chapter 22 63 Show how the acetoacetic ester synthesis is used to make 3-propylhex-5-en-2-one. The target compound is acetone with an n-propyl group and an allyl group as substituents: Solved Problem 7 Solution
  • 64. Chapter 22 64 Hydrolysis proceeds with decarboxylation to give the disubstituted acetone product. With an n-propyl halide and an allyl halide as the alkylating agents, the acetoacetic ester synthesis should produce 3-propyl-5-hexen-2-one. Two alkylation steps give the required substitution: Solved Problem 7 (Continued) Solution (Continued)
  • 65. Chapter 22 65 Conjugate Additions: The Michael Reaction ī‚§ Îą,β-unsaturated carbonyl compounds have unusually electrophilic double bonds. ī‚§ The β-carbon is electrophilic because it shares the partial positive charge of the carbonyl carbon through resonance.
  • 66. Chapter 22 66 1,2-Addition and 1,4-Addition ī‚§ When attack occurs at the carbonyl group, protonation of the oxygen leads to a 1,2-addition. ī‚§ When attack occurs at the β-position, the oxygen atom is the fourth atom counting from the nucleophile, and the addition is called a 1,4-addition.
  • 68. Chapter 22 68 1,4-Addition of an Enolate to Methyl Vinyl Ketone (MVK) ī‚§ An enolate will do a 1,4-attack on the Îą,β- unsaturated ketone (MVK).
  • 69. Chapter 22 69 Show how the following diketone might be synthesized using a Michael addition. A Michael addition would have formed a new bond at the β carbon of the acceptor. Therefore, we break this molecule apart at the β,Îŗ bond. Solved Problem 8 Solution
  • 70. Chapter 22 70 Robinson Annulation ī‚§ With enough base, the product of the Michael reaction undergoes a spontaneous intramolecular aldol condensation, usually with dehydration, to give a six-membered ring—a conjugated cyclohexenone.

Editor's Notes

  1. Copyright Š 2006 Pearson Prentice Hall, Inc.
  2. Copyright Š 2006 Pearson Prentice Hall, Inc.
  3. Copyright Š 2006 Pearson Prentice Hall, Inc.
  4. Copyright Š 2006 Pearson Prentice Hall, Inc.
  5. Copyright Š 2006 Pearson Prentice Hall, Inc.
  6. Copyright Š 2006 Pearson Prentice Hall, Inc.
  7. Copyright Š 2006 Pearson Prentice Hall, Inc.
  8. Copyright Š 2006 Pearson Prentice Hall, Inc.
  9. Copyright Š 2006 Pearson Prentice Hall, Inc.
  10. Copyright Š 2006 Pearson Prentice Hall, Inc.
  11. Figure: 22_04-22UNT.jpg Title: Michael Donors and Acceptors Caption: Some Common Michael Donors and Michael Acceptors Notes:
  12. Copyright Š 2006 Pearson Prentice Hall, Inc.