Chapter 22:
Carbonyl α-Substitutions
326
So far, all the reactions of carbonyl compounds
that we have seen were directly at the carbonyl.
But there is another possibility if the carbonyl has
α-hydrogens, α−substitution
327
Keto-Enol Tautomerism (22-2)
This substitution reaction can take place under
acidic condition from the enol tautomer acting
as the nucleophile.
Or under basic condition.
Tautomerism: Rapid conversion of two isomeric
chemical species
328
Under weakly basic (RO- or HO-)
conditions, only a small amount of the
enolate anion can be produced. In these
cases the nucleophile in the substitution
reaction is most likely the enol tautomer.
329
However, the use of stronger bases such as LDA
(Lithium DiisopropylAmide) generates the
enolate ion in quantitative amount, making this
species the nucleophile in the substitution
reaction.
330
Practice Questions
Draw the enol tautomers for each of the
following compounds. For compounds that have
more than one enol tautomer, indicate which is
more stable.
O
O
O
O
331
• Alkylation of Enolate Ion (22-3)
The reaction follows a SN2 mechanism where the
nucleophile is the enolate anion. Alkylation of
enolate anion is normally at the carbon, but some
side reaction at the oxygen is also possible.
332
333
This reaction has some serious limitations:
◦ Difficult at best with ketones, esters and nitriles
◦ Carbonyl with only one type of α-proton give
good results
◦ Only a strong base such as LDA can be used
◦ Condensation is often a side reaction.
◦ Cannot be done with aldehydes.
◦ To be successful, it requires small reactive alkyl
halides (R-X)
reactivity order: allylic > benzylic > methyl > 1o
◦ In practice, alkylation of enolate is rarely
attempted
334
The main reason for all of these
limitations is because similar conditions
are used to carry out condensation
reactions. Alkylation and condensation
always compete with one another.
335
In order to minimize this problem, one
must form the enolate ion quantitatively
and rapidly. The use of LDA is
recommended (yet condensation will still
take place).
336
• Formation and Alkylation of Enamines (22-4)
This is a milder alternative to direct
alkylation of carbonyl and is the best way to
alkylate aldehydes. Aldehydes cannot be
alkylated directly but using an enamine,
followed by an hydrolysis, alkylated
aldehydes can be obtained. Ketones can also
be obtained via this method.
337
Enamines are very similar to enolate ions in
their structures…and reactivity.
In the presence of an electrophile, they will act
as the nucleophile and produce the iminium ion.
This ion will generate the corresponding
carbonyl (ketone or aldehyde) on hydrolysis.
338
The synthesis of the enamine follows a
similar mechanism as the formation of the
imine (Chapter 18). However, here the
carbinolamine cannot eliminate a proton
from the amine function and the
elimination takes place at the α position
instead.
339
Acylation of enamines is also easy to do
by an acyl substitution mechanism with
acid chlorides, β diketones are obtained.
340
Practice Questions
How would you prepare the following compound
using an enamine intermediate?
341
How would you prepare the following
compounds from a ketone and an alkyl halide?
342
• α-Halogenation of Ketones (22-5)
This reaction if possible for any halogens (Cl2, Br2,
I2) under acidic conditions (acetic acid is often
used). Therefore, the mechanism will proceed via
the enol intermediate.
343
Under basic conditions it also possible carry
out this reaction. However, if more than
one α-hydrogens are present, they will all be
substituted.
Cl
Cl
H
Cl2
_
OH , H2O
O
O
O
O
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
344
This polyhalogenation under basic conditions is
the basis of the haloform reaction producing
carboxylic acid. Because the CX3 group on the
carbonyl is a good leaving group, acyl
substitution with the hydroxide ion takes place
leading to the formation of carboxylate anions.
345
• α Bromination of Acids:The HVZ Reaction (22-6)
The HVZ (Hell-Volhard-Zelinsky) reaction
is not one carried out on aldehydes/ketones
but rather is a reaction of a carboxylic acid,
and is very similar to the halogenation
reaction we just covered. This reaction
gives α-brominated carboxylic acids.
