$R_f$ 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). $R_f$ value for (X) in the same solvent system under identical condition will be :

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Alcohols Phenols and Ethers

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 :

(A)0.42
(B)0.82
(C)0.62
(D)0.12✓ Correct

Answer: (D)

Solution

Dilute sulphuric acid brings about Markovnikov hydration of 2-methylpropene: the proton adds to give the stable tertiary carbocation, which water then captures, so the major product is

tert

-butyl alcohol,

(C H_{3})_{3} COH

. The hydroxyl group makes this alcohol much more polar than the parent alkene, and on a polar adsorbent a more polar solute is held more strongly and moves less, giving a smaller

R_{f}

. The

R_{f}

of X must therefore lie well below

0.42

, and of the values offered only

0.12

qualifies.

The

R_{f}

value of X is

0.12

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Alcohols Phenols and Ethers →

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Alcohols Phenols and Ethers

Solution

More JEE Main questions on Alcohols Phenols and Ethers