Match the Compounds (List - I) with the appropriate Catalyst/Reagents (List - II) for their reduction into corresponding amines.

Choose the correct answer from the options given below :

(A) Amide $$\rightarrow$$ (III) $$LiAlH_4$$

• Why this reagent: $$LiAlH_4$$ is a powerful nucleophilic reducing agent capable of reducing the highly stable $$C=O$$ bond of an amide all the way to a $$CH_2$$ group to yield a primary amine.
• Why others fail: $$H_2/Ni$$ is usually not strong enough to reduce amides efficiently under standard conditions. $$Sn/HCl$$ is specifically used for nitro groups and won't touch the amide carbonyl. $$NaOH$$ would simply cause hydrolysis to a carboxylate salt and ammonia, not reduction.

(B) Nitrobenzene $$\rightarrow$$ (IV) $$Sn, HCl$$

• Why this reagent: This is a classic laboratory method for preparing aniline. The metal/acid combination ($$Sn/HCl$$ or $$Fe/HCl$$) provides the protons and electrons necessary to reduce $$-NO_2$$ to $$-NH_2$$.
• Why others fail: While $$H_2/Ni$$ could technically work, $$Sn/HCl$$ is the preferred "textbook" match for aromatic nitro compounds in this list. $$LiAlH_4$$ is avoided for aromatic nitro compounds because it can lead to azo compounds (coupling) rather than pure amines.

(C) Nitrile $$\rightarrow$$ (II) $$H_2/Ni$$

• Why this reagent: Catalytic hydrogenation is the most efficient way to add four hydrogen atoms across the $$C\equiv N$$ triple bond to produce a primary amine ($$R-CH_2NH_2$$).
• Why others fail: $$NaOH$$ would hydrolyze the nitrile to an amide and then a carboxylic acid. $$Sn/HCl$$ is not the standard reagent for full nitrile reduction (it is used in the Stephen reduction to stop at the aldehyde, but only with $$SnCl_2$$).

(D) Phthalimide $$\rightarrow$$ (I) $$NaOH$$

• Why this reagent: This refers to the final step of the Gabriel Phthalimide Synthesis. Once an alkyl group is attached to the nitrogen, base-catalyzed hydrolysis ($$NaOH$$) is used to break the cyclic imide bonds and release the primary amine.
• Why others fail: The goal here is "cleavage" to release the amine, not "reduction" of the aromatic ring or carbonyls. Reducing agents like $$LiAlH_4$$ or $$H_2/Ni$$ would attack the carbonyl groups of the phthalimide ring itself, destroying the starting material instead of releasing the desired amine.