A student has been given a compound " x " of molecular formula −C6H7N. ' x ' is sparingly soluble in water. However, on addition of dilute mineral acid , ' x ' becomes soluble in water. ' x ' when treated with CHCl3 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:-
(A)5
(B)8
(C)4
(D)7✓ Correct
Answer: (D)
Solution
The formula C6H7N for a base that is sparingly soluble in water, dissolves in dilute acid, and gives the foul-smelling isocyanide of the carbylamine test identifies a primary aromatic amine, aniline. With benzenesulphonyl chloride (Hinsberg reagent) a primary amine yields N-phenylbenzenesulphonamide, C6H5SO2NHC6H5, whose N--H is acidic enough to dissolve in alkali. Its distinct hydrogens are the N--H proton together with the ortho, meta and para hydrogens of each monosubstituted ring, giving 1+3+3=7 different hydrogen environments.
JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option