JEE Main 2026 — Apr 5, Shift 1 — Chemistry solution

JEE Main2026Apr 5, Shift 1Chemistry

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 _________ $\times 1 0^{- 1}$ %.

(Given molar mass in g mol $^{- 1}$ H:1, C:12, N:14, O:16)

Your answer

Solution

Nitration of phenol with dilute HNO $_{3}$ at 298 K gives a mixture of $o$ - and $p$ -nitrophenol. Only the $o$ -isomer is steam volatile (intramolecular H-bonding), so X = $o$ -nitrophenol, $C_{6} H_{5} NO_{3}$ , M = 139.

Mass percent of O in phenol ( $C_{6} H_{6} O$ , M = 94): $\frac{16}{94} \times 100 \approx 17.02%$ .

Mass percent of O in $o$ -nitrophenol: $\frac{3 \times 16}{139} \times 100 = \frac{48}{139} \times 100 \approx 34.53%$ .

Increase $\approx 34.53 - 17.02 = 17.51% = 175.1 \times 1 0^{- 1} %$ , so the requested integer is $175$ .

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