$R_f$ value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42. 2-methylpropene is treated with dilute $H_2SO_4$ to give major organic product (X). $R_f$ value for (X) in the same solvent system under identical condition will be:
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- 0.42
- 0.82
- 0.32
- 0.52
The Correct Option is C
Solution and Explanation
Step 1: Identify the chemical reaction.
When 2-methylpropene ($CH_2=C(CH_3)_2$) is treated with dilute sulfuric acid ($H_2SO_4$), it undergoes an acid-catalyzed hydration reaction following Markovnikov's rule. The water molecule adds across the double bond to form the more stable tertiary carbocation intermediate.
$$\text{CH}_2=\text{C}(\text{CH}_3)_2 + \text{H}_2\text{O} \xrightarrow{\text{dil. H}_2\text{SO}_4} (\text{CH}_3)_3\text{C}-\text{OH}$$
The major product (X) is 2-methylpropan-2-ol (tert-butyl alcohol).
Step 2: Compare polarities.
2-methylpropene is an alkene, which is a non-polar hydrocarbon. In contrast, 2-methylpropan-2-ol is an alcohol containing a highly polar hydroxyl (-OH) group capable of hydrogen bonding.
Step 3: Relate polarity to $R_f$ value.
In normal-phase thin-layer chromatography (TLC), which uses a polar stationary phase like silica gel, more polar molecules bind more strongly to the stationary phase. This strong interaction causes polar molecules to move more slowly up the plate, resulting in a lower $R_f$ value. Since the product (X) is much more polar than the starting alkene, its $R_f$ value must be significantly lower than 0.42.
Among the options, the value lower than 0.42 is 0.32. Therefore, the $R_f$ value for (X) is 0.32.
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