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 compound 'x' given in the problem has the molecular formula C_6H_7N, indicating that it is an aromatic amine, most likely aniline due to the structure fitting this formula. Aniline is known for being sparingly soluble in water, confirming that this is consistent with the compound properties described.

Let's examine the chemical reaction described:

• Aniline (or compound 'x') is treated with benzenesulphonyl chloride. This reaction is known to produce a sulphonamide.
• The sulphonamide formed is soluble in alkali (NaOH), which is characteristic of N-substituted benzenesulphonamides.

Now, to deduce the structure of 'z', the product of this reaction:

• The original compound 'x' (aniline) has the formula C_6H_5NH_2.
• When aniline reacts with benzenesulphonyl chloride (C_6H_5SO_2Cl), it forms N-phenylbenzenesulphonamide, a compound typically of the form C_6H_5SO_2NHC_6H_5.

Analyzing the hydrogen atoms in compound 'z':

• The introduction of the sulphonyl group from the reaction will result in a total of 5 different hydrogen types due to different chemical environments in the structure of N-phenylbenzenesulphonamide.

Therefore, the number of different types of hydrogen atoms present in compound 'z' is 5.

Thus, the correct answer is 5.