Q.Reduction of nitrobenzene by which of the following reagent gives aniline?
(Two or more options may be correct.)

The key idea is Electrophilic Aromatic Substitution — specifically, the reduction of the nitro group ($-NO_2$) to an amino group ($-NH_2$) on the benzene ring.

Reasoning steps:

1. Nitrobenzene is reduced to aniline by catalytic hydrogenation ($H_2$-Pd) or by chemical reduction using a metal and acid (Sn/HCl or Fe/HCl). …

The question is about the reduction of the nitro group ($-NO_2$) to the amino group ($-NH_2$) in an aromatic ring. This is a classic transformation in organic chemistry, and the key point is that the reduction requires a source of active hydrogen atoms, usually generated in situ by a metal-acid combination or by catalytic hydrogenation.

Let’s go through each option.

1. Sn/HCl — Tin and hydrochloric acid. This is a classic chemical reduction. Tin reacts with HCl to produce nascent hydrogen ($[H]$), which reduces the nitro group stepwise: nitrobenzene → nitrosobenzene → phenylhydroxylamine → aniline. The reaction is carried out in acidic medium, and aniline is obtained as the salt (anilinium chloride), which is then neutralized. This works perfectly.

2. Fe/HCl — Iron and hydrochloric acid. This is the Béchamp reduction, historically the most common industrial method. Iron also generates nascent hydrogen with HCl. The mechanism is similar, and it cleanly gives aniline. This works.

3. $H_2$-Pd — Catalytic hydrogenation using palladium on carbon. Molecular hydrogen in the presence of a metal catalyst (Pd, Pt, Ni) reduces the nitro group directly to the amino group. This is a heterogeneous catalytic reduction and works efficiently for nitrobenzene to aniline. No acidic medium is required here — the hydrogen adds across the N–O bonds. This works. …

Concept: Reduction of Nitrobenzene to Aniline

Nitrobenzene ($C_6H_5NO_2$) can be reduced to aniline ($C_6H_5NH_2$) using different reducing agents. The key is to identify which reagents provide active hydrogen in acidic or neutral medium to reduce the nitro group ($-NO_2$) to an amino group ($-NH_2$).

Method: Reagent-Specific Reduction Analysis

Steps:

1. Identify the functional group change

   Nitrobenzene → Aniline: $-NO_2$ is reduced to $-NH_2$ (6-electron reduction).

2. Check each reagent’s reducing capability

   • Sn/HCl → Strong acidic reducing agent. Sn + HCl produces nascent hydrogen ($[H]$) which reduces $-NO_2$ to $-NH_2$. ✓ Works
   • Fe/HCl → Same mechanism as Sn/HCl. Fe + HCl gives $[H]$ for reduction. ✓ Works
   • $H_2$-Pd → Catalytic hydrogenation. $H_2$ gas with Pd catalyst reduces $-NO_2$ to $-NH_2$. …

Common Mistakes Students Make on This Question

This question tests your understanding of reduction of nitrobenzene to aniline — a classic reaction in organic chemistry. The key is knowing which reducing agents work and which give different products.

✗ Mistake 1: Assuming all reducing agents give aniline

Many students think any reducing agent will convert nitrobenzene ($C_6H_5NO_2$) to aniline ($C_6H_5NH_2$). This is not true.

• Correct fact:
  • Sn/HCl and Fe/HCl reduce nitrobenzene to aniline.
  • $H_2$-Pd also reduces it to aniline (catalytic hydrogenation).
  • Sn/NH$_4$OH does not give aniline — it gives azoxybenzene or other intermediates (like phenylhydroxylamine) because the medium is alkaline.

How to avoid:

Memorise the medium dependence:

• Acidic medium (HCl with Sn or Fe) → aniline
• Neutral/alkaline medium (Sn/NH$_4$OH) → different products (azoxy, azo, hydrazo compounds)

✗ Mistake 2: Forgetting that catalytic hydrogenation works

Some students think only metal-acid reductions work, and ignore $H_2$-Pd.

• Correct fact: $H_2$ with Pd/C (or Pt, Ni) reduces nitrobenzene to aniline cleanly.

How to avoid:

Remember: Catalytic hydrogenation is a standard method for reducing nitro groups to amines. Include it in your list of valid reagents.

✗ Mistake 3: Confusing Sn/NH$_4$OH with Sn/HCl

Students often misread or misremember the reagent — Sn/NH$_4$OH looks similar to Sn/HCl but behaves very differently.

• Correct fact: Sn/NH$_4$OH is alkaline → gives azoxybenzene (not aniline). Sn/HCl is acidic → gives aniline.

How to avoid: …