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Question
Which would undergo SN2 reaction faster in the following pair and why ?
CH3 – CH2 – Br and CH3 – CH2 – I
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Solution
CH3 – CH2 – I would undergo an SN2 reaction faster than CH3 – CH2 – Br. Since iodine is a better leaving group because of its large size, it will be released at a faster rate in the presence of an incoming nucleophile as compared to bromine.
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\[\begin{array}{cc}\ce{CH3CH2CHCH3}\\
\phantom{.....}|\\
\phantom{......}\ce{Br}\\
\end{array}\] or \[\begin{array}{cc}\phantom{.......}\ce{CH3}\\
\phantom{...}|\\
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\phantom{...}|\\
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