'A' is a neutral organic compound (M. F : C_8H_9ON$C_{8}H_{9}ON$). On treatment with aqueous Br_2/HO^(-)$Br_{2}/HO^{(-)}$, 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous NaNO_2/HCl(0-5^circC)$NaNO_{2}/HCl(0-5^{\circ}C)$ 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$KMnO_{4}$ produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is

### Core Logic Let's trace the sequence: 1. A (C_8H_9ON$C_8H_9ON$) is neutral and reacts with Br_2/OH^-$Br_2/OH^-$ (Hofmann Bromamide Degradation). This means A is a primary amide. 2. Product B is soluble in dilute acid, meaning it is a primary amine (Ar-NH_2$Ar-NH_2$ or alkyl amine). 3. B reacts with NaNO_2/HCl$NaNO_2/HCl$ at 0-5^circC$0-5^{\circ}C$ to form C. Since C undergoes Sandmeyer with CuCN to form D, B must be an aromatic primary amine, and C is a diazonium salt. 4. D is an aryl cyanide (Ar-CN$Ar-CN$). Hydrolysis of D yields E (Ar-COOH$Ar-COOH$). 5. Crucially, E is also obtainable from the direct hydrolysis of A. This confirms A is an aryl amide of the form Ar-CONH_2$Ar-CONH_2$.

6. Let's analyze the formula C_8H_9ON$C_8H_9ON$. The amide group is -CONH_2$-CONH_2$. Removing -CONH_2$-CONH_2$ leaves C_7H_7$C_7H_7$. A benzene ring with one methyl group is a tolyl group. So A is a methylbenzamide (CH_3-C_6H_4-CONH_2$CH_3-C_6H_4-CONH_2$). 7. E is methylbenzoic acid (CH_3-C_6H_4-COOH$CH_3-C_6H_4-COOH$). 8. E is oxidized by acidified KMnO_4$KMnO_4$ to F. The methyl group on the benzene ring oxidizes to -COOH$-COOH$. Thus, F is a benzenedicarboxylic acid (HOOC-C_6H_4-COOH$HOOC-C_6H_4-COOH$).

9. The problem states that F contains two different types of hydrogen atoms. Let's check the isomers of benzenedicarboxylic acid: - Phthalic acid (ortho): Contains 2 types of aromatic hydrogens + 1 type of COOH hydrogen = 3 types. - Isophthalic acid (meta): Contains 3 types of aromatic hydrogens + 1 type of COOH hydrogen = 4 types. - Terephthalic acid (para): Due to symmetry, all 4 aromatic hydrogens are equivalent. So it contains 1 type of aromatic hydrogen + 1 type of COOH hydrogen = 2 types.

Since F has only two types of hydrogens, F must be terephthalic acid (para isomer). Thus, E is p-methylbenzoic acid, and A is p-methylbenzamide. ### Step 1: Final Identification Compound A is p-methylbenzamide. This corresponds to the structure in option (3). ### Pattern Recognition A classic sequence linking Hofmann bromamide, Sandmeyer, and side-chain oxidation. Symmetrical molecules like para-isomers minimize the number of unique proton environments (critical for NMR or simple counting). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids