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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?

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    $\begingroup$ You are close. Think about reordering the steps you've described to avoid any negatively charged intermediates. Under acidic conditions only positively charged and neutral species will be present. $\endgroup$
    – jerepierre
    Mar 17 '15 at 13:02
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    $\begingroup$ I am sorry could you tell me the first step please. I am really stuck. Why only positive charge in acidic conditions? $\endgroup$
    – RobChem
    Mar 17 '15 at 13:04
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    $\begingroup$ The Friedel-Crafts alkylation is reversible (see here under "Friedel–Crafts dealkylation"). So first protonate the ring and then dealkylate $\endgroup$
    – ron
    Mar 17 '15 at 13:07
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    $\begingroup$ This is really a reverse Friedel-Crafts followed by a normal Friedel-Crafts. So whatever the last step of a FC alkylation is will be the first step of this reaction. $\endgroup$
    – jerepierre
    Mar 17 '15 at 13:09

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.

Reverse Friedel-Crafts / forward Friedel-Crafts

  • $\begingroup$ Having come back to this question again, I realized that the second species that you have drawn in your mechanism seems massively unstable (cation on carbon with bromine substituent). I can't see another way of doing this so I don't doubt the mechanism. However, why is this possible? Also, is Acylation reversible too? $\endgroup$
    – RobChem
    Apr 21 '15 at 12:26
  • $\begingroup$ @RobChem Inductively bromine is electron withdrawing (destabilizing cation) but it's also electron donating by resonance (stabilizing cation), so I think it's a reasonable intermediate. Theoretically, FC acylation is reversible too. I'm not aware of any examples though. $\endgroup$
    – jerepierre
    Apr 21 '15 at 14:38
  • $\begingroup$ I guess the aluminium chloride is necessary so the chloride anion has a place to go. $\endgroup$
    – Karl
    May 6 '16 at 17:36

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