HydrocarbonsMedium
Question
$R_{f}$ 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 :
Options
A.0.42
B.0.82
C.0.62
D.0.12
Solution
Sol. [Key Concept: $2$-methylpropene undergoes Markovnikov hydration to tert-butanol, which is more polar; since $R_{f} \propto \frac{1}{polarity}$, its $R_{f}$ drops to $0.12$.]
Ter butanol ( X ) is more polar than 2-Methyl propene and $R_{f} \propto \frac{1}{\text{~}\text{polarity}\text{~}}$ so $R_{f}$ of ter.butanol is less than 2-Methylpropen
So. Ans is 0.12
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