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Author: Subject: Sodium Ethoxide and anhydrous EtOH
Organikum
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[*] posted on 9-10-2004 at 03:36
Sodium Ethoxide and anhydrous EtOH


A sep-funnel or anything else with an outlet (faucet) at the bottom is filled to 1/3rd with NaOH pellets and filled full with EtOH of 92% or of higher concentration (preferred). This sits for at least half an hour and then about 1/4th of the liquid at the bottom is withdraw very slowly. Then the rest is withdraw. This rest consists of EtOH which contains sodium ethoxide. Distilled to yield anhydrous EtOH, sodium ethoxide is left back which can be used to dry more EtOH so desired.
If the alcohol is pre-dried with anhydrous CuSO4 which was dehydrated at 300°C+ much less NaOH is needed. This pre-drying takes time though, the longer the better in special if no stirring is applied.
If the ethoxide is whats desired the alcohol should be left in the vessel for longer time, if bigger amounts are wanted it is favorable to withdraw alcohol/ethoxide from top and water/NaOH/alc from bottom and to refill alcohol. This is an almost continous process then.

Tried and true.




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[*] posted on 9-10-2004 at 06:57
A great procedure!


At the first look it sounds quite unbelievable, especially for those, who wasted their sodium metal for this reaction. :D

BTW: Did someone try this reaction with MeOH instead of EtOH?




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[*] posted on 9-10-2004 at 10:35


Interesting, but what makes the first quarter contain the water, does it form a seperate layer? Why are we removing 1/4 of the ethanol solution and not a visible product or anything related to the amount of water in the ethanol?

How much NaOH is dissolved in the liquid contaminating the ethoxide product?

This is a very unauthadox method and given how easy it is to mistake a solution of sodium hydroxide in ethanol with sodium ethoxide it does make me wonder.
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[*] posted on 9-10-2004 at 11:18


- Interesting, but what makes the first quarter contain the water, does it form a seperate layer?
- Yes it does. Salting out is the principle behind this.

- Why are we removing 1/4 of the ethanol solution and not a visible product or anything related to the amount of water in the ethanol?
- To make sure all water is removed. A rule of thumb not more not less, not optimized.

- How much NaOH is dissolved in the liquid contaminating the ethoxide product?
- This I cannot answer, not very much if the ethoxide/alcohol is withdrawn from top and not at the bottom. This is what the patent says at least. I guess the alkoxide salts the NaOH out of the alc. for its higher solubility.

- This is a very unauthadox method
- This a a patented method

As I posted it at the HIVE times ago:
"Patent US2796443 1956 to Dow Chemicals, describes a process for the production of alkali-alkoxides which requires no distillation. I basically works with an excess of hydroxide which saturates the water which can be withdrawn at the bottom, whereas the alcoholic anhydrous alcohol/alkoxide solution is withdrawn at top of a reactor."

The whole thread is interesting I want to say:
Alkoxides

There also a method for removing the water by azeotropic distillation with toluene is described. I believe the method as told by me is easier. And it works. Whereby I am not sure for the azeotropic distillation as salts fuck up almost every azeotrope. But thats an educated guess, not more. Benzene might work and would work for sure better than toluene, when toluene works at all.

- At the first look it sounds quite unbelievable, especially for those, who wasted their sodium metal for this reaction.
- You dont really want me comment on this?

- BTW: Did someone try this reaction with MeOH instead of EtOH?
- Works on MeOH too. No problem.


In short:
Adding excess NaOH to EtOH which contains water salts the water out. Separate and distill, all remaining water will by sure be removed by the alkoxide formed in step one.
Nor so hard at all.

Marvin: The process is so obvious and easy I thought it being ok to post it as rule of thumb. Withdrawing 1/4th works for sure and will give also the beginner a good result. Those who know the art may be able to distinguish the aqueous and the alcoholic layer by eyeballing though... :P
Just remove the aqueous layer as it is formed, shift the equilibrium this way and enjoy! Thats for pure alkoxides. For anhydrous alc. and alkoxides which maybe a small contamination of hydroxide the method as described first suffices.

