R f value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42 . 2-methylpropene is treated with dilute H 2 SO 4 to give major organic product (X). R f value for (X) in the same solvent system under identical condition will be :

Chemistry · Hydrocarbons

JEE Main 2026 — 6 April, Morning Shift — Question 57

$\mathrm{R}_{\mathrm{f}}$ value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42 . 2-methylpropene is treated with dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ to give major organic product (X). $\mathrm{R}_{\mathrm{f}}$ value for (X) in the same solvent system under identical condition will be :

Option A:
0.42
Option B:
0.82
Option C:
0.62
Option D:
0.12
Correct

Answer: D

Step-by-step solution

Ter butanol (X) is more polar than 2-Methyl propene and $R_{f} \propto \frac{1}{\text { polarity }}$ so $R_{f}$ of ter.butanol is less than 2-Methylpropen

So. Ans is 0.12

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Chemistry
Chapter: Hydrocarbons
Topic: Properties & Uses of Alkenes and Dienes

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Step-by-step solution

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