Aniline on nitration yield 47% m-nitroaniline also. This is due to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Formation of anilinium ion by protonation of aniline
The –\( \ce{NH2} \) group of aniline is ortho/para directing. But nitration is carried out in a strongly acidic medium (\( \ce{HNO3} \) + \( \ce{H2SO4} \)), where a large part of the aniline is protonated:
\( \displaystyle \ce{C6H5NH2 + H+ -> C6H5NH3+} \)
The anilinium ion has no lone pair on nitrogen and carries a positive charge, so it is an electron withdrawing, meta directing group. Nitration of this part gives m-nitroaniline (about 47%), alongside p- (51%) and o- (2%) products.
Why the other options are incorrect
Option (1) +M or +R effect of –NH2 group attached with benzene — Incorrect. This effect directs to ortho and para positions; it explains the o/p products, not the meta product.
Option (2) –M or –R effect of –NH2 group attached with benzene — Incorrect. The free –\( \ce{NH2} \) group has a +R effect, not a −R effect.
Option (4) Very strong reducing nature of HNO3 acid — Incorrect. \( \ce{HNO3} \) is an oxidising agent, not a reducing agent.
NCERT (2026-27 edition):
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.”