Nitration of aniline with nitric acid and sulphuric acid gives major product as
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → p-nitroaniline
The –\( \ce{NH2} \) group of aniline is a strong ortho/para director. In the strongly acidic nitrating mixture, however, much of the aniline is protonated to the anilinium ion, \( \ce{C6H5NH3+} \), which is meta directing:
\( \displaystyle \ce{C6H5NH2 + H+ -> C6H5NH3+} \)
So the major product is p-nitroaniline, with a large amount of the meta isomer.
Why the other options are incorrect
Option (1) m-nitroaniline — Incorrect. It is formed in a large amount (47%) because of the anilinium ion, but still less than the para isomer.
Option (3) o-nitroaniline — Incorrect. Only about 2% is formed.
Option (4) Anilinium hydrogen sulphate — Incorrect. This salt forms when aniline reacts with sulphuric acid as a base, \( \ce{C6H5NH2 + H2SO4 -> C6H5NH3+HSO4^-} \); it is an acid-base product, not a nitration product. On heating it gives sulphanilic acid.
NCERT (2026-27 edition):
Nitration
