This document summarizes Chapter 22 of an organic chemistry textbook. It discusses various reactions of carbonyl compounds including alpha substitution, condensations with aldehydes/ketones/esters, and keto-enol tautomerism. Specific reactions covered include halogenation, aldol condensation, Claisen condensation, malonic ester synthesis, and acetoacetic ester synthesis. Mechanisms are provided for reactions such as enolate formation, aldol addition, Claisen condensation and decarboxylation of alkylmalonic acids. Examples are given to illustrate reactions like the malonic ester synthesis and alkylation of acetoacetic ester.

1. Chapter 22
Copyright © 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
 Alpha substitution is the substitution of one of the
hydrogens attached to the alpha-carbon for an
electrophile.
 The reaction occurs through an enolate ion
intermediate.

3. Chapter 22 3
Condensation with an Aldehyde or
Ketone
 The enolate ion attacks the carbonyl group to form an
alkoxide.
 Protonation of the alkoxide gives the addition
product: a -hydroxy carbonyl compound.

4. Chapter 22 4
Condensation with Esters
 The enolate adds to the ester to form a tetrahedral
intermediate.
 Elimination of the leaving group (alkoxide) gives the
substitution product (a -carbonyl compound).

5. Chapter 22 5
Keto–Enol Tautomers
O
H
H
OH
H
keto form
(99.99%)
enol form
(0.01%)
 Tautomerization is an interconversion of
isomers that occur through the migration of a
proton and the movement of a double bond.
 Tautomers are not resonance form.

6. Chapter 22 6
Base–Catalyzed Tautomerism
 In the presence of strong bases, ketones and aldehydes act as
weak proton acids.
 A proton on the carbon is abstracted to form a resonance-
stabilized enolate ion with the negative charge spread over a
carbon atom and an oxygen atom.
 The equilibrium favors the keto form over the enolate ion.

7. Chapter 22 7
Acid-Catalyzed Tautomerism
 In acid, a proton is moved from the -carbon
to oxygen by first protonating oxygen and
then removing a proton from the carbon.

8. Chapter 22 8
Racemization
 For aldehydes and ketones, the keto form is greatly
favored at equilibrium.
 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 Hydrogens
 pKa for H of aldehyde or ketone ~20.
 Much more acidic than alkane or alkene
(pKa > 40) or alkyne (pKa = 25).
 Less acidic than water (pKa = 15.7) or
alcohol (pKa = 16–19).
 Only a small amount of enolate ion is
present at equilibrium.

10. Chapter 22 10
Formation and Stability of
Enolate Ions
 The equilibrium mixture contains only a small
fraction of the deprotonated, enolate form.

11. Chapter 22 11
Energy Diagram of Enolate
Reaction
 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)
 LDA is made by using an alkyllithium reagent
to deprotonate diisopropylamine.

13. Chapter 22 13
Enolate of Cyclohexanone
 When LDA reacts with a ketone, it abstracts
the -proton to form the lithium salt of the
enolate.

14. Chapter 22 14
The Halogenation of Ketones
 When a ketone is treated with a halogen and a base,
an halogenation reaction occurs.
 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
 The base-promoted halogenation takes place by a
nucleophilic attack of an enolate ion on the
electrophilic halogen molecule.
 The products are the halogenated ketone and a
halide ion.

16. Chapter 22 16
Multiple Halogenations
 The -haloketone produced is more reactive than
ketone because the enolate ion is stabilized by the
electron-withdrawing halogen.
 The second halogenation occurs faster than the first.
 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
 A methyl ketone reacts with a halogen under
strongly basic conditions to give a
carboxylate ion and a molecule of haloform.
 The trihalomethyl intermediate is not isolated.

18. Chapter 22 18
Mechanism of Haloform
Formation
 The trihalomethyl ketone reacts with hydroxide ion to
give a carboxylic acid.
 A fast proton exchange gives a carboxylate ion and a
haloform.
 When Cl2 is used, chloroform is formed; Br2 forms
bromoform ; and I2 forms iodoform.

19. Chapter 22 19
Positive Iodoform Test
for Alcohols
 The iodine oxidizes the alcohol to a methyl
ketone and it will give a positive iodoform test.
 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 α Halogenation
 Ketones also undergo acid-catalyzed halogenation.
 Acidic halogenation may replace one or more alpha
hydrogens depending on how much halogen is used.
 Acetic acid serves as both the solvent and the acid
catalyst.

