A student has been given a compound " $x$ " of molecular formula- $\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{~N}$. ' $x$ ' is sparingly soluble in water. However, on addition of dilute mineral acid, ' $x$ ' becomes soluble in water. ' $x$ ' when treated with $\mathrm{CHCl}_{3}$ and $\mathrm{KOH}($ alc $)$, ' $y$ ' is produced. ' y ' has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, ' $x$ ' gives a compound ' $z$ ' which is soluble in alkali. The number of different " $H$ " atoms present in ' $z$ ' is:-

• A

  8
• B

  4
• C

  5
• Correct answer
  D

  7

Solution

Compound X has molecular formula $C_6H_7N$, is sparingly soluble in water but soluble in dilute mineral acid → primary aromatic amine (aniline).
X with $\text{CHCl}_3 + \text{KOH (alc.)}$ gives Y with unpleasant smell → carbylamine reaction, confirms primary amine.
X with benzenesulphonyl chloride gives Z which is soluble in alkali → Hinsberg test confirms primary amine. Z is N-phenylbenzenesulfonamide.
Structure of Z: $\text{C}_6\text{H}_5\text{-NH-SO}_2\text{-C}_6\text{H}_5$

Hence, the correct answer is option: 'D'.