During Fries Rearrangement, oxygen donates its lone pair to $\ce{AlCl_3}$ forming $\ce{-O^+-AlCl3}$. Since this is a strong electron withdrawing group, shouldn't the final product be meta-substituted?
1 Answer
The above is the mechanism for Fries Rearrangement for a brominated ester( doesn't matter really, for any ester the mechanism essentially remains to be the same).As per what I infer from your question, I think you mean to ask why even in the presence of $\ce{O^+-AlCl3}$ which is a strongly deactivating group, the major product is not meta-substituted.
As you can clearly see, in the next step itself a stable acylium is formed and the formal charge on $\ce{O}$ is no longer positive.So, as phenoxide ion is an activating group and o,p-director, two possible products are expected which are ortho and para substituted.However, at higher temperatures, the major product is ortho substituted which is driven by chelation.
NOTE:
As far as I know, both intramolecular and intermolecular reaction mechanisms are able to explain the major products and it is still not confirmed which one is the actual mechanism.
Image source:
https://www.synarchive.com/named-reactions/fries-rearrangement
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$\begingroup$ Such ping doesn't work chemistry.meta.stackexchange.com/questions/3889/… $\endgroup$– MithoronJul 6, 2020 at 15:57
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$\begingroup$ @Mithoron Niescte has asked this question and I was not sure if I understood Niescte's question correctly.So, wanted to confirm with the author of the post, whether I interpreted in the correct way.If this is against the policy of the community, then I can delete that comment $\endgroup$ Jul 6, 2020 at 16:31
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$\begingroup$ sigh You'd have to comment under his question or comment, otherwise it just doesn't ping, that's my point - check out the meta stuff I linked for more info about pinging. BTW asking about clarification, before answering would be better... $\endgroup$– MithoronJul 6, 2020 at 22:48