Which compound in each of the following pairs will react faster in `S_(N)2` reaction with `OH^(-)`?
(a) `CH_(3)Br` or `CH_(3)I`
(b) `CH_(2)=CHBr` or `CH_(2)=CHCH_(2)Br`.

Explanation

(a) Since `I^(-)` ion is a better leaving group than `Br^(-)` ion, therefore, iodidies are more reactive than bromides. In other words, `CH_(3)I` is more reactive than `CH_(3)Br`.
(b) Reactiviity in `S_(N)2` reactions among other things depends upon the magnitude of the +ve charge on the carbon atom carrying the halogen. now the `sp^(2)`-carbon atom of the `CH_(2)` group carrrying the Br atom in `CH_(2)=CHCH_(2)Br` carries more +ve charge due to -I-effect of the Br atom and the `CH_(2)=CH`- group as compared to `sp^(2)`-carbon atom of `CH` group in `CH_(2)=CHBr` due to electron-donating resonance effect.

As a result, `CH_(2)=CHCH_(2)Br` will react faster than `CH_(2)=CHBr` with `OH^(-)` ions.