# Friedel–Crafts alkylation of benzene with tetrachloromethane

There was this question in an exam at our school which asked what the major product will be when excess of benzene is reacted with tetrachloromethane in the presence of aluminium chloride.

All of us got it wrong in the exam, but we realized that it's just Friedel–Crafts alkylation with extra steps. So later we came up with two competing answers: one group claimed it is tetraphenylmethane $(\ce{CPh4})$, while the other claimed it's triphenylchloromethane $(\ce{CClPh3})$ (aka trityl chloride).

Reason in favour of $\ce{CClPh3}$: After the formation of it, the negative inductive effect of the phenyl groups will supposedly be high enough not to allow the release of the last remaining Cl to form a carbocation. Also, there will be high steric hindrance in $\text{CPh}_4$.

Reason in favour of $\ce{CPh4}$: The phenyl groups are bonded to the central carbon through $\sigma$ bonds. So they can just rotate their planes to escape steric hindrance. Also, we do not know if the combined inductive effect of the phenyl groups is high enough not to allow the release of chloride. After all, the chlorine molecule is highly electronegative.

So, we're grappling with this problem and do not know the solution to it. Any help will be highly appreciated.