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Thinking about Friedel-Crafts acylation using carboxylic acid anhydride got me wondering about how exactly aluminium chloride ends up in that reaction.

Generally, despite the fact that Friedel-Crafts alkylation can be carried out using catalytic amounts of aluminium chloride - when chloride of a carboxylic acid is used as an acylating agent in Friedel-Crafts acylation at least 1 equivalent of aluminium chloride has to be used. This is due to the fact that aluminium chloride is complexed by resulting ketone (please see figure below).

enter image description here

Formation of such complex takes aluminium chloride out of action: 1 mole of aluminium chloride gets deactivated for every 1 mole of ketone produced, so it can not activate another molecule of acid chloride. And this is the reason why Friedel-Crafts acylation needs stochiometric amounts of aluminium chloride. (This is often omitted when Friedel-Crafts acylation is being introduced in organic chemistry textbooks, but it is sometimes mentioned later on - especially when it comes to actual lab work and real life conducting Friedel-Crafts acylation.) In the end formation of a ketone-aluminium chloride complex is not a big problem because when chemist who was running the reaction decides to isolate the product, the ketone can be liberated by hydrolysis of that complex using aqueous acid (note that water is unwanted in Friedel-Crafts reaction).

enter image description here

In instance of using carboxylic acid anhydride at least 2 equivalents of aluminium chloride are needed. One surely ends up bounded to the ketone - similarly as described above, but I am not sure what exactly happens with the second equivalent. My guess is that it reacts somehow with carboxylic acid that was produced from anhydride while the other part of anhydride got bounded to aromatic unit.

Hence my questions:

(1) What happens with that second equivalent of aluminium chloride?

(2) How does aluminium chloride react with acetic acid in nonaqueous medium? (My guess would be that this is something like in the figure below. However, I don't know if reaction goes all the way to chelate or not.)enter image description here

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    $\begingroup$ Does this answer your question? Mechanism of Friedel-Crafts acylation with succinic anhydride $\endgroup$
    – cngzz1
    Jan 2, 2021 at 1:03
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    $\begingroup$ @cngzz1 - thank you for the comment but it doesn't answer my question. What's under link provided makes it look like if AlCl3 was restored - at the end of therein presented mechanism H+ protonates carboxylate bonded to Al and AlCl3 is released. If that was the end then only catalytic amount of AlCl3 would be needed (as in F-C alkylation). Synthetic procedures for F-C acylation require stoichiometric amounts of AlCl3. At least 1 eq. for acid chlorides and 2 eq. for anhydrides. What exactly happens with 2nd eq. of AlCl3 is still unclear. Hence, I think this question needs to be reopened. $\endgroup$
    – Fractal
    Jan 2, 2021 at 13:42
  • $\begingroup$ For your first question: After aqueous work up, AlCl3 is transformed into aluminium hydroxide, which is insoluble, so can be filtered. This might help: organic-chemistry.org/namedreactions/… $\endgroup$
    – cngzz1
    Jan 3, 2021 at 1:03
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    $\begingroup$ @cngzz1 - Actually, from the very beginning my question is about what happens with 2nd eq. of AlCl3 in F-C acylation using carboxylic acid anhydride in reaction mixture (before work up). In other words: how exactly this 2nd eq. of AlCl3 is blocked from activating acylating agent (anhydride) while knowing that 1st eq. of AlCl3 is probably bounded by ketone carbonyl group - as in case of using acid chlorides. (This is why I put this suggestion at the end of my question that maybe AlCl3 reacts with carboxylic acid produced from anhydride in acylation process, but I am not sure if thats the case.) $\endgroup$
    – Fractal
    Jan 3, 2021 at 10:07

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