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JEE Main 20266 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Hydrocarbons

JEE Main 2026 · 6 April · Morning Shift · Chemistry · Hydrocarbons

R_f value for 2 -methylpropene in a solvent system (Ethyl acetate +...

RfR_f value for 22-methylpropene in a solvent system (Ethyl acetate + ether) is 0.420.42. 22-methylpropene is treated with dilute H2SO4\text{H}_2\text{SO}_4 to give major organic product (X). RfR_f value for (X) in the same solvent system under identical condition will be:

Options

  1. A0.420.42
  2. B0.820.82
  3. C0.620.62
  4. D0.120.12

Correct answer

D. 0.120.12

Step-by-step solution

22-methylpropene reacts with dilute H2SO4H_2SO_4 to undergo hydration, forming tert-butyl alcohol as the major product (X).

The reaction is:

CH2=C(CH3)2→dil. H2SO4(CH3)3C−OHCH_2=C(CH_3)_2 \xrightarrow{\text{dil. } H_2SO_4} (CH_3)_3C-OH

Product (X) is an alcohol, which is significantly more polar than the starting alkene due to the presence of the −OH-OH group. In normal phase chromatography, the stationary phase (such as silica gel) is highly polar. The polar −OH-OH group forms strong hydrogen bonds with the stationary phase, causing the alcohol to be adsorbed more strongly than the non-polar alkene.

Consequently, the alcohol travels a shorter distance along the chromatography plate, resulting in a lower RfR_f value compared to the alkene.

Since the RfR_f value of 22-methylpropene is 0.420.42, the RfR_f value of product (X) must be less than 0.420.42. Among the given choices, only 0.120.12 satisfies this condition.

Answer: 0.120.12
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