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Jackson
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[*] posted on 11-3-2019 at 07:35
Disodium 1,2-ethanediolate Production

Hi, Could I produce Disodium 1,2-ethanediolate by heating a mixture of ethylene glycol and sodium hydroxide. My idea is that the sodium hydroxide deprotonates the ethylene glycol like (CH2OH)2 + NaOH —> Na-O-CH2-CH2-OH + H2O. Then that reaction happens twice. The reaction would be driven forward by the water being removed by boiling as it has a lower boiling point than sodium hydroxide, ethylene glycol, and Disodium 1,2-ethanediolate. I need this for a Fischer esterification.
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[*] posted on 11-3-2019 at 07:54


I would use a vacuum, could very well work.
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[*] posted on 11-3-2019 at 14:22


Just make sure it's protected from air, because alkoxides get oxidized pretty quickly -- and the rxn mixture may attack glass



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[*] posted on 11-3-2019 at 14:47


Could toluene azeotrope with the water and boil at lower T in this case?



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12thealchemist
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[*] posted on 12-3-2019 at 01:18


Quote: Originally posted by fusso  
Could toluene azeotrope with the water and boil at lower T in this case?


Unfortunately toluene forms an azeotrope with ethylene glycol consisting of 93.5% ethylene glycol boiling at 110.2°C. (toluene bp 110.8°C). While you could distil off the water-toluene azeotrope (bp 84.1°C), unless the reaction was fairly swift, you'd run the risk of losing a lot of your reagent - I believe ethylene glycol and toluene are miscible.



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Jackson
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[*] posted on 12-3-2019 at 07:40


I know that NaOH does deprotanate ethylene glycol, so this could mean that diols form more stable alkoxides. Does anyone know how it behaves?
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[*] posted on 12-3-2019 at 08:14


NaOH also deprotonates ethanol, the questions is to what extend it does so. You can have a look at the pKa of the two alcohols, I'm pretty sure that will give a good indication for your question.

A quick search shows 15.1 and 15.9 for ethanediol and ethanol respectively, which means the first is about 7 times more acidic. So I think your reasoning is correct.
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