Aldol Condensation: Mechanism & Examples | Chemca

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Aldol Condensation: Mechanism, Cross Aldol & Examples

By Chemca Editorial Team • Last Updated: January 2026 • 12 min read

Aldol Condensation is a fundamental carbon-carbon bond-forming reaction where two molecules of an aldehyde or ketone having at least one $\alpha$-hydrogen react in the presence of a dilute base to form a $\beta$-hydroxy aldehyde (Aldol) or $\beta$-hydroxy ketone (Ketol). Upon heating, this product undergoes dehydration to form an $\alpha,\beta$-unsaturated carbonyl compound.

1. General Reaction

The reaction occurs in two stages:

Aldol Addition: Formation of $\beta$-hydroxy aldehyde.
Condensation: Elimination of water molecule (Dehydration) to form a double bond.

$$ 2CH_3CHO \xrightarrow{\text{Dil. NaOH}} \underbrace{CH_3CH(OH)CH_2CHO}_{\text{3-Hydroxybutanal (Aldol)}} \xrightarrow{\Delta, -H_2O} \underbrace{CH_3CH=CHCHO}_{\text{But-2-enal (Crotonaldehyde)}} $$

Conditions:

Substrate: Aldehydes/Ketones with $\alpha$-hydrogen (e.g., Acetaldehyde, Acetone).
Reagent: Dilute Base ($NaOH$, $KOH$, $Na_2CO_3$, or $Ba(OH)_2$).
Final Product: $\alpha,\beta$-Unsaturated Carbonyl.

2. Detailed Mechanism

The reaction proceeds via the formation of a resonance-stabilized Enolate Ion.

Step 1: Formation of Enolate Ion

The base ($OH^-$) abstracts an acidic $\alpha$-proton from the aldehyde to generate a nucleophilic enolate ion.

$$ OH^- + H-CH_2CHO \rightleftharpoons H_2O + [ \bar{C}H_2-CHO \leftrightarrow CH_2=C(O^-)H ] $$

Step 2: Nucleophilic Attack

The enolate ion attacks the electrophilic carbonyl carbon of a second aldehyde molecule.

$$ CH_3CHO + \bar{C}H_2CHO \rightarrow CH_3-CH(O^-)-CH_2CHO $$

Step 3: Protonation (Aldol Formation)

The alkoxide ion accepts a proton from water to form the Aldol.

$$ CH_3-CH(O^-)-CH_2CHO + H_2O \rightarrow CH_3-CH(OH)-CH_2CHO + OH^- $$

Step 4: Dehydration (Condensation)

Upon heating, the aldol undergoes elimination of water (usually via $E1cB$ mechanism) to form a stable conjugated system.

3. Cross Aldol Condensation

When condensation is carried out between two different aldehydes or ketones, it is called Cross Aldol Condensation.

Mixture of Products

If both reactants have $\alpha$-hydrogens, a mixture of four products is obtained (two self-aldol products and two cross-aldol products).

Example: Ethanal + Propanal

Reacting $CH_3CHO$ and $CH_3CH_2CHO$ yields:

But-2-enal (Self: Ethanal)
2-Methylpent-2-enal (Self: Propanal)
2-Methylbut-2-enal (Cross: Ethanal Enolate + Propanal)
Pent-2-enal (Cross: Propanal Enolate + Ethanal)

4. Aldol vs. Cannizzaro

Feature	Aldol Condensation	Cannizzaro Reaction
Requirement	Must have $\alpha$-Hydrogen	Must NOT have $\alpha$-Hydrogen
Reagent	Dilute Base	Concentrated Base (50% NaOH)
Product Type	Coupling (Bond formation)	Disproportionation (Redox)
Examples	Acetaldehyde, Acetone	Formaldehyde, Benzaldehyde

Aldol Condensation Quiz

Test your concepts on Aldehydes and Ketones. 10 MCQs with explanations.

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