Reverse Friedel-Crafts mechanism I need to find a mechanism for the above reaction but I cannot make much progress. I do have something in mind but it doesn't seem entirely plausible: First the tert-butyl group leaves (this seems wrong but tert-butyl cation is stable and thus a reasonable leaving group maybe? Also the negatively charged ring that it would leave behind would be stabilised by the EWG Bromines). The cation is then attacked by Cl and the ring is attacked by a proton (one product formed). Then a Friedel-Craft alkylation to make the other product (hence the excess of benzene to promote only mono-substitution). I doubt that is right, purely because of the first step. Can you help?
This reaction proceeds through a reverse Friedel-Crafts alkylation followed by a forward Friedel-Crafts alkylation. The last step of Friedel-Crafts alkylation is deprotonation, which means that the first step of the reverse process will be protonation. Next, the carbon-carbon bond breaks, with 1,2,4,5,-tetrabromobenzene acting as a leaving group. The resulting t-butyl cation can undergo the typical electrophilic aromatic substitution steps, attack by an aromatic ring (benzene here) and deprotonation to regain aromaticity.

One might wonder why aluminum chloride is present in the reaction mixture, since we can write a reasonable mechanism that excludes it. Likely it is present in case chloride (from HCl) traps it to form t-butyl chloride. Aluminum chloride will catalyze the reformation of t-butyl cation.