Identify the final major product [Q] in the following reactions.

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

Two reactions are used here.

(i) DIBAL-H reduction of an ester. Diisobutylaluminium hydride is a mild, bulky reducing agent. At low temperature it reduces an ester only as far as the aldehyde. So ethyl benzoate gives benzaldehyde, P:

    \( \displaystyle \ce{C6H5COOC2H5} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) DIBAL-H}} \ce{C6H5CHO + C2H5OH} \)

(ii) Claisen–Schmidt (cross aldol) condensation. Benzaldehyde has no α-hydrogen, so it cannot form an enolate and acts only as the carbonyl acceptor. The base removes an α-hydrogen from acetophenone, and the carbanion adds to the C=O of benzaldehyde. The β-hydroxy ketone formed loses water at once, because the new C=C is conjugated with both the ring and the C=O group:

    \( \displaystyle \ce{C6H5CHO + CH3COC6H5} \xrightarrow[\text{293 K}]{\ce{OH^-}} \ce{C6H5CH=CHCOC6H5 + H2O} \)

So Q is 1,3-diphenylprop-2-en-1-one (benzalacetophenone, or chalcone).

Why the other options are incorrect

• Option (1) 1,3-diphenylprop-2-en-1-ol — Incorrect. This alcohol would need the C=O group to be reduced (for example by \( \ce{NaBH4} \)); a base does not reduce it.

• Option (3) 1,3-diphenylpropene — Incorrect. It has no oxygen at all. Removing the C=O group completely needs a Clemmensen or Wolff–Kishner reduction.

• Option (4) 2-hydroxy-1,3-diphenylpropan-1-one — Incorrect. Its –OH is on the α-carbon. An aldol product has the –OH on the β-carbon, and here it dehydrates anyway.

NCERT (2026-27 edition):
Preparation of Aldehydes – “… Alternatively, nitriles are selectively reduced by diisobutylaluminium hydride, (DIBAL-H) to imines followed by hydrolysis to aldehydes: … Similarly, esters are also reduced to aldehydes with DIBAL-H.”

Cross Aldol Condensation – …