Predict the order of reactivity of the following compounds in $S_{N} 1$ and $S_{N} 2$ reactions :
The four isomeric bromobutanes
$C_{6} H_{5} C H_{2} B r, C_{6} H_{5} C H (C_{6} H_{5}) B r, C_{6} H_{5} C H (C H_{3}) B r, C_{6} H_{5} C (C H_{3}) (C_{6} H_{5}) B r$

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Solution

$S N^{1}$ reaction order
$C_{6} H_{5} C (C H_{3}) (C_{6} H_{5}) B r > C_{6} H_{5} C H (C_{6} H_{5}) B r > C_{6} H_{5} C H (C H_{3}) B r > C_{6} H_{5} C H_{2} B r$
$S N^{2}$ reaction order ,
$C_{6} H_{5} C H_{2} B r > C_{6} H_{5} C H (C H_{3}) B r > C_{6} H_{5} C H (C_{6} H_{5}) B r > C_{6} H_{5} C (C H_{3}) (C_{6} H_{5}) B r$
of the two secondary bromides , the carbocation intermediate obtained from $C_{6} H_{5} C H (C_{6} H_{5}) B r$ is more stable than obtained from $C_{6} H_{5} C H (C_{6} H_{5}) B r$ because it is stablised by two phenyl group 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 , $C_{6} H_{5} C H (C H_{3}) B r$ in $S N^{2}$ reactions.

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