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Jacob
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[*] posted on 7-4-2019 at 05:35
Potassium ethoxide


Hi there! I'm a noob please bear with me.

Is it possible to make potassium ethoxide by mixing KOH and EtOH in the presence of a dehydrating agent like MgSO4 or CaO?

The easy way is adding metallic potassium to ethanol. But nobody likes to deal with potassium or hydrogen.

My understanding is that added KOH reacts with EtOH in a reversible way. But if there is a water absorber the product water is gone so the reaction moves to the right side.

KOH + C2H5OH <-> C2H5KO + H2O
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Tsjerk
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[*] posted on 7-4-2019 at 06:27


This has been done for sodium ethoxide, there is a very old thread about it somewhere here.
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[*] posted on 7-4-2019 at 08:11


Quote: Originally posted by Tsjerk  
This has been done for sodium ethoxide, there is a very old thread about it somewhere here.

Not with a drying agent that reacts with potassium hydroxide.
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Tsjerk
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[*] posted on 7-4-2019 at 08:27


That is true, but I guess if it works with NaOH and ethanol it could work with KOH. Although I think most uses of K-ethoxide should work with the sodium salt.
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[*] posted on 7-4-2019 at 08:57


I think commercial KOH may have more water in than commercial NaOH.
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Jacob
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[*] posted on 7-4-2019 at 10:00


Quote: Originally posted by Tsjerk  
This has been done for sodium ethoxide, there is a very old thread about it somewhere here.


Still trying to find it. It's a mess!
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Jacob
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[*] posted on 7-4-2019 at 10:02


Quote: Originally posted by unionised  
Quote: Originally posted by Tsjerk  
This has been done for sodium ethoxide, there is a very old thread about it somewhere here.

Not with a drying agent that reacts with potassium hydroxide.


Yes, my bad. :P MgSO4 reacts with KOH. K2CO3 maybe?
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brubei
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[*] posted on 7-4-2019 at 10:39


Maybe Lithium or powdered Al will form M(OH)x + H2 ?



[Edited on 7-4-2019 by brubei]




I'm French so excuse my language
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[*] posted on 7-4-2019 at 12:52


Azeotropic distillation must be easier than all that other hassle.
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Chemi Pharma
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[*] posted on 7-4-2019 at 13:54


You can produce it from KOH, anhydrous Ethanol and hexane, refluxing the mixture using a destilation column and a Dean Stark trap adapter, purging the water while it's produced. When no more water comes over the reaction is finished and you will have an alcoholic solution of potasium ethoxide. Then, distill the excess of alcohol to dryness, leaving the potassium ethoxide at the flask as a white powder.

You can follow the same directions given in the Patent I'm attaching, teaching how to produce the so expensive potassium tert-butoxide from the OTC KOH, T-Butil alcohol and Hexane.

Attachment: potassium tertbutoxide from KOH, Tertbutil alcohol and Hexane.PDF (411kB)
This file has been downloaded 62 times

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clearly_not_atara
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[*] posted on 7-4-2019 at 14:38


I think that you could probably get a pretty decent solution of KOEt by the rxn of K2CO3 with Mg(OEt)2, the latter being produced by dissolving magnesium in refluxing ethanol with a little iodine (or similar?) as an activator. It's not quite a "dehydrating agent method"--it's double displacement--but it is technically a dehydrating agent!

Chemi: in the above patent, do you think you could use e.g. heptane, cyclohexane, toluene, or... pretty much anything that isn't hexane? :p

[Edited on 7-4-2019 by clearly_not_atara]




[Edited on 04-20-1969 by clearly_not_atara]
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[*] posted on 8-4-2019 at 05:09


@Clearly, although the Patent say to use hexane or cyclohexane, I think Toluene, Benzene, hydrocarbons like Heptane as you point, or other non polar solvent with low boiling point that forms a tertiary azeotropic mixture with water and the alcohol may be useful too. All of them are less dense than water, making easy to extract the generated water by a Dean Stark/refluxing column system apparatus.

