Benzaldehyde, lacking any alpha-hydrogen, condenses with an aliphatic aldehyde or a methyl ketone (which DOES have an alpha-hydrogen) in dilute alkali at room temperature, giving an alpha,beta-unsaturated aldehyde/ketone -- a crossed aldol condensation that runs cleanly precisely because benzaldehyde cannot self-condense. Benzaldehyde + acetaldehyde gives cinnamaldehyde (C6H5-CH=CH-CHO); benzaldehyde + acetone gives benzylideneaceto …
Step 1. Benzaldehyde has NO alpha-hydrogen of its own (its 'alpha-position' is the aromatic ring, which cannot be deprotonated by dilute base the way an aliphatic methyl group can), so it cannot self-condense.
Step 2. Acetone, CH3-CO-CH3, DOES have alpha-hydrogens on both of its methyl groups; in dilute NaOH, one of those alpha-carbons is deprotonated to give the resonance-stabilised enolate carbanion, CH2=C(O-)-CH3 (equivalently -:CH2-CO-CH3).
Step 3. This acetone-derived carbanion attacks benzaldehyde's electrophilic carbonyl carbon, forming the new C-C bond and, after protonation, the beta-hydroxy ketone (aldol) intermediate, C6H5-CH(OH)-CH2-CO-CH3. …
Recognise that benzaldehyde (no alpha-hydrogen) can only act as the electrophile, while acetone (which has alpha-hydrogens) supplies the attacking enolate carbanion -- this specific pairing is the Cl …
Drawing the SELF-aldol product of acetone alone (mesityl oxide-type structure), forgetting that benzaldehyde is present specifically to act as the el …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2018Set ANNUAL5 marks
Q.Explain the mechanism of Claisen Schmidt reaction.
›Reveal solutionSolution
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 (C6H5CHO) reacts with acetophenone (C6H5COCH3) in the presence of dilute NaOH to give benzalacetophenone (a chalcone):
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).
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.
C6H5COC−H2+C6H5CHO→C6H5COCH2CH(O−)C6H5
Protonation: The alkoxide picks up a proton from water/solvent to give the neutral β-hydroxy ketone (the aldol product).
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.
C6H5COCH2CH(OH)C6H5OH−,−H2OC6H5COCH=CHC6H5 (benzalacetophenone / a chalcone)