Reaction of aniline with \(HNO_3\) and conc. \(H_2SO_4\) at \(298 K\) will produce \(47\%\) of

m-Nitroaniline

The correct answer is option 3. m-Nitroaniline.

The nitration of aniline with conc.nitric acid (\(HNO_3\)) and conc. sulfuric acid (\(H_2SO_4\)) at \(298 \, \text{K}\) involves the following considerations:

1. Aniline and Acids Reaction:

Aniline (\(C_6H_5NH_2\)) is a basic compound. When treated with strong acids like concentrated sulfuric acid, the amino group (\(-NH_2\)) gets protonated to form the anilinium ion (\(C_6H_5NH_3^n+\)).

2. Electrophilic Aromatic Substitution:

The nitration reaction is an electrophilic aromatic substitution where a nitro group (\(-NO_2\)) is introduced to the benzene ring. The presence of the protonated amino group (\(C_6H_5NH_3^+\)) significantly influences the position where the nitro group is introduced.

The amino group in aniline (\(-NH_2\)) is an electron-donating group and typically directs electrophiles to the ortho and para positions relative to itself. However, when it is protonated to form the anilinium ion (\(-NH_3^+\)), it becomes an electron-withdrawing group. Electron-withdrawing groups direct electrophiles to the meta position on the benzene ring.

Under the reaction conditions (strong acids and 298 K), the anilinium ion is the major species. The electron-withdrawing nature of the anilinium ion directs the nitro group to the meta position, resulting predominantly in the formation of m-nitroaniline.

Therefore, the major product of the nitration of aniline under these conditions is m-nitroaniline, which accounts for the \(47\%\) mentioned in the problem. The protonated amino group directs the incoming nitro group to the meta position due to its electron-withdrawing effect.