One mole of phenol is treated with dilute HNO3 at 298 K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound (X) is separated. The increase in percentage of oxygen in (X) with respect to phenol is 10-1% (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16)
Answer
175
Solution
Phenol + dil. HNO3 (298 K) → o-nitrophenol + p-nitrophenol.
o-Nitrophenol has intramolecular H-bonding (no association), so it is steam volatile ⇒ X = o-nitrophenol [figure] , C6H5NO3
M(X) = 72 + 5 + 14 + 48 = 139 g mol-1 ⇒ % O in X = 48139 100 = 34.53\%
Phenol C6H6O: M = 94 ⇒ % O = 1694 100 = 17.02\%
∴ Increase = 34.53 − 17.02 = 17.51% = 175.1 × 10-1% ≈ 175 × 10-1%
Answer: 175