Aniline forms meta nitro derivative in strongly acidic medium. The reason is-
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Anilinium ion is meta directing
The −\( \ce{NH2} \) group of aniline is a strongly activating, ortho- and para-directing group, because its lone pair is delocalised into the ring. In the strongly acidic nitrating mixture (conc. \( \ce{HNO3} \) + conc. \( \ce{H2SO4} \)), however, aniline is protonated:
\( \displaystyle \ce{C6H5NH2 + H^+ -> C6H5NH3^+}\ \ (\text{anilinium ion}) \)
In the anilinium ion the nitrogen has no free lone pair, and the positive −\( \ce{NH3^{+}} \) group withdraws electrons from the ring. It is a deactivating, meta-directing group, so a significant amount of m-nitroaniline is formed along with the o- and p-isomers:
\( \displaystyle \ce{C6H5NH2} \xrightarrow[\text{288 K}]{\text{conc. }\ce{HNO3},\ \text{conc. }\ce{H2SO4}} \text{p-nitroaniline (51\%)} + \text{m-nitroaniline (47\%)} + \text{o-nitroaniline (2\%)} \)
Why the other options are incorrect
Option (1) −NH2 group is meta directing — Incorrect. Free −\( \ce{NH2} \) directs to ortho and para positions; the meta product comes from the protonated form.
Option (2) −NH2 group is o-p directing — Incorrect. This is true, but it explains the ortho and para products, not the meta derivative.
Option (4) −NH2 group is highly reactive — Incorrect. High reactivity explains the tarry oxidation products and polysubstitution, not the formation of a meta derivative.
NCERT (2026-27 edition):
Electrophilic Substitution – “… (b) Nitration: Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.”