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Predict the order of reactivity of the c...

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`

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`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))Br gt 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 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 .
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