JEE Main 2026 — Jan 22, Shift 2 — Chemistry solution

JEE Main2026Jan 22, Shift 2Chemistry

The compound A, $C_{8} H_{8} O_{2}$ reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as :

A
2-hydroxy acetophenone
B
4-methoxy benzaldehyde
C
4-hydroxy benzylaldehyde
D
4-methyl benzoic acid

Solution

For a clean cross-aldol with acetophenone to give a single product, A must be a non-enolisable aldehyde (no $α$ -hydrogen).

Among the choices with formula $C_{8} H_{8} O_{2}$ , 4-methoxybenzaldehyde (anisaldehyde) fits - it lacks $α$ -hydrogens.

With conc. NaOH, 4-methoxybenzaldehyde undergoes the Cannizzaro reaction (because it has no $α$ -H) to give a substituted benzyl alcohol (4-methoxybenzyl alcohol) and the corresponding benzoate, confirming the choice.

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