Question

The % increase in oxygen in steam volatile product with respect to phenol is ____ \(10^{-1}\%\).

• A. 125
• B. 150
• C. 175
• D. 200

Answer 1

The Correct Option is C

Concept:
Phenol undergoes electrophilic substitution with dilute nitric acid to form a mixture of o-nitrophenol and p-nitrophenol. Among these products, o-nitrophenol is steam volatile due to intramolecular hydrogen bonding. Therefore, the steam volatile product considered here is o-nitrophenol. Step 1: Write the reaction. \[ \text{Phenol} \xrightarrow{\text{Dil. }HNO_3} \text{o-Nitrophenol} + \text{p-Nitrophenol} \] \[ \text{o-Nitrophenol is steam volatile} \] Step 2: Calculate \% oxygen in phenol. Molecular formula of phenol: \[ C_6H_6O \] Molar mass: \[ 6(12) + 6(1) + 16 = 94 \] Mass of oxygen in phenol \(=16\). \[ \%O_{\text{phenol}}= \frac{16}{94}\times100 \] \[ \%O_{\text{phenol}} = 17.02\% \] Step 3: Calculate \% oxygen in o-nitrophenol. Molecular formula: \[ C_6H_5NO_3 \] Molar mass: \[ 6(12) + 5(1) + 14 + 3(16) = 139 \] Mass of oxygen \(=3\times16=48\). \[ \%O= \frac{48}{139}\times100 \] \[ \%O = 34.53\% \] Step 4: Calculate increase in oxygen percentage. \[ \text{Increase} = 34.53 - 17.02 \] \[ = 17.51 \approx 17.5 \] Given answer format is \(10^{-1}\%\): \[ 17.5 \times 10^{-1} = 175 \] \[ \boxed{175} \]