Claisen–Schmidt condensation: practice and mechanism
The ketone enolate attacks benzaldehyde (which has no α-H itself). After loss of water a conjugated enone forms with the C=C next to the ring.
Example
A crossed aldol condensation: the ketone forms the enolate and benzaldehyde (no α H, cannot form an enolate itself) is the electrophile. After loss of water an α,β-unsaturated ketone forms; acetophenone gives chalcone.
Mechanism step by step
- 1
NaOH forms the ketone enolate (benzaldehyde has no α-H and cannot form one).
- 2
The enolate attacks the more reactive aldehyde: a β-hydroxy ketone forms.
- 3
Water is eliminated via an enolate (E1cB): an enone conjugated with both the ring and the C=O forms.
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 condensation · Wittig reaction
Frequently asked questions
Why no mixture as in crossed aldol reactions?
Benzaldehyde can only be the electrophile, and aldehydes are more reactive than ketones. So almost only the ketone enolate reacts with the aldehyde.
Why is water lost so easily?
The product is conjugated through the C=C with both the benzene ring and the C=O. This stability drives the elimination.