\('A' + H_2O/H^+ \longrightarrow 'B' + NaOH + CaO \longrightarrow \text{ Benzene}\)
The compound \('A'\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Benzamide
Work backwards from benzene. Heating the sodium salt of a carboxylic acid with soda lime (NaOH + CaO) removes \( \ce{CO2} \) and gives a hydrocarbon with one carbon less; this is decarboxylation. Benzene is obtained this way from benzoic acid, so B is benzoic acid:
\( \ce{C6H5COOH + NaOH -> C6H5COONa + H2O} \); \( \ce{C6H5COONa} \xrightarrow[\Delta]{\ce{NaOH + CaO}} \ce{C6H6 + Na2CO3} \)
A gives benzoic acid on acid hydrolysis; an amide is hydrolysed to the acid:
\( \ce{C6H5CONH2 + H2O + H^+ -> C6H5COOH + NH4^+} \)
So A is benzamide.
Why the other options are incorrect
Option (1) Aniline — Incorrect. Aniline is not hydrolysed by \( \ce{H2O} \)/\( \ce{H^{+}} \) (it only forms the anilinium ion), and it has no carboxyl group to lose.
Option (2) Acetanilide — Incorrect. Hydrolysis gives aniline and acetic acid; neither gives benzene with soda lime.
Option (3) Benzylchloride — Incorrect. Hydrolysis gives benzyl alcohol, \( \ce{C6H5CH2OH} \), which is not decarboxylated by soda lime.
NCERT (2026-27 edition):
Decarboxylation
