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Organic Chemistry

23.1 Carbonyl Condensations: The Aldol Reaction

Organic Chemistry23.1 Carbonyl Condensations: The Aldol Reaction

23.1 • Carbonyl Condensations: The Aldol Reaction

Carbonyl condensation reactions take place between two carbonyl partners and involve a combination of nucleophilic addition and α-substitution steps. One partner is converted into an enolate-ion nucleophile and adds to the electrophilic carbonyl group of the second partner. In so doing, the nucleophilic partner undergoes an α-substitution reaction and the electrophilic partner undergoes a nucleophilic addition. The general mechanism of the process is shown in Figure 23.2.

Figure 23.2 MECHANISM
The general mechanism of a carbonyl condensation reaction. One partner becomes a nucleophilic donor and adds to the second partner as an electrophilic acceptor. After protonation, the final product is a β-hydroxy carbonyl compound.
The three-step reaction is involved in the formation of beta-hydroxy carbonyl compound on the reaction of a carbonyl compound with base.

Aldehydes and ketones with an α hydrogen atom undergo a base-catalyzed carbonyl condensation reaction called the aldol reaction. For example, treatment of acetaldehyde with a base such as sodium ethoxide or sodium hydroxide in a protic solvent leads to rapid and reversible formation of 3-hydroxybutanal, known commonly as aldol (aldehyde + alcohol), hence the general name of the reaction.

Acetaldehyde reacts with sodium hydroxide and ethanol to yield an enolate ion, which further reacts with ethanal and forms 3-hydroxybutanal (aldol compound) with the release of hydroxide ion.

The exact position of the aldol equilibrium depends both on reaction conditions and on substrate structure. The equilibrium generally favors the condensation product in the case of aldehydes with no α substituent (RCH2CHO) but favors the reactant for disubstituted aldehydes (R2CHCHO) and for most ketones. Steric factors are probably responsible for these trends, since increased substitution near the reaction site increases steric congestion in the aldol product.

2 equivalents of phenylacetaldehyde react reversibly in sodium hydroxide and ethanol to produce beta-hydroxyaldehyde (90%); 2 equivalents of cyclohexanone react reversibly in sodium hydroxide and ethanol to produce beta-hydroxyketone (22%).

Worked Example 23.1

Predicting the Product of an Aldol Reaction

What is the structure of the aldol product from propanal?


An aldol reaction combines two molecules of reactant by forming a bond between the α carbon of one partner and the carbonyl carbon of the second partner. The product is a β-hydroxy aldehyde or ketone, meaning that the two oxygen atoms in the product have a 1,3 relationship.


Two molecules of ethanaldehyde react with sodium hydroxide to yield beta-hydroxy aldehyde (3-hydroxy-2-methylpentanal). New bond formed between C 2 and C 3.
Problem 23-1
Predict the aldol reaction product of the following compounds:
The structure of butyraldehyde shows a four-carbon chain carrying a carbonyl group at first carbon and is single-bonded to H and propyl group.
The structure of acetophenone shows a benzene ring single-bonded to the carbonyl group at the first carbon. The carbonyl carbon is single-bonded to a methyl group.
The structure of cyclopentanone shows a five-membered cyclic ring. The first carbon of the ring is double-bonded to an oxygen atom. Other carbon-carbon bonds are single-bonded.
Problem 23-2
Using curved arrows to indicate the electron flow in each step, show how the base-catalyzed reverse-aldol reaction of 4-hydroxy-4-methyl-2-pentanone takes place to yield 2 equivalents of acetone.
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