Predict the order of reactivity of the following compounds in `S _(N)1` and `S_(N)2` reactions:
(i) The four isomeric bromobutanes
(ii) `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

(a) The four isomeric bromobutanes are:
(i) `CH_(3)CH_(2)CH_(2)CH_(2)Br` (ii) `CH_(3)underset(CH_(3))underset(|)CHCH_(2)Br` (iii) `CH_(3)CH_(2)underset(Br)underset(|)CHCH_(3)` (iv) `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Br`
In `S_(N)1` reactions, the order of reactivity depends upon the stability of the intermediate carbocations. Since the stability of carbocation decreases as `3^(@) gt 2^(@) gt 1^(@)`, the order of reactivity also decreases in the same order. (i) and (ii) are primary alkyl halides, but the carbocation intermediate derived from `(CH_(3))_(2)CHCH_(2)Br` is more stable than that 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. (iii) in a secondary, and (iv) is tertiary bromide. Hence, the order of reactivity in `S_(N)1` reaction is:
`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))_(2)CBr`
The reactivity in `S_(N)2` reactions follows the reverse order as the steric hindrance around the electrophilic carbon increases in that order. Thus, the order of reactivity in `S_(N)2` reactions is :
`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))_(2)CBr`
(b) In `S_(N)1` reactions, the reactivity increases with increase in stability of the intermediate carbocations formed as `3^(@) gt 2^(@) gt 1^(@)`. Therefore, `C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br` which gives `3^(@)` carbocation [i.e, `C_(6)H_(5)C^(+)(CH_(3))C_(6)H_(5)`] is the most reactive. Of the two secondary bromides, `C_(6)H_(5)-CH(C_(6)H_(5)) Br` and `C_(6)H_(5)-CH(CH_(3))Br`, the carbocation intermediate obtained from `C_(6)H_(5)-CH(C_(6)H_(5)) Br` [i.e., `C_(6)H_(5)-CH(C_(6)H_(5))`] is more stable than that obtained from `C_(6)H_(5)CH(CH_(3))Br` [i.e., `C_(6)H_(5) CH(CH_(3))`] because it is stabilised by two phenyl groups due to resonance. Therefore, `C_(6)H_(5)CH(C_(6)H_(5))Br` is more reactive than `C_(6)H_(5)CH(CH_(3)) Br`.
The fourth alkyl bromide, i.e., `C_(6)H_(5)CH_(2)Br` which gives primary carbocation i.e., `C_(6)H_(5)CH_(2)^(+)` is least reactive. Hence, the overall reactivity of these alkyl bromides towards `S_(N)1` reactions decreases as:
For `S_(N)1` reaction : `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`
In `S_(N)2` reactions, the reactivity depends upon the steric hindrance and decreases as `3^(@) lt 2^(@) lt 1^(@)`. Therefore, primary bromide, `C_(6)H_(6)CH_(2)Br` is most reactive and tertiary bromide `C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br` is least reactive. Among two secondary bromides, `C_(6)H_(5)CH(C_(6)H_(5)) Br` is less reactive than `C_(6)H_(5)CH(CH_(3))Br` because a phenyl group is bulkier than a methyl group and causes more steric hindrance. So, the correct order is
For `S_(N)2` reaction: `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`