Correct Option: (B) Reaction of Nitrobenzene with Hydrogen gas in presence of finely divided Nickel

Step 1 (Governing Reaction Principle): [0.5 Mark]

The preparation of aniline from nitrobenzene is a classic reduction reaction. Nitrobenzene ($\text{C}_6\text{H}_5\text{NO}_2$) undergoes catalytic hydrogenation in the presence of metal catalysts like Nickel (Ni), Palladium (Pd), or Platinum (Pt) to yield aniline ($\text{C}_6\text{H}_5\text{NH}_2$).

The chemical equation is represented as:

Step 2 (Step-by-Step Analysis of Options): [0.25 Mark]

• (A) $\text{C}_6\text{H}_5\text{Cl} + \text{NH}_3$ requires high pressure and temperature (ammonolysis of aryl halides is difficult due to resonance).
• (B) Nitrobenzene reduction is the standard industrial method for producing aniline.
• (C) Reduction of Benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$) with $\text{LiAlH}_4$ yields Benzylamine ($\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$), not aniline.
• (D) Reduction of Benzonitrile ($\text{C}_6\text{H}_5\text{CN}$) with $\text{LiAlH}_4$ also yields Benzylamine.

Step 3 (Final Concluded Answer): [0.25 Mark]

Since the reduction of the nitro group ($-\text{NO}_2$) directly attached to the benzene ring using $\text{H}_2/\text{Ni}$ produces the primary aromatic amine (aniline), Option (B) is the correct choice.

💡 EL10 Concept Insight: Think of this like a "swap" game: you are simply replacing the two oxygen atoms attached to the nitrogen with two hydrogen atoms. Because the nitrogen is already stuck to the benzene ring, swapping those atoms turns the nitrobenzene directly into aniline!