Chemistry Q73 – JEE-MAIN 2026

Home Chemistry Basic Organic ChemistryQuestion #118

One mole of phenol is treated with dilute

H N O_{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 10^{- 1} %

(Given molar mass in g

m o l^{- 1}

H:1, C:12, N:14, O:16)

Enter Calculated Numerical Value:

Clue 1 of 3: Organic Reactions and Isomerism

Identify the products formed when phenol reacts with dilute nitric acid and determine which isomer is steam volatile.

🧠 Full Solution Path

Step 1: Identify the steam volatile product✦ Active

Phenol reacts with dilute $H N O_{3}$ to form ortho-nitrophenol and para-nitrophenol. Ortho-nitrophenol exhibits intramolecular hydrogen bonding, making it steam volatile, while para-nitrophenol exhibits intermolecular hydrogen bonding, making it less steam volatile. Therefore, the steam volatile compound (X) is ortho-nitrophenol.

Molecular formula of Phenol: $C_{6} H_{6} O$

Molecular formula of Ortho-nitrophenol: $C_{6} H_{5} N O_{3}$

Step 2: Calculate the percentage of oxygen in phenol and ortho-nitrophenol○ Expand

Given atomic masses: H=1, C=12, N=14, O=16.

For Phenol ( $C_{6} H_{6} O$ ):

Molar mass = 6 (12) + 6 (1) + 1 (16) = 72 + 6 + 16 = 94 g/mol

Mass of oxygen = 1 \times 16 = 16 g

Percentage of oxygen in phenol = \frac{16}{94} \times 100 \approx 17.021276 %

For Ortho-nitrophenol ( $C_{6} H_{5} N O_{3}$ ):

Molar mass = 6 (12) + 5 (1) + 1 (14) + 3 (16) = 72 + 5 + 14 + 48 = 139 g/mol

Mass of oxygen = 3 \times 16 = 48 g

Percentage of oxygen in ortho-nitrophenol = \frac{48}{139} \times 100 \approx 34.532374 %

Step 3: Calculate the increase in percentage of oxygen and express in the required format○ Expand

Increase in percentage of oxygen = (Percentage of oxygen in ortho-nitrophenol) - (Percentage of oxygen in phenol)

Increase = 34.532374 % - 17.021276 % = 17.511098 %

The question asks for the value in the format _______ $\times 10^{- 1} %$ . Let the value in the blank be $Y$ .

Y \times 10^{- 1} = 17.511098

Y = 17.511098 \times 10 = 175.11098

Rounding to two decimal places, the answer is $175.11$ .

💡 Teacher's Secret Hint

Pay attention to the required format for the final answer, which involves a factor of $10^{- 1}$ .

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Chemistry Question 73 – JEE-MAIN 2026

Clue 1 of 3: Organic Reactions and Isomerism

🧠 Full Solution Path