Question

Observe the following reaction sequence: \[ (P) \xrightarrow[\Delta]{NH_3} (Q) \xrightarrow{Br_2/KOH} (R) \] Which of the following is the correct structure for \( P \), \( Q \) and \( R \)?

• A. \( P = \) Benzoic acid (\({-COOH}\)), \( Q = \) Benzamide (\({-CONH2}\)), \( R = \) Aniline (\({-NH2}\))
• B. \( P = \) Benzoic acid, \( Q = \) Ammonium benzoate, \( R = \) Benzene
• C. \( P = \) Ammonium benzoate, \( Q = \) Benzamide, \( R = \) Benzene
• D. \( P = \) Benzoic acid, \( Q = \) Aniline, \( R = \) Benzene

Hint:

Key organic reactions to remember:
Carboxylic acid + \(NH_3\) (heat) \(\rightarrow\) Amide
Hofmann bromamide reaction converts amides to amines with one carbon less
Aromatic amides give aromatic amines

Answer 1

The Correct Option is A

Step 1: Benzoic acid reacts with ammonia on heating to form benzamide: \[ {C6H5COOH + NH3 ->[\Delta] C6H5CONH2 + H2O} \] Thus, \[ P = \text{Benzoic acid}, Q = \text{Benzamide} \]
Step 2: Benzamide undergoes Hofmann bromamide reaction with \( {Br2/KOH} \), forming aniline: \[ {C6H5CONH2 ->[Br2/KOH] C6H5NH2} \] Thus, \[ R = \text{Aniline} \]