Understand the assertion

The assertion says:

Hydrolysis of an alkyl chloride is slow, but in the presence of $\mathrm{NaI}$, the rate of hydrolysis increases.

This is true.

Why? Alkyl chlorides generally hydrolyze slowly because $\mathrm{Cl^-}$ is not as good a leaving group as $\mathrm{I^-}$, and the carbon–chlorine bond is stronger than the carbon–iodine bond.

In presence of $\mathrm{NaI}$, iodide ion first converts alkyl chloride into alkyl iodide: $\mathrm{R{-}Cl + I^- \rightarrow R{-}I + Cl^-}$

This is a halogen exchange (Finkelstein-type process). The alkyl iodide formed is more reactive toward hydrolysis: $\mathrm{R{-}I + H_2O \rightarrow R{-}OH + HI}$

Since $\mathrm{I^-}$ is a much better leaving group, hydrolysis becomes faster.

Understand the reason

The reason says:

$\mathrm{I^-}$ is a good nucleophile as well as a good leaving group.

This is also true.

• As a good nucleophile, $\mathrm{I^-}$ can attack $\mathrm{R{-}Cl}$ and form $\mathrm{R{-}I}$.
• As a good leaving group, once $\mathrm{R{-}I}$ is formed, it hydrolyzes more easily.

Check whether R explains A

Yes, it does.

The increase in rate of hydrolysis in presence of $\mathrm{NaI}$ is precisely because iodide ion helps form the more reactive alkyl iodide intermediate, and $\mathrm{I^-}$ is both:

• a good nucleophile (for substitution of $\mathrm{Cl^-}$), and
• a good leaving group (making hydrolysis of $\mathrm{R{-}I}$ faster).

Evaluate the options

• A: A is true but R is false $\rightarrow$ Incorrect
• B: A is false but R is true $\rightarrow$ Incorrect
• C: Both A and R are true and R is the correct explanation of A $\rightarrow$ Correct
• D: Both A and R are true but R is NOT the correct explanation of A $\rightarrow$ Incorrect

Final answer

The correct option is: $\boxed{\text{C}}$

Comparison with stored correct answer

Stored correct answer = C

My derived answer = C

Hence, they agree.