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Williamson Ether and Carboxylic Acids
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Mercy Gina
Williamson Ether and Carboxylic Acids
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.
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.
No, I dont think so otherwise Williamson synthesis of methyl ethers would not workMore
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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.
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.
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.
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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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.
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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