Reaction of benzaldehyde with acetophenone in the presence of methanolic NaOH yields a major stable product as
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → trans- 1,3-diphenylprop-2-en-1-one
In a cross aldol condensation two different carbonyl compounds react in the presence of a base. Here acetophenone, \( \ce{C6H5COCH3} \), has α-hydrogens and benzaldehyde has none, so the base removes an α-hydrogen from acetophenone only. The carbanion (enolate) formed attacks the carbonyl carbon of benzaldehyde, which cannot condense with itself, so one main product is formed. This base-catalysed cross aldol condensation of an aromatic aldehyde with a ketone is also called the Claisen–Schmidt condensation.
The β-hydroxy ketone first formed loses water readily, because the new C=C bond is conjugated with both the benzene ring and the C=O group:
\( \displaystyle \ce{C6H5CHO + CH3COC6H5} \xrightarrow{\ce{NaOH}/\ce{CH3OH}} \ce{C6H5-CH(OH)-CH2-CO-C6H5} \)
\( \displaystyle \ce{C6H5-CH(OH)-CH2-CO-C6H5} \xrightarrow{-\ce{H2O}} \ce{C6H5-CH=CH-CO-C6H5} \)
The product, 1,3-diphenylprop-2-en-1-one (benzalacetophenone, chalcone), can exist as cis and trans isomers. In the trans isomer the phenyl group and the benzoyl group lie on opposite sides of the double bond, away from each other, so it is the more stable one and the major product.
Why the other options are incorrect
Option (1) cis-1,3-diphenylprop-2-en-1-one — Incorrect. In the cis isomer the two bulky groups are on the same side of the C=C bond and crowd each other, so it is less stable than the trans isomer.
Option (2) 1,3-diphenyl propanone — Incorrect. A saturated ketone with no C=C bond would need hydrogenation of the condensation product; the base does not supply hydrogen.
Option (4) Benzophenone — Incorrect. Benzophenone, \( \ce{C6H5COC6H5} \), has no new carbon–carbon bond between the two reactants; it is not formed in an aldol condensation.
NCERT (2026-27 edition):
Cross Aldol Condensation – …
