Predict the order of reactivity of the following 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 and C_6 H_5 C(CH_3)(C_6H_5)Br`

Explanation

a.
Out of `1^(@) C^(o+) (A)` and `(C),(C)` is more stable due to `+I` effect of two `(Me)` groups. Therefore, `(III)` is more reactive than `(I)` in `SN^(1)` reaction.
`SN^(1) implies )I) gt (III) gt (II) gt (IV)`
`SN^(2) implies (I) gt (III) gt (II) gt (IV)`
b. `SN^(1) implies C_(6) H_(5) C(CH_(3)) (C_(6)H_(5))Br (IV)`
`gt C_(6) H_(5) CH (C_(96) H_(5)) - Br gt C_(6) H_(5) CH(CH_(3)) Br gt C_(6) H_(5) CH_(2)Br`
`gt (II) gt (III) gt (I)`
`SN^(2) implies C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br lt C_(6) H_(5) CH(C_(6) H_(5)()-Br lt`
`(IV) lt (II)`
`C_(6)H_(5)CH(CH_(3))Br lt C_(6)H_(5)H_(2)Br`
`(III) lt (I)`
Of the two secondary bromides, the carbocation intermediate obtained from `(II)` is more stable than that obtained form `(III)` becuase it is stablised byu two phenyl groups due to resonacne. Therefore the foremer bormide is more reactive than the latter in `SN^(-1)` reacations. A phenyl group is bulkier than a methuyl group . THerefore `(II0` is less reactive than `(III)` in `SN^(2)` reacations.