Predict the % of nitration products when aniline is subjected to nitration with Conc. $HNO_3$/Conc. $H_2SO_4$.

o-nitroaniline 2%; m-nitroaniline 47%; p-nitroaniline 51%

The correct answer is Option (2) → o-nitroaniline 2%; m-nitroaniline 47%; p-nitroaniline 51%
When aniline is subjected to nitration using concentrated nitric acid and concentrated sulfuric acid, the major products are meta (47%) and para (51%), while ortho is very less (2%).

• In strongly acidic medium (conc. HNO₃ / conc. H₂SO₄), aniline gets protonated to form the anilinium ion (–NH₃⁺).

• This group is electron-withdrawing and deactivating, and it acts as a meta-directing group.

• Hence, meta substitution becomes significant (~47%).

• However, a small amount of free (unprotonated) aniline still exists, which is ortho/para directing:

  • Para product is favored due to less steric hindrance → 51%

  • Ortho is very less due to steric hindrance → 2%

Key idea 
Strong acidic medium → –NH₂ becomes –NH₃⁺ → meta directing dominates, but para still major due to steric + partial free base

To memorise

In strong acidic conditions, aniline behaves as a meta director; otherwise, it is an ortho/para director.Trick line: “Acid bana de meta, free amine de ortho-para.”