Observe the following reaction sequence:

(P) —NH₃, Δ→ (Q) —Br₂/KOH→ (R)

Which of the following is the correct structure for P, Q and R?

Correct Option: The first option (corresponding to the structures shown in the first image, where P is benzoic acid, Q is benzamide, and R is aniline).

Step-by-Step Explanation:

1. Conversion of P to Q:
When a carboxylic acid is treated with ammonia (NH₃) followed by heating (represented by Δ), it undergoes a nucleophilic acyl substitution followed by thermal dehydration.
Initially, the carboxylic acid reacts with ammonia in an acid-base reaction to form an ammonium carboxylate salt:
C₆H₅COOH + NH₃ → C₆H₅COO^-NH₄⁺
Upon heating (Δ), this salt undergoes dehydration (loss of a water molecule, H₂O) to yield a primary amide (benzamide):
C₆H₅COO^-NH₄⁺ —Δ→ C₆H₅CONH₂ + H₂O
Therefore, starting material P is benzoic acid (C₆H₅COOH) and product Q is benzamide (C₆H₅CONH₂).

2. Conversion of Q to R:
When benzamide (Q) is treated with bromine in the presence of potassium hydroxide (Br₂/KOH), it undergoes the Hoffmann Bromamide Degradation reaction.
In this reaction, the primary amide is converted into a primary amine containing one carbon atom less than the starting amide via an isocyanate intermediate:
C₆H₅CONH₂ + Br₂ + 4KOH → C₆H₅NH₂ + K₂CO₃ + 2KBr + 2H₂O
Thus, the primary amide group (-CONH₂) is degraded to an amine group (-NH₂). The resulting product R is aniline (C₆H₅NH₂).

Conclusion:
Analyzing the structures shown in the provided images:
• P: Benzoic acid (benzene ring attached to -COOH)
• Q: Benzamide (benzene ring attached to -CONH₂)
• R: Aniline (benzene ring attached to -NH₂)
These structures correspond exactly to the first option.