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What would happen if you tried a Williamson Ether Synthesis on something that has both an alcohol and a carboxylic acid? Would both -OH groups be functionalized, where the alcohol would become an ether and the carboxylic acid become an ester?

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  • $\begingroup$ Hi Tbl, welcome to Chemistry SE. Have you tried to do any research on this on your own? The products formed will likely depend heavily on the nature of reactants and the reaction conditions. $\endgroup$ – Michael Lautman Jun 6 at 21:34
  • $\begingroup$ I’ve been trying to find the answer to this question for a while, even pulled out my old organic textbook to try and answer this question. The only things I’ve found have basically said that for Williamson Ether Synthesis, turn the carboxylic acid into an alcohol then run the reaction, but I don’t actually want the carboxylic acid to react. It would be awesome if it didn’t, and only the alcohol reacted. $\endgroup$ – Tbl Jun 6 at 22:33
  • $\begingroup$ If you make the ether-ester from the dianion, then saponify the ester to get the ester-carboxylic acid if that is what you want. $\endgroup$ – user55119 Jul 29 at 23:24
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In priciple this should work. If you dissolve your omega-hydroxycarboxylic acid in excess aqueous hydroxide (or hydrolyse the lactone) then add a reactive electrophile like MeI, I think you will get the ether you want. Any ester that forms will quickly hydrolyse back. I do not have database access to search for this, but perhaps someone else can find a reference.

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  • $\begingroup$ Won’t this just give lots of methanol? $\endgroup$ – PCK Jun 13 at 19:28
  • $\begingroup$ No, I don't think so otherwise Williamson synthesis of methyl ethers would not work $\endgroup$ – Waylander Jun 13 at 22:23
  • $\begingroup$ True, I wonder why? Heavier solvation of hydroxide vs alkoxide? $\endgroup$ – PCK Jun 14 at 6:21
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There is a general principle known as “last in-first out” that I think applies here. If a molecule contains multiple acidic positions with significantly different pKa, you can selectively react on the least acidic position (in your case, the alcohol).

Specifically, I would try treating your molecule with 2 eq. Of strong base such as NaH or LDA. The first equivalent will deprotonate all of your acid to form the quite stable metal acetate. The second equivalent removes the alcohol hydrogen to generate a less stable alkoxide. If you add 1.0 eq. Of the electrophile it ought to react selectively on the more reactive alkoxide nucleophile, leaving the metal acetate unaffected, then you can just neutralise in work up.

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  • $\begingroup$ Certainly true that the alkoxide is a better nucleophile than the carboxylate $\endgroup$ – Waylander Jun 13 at 22:25

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