Question

$R_f$ 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:

• A. 0.42
• B. 0.82
• C. 0.32
• D. 0.52

Hint:

Identify the product of the hydration of 2-methylpropene and recall that more polar compounds have lower Rf values in standard TLC.

Answer 1

The Correct Option is C

The $R_f$ value is a physical constant that indicates how far a compound travels in a particular solvent system relative to the solvent front. A higher polarity usually corresponds to a lower $R_f$ value on a polar stationary phase like silica.

1. Starting material: 2-methylpropene (Alkene, Non-polar). $R_f = 0.42$.
2. Reaction: Dilute $H_2SO_4$ converts an alkene to an alcohol via hydration.
$$CH_2=C(CH_3)_2 \xrightarrow{H_2O/H^+} (CH_3)_3COH$$
3. Product (X): 2-methylpropan-2-ol (Alcohol, Polar).

Comparison: Alcohols are more polar than alkenes because of the dipole moment of the C-O and O-H bonds. Because (X) is more polar, it will be more strongly adsorbed by the stationary phase in the chromatography experiment. This increased adsorption reduces the distance the compound travels with the solvent.
Consequently, $R_f(X)<R_f(\text{alkene})$.
$R_f(X)<0.42$.
Comparing the options (0.42, 0.82, 0.32, 0.52), 0.32 is the only option that reflects the expected decrease in $R_f$ due to increased polarity.