JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Compounds Containing Nitrogen

Sulphanilic acid reacts with $\mathrm{NaN}{\mathrm{O}}_2$ and $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ to form its diazonium salt (X); since sulphanilic acid exists as a zwitterion (${}^{-}{\mathrm{O}}_3\mathrm{S}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{N}{\mathrm{H}}_3^{+}$), the diazotization gives an internal diazonium salt, ${}^{-}{\mathrm{O}}_3\mathrm{S}-{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{N}}_2^{+}$.

X couples with 1-naphthylamine at the ring position para to the amino group — the position of highest electron density on the naphthalene ring — forming the azo linkage $-\mathrm{N}=\mathrm{N}-$.

The resulting azo dye (Y) is $\mathrm{H}{\mathrm{O}}_3\mathrm{S}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{N}=\mathrm{N}-{\mathrm{C}}_{10}{\mathrm{H}}_6-\mathrm{N}{\mathrm{H}}_2$, a red dye formed by this Griess–Ilosvay coupling.