Which compound in each of the following pairs will react faster in $\ce{S_N2}$ reaction with $\ce{–OH}$?
$\ce{CH_3Br or CH_3I}.$
$\ce{CH_3Br or CH_3I}.$
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In the $\ce{S_N2}$ mechanism, the reactivity of halides for same halides group increase down the group. Because increase in size increase, the halide becomes a better leaving group. Therefore, $CH_3I$ will react faster than $\ce{CH_3Br}$ in $\ce{S_N2}$ reactions with $\ce{OH^-}$.
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