346
Because acids, esters and amides do not
enolize enough, they cannot be brominated
directly. Hence, the acid is transformed
first into the acid bromide, α-brominated
and hydrolyzed back to a carboxylic acid.
The mechanism is shown on the next slide.
The formation of the intermediate acid
bromide (labelled A on the next slide)
follow the same mechanism we have seen
for alcohols (Chapter 11 Alcohols/phenols)
347
Mechanism of the HVZ Reaction
A
348
• The Aldol Condensation (22-7)
In the aldol condensation, the same carbonyl can
be used as the nucleophile (enolate) and as the
electrophile (carbonyl). The result is the
formation of a β-hydroxycarbonyl. This is the
reason why alkylation was so difficult to perform,
this reaction always competes with others
whenever an enolate anion is produced.
enolate
O
R
CH2-
ketone
O
R
CH3
nucleophile electrophile
349
Ketones and aldehydes will give aldol
condensation.
2
3
2
3
1 H
2
3
1 H
1 OH
OH3 2
O
O
O
3
3
2 1
O
3
2 1
O
1 H
OH-
2 OH
1
3
2 1
O
350
Based Catalyzed Aldol Addition
351
Practice Question
What is the aldol addition products are formed
from the following carbonyls?
352
The aldol condensation of aldehydes and ketones
can also be catalyzed by acid. The enol tautomer
is acting as the nucleophile in these cases.
353
Dehydration of the Aldol Product (22-8)
◦ Aldol products, β-hydroxycarbonyls are
very easy to dehydrate because the
resulting product is very stable
(unsaturated carbonyls). Since the aldol
addition is at equilibrium, this step is
usually necessary to complete the
reaction. These two steps together
(aldol addition + dehydration) will
generate α,β
β-unsaturated carbonyl.
354
The dehydration of the aldol addition
product can be done under acidic or basic
conditions.
O
O
H2SO4
O
O
NaOH
H
H
355
Base--Catalyzed Dehydration of Aldol
Base
Product
356
Acid--Catalyzed Dehydration of Aldol
Acid
Product
H+
O
-
O
B
H
+
OH OH
OH
H
OH OH
H
H +
H
H
B-
-
B
H
+
OH OH2
O+
H
H+
H
OH OH
H
O
H
357
Competition Between
Condensation and α-Substitution.
We have seen that for both of these reactions an
enolate anion must be formed. Different
conditions can be used to control the reaction
outcome depending on the type of reaction
needed:
358
• Crossed Aldol Condensations (22-9)
The condensation reaction between two
structurally different aldehydes/ketones.
For example, what would be the product
of this reaction? How many products can
be formed?
359
These mixed aldol (cross aldol) reactions are very
difficult to control and more than one product is
usually obtained.
However, it is possible to limit the possibilities by
choosing the correct reaction conditions:
A- One carbonyl has no α-hydrogen (cannot enolize)
B- Form the enolate first, then add the second
carbonyl
360
A- One carbonyl has no α-hydrogen (cannot enolize)
361
B- Form the enolate first, then add the second
carbonyl
O-
O
LDA
O
1)
O
OH
H
2) H2O
fformation
i off the
h enolate
l
should be quantitative
362
Practice Question
What are the products of the following crossed
aldol additions?
363
• Aldol Cyclizations (22-10)
When a molecule contains two carbonyl
groups within its structure, it is possible for
this molecule to cyclize via an aldol
addition. The stability of cyclic structures
is important, hence 5 and 6-membered
rings are favoured.
364
365
Practice Question
What is the major product of the
following reactions?
366
Planning Syntheses using Aldol Condensations
(22-11)
When planning to carry out a synthesis that will
involved an aldol condensation, always look at the
bond between the α and β carbons since this is the
bond created during the aldol condensation.
367
• Claisen Condensation (22-12)
This reaction is identical to the aldol condensation
but takes place between 2 esters. A small
difference exists in the mechanism to explain the
formation of the final product, a β-ketoester. You
should recognize in the second step the acyl
substitution mechanism that we have seen in a
previous chapter.
