Nitration of aniline with a mixture of nitric acid and sulphuric acid gives a mixture of products. This is because
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → anilinium ion is formed and NH2 group is ortho and para directing.
The –\( \ce{NH2} \) group of aniline is a strongly activating, ortho and para directing group: its lone pair is pushed into the ring by resonance. So aniline on its own would be nitrated at the ortho and para positions.
But nitration is done in a strongly acidic mixture (\( \ce{HNO3} \) + \( \ce{H2SO4} \)). Being a base, much of the aniline is protonated to the anilinium ion:
\( \displaystyle \ce{C6H5NH2 + H^+ -> C6H5NH3^+} \)
The –\( \ce{NH3^{+}} \) group has no lone pair and pulls electrons out of the ring, so it is deactivating and meta directing.
The reaction mixture therefore contains both free aniline (o/p directing) and the anilinium ion (meta directing). The meta product comes from the anilinium ion.
Why the other options are incorrect
Option (1) NH2 group is para directing — Incorrect. True for free aniline, but it cannot explain why almost half the product is the meta isomer.
Option (2) NH2 group is meta directing — Incorrect. –\( \ce{NH2} \) is ortho and para directing; it is the –NH3+ group that directs meta.
Option (4) NH2 group is a deactivating group — Incorrect. –\( \ce{NH2} \) is a strongly activating group.
NCERT (2026-27 edition):
Nitration
