Question

In the following pair of compounds, which compound undergoes an $\text{S}_{N}$2 reaction faster and why?

Hint:

In SN2 reactions, a better leaving group (I\(^-\)>Br\(^-\)) facilitates a faster substitution reaction.

Answer 1

- The SN2 reaction rate depends on the leaving group. The better the leaving group, the faster the SN2 substitution. - Iodine (I\(^-\)) is a better leaving group than bromine (Br\(^-\)), due to the larger size and weaker bond of I\(^-\) with the carbon atom. - Therefore, ethyl iodide (CH\(_3\)CH\(_2\)I) will undergo the SN2 reaction faster than ethyl bromide (CH\(_3\)CH\(_2\)Br).
\

\end{center}