The major product obtained in the given reaction is:

We know that ${\mathrm{AlCl}}_3$ is a lewis acid. It can give Friedel Crafts reactions. Here, we can see that ${\mathrm{AlCl}}_3$ acts as a lewis acid and accepts the electron pair from the chlorine atom. This forms$-{\mathrm{AlCl}}_4.$ Alongside this, a carbocation is also formed.

Now, this carbocation is an electrophilic carbon. So, benzene rings can give electrophilic substitution reactions here. So, as a result, the double bond of benzene will attack on the carbocation and a C-C bond will be formed there. Then after losing a proton, the carbon ring becomes aromatic. This, we obtain a compound which has two ring structures. One ring is aromatic and the other is not aromatic.

Friedel-Craft alkylation takes place to give a six membered ring.