Assertion A: Hydrolysis of an alkyl chloride is a slow reaction but in the presence of NaI, the rate of hydrolysis increases.
Reason R: I$$^-$$ is a good nucleophile as well as a good leaving group.
In the light of the above statements, choose the correct answer from the options given below:

We need to evaluate the Assertion and Reason about the hydrolysis of alkyl chlorides and the role of NaI.

The Assertion A states that hydrolysis of an alkyl chloride is a slow reaction but in the presence of NaI, the rate of hydrolysis increases. This assertion is TRUE because hydrolysis of an alkyl chloride ($$RCl$$) is typically slow due to the relatively strong $$C-Cl$$ bond and the fact that water is a weak nucleophile. However, when NaI is added, the reaction rate increases dramatically through a two-step mechanism known as the Finkelstein reaction followed by hydrolysis.

Initially, $$I^-$$ from NaI attacks the carbon bearing the chlorine via an $$S_N2$$ mechanism, displacing $$Cl^-$$:
$$R-Cl + I^- \xrightarrow{S_N2} R-I + Cl^-$$
This reaction proceeds quickly because $$I^-$$ is a strong nucleophile, being a large, highly polarizable anion.

Next, the resulting alkyl iodide ($$R-I$$) undergoes hydrolysis much faster than the original alkyl chloride:
$$R-I + H_2O \rightarrow R-OH + HI$$
This step is fast because $$I^-$$ is an excellent leaving group—it is a weak base and a large, stable anion that easily departs with the bonding electrons.

Therefore, the overall effect of adding NaI is to convert the slow hydrolysis of $$RCl$$ into a faster two-step process via $$RI$$.

The Reason R asserts that $$I^-$$ is a good nucleophile as well as a good leaving group, which is also TRUE. On one hand, $$I^-$$ is a good nucleophile because it is large and highly polarizable, making it very effective in $$S_N2$$ reactions. The nucleophilicity order is $$I^- > Br^- > Cl^- > F^-$$. On the other hand, $$I^-$$ is a good leaving group because its conjugate acid, HI, is very strong ($$pK_a \approx -10$$), so $$I^-$$ is a very weak base. Accordingly, leaving group ability follows the same order: $$I^- > Br^- > Cl^- > F^-$$.

It follows that R explains A, since the acceleration of $$RCl$$ hydrolysis by NaI relies precisely on the dual role of $$I^-$$ as both a strong nucleophile (to displace $$Cl^-$$ in the first step) and a good leaving group (to facilitate the second step of hydrolysis).

The correct answer is Option 3: Both A and R are true and R is the correct explanation of A.