Aniline is treated with concentrated sulphuric acid. Which compound is formed first, before the mixture is heated at 453–473 K?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Anilinium hydrogensulphate
Aniline is a base. With concentrated sulphuric acid it first accepts a proton and forms a salt, anilinium hydrogensulphate:
\( \displaystyle \ce{C6H5NH2 + H2SO4 -> C6H5NH3^+HSO4^-} \)
Only when this salt is heated with sulphuric acid at 453–473 K does sulphonation take place, giving p-aminobenzene sulphonic acid (sulphanilic acid) as the major product:
\( \displaystyle \ce{C6H5NH3^+HSO4^-} \xrightarrow{\text{453-473 K}} \ce{H2N-C6H4-SO3H + H2O} \)
Why the other options are incorrect
Option (1) Sulphanilic acid — Incorrect. It is the product of the second step, formed only on heating at 453–473 K.
Option (3) Benzenediazonium hydrogensulphate — Incorrect. A diazonium salt needs nitrous acid (sodium nitrite and acid) in the cold; sulphuric acid alone does not give it.
Option (4) o-Aminobenzenesulphonic acid — Incorrect. No sulphonation happens before heating, and when it does, the major product is the para acid, not the ortho acid.
NCERT (2026-27 edition):
9.6 Chemical Reactions – “(c) Sulphonation: Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid, as the major product.”