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 cone. NaOH forms a substituted benzyl alcohol as one of the two products. The compound A is :

We need to identify compound A ($$C_8H_8O_2$$) that undergoes cross-Aldol condensation with acetophenone and gives a substituted benzyl alcohol with conc. NaOH.

Clue 1: Cross-Aldol condensation with acetophenone ($$C_6H_5COCH_3$$) gives a single product. This means A must be an aldehyde without alpha-hydrogens (so it acts only as the electrophilic component, avoiding multiple products).

Clue 2: A with conc. NaOH gives a substituted benzyl alcohol. This is the Cannizzaro reaction, which occurs only with aldehydes lacking alpha-hydrogens. In this reaction, one molecule is oxidised to carboxylic acid and another is reduced to alcohol (benzyl alcohol).

Clue 3: Molecular formula $$C_8H_8O_2$$.

Checking Option 4: 4-Methoxybenzaldehyde ($$p\text{-}CH_3OC_6H_4CHO$$)

Formula: $$C_8H_8O_2$$ ✓ (CH$$_3$$O = OCH$$_3$$ on benzene ring + CHO)

It is an aldehyde with no alpha-hydrogen ✓

It can undergo cross-Aldol with acetophenone ✓

It undergoes Cannizzaro reaction with NaOH to give 4-methoxybenzyl alcohol ✓

The correct answer is Option 4: 4-methoxy benzaldehyde.