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 :
- 8
- 4
- 5
- 7
Answer
(4) 7
Solution
C6H7N, basic (dissolves in dilute acid), gives carbylamine reaction (offensive smelling isocyanide 'y' = C6H5NC) 'x' is aniline, C6H5NH2.
C6H5NH2 + C6H5SO2Cl C6H5SO2NHC6H5 + HCl
'z' = N-phenylbenzenesulphonamide (alkali-soluble because of the acidic N–H):
[figure]
Different H atoms:
Ring attached to SO2: ortho, meta, para = 3
Ring attached to N: ortho, meta, para = 3
N–H = 1
Total = 3 + 3 + 1 = 7
Answer: (4)