368
In this reaction a second step (acidic workup) is
necessary since the product of the claisen
condensation is more acidic than the original
esters and is deprotonated with the base that is
expelled.
pKa~ 11
O
R CH2 C
R'O C C
H
O
H3O+
R
O
R CH2 C
R'O C C
O
R
369
The Dieckmann Condensation (22-13)
This is similar to the intramolecular Aldol
reaction, 5 and 6-membered rings are
preferred.
370
• Crossed Claisen Condensation (22-14)
Similar to the Aldol condensation, when
two different esters are used, cross
condensation results.
371
Other carbonyl can be used to replace one
of the ester. When a ketone is used, it always
forms the enolate since the protons are
more acidic. Reaction with an ester gives a
β-diketone.
372
Reaction of a ketone with diethyl
carbonate is another good method to get
β-ketoesters.
373
Practice Questions
Which of the following cannot give
Claisen condensation?
374
What are the products of the following
reactions?
O
O
OCH3
1) CH3O2) HCl
375
• Malonic ester synthesis (22-16)
In this reaction the α-hydrogen are very
acidic because they are located between 2
carbonyl groups…hence, we say that the
carbonyl is activated. The deprotonation is
quantitative and unambiguous. Carboxylic
acids are obtained after an hydrolysis and
decarboxylation steps.
376
Simple bases are used because of the increased
acidity of the β-dicarbonyl compounds.
377
Examples of the Malonic Ester Synthesis
378
We have seen how to hydrolyze (undeer
basic or acidic conditions) esters to acids
(section on carboxylic acid derivatives) .
Because of the structure of this diacid, it
is very easy (just requires heat) to remove
CO2 and obtain the product, a carboxylic
acid.
379
Practice Question
What are the products of the following
reactions?
380
• Acetoacetic ester synthesis (22-17)
The principles seen in the malonic ester
synthesis also apply here…except, because
there is only one possible carboxyl group to
eliminate, a ketone will be obtained. This is
an excellent method to get an alkylated
ketone and avoid condensation. This
reaction will work for any β-ketoester.
381
Decarboxylation follows the same
mechanism, but only one carboxyl group is
present, hence a ketone is produced.
382
Example of Acetoacetic Ester Synthesis
383
Practice Questions
What alkyl bromide would do you need in the
malonic ester synthesis of the following acids?
384
What alkyl bromide would you use to
carry out the acetoacetic ester synthesis
of the following ketones?
385
• Michael Addition (22-18)
Enolate ion are not as nucleophilic as
other ion such as RLi or RMgX. This is due
to the resonance stabilization. Because of
this, enolate ion will add in a conjugate
β-unsaturated
fashion in the presence of α,β
carbonyl. This reaction is called the
Michael Addition.
386
The Michael addition will work with any enolate
ion, but the results are better with a stable
enolate is used. Enolates from:
◦ β-ketoesters
◦ β-diketones
◦ Malonic ester
387
Normally the reactant forming the enolate
is called: Michael Donor (because it
provides the electrons for the new C-C
bond)
The reactant that is attacked, the α,β
βunsaturated compound is called the
Michael Acceptor (because it accepts the
electron pair).
These reactants are not limited to carbonyl
compounds (see next slide)
388
389
Mechanism of the Michael Addition
Example of Michael Addition
390
Practice Question
How would you prepare the following
compound using a Michael addition in
your synthesis?
391
• Robinson Annulation (18.17)
qAnnulation: Formation of a ring structure
from acyclic structure.
qThe Robinson Annulation is a two step
process.
q(1) Michael Addition
q(2) Intramolecular Aldol Condensation
392
Mechanism of Robinson Annulation
393
How would you prepare the following
compound using a Robinson Annulation?
The key questions are, which part of this
molecule came from the Michael donor,
which one came from the Michael acceptor?
394
First identify the β and γ carbon on the
target molecule…count going away from
the double bond.
Draw a line going through the β and γ
carbon and splitting the double bond in
O
half
acceptor
donor
395
Therefore, in order to carry out this
synthesis, we need:
396
Practice Questions
Propose a synthesis for the following
compound, using a Robinson Annulation?
397
What are the product of the following
Robinson Annulations?
398