Q.Explain the mechanism of Claisen Schmidt reaction.

Claisen–Schmidt reaction: dilute-base-catalysed crossed aldol condensation between an aromatic aldehyde (no α-H) and a ketone/aldehyde with α-H, giving an α,β-unsaturated carbonyl compound (chalcone) after dehydration.

Example reaction

Benzaldehyde ($C_6H_5CHO$) reacts with acetophenone ($C_6H_5COCH_3$) in the presence of dilute NaOH to give benzalacetophenone (a chalcone):

$C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{\text{dil. NaOH}} C_6H_5CH=CHCOC_6H_5 + H_2O$

Mechanism (step-by-step)

1. Formation of carbanion (enolate): Hydroxide ion abstracts an α-hydrogen from the ketone (acetophenone), since this hydrogen is acidic due to the adjacent carbonyl group. This generates a resonance-stabilised carbanion (enolate ion).

$C_6H_5COCH_3 + OH^- \rightarrow C_6H_5CO\overset{-}{C}H_2 \; (\leftrightarrow C_6H_5C(O^-)=CH_2) + H_2O$

2. Nucleophilic addition (aldol step): The nucleophilic carbanion attacks the electrophilic carbonyl carbon of benzaldehyde (which has NO α-hydrogen and so cannot itself form a carbanion). This gives an alkoxide intermediate.

$C_6H_5CO\overset{-}{C}H_2 + C_6H_5CHO \rightarrow C_6H_5COCH_2CH(O^-)C_6H_5$

3. Protonation: The alkoxide picks up a proton from water/solvent to give the neutral β-hydroxy ketone (the aldol product).

$C_6H_5COCH_2CH(O^-)C_6H_5 + H_2O \rightarrow C_6H_5COCH_2CH(OH)C_6H_5 + OH^-$

4. Base-catalysed dehydration (E1cb): Base removes the acidic α-hydrogen (adjacent to the carbonyl) once more; the resulting carbanion expels the β-hydroxide as $OH^-$, forming the carbon–carbon double bond. Elimination is strongly favoured here because the resulting alkene is conjugated with both the carbonyl group and the aromatic ring, giving extra stabilisation.

$C_6H_5COCH_2CH(OH)C_6H_5 \xrightarrow{OH^-,\;-H_2O} C_6H_5COCH=CHC_6H_5$ (benzalacetophenone / a chalcone)

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