Yes sometimes this is not really hard.
And sometimes its much much harder as it sounds I want to add. :o


[Edited on 9-10-2004 by Organikum]




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[*] posted on 9-10-2004 at 12:25


I have a complaint: Edit your post instead of doubleposting! :P Hah! Got you there! :P

Now it's very possible I was wrong there. I haven't heard of any method for making sodiumethoxide this way untill today. I can't for the life of me remember what I said in that thread though. Tzzz, Alzheimer and still so young.




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[*] posted on 9-12-2004 at 15:50


Isn't distilling off the ethanol from the sodium ethoxide dangerous? Sodium ethoxide is extremely flammable.

Edit: By the way this doesn't work with methanol, the patent says it requires aliphatic alcohols with two carbons or more. It must be heated to at least 70 degrees C. The resulting alcohol/alkoxide solution has about 2-3% dissolved hydroxide. This dissolved hydroxide won't harm alkoxide catalyzed condensations will it?

[Edited on 10-12-2004 by Mendeleev]




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[*] posted on 9-12-2004 at 17:36


There are posts missing from this thread and I'm almost sure one is mine. I remeber reading that patent as part of this thread and posting a bunch of reasons why the process would work badly for ethoxide, why counterflow exchange was so important to the process in the patent and that a solution of sodium hydroxide in ethanol or methanol is very difficult to tell apart from methoxide and ethoxide. I remeber a doublepost by Organikum seperated by about an hour (hense vulture's comments about a doublepost) as well as a sizable argument between them. I also think I remeber Organikum posting a low yeilding reaction as evidence the process did produce ethoxide or methoxide.

My conclusion I remeber is that I wouldnt trust the process at all without a reliable testing. It looks highly broken to me.

Can anyone confirm or am I going insane?
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[*] posted on 9-12-2004 at 17:42


No, I noticed the same. Hmmm.... very peculiar. Although I am not sure about the specifics.
I can't remember when this happened elsewhere, except pictures disappearing.

[Edited on 10-12-2004 by chemoleo]




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[*] posted on 9-12-2004 at 17:54


Yes, something has changed. Couldn't tell you exactly what, but it looked wrong to me -- and I followed the thread closely, since it seemed like a useful idea.... assuming it works.

Edit: Sorry, this post is pretty useless, Chemoleo submitted his confirming post while I was making this one. Can someone either recover it (from archive etc) or make a summary post from memory? I hate the idea of us losing posts (and what they contain).

There is evidently criticism/scepticism of the method but I was under the impression Org had actually tested it? In which case I trust his experiment more than others' theories.

[Edited on 10-12-2004 by Oxydro]
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[*] posted on 10-12-2004 at 01:39


I deleted no information but solely a post with an unnecessary argument between me and vulture.

I never saw a post of Marvin as he describes, I have no possibility to delete posts which are not mine either.
Sorry Marvin, I vote for the "going insane" option.

I want to point out that I posted the method mainly for making anhydrous alcohol, not ethoxide, but it works for this purpose too. It may or maynot work for methoxide, I apologize if I was wrong on this, I thought I read somebody doing it with success.
I never posted more information than the one above and no "low yielding reaction" blabla.

The patent was found in the Ullmanns IIRC, which names usually only patents which methods are actually used in industry. The working principle is solubility, density and salting out - counterflow makes it continous, withdrawing from the bottom only makes it batch.

[Edited on 10-12-2004 by Organikum]




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[*] posted on 10-12-2004 at 02:27


I am currently trying to turn a bromo benzaldehyde into a methoxy benzadehyde by leaving the bromo benzaldehyde in a methoxide solution made as Organikum described for many weeks (two so far) without copper catalyst. The temperature in my attic now may reach 40ºC easy during the day. The thing has already turned from “crystals in clear alcohol” to “thick white paint” look.

If it works, would it be a test for methoxide or could hidroxide in EtOH do this trick? I intent to test the water extracts for NaBr to check if it works.

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[*] posted on 8-6-2005 at 00:07


This convenient method for sodium ethoxide would likely have value
with regards to production of ethanol
solutions of anhydrous hydrazine .
If a 50% in excess of the theoretical
amount of sodium ethoxide was used
for freebasing hydrazine from hydrazine sulfate , the result would be an equimolar
amount of anhydrous hydrazine and sodium ethoxide in anhydrous ethanol ,
which could be decanted from the solid residue of sodium sulfate . Such a solution would be ideal for the production of sodium azide which would precipitate
upon addition of isopropyl nitrite , ethylene glycol dinitrite , or perhaps upon
bubbling through the solution N2O3 .

