'A' is a neutral organic compound (M. F : C₈H₉ON ). On treatment with aqueous Br₂/HO⁽⁻⁾, ' A ' forms a compound ' B ' which is soluble in dilute acid. ' B ' on treatment with aqueous NaNO₂/HCl(0-5°C) produces a compound ' C ' which on treatment with CuCN/NaCN produces ' D '. Hydrolysis of ' D ' produces ' E ' which is also obtainable from the hydrolysis of ' A '. ' E ' on treatment with acidified KMnO₄ produces ' F '. ' F ' contains two different types of hydrogen atoms. The structure of ' A ' is

Solution

Chemistry ma’am · 3 steps

Read the chain backwards from F: only two kinds of H means a para diacid, so the amide in A must be para.

1. Hofmann step. Br₂/OH⁻ on a primary amide ArCONH₂ gives the amine ArNH₂, with one carbon less. B is basic, so it dissolves in dilute acid. With C₈H₉NO, A must be a toluamide, CH₃C₆H₄CONH₂.

   CH₃C₆H₄CONH₂ →(Br₂/OH⁻) CH₃C₆H₄NH₂ →(NaNO₂/HCl) CH₃C₆H₄N₂⁺Cl⁻ →(CuCN) CH₃C₆H₄CN →(H₃O⁺) CH₃C₆H₄COOH

2. Carbon comes back. Diazotisation and the Sandmeyer reaction put -CN exactly where -CONH₂ was. Hydrolysis then gives toluic acid E, the same acid that hydrolysis of A gives. Acidified KMnO₄ turns the ring -CH₃ into -COOH, so F is a benzenedicarboxylic acid.

   Option	F after the sequence	Types of H in F
   (a) para-toluamide	terephthalic acid (1,4)	2: four equal ring H, two equal COOH
   (b) meta-toluamide	isophthalic acid (1,3)	4: H-2, H-4/H-6, H-5, COOH
   (c) ortho-toluamide	phthalic acid (1,2)	3: H-3/H-6, H-4/H-5, COOH
   (d) N-methylbenzamide	no Hofmann reaction (no -NH₂)	chain fails at B

3. So. Only the para isomer gives a fully symmetric diacid with two types of hydrogen. A is para-toluamide, option (a).

The idea to keep. When a question ends with how many types of H a product has, use symmetry: a 1,4-disubstituted benzene with identical groups is the most symmetric. Then trace the substituent positions back to the start.

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