Complete the following reaction given below

The correct answer is Option (1) → 1,3-diphenylprop-2-en-1-one (benzalacetophenone)

When an aldehyde or ketone that has α-hydrogens is treated with dilute alkali, it forms a carbanion (enolate) that adds to the C=O group of another carbonyl compound. This is the aldol reaction; it gives a β-hydroxy carbonyl compound, which on heating loses water to give an α,β-unsaturated carbonyl compound (aldol condensation).

When an aromatic aldehyde without α-hydrogens reacts with a ketone that has them, the crossed product is formed cleanly. This is the Claisen–Schmidt condensation:

• Acetophenone (\( \ce{C6H5COCH3} \)) has α-hydrogens; \( \ce{OH^{-}} \) removes one to form the carbanion.

• Benzaldehyde has no α-hydrogen, so it can only accept the carbanion at its C=O carbon.

    \( \displaystyle \ce{C6H5CHO + CH3COC6H5} \xrightarrow[293\ \mathrm{K}]{\text{dil. }\ce{NaOH}} \ce{C6H5CH(OH)CH2COC6H5} \)

    \( \displaystyle \ce{C6H5CH(OH)CH2COC6H5} \xrightarrow{-\ce{H2O}} \ce{C6H5CH=CH-CO-C6H5} \)

The double bond forms in conjugation with both the benzene ring and the C=O group, which makes the product very stable. So A is C₆H₅CH=CH–CO–C₆H₅, benzalacetophenone (chalcone).

Why the other options are incorrect

• Option (2) 1,3-diphenylprop-2-en-1-ol — Incorrect. Only the C=O group has been reduced to CH–OH; NaOH is not a reducing agent, so this alcohol is not formed.

• Option (3) 1,3-diphenylprop-1-ene-1,3-diol — Incorrect. The C=O group has been reduced to CH–OH and an extra –OH sits on a C=C carbon. NaOH reduces nothing, and the condensation removes water; it does not add an –OH group.

• Option (4) 1,3-diphenylprop-1-en-1-ol — Incorrect. This is the enol form of the saturated ketone 1,3-diphenylpropan-1-one. It would need the C=C bond of the product to be hydrogenated, which NaOH cannot do.

NCERT (2026-27 edition):
Cross Aldol Condensation

Note: The same/similar question has been asked in different years by NTA as per the details given below: