Chapter 21
Reactions of a-Hydrogens:
Condensation Reactions
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Assignment for Chapter 21
DO:
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


Sections 21.0 through 21.4
Sections 21.7 through 21.9
Sections 21.12 through 21.17
Section 21.22
SKIP:



Sections 21.5 through 21.6
Sections 21.10 through 21.11
Sections 21.18 through 21.21
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ALSO DO:
Summary
Problems
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Problem Assignment
In-Text Problems:


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21-1 through 21-9
21-12 through 21-17
21-25 through 21-45
End-of-Chapter Problems:


1 through 11
13 through 24
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Tautomerism
Compounds whose structures differ markedly in
the arrangement of atoms, but which exist in
equilibrium, are called tautomers.
Most often, tautomers are species that differ by
the position of a hydrogen atom and which exist
in equilibrium.
The best-known example is keto-enol
tautomerism.
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Keto-Enol Tautomerism
H
O
C
C
O
K
C
C
H
keto
enol
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Keto-Enol Tautomerism
In general, the equilibrium favors the
keto form very dramatically.
If one calculates the relative energies
of the keto and enol forms, one
concludes that the formation of enol
from keto should be endothermic by
about 75 kJ/mole.
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Keto-Enol Tautomerism
From this energy, one calculates an
equilibrium constant for enolization:
6.3 x 10-14
Clearly, for most aldehydes and
ketones, the ability to form an enol is
an extremely minor property.
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Keto-Enol Tautomerism
In the case of 1,3-dicarbonyl
compounds, however, the equilibrium
may shift to favor the enol form, since a
stabilized, hydrogen-bonded structure is
now possible.
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Keto-Enol Tautomerism in 1,3Dicarbonyl Compounds
H
O
O
O
O
C
C
K
C
C
CH3
C
H
CH3
H
keto
CH3
C
CH3
H
enol
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Keto-Enol Tautomerism in 1,3Dicarbonyl Compounds
The equilibrium lies substantially to the right.
In simple ketones, such a hydrogen-bonded
structure cannot form, and the percentage of
enol found in an equilibrium mixture is very
small.
The following tables illustrate some typical enol
percentages. Notice the difference between
simple ketones and dicarbonyl compounds.
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Some Representative Enol
Percents
O
O
CH3 C
CH2 C
OH
H
CH3 C
O
CH C
H
% enol = 98
O
O
CH3 C
CH2 C
OH
CH3
CH3 C
O
CH C
CH3
% enol = 80
O
O
CH3 C
CH2 C
OH
OC2H5
CH3 C
O
CH C
OC2H5
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% enol = WWU
8
More Representative Enol
Percents
O
OH
-4
% enol = 4.1 x 10
O
OH
CH3 C CH3
CH2 C
O
CH2 CH C
CH3
-4
% enol = < 2 x 10
OH
CH3
CH2
CH C
CH2
-3
% enol = 2.5 x 10
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One last comment on this...
You may recognize some structural
similarities between enols and
enamines.
Whenever an enol form can exist, it
has the potential to be a
nucleophile!
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Acidity of a-Hydrogens
Review material in Chapter 7, Section
7.7
The acidity of a hydrogen attached to
the a-carbon of a carbonyl compound is
much higher than the acidity of a
typical C-H hydrogen.
pKa values range from about 19 to 20
(compared with 48 to 50)
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Acidity of a-Hydrogens:
The Reason
R
H
O
C
C
O
CH3
R
R
C
C
CH3
R
+
+
B
B
H
O
R
C
C
CH3
R
an enolate ion
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Acidity of a-Hydrogens
Resonance stabilization of the enolate
ion shifts the equilibrium to the right,
thereby making the C-H bond more
acidic.
Once formed, the enolate ion is capable
of reacting as a nucleophile. The
a-carbon of the enolate ion bears
substantial negative charge.
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Base-Promoted Halogenation
of Ketones
O
R
CH2 C
+
O
R
R
CH
C
R
Br
Br2
+
+
Br-
OH-
+
O
H
H
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Base-Promoted Halogenation
of Ketones
The experimental rate law is:
Rate = k[ketone][OH-]
Note that the rate law does not contain
bromine!
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Mechanism
1)
O
O
R
C
R
CH2 H
C
slow
CH2
+
O
H
C
H
Note that the
first step is ratedetermining
CH2
2)
O
C
H
O
R
R
O
+
O
CH2
+
Br
Br
fast
R
C
CH2 Br
+
Br
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Example
O
C
O
CH3
C
CH2 Br
NaOH
+
+
Br2
Br-
(one equivalent)
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But...
The halogenation is difficult to stop at
the mono-substitution stage.
Often, poly-halogenated products are
formed in this reaction.
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With an excess of bromine:
O
C
CH3
NaOH
+
Br2
(excess)
O
Br
C
C
Br
Br
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•There is also an acid-catalyzed halogenation
reaction, which operates through the
formation of the enol form of the ketone
(recall that the enol is nucleophilic).
•Once formed, the enol displaces bromide
ion from Br2, forming the brominated
product.
•In the acid-catalyzed mechanism, monosubstitution is the predominant result.
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Example
O
O
Br
CH3COOH
+ Br2
+ HBr
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Alkylation of Enolate Ions
In the presence of a very strong base, the
a-hydrogen of an aldehyde or ketone can be
replaced by an alkyl group.
Once again, the strong base removes an
a-hydrogen to form an enolate ion.
The enolate ion, acting as a nucleophile,
participates in an SN2 substitution with an
alkyl halide.
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Alkylation of a Ketone
O
C
O
C
O
CH3
.. _
CH2
strong base
C
THF
.. _
CH2
O
CH3 I
C
CH2 CH3
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… and the “strong base” is:
CH3
CH3 C
O
K
CH3
potassium tert-butoxide
NaNH 2
sodium amide
KH
potassium hydride
CH3
N
CH3
CH
CH
CH3
CH3
Li
lithium diisopropylamide
"LDA"
Best Choice
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Mechanism
1)
O
O
R
C
R
CH2 H
C
+
+
Base
2)
CH2
Base-H
O
R
C
O
CH2
+
R
X
slow
R
C
CH2 R
S N2
+
X
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Alkylation of Enolate Ions
Remember that enamines can also react
with alkyl halides to give similar
products.
Review Chapter 16, Section 16.13.
See also Chapter 21, Section 21.8.
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Example
O
O
LDA
CH3 CH2 Br
CH2 CH3
THF
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Here’s something different:
O
Se
Br
LDA
THF
O
Se
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