Explanation

The compound x has molecular formula C₆H₇N and becomes soluble in dilute mineral acid, indicating that it is a basic amine.

On treatment with CHCl₃ and alcoholic KOH, compound x gives a foul-smelling product y. This is the carbylamine test, which is given only by primary amines.

Hence, x is a primary amine. The only aromatic primary amine with molecular formula C₆H₇N is aniline (C₆H₅NH₂).

Aniline reacts with benzenesulphonyl chloride in the Hinsberg test to form a sulphonamide z, which is soluble in alkali due to the presence of acidic N–H hydrogen.

The structure of compound z is:

C₆H₅–SO₂–NH–C₆H₅

Now, count the number of different (chemically non-equivalent) hydrogen atoms in compound z.

Each benzene ring in z is a monosubstituted phenyl ring. In a monosubstituted benzene ring, aromatic hydrogens fall into three distinct environments:

• Ortho hydrogens (2 equivalent)
• Meta hydrogens (2 equivalent)
• Para hydrogen (1)

Thus, each monosubstituted phenyl ring contributes 3 different aromatic H types.

Since there are two monosubstituted phenyl rings:

$$ 3 + 3 = 6 $$

Additionally, there is one N–H hydrogen.

Therefore, total number of different hydrogen atoms:

$$ 6 + 1 = 7 $$

Hence, the correct answer is 7.

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