The major product in the above reaction is

The correct answer is Option (2) → \(C_6H_5CH=CHCOC_6H_5\)

An aldehyde or ketone with at least one α-hydrogen forms a carbanion in dilute alkali, which adds to the carbonyl group of another molecule (aldol reaction). On heating, the β-hydroxy carbonyl compound loses water and gives an α,β-unsaturated carbonyl compound (aldol condensation). When the two partners are different, it is called cross aldol condensation.

Here benzaldehyde has no α-hydrogen, so it can only accept the carbanion; acetophenone has α-hydrogens on its \( \ce{CH3} \) group and supplies the carbanion. Only one cross product is therefore formed:

    \( \displaystyle \ce{C6H5CHO + CH3COC6H5} \xrightarrow{\text{dil. NaOH}} \ce{C6H5CH(OH)CH2COC6H5} \)

    \( \displaystyle \ce{C6H5CH(OH)CH2COC6H5} \xrightarrow{\Delta} \ce{C6H5CH=CHCOC6H5 + H2O} \)

So the major product is C₆H₅CH=CHCOC₆H₅ (1,3-diphenylprop-2-en-1-one, benzalacetophenone).

Why the other options are incorrect

• Option (1) \(C_6H_5CH=CHC_6H_5\) — Incorrect. Stilbene has no C=O group at all; an aldol condensation keeps the carbonyl group of the ketone.

• Option (3) \(C_6H_5CH_2OH + C_6H_5CH(OH)C_6H_5\) — Incorrect. These are reduction products (alcohols); no reducing agent is present.

• Option (4) \(C_6H_5CH_2OH + C_6H_5COONa\) — Incorrect. These are the products of the Cannizzaro reaction of benzaldehyde, which needs concentrated alkali and no partner with α-hydrogens:

      \( \displaystyle \ce{2C6H5CHO + NaOH(conc.) -> C6H5CH2OH + C6H5COONa} \)

NCERT (2026-27 edition):
Cross Aldol Condensation