An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br_2$Br_{2}$ and KOH forms compound (R) having molecular formula C_9H_7N$C_{9}H_{7}N$. Names of P, Q and R respectively are.

### Core Logic The reaction of an amide with Br_2$Br_2$ and KOH is Hoffmann bromamide degradation. It steps down the carbon chain by one carbonyl carbon to form a primary amine. Let's analyze the options and molecular formula. The question says (R) has molecular formula C_9H_7N$C_9H_7N$. Wait, looking at the standard solutions for this type of problem, aniline is C_6H_7N$C_6H_7N$. The PDF says C_9H_7N$C_9H_7N$ which is likely a typo in the original paper for C_6H_7N$C_6H_7N$, since option (1) gives Aniline (C_6H_7N$C_6H_7N$). Let's assume the standard sequence: 1. mathrmPh-COOH xrightarrowmathrmNH_3, Delta mathrmPh-CO-NH_2$\mathrm{Ph-COOH} \xrightarrow{\mathrm{NH}_3, \Delta} \mathrm{Ph-CO-NH}_2$ (Benzoic acid to Benzamide) 2. mathrmPh-CO-NH_2 xrightarrowmathrmBr_2/mathrmKOH mathrmPh-NH_2$\mathrm{Ph-CO-NH}_2 \xrightarrow{\mathrm{Br}_2/\mathrm{KOH}} \mathrm{Ph-NH}_2$ (Benzamide to Aniline) Aniline is C_6H_5NH_2 = C_6H_7N$C_6H_5NH_2 = C_6H_7N$. So P is Benzoic acid, Q is Benzamide, R is Aniline. ### Pattern Recognition Reaction sequence: Carboxylic\ Acid xrightarrowNH_3, Delta Amide xrightarrowBr_2/KOH Amine$Carboxylic\ Acid \xrightarrow{NH_3, \Delta} Amide \xrightarrow{Br_2/KOH} Amine$. The Br_2/KOH$Br_2/KOH$ step is Hoffmann bromamide reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids