For the nitration of aniline, select the incorrect option.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The yield of para isomer product is only 4% less than meta-isomer product
The –\( \ce{NH2} \) group of aniline is a strong activating, ortho and para directing group. But direct nitration is done in a strongly acidic mixture (conc. \( \ce{HNO3} \) + conc. \( \ce{H2SO4} \)), where aniline is protonated to the anilinium ion:
\( \displaystyle \ce{C6H5NH2 + H^+ -> C6H5NH3^+} \)
The –\( \ce{NH3^{+}} \) group is deactivating and meta directing. As a result, a large amount of the meta product forms along with the para product, and tarry oxidation products are also formed.
Option (1) Meta-isomer product yield is more than ortho-isomer product yield. — A correct statement. The meta isomer is 47%, while the ortho isomer is only 2%.
Option (2) Anilinium ion is produced. — A correct statement. The basic –\( \ce{NH2} \) group takes a proton from the strongly acidic nitrating mixture.
Option (3) -NH2 acts as activating group. — A correct statement. Its lone pair is delocalised into the ring and increases the electron density at the ortho and para positions.
Option (4) The yield of para isomer product is only 4% less than meta-isomer product — The incorrect statement. The para isomer (51%) is 4% more than the meta isomer (47%), not 4% less.
NCERT (2026-27 edition):
Nitration
