What is the major product (X) in the following series of reactions?

The correct answer is Option (3) → the para zwitterion of sulphanilic acid

Aniline is a base, so it first reacts with sulphuric acid to form a salt, anilinium hydrogensulphate (A). When this salt is heated at 453–473 K (180–200 °C), a –\( \ce{SO3H} \) group enters the ring. The –\( \ce{NH2} \) group directs it to the ortho and para positions, and the bulky group goes mainly to the less crowded para position. The product is p-aminobenzenesulphonic acid, called sulphanilic acid (B).

    \( \displaystyle \ce{C6H5NH2 + H2SO4 -> C6H5NH3^{+}HSO4^{-}}\quad(\text{A}) \)

    \( \displaystyle \ce{C6H5NH3^{+}HSO4^{-}} \xrightarrow{\text{453-473 K}} \ce{H2N-C6H4-SO3H + H2O}\quad(\text{B, para}) \)

Sulphanilic acid has a basic –\( \ce{NH2} \) group and a strongly acidic –\( \ce{SO3H} \) group in the same molecule. The acidic proton moves to the nitrogen, forming a dipolar ion, or zwitterion. This is X in the equilibrium:

    \( \displaystyle \ce{H2N-C6H4-SO3H <=> H3N^{+}-C6H4-SO3^{-}}\quad(\text{X}) \)

So X is the para zwitterion, with –\( \ce{NH3^{+}} \) and –\( \ce{SO3^{-}} \) opposite each other on the ring.

Why the other options are incorrect

• Option (1) the ortho zwitterion — Incorrect. Sulphonation at high temperature gives the para acid; the bulky –\( \ce{SO3H} \) group avoids the crowded ortho position.

• Option (2) the meta zwitterion — Incorrect. –\( \ce{NH2} \) is an ortho/para director. Meta substitution needs a meta-directing group, which is not what drives this baking process.

• Option (4) the meta disulphonated zwitterion — Incorrect. Only one –\( \ce{SO3H} \) group enters, and it does not go to the meta positions.

NCERT (2026-27 edition):
Sulphonation