Question

An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with \( \mathrm{Br_2} \) and KOH forms compound (R) having molecular formula \( \mathrm{C_6H_7N} \). Name P, Q and R respectively are :

• A. Benzoic acid, benzamide, aniline
• B. Toluic acid, methylbenzamide, 2-methylaniline
• C. Benzoic acid, 4-methylbenzamide, 4-methylaniline
• D. Phenylethanoic acid, phenylethanamide, benzamine

Hint:

In Hoffmann bromamide degradation: \[ \mathrm{RCONH_2 \rightarrow RNH_2} \] the product amine always contains one carbon atom less than the parent amide.

Answer 1

The Correct Option is A

Concept: Carboxylic acids react with ammonia under heating to form amides. Further, amides on treatment with: \[ \mathrm{Br_2/KOH} \] undergo Hoffmann bromamide degradation to produce primary amines having one carbon atom less. General reaction: \[ \mathrm{RCONH_2 \xrightarrow{Br_2/KOH} RNH_2} \]

Step 1: Identifying compound \(R\). Given: \[ \mathrm{R} \] has molecular formula: \[ \mathrm{C_6H_7N} \] This corresponds to: \[ \mathrm{C_6H_5NH_2} \] which is aniline. Thus: \[ R = \text{aniline} \]

Step 2: Finding compound \(Q\). In Hoffmann degradation: \[ \mathrm{RCONH_2 \rightarrow RNH_2} \] If the product is aniline: \[ \mathrm{C_6H_5NH_2} \] then the amide must be: \[ \mathrm{C_6H_5CONH_2} \] which is benzamide. Thus: \[ Q = \text{benzamide} \]

Step 3: Finding compound \(P\). Benzamide is formed from benzoic acid by treatment with ammonia and heating. Reaction: \[ \mathrm{C_6H_5COOH \xrightarrow{NH_3,\ heat} C_6H_5CONH_2} \] Thus: \[ P = \text{benzoic acid} \] Hence: \[ P = \text{Benzoic acid} \] \[ Q = \text{Benzamide} \] \[ R = \text{Aniline} \] Therefore, correct option is: \[ \boxed{(1)} \]