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

- SN2 reaction rate is influenced by the leaving group; a superior leaving group results in a faster SN2 substitution. - Iodine (I\(^-\)) functions as a superior leaving group compared to bromine (Br\(^-\)), attributable to iodine's larger atomic size and the weaker bond it forms with the carbon atom.- Consequently, ethyl iodide (CH\(_3\)CH\(_2\)I) will experience a faster SN2 reaction than ethyl bromide (CH\(_3\)CH\(_2\)Br).
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