In the following reaction, identify A and B.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A = Benzaldehyde, B = Acetophenone
An aldehyde or ketone with an α-hydrogen forms a carbanion (enolate) in dilute alkali. This carbanion adds to the C=O group of another carbonyl compound (aldol addition), and the product then loses water to give an α,β-unsaturated carbonyl compound (aldol condensation). When the two partners are different, it is a cross aldol condensation; with an aromatic aldehyde and a ketone it is called the Claisen–Schmidt reaction.
The product Ph–CH=CH–CO–Ph can be split at the C=C bond into its two parts:
Ph–CH= comes from benzaldehyde, \( \ce{C6H5CHO} \), which has no α-hydrogen and can only accept the carbanion.
=CH–CO–Ph comes from acetophenone, \( \ce{C6H5COCH3} \), whose \( \ce{CH3} \) group supplies the α-hydrogens and forms the carbanion.
\( \displaystyle \ce{C6H5CHO + CH3COC6H5} \xrightarrow[\text{293 K}]{\ce{OH^-}} \ce{C6H5CH=CHCOC6H5 + H2O} \)
The product is benzalacetophenone (chalcone).
Why the other options are incorrect
Option (2) A = Benzaldehyde, B = Acetone — Incorrect. Acetone would give \( \ce{Ph-CH=CH-CO-CH3} \) (benzalacetone), which has a methyl group, not a phenyl group, on the C=O.
Option (3) A = Benzaldehyde, B = Benzophenone — Incorrect. Benzophenone (Ph–CO–Ph) has no α-hydrogen, so it cannot form the carbanion; with benzaldehyde no aldol condensation is possible.
Option (4) A = Acetophenone, B = Acetophenone — Incorrect. Self-condensation of acetophenone gives \( \ce{Ph-C(CH3)=CH-CO-Ph} \), which carries an extra methyl group on the double bond.
NCERT (2026-27 edition):
Cross Aldol Condensation
