Predict the order of reactivity of the compounds in `S_(N)1` and `S_(N) 2` reactions .
`C_(6)H_(5)CH_(2) Br, C_(6)H_(5) CH (C_(6)H_(5)) Br , C_(6) H_(5) CH(CH_(3)) Br , C_(6)H_(5) C (CH_(3)) (C_(6)H_(5)) Br`

Explanation

(i) `(CH_(3)CH_(2)CH_(2)CH_(2)Br lt (CH_(3))_(2)CHCH_(2)Br lt CH_(3)CH_(2)CH(Br)CH_(3) lt (CH_(3))_(3)CBr(S_(N)1)`
`CH_(3)CH_(2)CH_(2)CH_(2)Br gt (CH_(3))_(2)CHCH_(2)Br gt CH_(3)CH_(2)CH(Br)CH_(3) gt (CH_(3))_(3)CBr(S_(N)2)`
Of the two primary bromides, the carbocation intermediate derived from `(CH_(3))_(2)CHCH_(2)Br` is more stable than derived from `CH_(3)CH_(2)CH_(2)CH_(2)Br` because of greater electron donating inductive effect of `(CH_(3))_(2)CH-` group. Therefore, `(CH_(3))_(2)CHCH_(2)Br` is more reactive than `CH_(3)CH_(2)CH_(2)CH_(2)Br` in `S_(N)1` reactions. `CH_(3)CH_(2)CH(Br)CH_(3)` is a secondary bromide and `(CH_(3))_(3)CBr` is a tertiary bromide. Hence the above order is followed in `S_(N)1`. The reactivity in `S_(N)2` reactions follows the reverse order as the steric hinderance around the electrophilic carbon increases in that order.
`C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br gt C_(6)H_(5)CH(C_(6)H_(5))Br gt C_(6)H_(5)CH(CH_(3))Brgt C_(6)H_(5)CH_(2)Br (S_(N)1)`
`C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br lt C_(6)H_(5)CH(C_(6)H_(5))Br lt C_(6)H_(5)CH(CH_(3))Br lt C_(6)H_(5)CH_(2)Br (S_(N)2)`
Of the two secondary bromides, the carbocation intermediate obtained from `C_(6)H_(5)CH(C_(6)H_(5))Br` is more stable than obtained from `C_(6)H_(5)CH(CH_(3))Br` because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in `S_(N)1` reactions. A phenyl group is bulkier than a methyl group. Therefore, `C_(6)H_(5)CH(C_(6)H_(5))Br` is less reactive than `C_(6)H_(5)CH(CH_(3))Br` in `S_(N)2` reactions.