To understand which reducing agent is used for the conversion of nitrobenzene (C\(_6\)H\(_5\)-NO\(_2\)) to aniline (C\(_6\)H\(_5\)-NH\(_2\)) in an acidic medium, we must analyze the role of each option provided.
- \(LiAlH_4\): Lithium aluminium hydride is a strong reducing agent that is generally used for the reduction of carboxylic acids, esters, and amides to alcohols. It is not commonly used for the reduction of nitro compounds to anilines.
- \(Sn/HCl\): This combination of tin metal with hydrochloric acid (HCl) acts as a reducing agent in an acidic medium. It converts nitrobenzene to aniline through a multi-step reaction involving the formation of a tin chloride complex that facilitates electron transfer. This is the correct answer for the reduction of nitrobenzene to aniline in acidic medium.
- \(Na/Alcohol\): Sodium in alcohol is typically used as a reducing agent for the reduction of carbonyl compounds. It is not utilized for reducing nitrobenzene to aniline in an acidic medium.
- \(H_2/Ni\): This combination is used in catalytic hydrogenation reactions, often requiring high temperatures. While it can reduce nitro compounds, it is not the conventional choice for laboratory preparation of aniline from nitrobenzene in an acidic medium.
Given the above explanations, \(Sn/HCl\) is the most appropriate reducing agent for converting nitrobenzene to aniline in acidic conditions. The reaction mechanism involves an initial reduction to a hydroxylamine intermediate, which further reduces to form aniline.
Thus, the correct answer is Sn/HCl.