Question

\(A\) is a neutral organic compound (M.F.: \(C_8H_9ON\)). On treatment with aqueous \(Br_2/HO^-\), \(A\) forms a compound \(B\) which is soluble in dilute acid. \(B\) on treatment with aqueous \(NaNO_2/HCl\) (0--5\(^\circ\)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 \(KMnO_4\) produces \(F\). \(F\) contains two different types of hydrogen atoms. The structure of \(A\) is 

• A. C
• B. A
• C. B
• D. D

Answer 1

The Correct Option is A

Step 1: Analyze molecular formula.
The molecular formula of \(A\) is \(C_8H_9ON\), which corresponds to a methyl-substituted benzamide.
Step 2: Reaction with \(Br_2/HO^-\).
Aqueous \(Br_2/HO^-\) indicates the Hofmann bromamide reaction, converting an amide into an amine with one fewer carbon atom. Thus, \(A\) must be a benzamide derivative, and \(B\) is an aromatic amine, soluble in dilute acid due to salt formation.
Step 3: Diazotization and Sandmeyer reaction.
Treatment of aromatic amine \(B\) with \(NaNO_2/HCl\) at 0–5\(^\circ\)C forms a diazonium salt \(C\). Reaction of \(C\) with \(CuCN/NaCN\) replaces the diazonium group with \(-CN\), forming nitrile \(D\).
Step 4: Hydrolysis of nitrile.
Hydrolysis of nitrile \(D\) gives carboxylic acid \(E\). Since \(E\) is also obtained from hydrolysis of \(A\), \(A\) must be a benzamide corresponding to that acid.
Step 5: Oxidation with acidified \(KMnO_4\).
Oxidation of \(E\) converts the methyl group into a carboxylic acid group. The product \(F\) having two different types of hydrogen atoms confirms a meta-substituted benzenedicarboxylic acid.
Step 6: Final identification.
Among the given options, only m-methyl benzamide satisfies all the reaction conditions and structural requirements.
Final Answer: \[ \boxed{\text{m-methyl benzamide}} \]