Which would undergo S_N2 reaction faster in the following pair and why?

CH₃-CH₂-Br and CH₃-CH₂-I

Ethyl iodide would undergo S_N2 reaction faster because Iodide ion is a good leaving group. For a given alkyl group the reactivity is of the order R-I> R-Br> R-Cl≫ R-F.

Explanation: S_N2 reaction occurs easily when the substrate has a good leaving group and is attacked by a strong nucleophile. In the above, Iodide is larger in size and forms a stable anion therefore, it is considered as a good leaving group and hence ethyl iodide undergoes reaction faster.