Aldol Condensation
Aldol Condensation: β-hydroxy carbonyl, key C–C bond forming reaction. Step-by-step mechanism, reagents and practice.
What happens
β-hydroxy carbonyl, key C–C bond forming reaction. Base generates an enolate from one aldehyde molecule, which then attacks the carbonyl carbon of a second, unreacted aldehyde molecule.
Reaction summary
2CH₃CHO → CH₃CH(OH)CH₂CHO
Reagents and conditions
- Starting material: Acetaldehyde
- Reagents: Dilute NaOH
- Conditions: Dilute NaOH, 0°C
- Products: 3-Hydroxybutanal
Mechanism
- Enolate formation: Dilute NaOH removes an α-proton from one acetaldehyde molecule, forming a resonance-stabilized enolate.
- Nucleophilic addition: The enolate carbon attacks the carbonyl carbon of a second, unreacted acetaldehyde molecule.
- Protonation: Protonation of the resulting alkoxide gives the neutral β-hydroxy aldehyde (aldol) product.
Common mistakes
- Confusing 'aldol addition' (stops at the β-hydroxy carbonyl, shown here) with 'aldol condensation' proper (goes on to eliminate water, giving an α,β-unsaturated carbonyl), the two terms are often used loosely.
- Forgetting only a catalytic amount of base is needed here (unlike Claisen), since the product isn't acidic enough to need a full equivalent to drive the equilibrium.
Memory trick
"Aldol" = ALcohol + aldehyde in one molecule, the product carries both a new OH and the original carbonyl.