What will be the product obtained as a result of the following reaction and why?

This reaction is an example of Friedel Crafts alkylation using a lewis acid. AlCl₃ is a lewis acid based on the fact that the central atom Al does not have a complete octate so it behaves as an acid. This property of lewis acid forms an important base for Friedel Craft reactions. It can be broken down into following steps.

Step 1: Formation of carbocation

The electron deficient centre of AlCl₃ is attacked by chlorine from the alkyl chloride which results into formation of 1° carbocation. This carbocation is less stable as compared to a possible 2° carbocation which can be formed by hydride shift.

Step 2: Formation of more stable carbocation

1° carbocation is less stable as compared to a 2° carbocation. When there is a possibility of formation of a higher ° cation, we observe a phenomenon called as hydride shift (in some cases even a methyl shift). The hydrogen from the β carbon shifts along with its electron pair to the positive carbon to give a more stable 2° cation.

Step 3: Nucleophilic attack of benzene ring.

This 2° cation is attacked by the double bond from benzene ring; or rather the hydrogen from the ring is lost and undergoes electrophilic substitution to give cumene as the major product. A minor product is also formed by attack of 1° carbocation as propyl benzene.