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Chemistry Question 63 – JEE-MAIN 2026

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Rf value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42. 2-methylpropene is treated with dilute H2SO4 to give major organic product (X). Rf value for (X) in the same solvent system under identical condition will be:

Determine the major organic product formed when 2-methylpropene reacts with dilute sulfuric acid.

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Ninja Strategy•Polarity Shift Prediction

Identify the change in polarity from reactant to product. A more polar product will have a lower Rf value in normal phase chromatography.

Step 1: Determine the major product (X)✦ Active

2-methylpropene undergoes acid-catalyzed hydration (reaction with dilute H2SO4) following Markovnikov's rule. The major product (X) is 2-methylpropan-2-ol (tert-butyl alcohol).

(CH3)2C=CH2→dil. H2SO4(CH3)3COH
Step 2: Compare the polarities of 2-methylpropene and product (X)○ Expand

2-methylpropene is an alkene, which is relatively nonpolar. Product (X), 2-methylpropan-2-ol, is an alcohol. The presence of the hydroxyl (-OH) group makes it significantly more polar than the alkene, as it can form hydrogen bonds.

Step 3: Relate polarity to Rf value in chromatography○ Expand

In normal phase chromatography (polar stationary phase, less polar mobile phase), more polar compounds interact more strongly with the stationary phase and thus travel shorter distances, resulting in lower Rf values. Since (X) is more polar than 2-methylpropene, its Rf value will be lower than 0.42. Among the given options, 0.12 is the only value less than 0.42.

💡 Teacher's Secret Hint

Remember that Rf values are always between 0 and 1.

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