m-chlorobenzaldehyde on treatment with 50% KOH solution yields :

Solution

Chemistry ma’am · 4 steps

No α-hydrogen plus strong alkali means Cannizzaro: one aldehyde is oxidised, the other reduced.

1. Spot the reaction. The -CHO group sits directly on the ring, so there is no α-hydrogen and aldol condensation is impossible. In concentrated (50%) KOH, such an aldehyde undergoes the Cannizzaro reaction.
2. Hydride transfer. OH⁻ adds to the carbonyl carbon of one molecule. That adduct passes a hydride ion (H⁻) to the carbonyl carbon of a second molecule. The first becomes a carboxylate (oxidised), and the second becomes an alkoxide that picks up a proton to give an alcohol (reduced).

   2 ClC₆H₄CHO + KOH → ClC₆H₄COO⁻ K⁺ + ClC₆H₄CH₂OH

3. Chlorine stays put. The aryl C-Cl bond has partial double-bond character from resonance with the ring. Hydroxide cannot replace it under these mild conditions, so the Cl remains at the meta position in both products.
4. So. The products are the m-chlorobenzoate ion and m-chlorobenzyl alcohol, which is option (a).

The idea to keep. Aldehyde with no α-H in concentrated NaOH or KOH: think Cannizzaro, giving a carboxylate salt plus an alcohol. Ring substituents like an aryl Cl just ride along unchanged.

Answer from the NTA final answer key. Solution by PrepWiser.