Question

A student has been given a compound "x" of molecular formula- $C_6H_7N$. 'x' is sparingly soluble in water... 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. 5
• B. 4
• C. 8
• D. 7

Hint:

The Hinsberg test is crucial: $1^\circ$ amine forms alkali-soluble sulfonamide; $2^\circ$ amine forms alkali-insoluble sulfonamide; $3^\circ$ amine does not react.

Answer 1

The Correct Option is A

The molecular formula of $X$ is $C_6H_7N$, which corresponds to aniline ($C_6H_5NH_2$), a primary aromatic amine. \[ X = \mathrm{C_6H_5NH_2} \] On treatment with benzenesulphonyl chloride ($\mathrm{C_6H_5SO_2Cl}$), aniline undergoes the Hinsberg reaction to form a sulphonamide: \[ Z = \mathrm{C_6H_5SO_2NH C_6H_5} \] The product $Z$ is soluble in alkali due to the presence of an acidic $\mathrm{N\!-\!H}$ proton, confirming that $X$ is a primary amine. Now, determine the number of different types of hydrogen atoms present in $Z$. \begin{itemize} \item One type of hydrogen due to the N--H proton. \item The aromatic protons in the phenyl rings appear in four distinct chemical environments due to symmetry and substitution effects. \end{itemize} \[ \text{Total different hydrogen atoms} = 1 + 4 = 5 \] \[ \therefore \text{Number of different hydrogen atoms in } Z = \boxed{5} \]