Among the following, which is the correct set of reducing agents for conversion of nitrobenzene into aniline?

(A) $H_2/Pd$
(B) $Sn+HCl$
(C) $Fe +HCl$
(D) $NaBH_4$

Choose the correct answer from the options given below:

(A), (B) and (C) only

The correct answer is Option (3) → (A), (B) and (C) only

Conversion of nitrobenzene to aniline involves reduction of the nitro group (-NO₂) to an amino group (-NH₂).

C₆H₅NO₂ → C₆H₅NH₂

This requires complete reduction of nitrogen from oxidation state +5 in NO₂ to -3 in NH₂.

Let us examine each reagent.     

Option (A) H₂ / Pd

Catalytic hydrogenation using hydrogen gas in presence of palladium catalyst effectively reduces nitrobenzene to aniline.

C₆H₅NO₂ + 3H₂ → C₆H₅NH₂ + 2H₂O

This is a standard industrial and laboratory method.

Hence, it is a correct reducing agent.

Option (B) Sn + HCl

Tin in presence of hydrochloric acid produces nascent hydrogen and acts as a strong reducing system.

Sn + 2HCl → SnCl₂ + H₂ (nascent hydrogen)

This reduces the nitro group stepwise:

Nitro → Nitroso → Hydroxylamine → Amine

Thus nitrobenzene is converted to aniline after neutralisation.

Hence, this is also correct.

Option (C) Fe + HCl

Iron with hydrochloric acid behaves similarly to tin.

Fe + HCl generates reducing conditions that convert nitrobenzene into aniline via intermediate stages.

This method is widely used in laboratory preparations.

Hence, this is also correct.

Option (D) NaBH₄

Sodium borohydride is a mild reducing agent.

It can reduce:

Aldehydes
Ketones
Acid chlorides

But it is generally not strong enough to reduce nitro groups attached to aromatic rings.

Reduction of nitrobenzene requires stronger hydride donors like LiAlH₄ or catalytic hydrogenation.

Therefore, NaBH₄ does not convert nitrobenzene to aniline.

Hence, it is incorrect.