The work up I have proposed to KEtO is only based on the Patent and don't represent the original work up, but I think it will be work as proposed, based on my personal experience synthesizing potassium tert-butoxy with KOH, Tert-butil alcohol and hexane. If I remember right the yield was almost the Patent claims.
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Dr.Bob
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[*] posted on 8-4-2019 at 05:57


It is very tough to make useful KOEt from KOH and water, as even though the equilibrium reaction seems like it should balance between the two sides, it will really be shifted far to the left, as water is very difficult to remove from this mixture. Someone posted a way to precipitate NaOEt from this type of reaction using acetone a while back as a way to remove the water and drive the reaction to the right, but not sure if that works for KOEt.

KOH + C2H5OH <<<<<-------> C2H5KO + H2O

Also, the problem is that KOH absorbs water very well, so it is often full of water of hydration, to the point of dissolving itself if not kept well sealed.

Lasly, for many reaction that need KOEt, any trace of water will quench the anions produced, so again, water is faster at reacting with anions than most bases are at making them, so there is a kinetic effect which means that one equivalent of water will all but kill a reaction, and even .10 eq of water may reduce the amount of anion present by 80-90%, in my experience. I recently tried some challenging reactions with various bases, and even the smallest traces of water were enough to all but kill the reaction,even with freshly distilled solvents, and well dried substrates. That is why is is useful to understand if a reaction is water sensitive in a minor way or a catalytic way. The first can be overcome by using more base to react with the traces of water (eg, in making biodiesel, you can add enough NaOMe to eventually shift the reaction to product, even with traces of water.) But in many strong base reactions, such as many using LDA, any trace of water will cause side reactions which can shift the reaction mostly to side products of hydroysis or decomposition.
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[*] posted on 8-4-2019 at 05:59


KOH can be made anhydrous by fusing it, but it should also work with the 85% stuff.
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Jacob
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[*] posted on 8-4-2019 at 06:36


Quote: Originally posted by brubei  
Maybe Lithium or powdered Al will form M(OH)x + H2 ?



[Edited on 7-4-2019 by brubei]


To remove water?

Would have been great if it made the insoluble Al(OH)3. But Al and KOH and water react to make H2 and KAl(OH)4 which is water soluble. Don't know if it's alcohol soluble too.

No idea about lithium either.

CaO and MgO might be just better.
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Jacob
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[*] posted on 8-4-2019 at 09:23


Quote: Originally posted by Chemi Pharma  
You can produce it from KOH, anhydrous Ethanol and hexane, refluxing the mixture using a destilation column and a Dean Stark trap adapter, purging the water while it's produced. When no more water comes over the reaction is finished and you will have an alcoholic solution of potasium ethoxide. Then, distill the excess of alcohol to dryness, leaving the potassium ethoxide at the flask as a white powder.

You can follow the same directions given in the Patent I'm attaching, teaching how to produce the so expensive potassium tert-butoxide from the OTC KOH, T-Butil alcohol and Hexane.


Thank you very much. What is the purpose of the Dean Stark trap anyways? If your liquids are immiscible just let them sit in a burette, why distill first?

And what role does hexane play in the patent? Removing water?
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[*] posted on 8-4-2019 at 14:16


Quote: Originally posted by Jacob  
Thank you very much. What is the purpose of the Dean Stark trap anyways? If your liquids are immiscible just let them sit in a burette, why distill first?

And what role does hexane play in the patent? Removing water?


Dean Stark is used generally to extract water from the condensed liquid formed inside a reflux column, without the need of stop the heat, reflux or open the apparatus. Look at wikipedia to understand better it's function: https://en.wikipedia.org/wiki/Dean%E2%80%93Stark_apparatus

The role hexane plays is acting as a withdrawing agent. In really, the Patent attempts to the fact you need to form an azeotropic tertiary mixture, between the alcohol, the water and the withdrawing agent to reflux it. When the vapors condense inside the column the amount of water sits at the bottom of the Dean Stark trap and could be trapped out, while the alcohol and hexane mixture floats at the top of the Dean Stark trap and return to the reaction vessel directing the reaction to the right (toward the formation of the alkoxy salt).

You must distill not at the begining, but just when no more water sits down at the Dean Stark trap, as I've said. At the beginning you must reflux, following the Patent direction. This is an adaptation of the work up described at the patent I have already done in my Lab with good results.
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