JEE Main 2026 — Jan 23, Shift 2 — Chemistry solution

JEE Main2026Jan 23, Shift 2Chemistry

A student has been given a compound "x" of molecular formula $C_{6} H_{7} N$ . 'x' is sparingly soluble in water. However, on addition of dilute mineral acid, 'x' becomes soluble in water. 'x' when treated with $C H C l_{3}$ and 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:-

Solution

$C_{6} H_{7} N$ with the given behaviour is aniline ( $C_{6} H_{5} N H_{2}$ ):

- Sparingly soluble in water but soluble in dilute $H C l$ (forms anilinium salt).

- With $C H C l_{3} / a l c . K O H$ (carbylamine reaction), aniline gives phenyl isocyanide (offensive smell) → this confirms a primary amine.

- With benzenesulphonyl chloride (Hinsberg test), aniline gives $N$ -phenyl benzenesulphonamide ( $C_{6} H_{5} - N H - S O_{2} - C_{6} H_{5}$ ), which is soluble in alkali because of the acidic $N - H$ .

Counting distinct H environments in $C_{6} H_{5} - N H - S O_{2} - C_{6} H_{5}$ (compound z):

- $N - H$ : 1 type

- Aniline ring (attached to NH): ortho, meta, para = 3 types

- Sulphonyl ring (attached to $S O_{2}$ ): ortho, meta, para = 3 types

Total = $1 + 3 + 3 = 7$ different hydrogen environments.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →