Chapter 22 – Enols and Enolates

Acidity of the α hydrogen
o The position next door to a carbonyl is called the α position
o When an α proton is abstracted, the resulting carbanion is resonance-stabilized.
-
This is called an enolate ion.
o The pKa of most aldehydes and ketones is around 20.
 So why do we show hydroxide deprotonating them?
 Sometimes we want to use the enolate to attack the starting material.
 See the aldol condensation below.
 The resulting enolate ion is so nucleophilic, that it will react completely,
driving the acid-base equilibrium forward.
 LeChatelier just won’t leave us alone!
 Often, we don’t want to use hydroxide.
 In mixed condensations we might want one aldehyde to
completely deprotonate so that it doesn’t attack its protonated
form instead of the intended electrophile.
 It may compete to react with the substrate instead of the enolate.
 In this case, we need a base strong enough to quantitatively
deprotonate the aldehyde or ketone, but which is less
nucleophilic.
o Specifically, we often see LDA (Lithium diisopropylamide)

o Because of the bulky alkyl groups, LDA is too hindered to
be a competitive nucleophile.
Alpha halogenations of ketones
o Base-promoted α halogenations
 Step 1: Deprotonation
LDA
-

Step 2: Enolate attacks the halogen
LDA
-


This reaction keeps going, because any additional α protons are even
more acidic than they were to begin with.
o Haloform reaction
 A test for methyl ketones
 When a ketone has a methyl group, the resulting –CX3 is a good enough
leaving group to be replaced by the hydroxide.
 Overall, a methyl ketone reacts with X2 and hydroxide to give a
carboxylate and a haloform (CHX3)
o Acid-catalyzed α halogenations
 Easier to add just one halogen
Alkylation of enolate ions
o SN2 is back again!
NaOH
-
-

Aldol condensation
o Enolate ion attacks a carbonyl
H+
-
o If you heat the product up, it loses water to form a new carbon-carbon double
bond.

A common question in this material is when you’ll lose water and when
you won’t.
 First, calm down. This material is on the take-home quiz and a
few questions on the ACS, so it’s less tested than the rest of the
semesters.
 Next, Mr. Baker’s take-home will make it clear when to do this.
o Either heat will be indicated with the arrow (as below).
o Or the resulting product will be highly conjugated, driving
the reaction forward.
 The ACS is multiple choice!
o Most likely, both products will not be shown.
o If both products are shown, I personally would choose the
dehydrated product, but as I don’t look at this test I can’t
tell you what to do here.
heat

o Crossed or mixed aldol condensations
 This just means that you started with two different ketones or aldehydes.
Claisen Condensations
o Similar to aldol, but with esters you get nucleophilic substitution at the carbonyl
instead of nucleophlic addition at the carbonyl.
o The drawing below shows the overall reaction, but remember that with
nucleophilic substitution at the carbonyl there is always a tetrahedral
intermediate.
NaOEt
EtOH
o Why does this become favorable?
-


The product deprotonates because the hydrogens which are α to both
carbonyls are even more acidic (pKa about 12)
Malonic ester synthesis
o

Malonic ester
A hydrogen α to two carbonyls will be even more acidic than a regular α
hydrogen.
In this case, the pKa is 13

o First, the malonic ester is deprotonated.
NaOEt
-
o Then, the enolate ion attacks an alkyl halide in an SN2 reaction.
-
o The esters undergo hydrolysis.
H3O+
o When heated, any carboxyl group that is β to a carbonyl will decarboxylate.
heat
2
1

 This is called decarboxylation of β-keto acids
 Overall, the COOH turns into a H.
o Again, be flexible in your thinking. This is just one example of a reaction, but the
individual steps can all happen with different chemicals.
Michael Addition
o When a pi bond is conjugated with a carbonyl, there are two electrophilic sites.
-
-
o Because of this, nucleophiles can add to either partially positive site.
o When a nucleophile adds to the carbon-carbon double bond, this is called
Michael Addition.
o How do you know whether a nucleophile will add to the alkene or to the
carbonyl?
 The really strong nucleophiles add to the carbonyl.
 Grignards and alkyl lithiums

The other nucleophiles add to the carbon-carbon double bond.
 Enolate ions
 Dialkyl cuprates (R2CuLi)
(CH3)2CuLi

Robinson annulations
o First, you do a Michael addition using an enolate ion as your nucleophile.
-
o If a cyclic aldol reaction would yield a six-membered ring, then do it!
OH
Base
1
2
2
1
1
3
3
3
4
4
-
6
6
7
7
o Finish it off with a dehydration.
OH
1
6
2
3
5
4
7
5
4
5
5
6
2
1
6
2
3
5
4
7
7