Q.Benzoic acid, treated with

Step 1. Benzoic acid + NH3 (i), then heat (ii), gives the ammonium benzoate salt, which loses water on heating to give A = benzamide, C6H5-CO-NH2 (Section 12.12.C.4).

Step 2. A (benzamide) + NaOBr undergoes Hofmann's bromamide degradation (Section 12.14.5.2, method 4): the amide loses its carbonyl carbon entirely and gives a primary amine with one carbon fewer -- since benzamide's 'alkyl' group is the ring itself, this gives B = aniline, C6H5-NH2 (the nitrogen migrates directly onto the ring, with loss of the carbonyl carbon as carbonate). …

Recognise NaOBr specifically as the reagent for Hofmann's bromamide degradation (amide to amine, minus one carbon), then recognise NaNO2/HCl at low tem …

• Forgetting that Hofmann degradation removes the carbonyl carbon entirely (benzamide's ring-attached carbon is lost as CO2/carbonate) -- B is aniline (a simple ring-N …