Which compound will react faster in $S N_{2}$ reaction with $-$ OH group?
1. $C H_{3} B r$ or
2. $C H_{3} I$
Give your answer in terms of 1 and 2.

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Solution

Methyl iodide will react faster than methyl bromide towards bimolecular nucleophilic substitution reaction with hydroxide ion. This is because iodide ion is a better leaving group than bromide ion as bigger iodide ion can easily accommodate the negative charge. Also, the $C - I$ bond is weaker than the $C - B r$ bond.

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