22. Chapter 22 22
Mechanism of Acid-Catalyzed
α Halogenation
 The mechanism of acid-catalyzed halogenation
involves attack of the enol form of the ketone on the
electrophile halogen molecule.
 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
 The HVZ reaction replaces a hydrogen atom with a
bromine atom on the alpha-carbon of a carboxylic
acid ( -bromoacid).
 The acid is treated with bromine and phosphorus
tribromide, followed by hydrolysis.

25. Chapter 22 25
Hell–Volhard–Zelinski Reaction:
Step 1
 The enol form of the acyl bromide serves as a
nucleophilic intermediate.
 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
 The enol is nucleophilic, so it attacks bromine
to give the alpha-brominated acyl bromide.
 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
 Because the enolate has two nucleophilic sites (the
oxygen and the carbon), it can react at either of
these sites.
 The reaction usually takes place primarily at the
carbon, forming a new C—C bond.

28. Chapter 22 28
Alkylation of Enolate Ions
 LDA forms the enolate.
 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
 Ketones or aldehydes react with a secondary
amine to form enamines.
 The enamine has a nucleophilic -carbon,
which can be used to attack electrophiles.

30. Chapter 22 30
Mechanism of Enolate Formation
 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
 The electrostatic potential map (EPM) of a simple
enamine shows a high negative electrostatic potential
(red) near the -carbon atom of the double bond.
 This is the nucleophilic carbon atom of the enamine.

32. Chapter 22 32
Alkylation of an Enamine
 Enamines displace halides from reactive alkyl
halides, giving alkylated iminium salts.
 The alkylated iminium salt can be hydrolyzed
to the ketone under acidic conditions.

33. Chapter 22 33
Acylation of Enamines
 The enamine attacks the acyl halide, forming an acyl
iminium salt.
 Hydrolysis of the iminium salt produces the -
diketone as the final product.

34. Chapter 22 34
Aldol Condensation
 Under basic conditions, the aldol condensation involves the
nucleophilic addition of an enolate ion to another carbonyl
group.
 When the reaction is carried out at low temperatures, the -
hydroxy carbonyl compound can be isolated.
 Heating will dehydrate the aldol product to the unsaturated
compound.

35. Chapter 22 35
Base-Catalyzed Aldol
Condensation: Step 1
 During Step 1, the base removes the -
proton, forming the enolate ion.
 The enolate ion has a nucleophilic -carbon.

36. Chapter 22 36
Base-Catalyzed Aldol
Condensation: Step 2
 The enolate attacks the carbonyl carbon of a
second molecule of carbonyl compound.

37. Chapter 22 37
Base-Catalyzed Aldol
Condensation: Step 3
 Protonation of the alkoxide gives the aldol
product.

38. Chapter 22 38
Dehydration of Aldol Products
 Heating a basic or acidic aldol dehydration of the
alcohol functional group.
 The product is a , -unsaturated conjugated
aldehyde or ketone.
 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
 Intramolecular aldol reactions of diketones are often
used for making five- and six-membered rings.
 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
 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
 Two different esters can be used, but
one ester should have no hydrogens.
 Useful esters are benzoates, formates,
carbonates, and oxalates.
 Ketones (pKa = 20) may also react with
an ester to form a -diketone.

47. Chapter 22 47
Crossed Claisen Condensation
 In a crossed Claisen condensation, an ester
without hydrogens serves as the
electrophilic component.

48. Chapter 22 48
Crossed Claisen Condensation
with Ketones and Esters
 Crossed Claisen condensation between ketones and
esters are also possible.
 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
 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 -keto ester.
First, break the structure apart at the bond ( to the ester carbonyl). This is the bond formed in
the Claisen condensation.
Solved Problem 5
Solution

53. Chapter 22 53
Next, replace the 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)

54. Chapter 22 54

55. Chapter 22 55
Malonic Ester Synthesis
 The malonic ester synthesis makes substituted
derivatives of acetic acids.
 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
 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
 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
 Ethoxide ion completely deprotonates acetoacetic
ester.
 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
 Acidic hydrolysis of the alkylacetoacetic ester initially
gives an alkylacetoacetic acid, which is a -keto acid.
 The keto group in the -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
 , -unsaturated carbonyl compounds have unusually
electrophilic double bonds.
 The -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
 When attack occurs at the carbonyl group,
protonation of the oxygen leads to a 1,2-addition.
 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.

67. Donors and Acceptors
Chapter 22 67

68. Chapter 22 68
1,4-Addition of an Enolate to
Methyl Vinyl Ketone (MVK)
 An enolate will do a 1,4-attack on the -
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 carbon of the acceptor. Therefore,
we break this molecule apart at the bond.
Solved Problem 8
Solution

70. Chapter 22 70
Robinson Annulation
 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.

71. Chapter 22 71
Robinson Mechanism