Reaction of aniline with conc. HNO3 and conc. H2SO4 at 298 K will produce 47% of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → m-Nitroaniline
The –\( \ce{NH2} \) group is ortho and para directing. But the nitrating mixture is strongly acidic, and the basic amino group is protonated to the anilinium ion, –\( \ce{NH3^{+}} \). This positively charged group withdraws electrons and is meta directing. Nitration therefore gives a large amount of the meta product besides the para and ortho products:
\( \displaystyle \ce{C6H5NH2 + H+ -> C6H5NH3+} \)
So the 47% product is m-nitroaniline.
Why the other options are incorrect
Option (1) p-Nitroaniline — Incorrect. It is the main product, about 51%; it is the major product (with almost no meta isomer) only when the –\( \ce{NH2} \) group is first protected by acetylation.
Option (2) o-Nitroaniline — Incorrect. Only about 2% is formed.
Option (4) 2,4-Dinitroaniline — Incorrect. It is not a product of this reaction; the protonated ring is deactivated, so a second nitro group does not enter under these conditions.
NCERT (2026-27 edition):
Nitration
