Explain the reaction of benzene with $n$-propyl chloride.

(N/A) The reaction of benzene with $n$-propyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
The reaction proceeds via the formation of a carbocation intermediate.
$CH_3CH_2CH_2Cl + AlCl_3 \rightarrow CH_3CH_2CH_2^+ + AlCl_4^-$
The initially formed $1^{\circ}$-carbocation $(CH_3CH_2CH_2^+)$ is less stable and undergoes a $1,2$-hydride shift to form a more stable $2^{\circ}$-carbocation $(CH_3CH^+CH_3)$.
Consequently,the major product formed is isopropylbenzene (cumene),while $n$-propylbenzene is formed as a minor product due to the direct attack of the $1^{\circ}$-carbocation before rearrangement.