Knoevenagel condensation (Doebner): practice and mechanism
Malonic acid condenses with the aldehyde to the alkylidene malonic acid, which loses CO₂ in hot pyridine. An (E)-α,β-unsaturated carboxylic acid forms (cinnamic acid from benzaldehyde).
Example
Pyridine deprotonates the very acidic malonic acid and the enolate attacks the aldehyde. After loss of water an alkylidene malonic acid forms, which decarboxylates on heating. Result: an (E)-cinnamic acid or α,β-unsaturated carboxylic acid.
Mechanism step by step
- 1
Pyridine (with a little piperidine) deprotonates the CH₂ group of malonic acid.
- 2
The enolate attacks the aldehyde; after loss of water the alkylidene malonic acid forms.
- 3
On heating one COOH group decarboxylates, giving the (E)-unsaturated acid.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The alkylidene malonic acid is a β-dicarboxylic acid and decarboxylates on heating. Only one COOH group remains.
- The addition product loses water because the product is conjugated with the ring and the C=O. The C=C double bond forms.
Often confused with: Horner–Wadsworth–Emmons reaction · Claisen–Schmidt condensation · Cannizzaro reaction
Frequently asked questions
How does it differ from the aldol condensation?
The nucleophilic partner is a very acidic "active methylene" (between two acceptors), so a weak base such as pyridine suffices.
What is the Doebner variant?
Malonic acid (instead of the ester) in pyridine: condensation and decarboxylation run in one pot, straight to cinnamic acid.