Benzaldehyde reacts with semicarbazide to give

The correct answer is Option (2) → benzaldehyde semicarbazone

Aldehydes and ketones react with ammonia derivatives (Z–NH₂) by nucleophilic addition followed by elimination of water: the –\( \ce{NH2} \) group adds to the C=O group, and the C=O becomes C=N–Z. With hydroxylamine the product is an oxime, with hydrazine a hydrazone, and with semicarbazide (\( \ce{H2N-NH-CO-NH2} \)) a semicarbazone:

    \( \displaystyle \ce{C6H5CHO + H2N-NH-CO-NH2 -> C6H5CH=N-NH-CO-NH2 + H2O} \)

Semicarbazide has two –\( \ce{NH2} \) groups, but only the one attached to NH reacts. The lone pair of the other –\( \ce{NH2} \), next to C=O, is delocalised onto the carbonyl oxygen by resonance (as in an amide), so it is not nucleophilic. The product is therefore C₆H₅CH=N–NH–CO–NH₂, the structure in Option (2).

Why the other options are incorrect

• Option (1) benzenediazonium chloride — Incorrect. It is made from aniline with \( \ce{NaNO2} \) and HCl at 273–278 K (diazotisation); it has no link with an aldehyde.

• Option (3) C₆H₅CH=N–CO–NHNH₂ — Incorrect. Here the aldehyde would have condensed with the amide –\( \ce{NH2} \), which is not nucleophilic enough to react.

• Option (4) benzamide — Incorrect. \( \ce{C6H5CONH2} \) is an amide of benzoic acid; it contains no C=N bond.