1. Identify the reaction type

The reaction is:

$\mathrm{CH_3CH_2CH_2CH_2Cl} + \mathrm{HO^-} \rightarrow \mathrm{CH_3CH_2CH_2CH_2OH} + \mathrm{Cl^-}$

This is the conversion of a primary alkyl halide (1-chlorobutane) to an alcohol by nucleophilic substitution.

For a primary haloalkane, the reaction with $\mathrm{HO^-}$ generally proceeds by the $S_N2$ mechanism.

2. Effect of solvent polarity on $S_N2$ reaction

In an $S_N2$ reaction, the nucleophile must attack the carbon directly. The reactivity depends strongly on the solvent.

• In polar protic solvents (more polar medium like water/alcohol), $\mathrm{HO^-}$ is heavily solvated by hydrogen bonding.
• Heavy solvation reduces the nucleophilicity of $\mathrm{HO^-}$.
• Therefore, $S_N2$ reactions are generally slower in such more polar protic media.

Hence, the conversion proceeds better in a less polar medium (more precisely, less strongly solvating medium).

So, Statement I is true.

3. Check Statement II

Statement II says the conversion proceeds well in the more polar medium.

For this particular reaction involving $\mathrm{HO^-}$ and a primary haloalkane, a more polar protic medium strongly solvates the nucleophile and lowers its nucleophilicity, so the substitution does not proceed better there.

Thus, Statement II is false.

4. Choose the correct option

• Statement I: True
• Statement II: False

Therefore, the correct option is:

$\boxed{\text{B}}$

5. Comparison with stored correct answer

The stored correct answer is D.

My derived answer is B. These do not match.

The likely issue is that the stored answer treats “more polar medium” too generally, but for an $S_N2$ reaction with $\mathrm{HO^-}$, increased protic polarity usually decreases nucleophilic strength due to solvation. Hence Statement II should not be true in the standard JEE context.