Claisen Condensation
Claisen Condensation: Ethoxide deprotonates the α-carbon of one ester molecule to form an ester enolate; this enolate attacks the carbonyl carbon of a…
What happens
Ethoxide deprotonates the α-carbon of one ester molecule to form an ester enolate; this enolate attacks the carbonyl carbon of a second ester molecule, and loss of ethoxide (the leaving group) regenerates a carbonyl, giving a β-ketoester after a final, essentially irreversible deprotonation by excess base.
Reaction summary
2 CH₃CO₂Et --NaOEt--> CH₃COCH₂CO₂Et + EtOH
Reagents and conditions
- Starting material: Ethyl acetate
- Reagents: Sodium ethoxide (NaOEt), full equivalent
- Products: Ethyl acetoacetate
Mechanism
- Enolate formation: Ethoxide removes an α-proton from one ester molecule, generating a resonance-stabilized ester enolate.
- Nucleophilic acyl substitution: The enolate attacks the carbonyl carbon of a second ester molecule; ethoxide is expelled as the tetrahedral intermediate collapses, forming the new C–C bond.
- Driving deprotonation: The β-ketoester product is far more acidic than a normal ester (flanked by two carbonyls); a full equivalent of ethoxide deprotonates it, pulling the equilibrium toward product until an acidic workup releases the neutral final product.
Common mistakes
- Forgetting a FULL equivalent of base is required (not catalytic), the final deprotonation of the acidic product is what drives the equilibrium to completion.
- Confusing this with the Aldol condensation, Claisen combines two esters (losing an alkoxide leaving group), Aldol combines two carbonyls with no leaving group and instead loses water on dehydration.
Memory trick
"Claisen needs a leaving group, Aldol doesn't": Claisen is on esters (which have OR to kick out), Aldol is on aldehydes/ketones (which don't).