CBSE Class 12 Chemistry 2023: question 24(a)

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Step-by-step solution

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1. (i) Sulphanilic acid is p-aminobenzenesulphonic acid. The acidic $-\mathrm{S}{\mathrm{O}}_3\mathrm{H}$ protonates the basic $-\mathrm{N}{\mathrm{H}}_2$, giving the zwitter ion:
   ${}^{+}{\mathrm{H}}_3\mathrm{N}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{S}{\mathrm{O}}_3^{-}$ (benzene ring with $-\mathrm{N}{\mathrm{H}}_3^{+}$ and $-\mathrm{S}{\mathrm{O}}_3^{-}$ para to each other).

   1 mark

2. (ii) By acetylation of aniline with acetic anhydride (or acetyl chloride) in pyridine to form acetanilide, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2+(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{)}_2\mathrm{O}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{NHCOC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$. The N lone pair is now also delocalised onto the C=O group, so the ring is less activated; after the desired substitution, hydrolysis regenerates the $-\mathrm{N}{\mathrm{H}}_2$ group.

   1 mark

Answer: (i) ${}^{+}{\mathrm{H}}_3\mathrm{N}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{S}{\mathrm{O}}_3^{-}$ (para); (ii) protect –NH₂ by acetylation (acetanilide)

Revise the chapter

• Chapter: Amines (notes, formulas, where marks are lost and every past question)