One mole of phenol is treated with dilute HNO_3 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)
A. A
When phenol is treated with dilute HNO _3 at 298 K , it undergoes electrophilic aromatic substitution to give a mixture of ortho-nitrophenol and para-nitrophenol. The mixture is separated by steam distillation. Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bonding, whereas para-nitrophenol is not steam volatile due to intermolecular hydrogen bonding. Thus, the steam volatile compound (X) is ortho-nitrophenol. Molecular formula of phenol: C _6 H _6 O Molar mass of phenol = (6 12) + (6 1) + 16 = 94 g mol ^ -1 Percentage of oxygen in phenol = 16 94 100 17.02\% Molecular formula of ortho-nitrophenol (X): C _6 H _5 NO _3 Molar mass of ortho-nitrophenol = (6 12) + (5 1) + 14 + (3 16) = 139 g mol ^ -1 Percentage of oxygen in ortho-nitrophenol = 48 139 100 34.53\% Increase in percentage of oxygen = 34.53\% - 17.02\% = 17.51\% We need to express this as y 10^ -1 \%. 17.51\% = 175.1 10^ -1 \% Rounding to the nearest integer, we get 175. Answer: 175
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