Claisen condensation: practice and mechanism
The ester enolate attacks a second ester molecule and ethoxide leaves. A β-keto ester forms; deprotonation of the acidic β-keto ester drives the reaction.
Example
The ester enolate attacks the carbonyl group of a second ester molecule and ethoxide is released. A β-keto ester forms. The driving force is deprotonation of the very acidic β-keto ester, which is why a full equivalent of base is needed.
Mechanism step by step
- 1
Ethoxide deprotonates an α-carbon of the ester to the ester enolate.
- 2
The enolate attacks the C=O group of a second ester molecule and ethoxide leaves: β-keto ester.
- 3
Ethoxide deprotonates the acidic β-keto ester (driving the equilibrium); acidic workup gives the neutral product.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Aldol addition · Ester hydrolysis (saponification)
Frequently asked questions
How does it differ from the aldol reaction?
In the aldol reaction an OH group remains (addition). In the Claisen condensation an alkoxy group leaves (substitution) and a second C=O group forms.
Why use NaOEt with ethyl esters?
Other alkoxides would transesterify and NaOH would hydrolyse. Ethoxide only swaps EtO for EtO.