I am uncertain as to whether the freebasing of the hydrazine would proceed under anhydrous conditions
from the very start . But at the very least
the use of the sodium ethoxide in some
part substituted for the NaOH which would
ordinarily be used , would reduce the water content and improve yields of
sodium azide made subsequently by
nitrosation of the mixture .
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[*] posted on 8-6-2005 at 20:10


Quote:
Originally posted by Tacho
I am currently trying to turn a bromo benzaldehyde into a methoxy benzadehyde by leaving the bromo benzaldehyde in a methoxide solution made as Organikum described for many weeks (two so far) without copper catalyst. The temperature in my attic now may reach 40ºC easy during the day. The thing has already turned from “crystals in clear alcohol” to “thick white paint” look.

If it works, would it be a test for methoxide or could hidroxide in EtOH do this trick? I intent to test the water extracts for NaBr to check if it works.



Unfortunately NaOH will react with the bromobenzaldehyde the same way as methoxybenzaldehyde does to give a hydroxy group rather than a methoxide group.

Is there any reason why CaO can't just be used to shift the equilibrium of the NaOH + methanol reaction to forming more methoxide? The Ca(OH)2 formed shouldn't be very soluble in methanol or ethanol I don't think.
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[*] posted on 9-6-2005 at 09:02


It won't work without the copper catalyst, to my knowledge metallic copper isn't the active species anyway. But I think the Cannizzaro reaction works with ethoxide instead of hydroxide too, giving sodium benzylate (sodium salt of benzyl alcohol) and the ethyl ester of the benzoic acid instead of sodium benzoate and benzyl alcohol. It might be what's happening.

[Edited on 9-6-2005 by trilobite]

[Edited on 9-6-2005 by trilobite]
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[*] posted on 9-6-2005 at 09:18


Googling for +Cannizzaro +alkoxide returns no relevant hits for me, so I can't verify the veracity of ethoxide replacing hydroxide in the Cannizzaro reaction; having just reviewed the mech, however, it seems theoretically possible.

Someone ought to try it out, then... ;)

But wouldn't the water generated by the reaction consume the ethoxide as well? (scratches head)

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[*] posted on 9-6-2005 at 10:02


What water? ;) Oh, I should have been talking about methoxides and methyl esters instead of ethoxides and ethyl ester, well anyway:

PhCHO + PhCHO + NaOMe --> PhCOOMe + PhCH2ONa

[Edited on 9-6-2005 by trilobite]
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[*] posted on 10-6-2005 at 06:20


When I wrote my previous post, my eyes were trying to shut themselves out of their own accord. :( Sorry about that, you're right, if it's just alkoxide and aldehyde in the reaction vessel, there's no water. :)

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[*] posted on 8-8-2005 at 11:07
alkoxide exchange


I don't know if there is already a thread about thi, but I coulnt find it with the search function. Please let me know if I missed one.
Anyway, a synthesis I'm thinking about needs sodium ethoxide. In the official write-up that's easy; they just dissolved sodium in dry alcohol.
My problem is the lack of sodium. On the other hand, I have magnesium and (even better) aluminium. Am I missing something or can I react the Mg with EtOH to get Mg(OEt)2 then add a solution of NaOH in EtOH to get Mg(OH)2 as a ppt and the NaOEt in solution?
Better yet can I use (cheap) Al as the metal.
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[*] posted on 8-8-2005 at 12:41


The problems I can see are 1) that Mg/Al doesn't dissolve all that well in water, let alone EtOH and 2) Mg(OEt)2 may be insoluble itself, so that you'd interconvert one insoluble compound to another. May not be a fast reaction at all.

PS merged thread with existing alkoxide thread.




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[*] posted on 8-8-2005 at 16:37


IIRC, there was something like this in Vogel. Magnesium is refluxed with EtOH, forming the ethoxide. The ethoxide reacts with residual water to form Mg(OH)<sub>2</sub>. I think this is finally distilled to yield ultradry ethanol.
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[*] posted on 8-8-2005 at 18:07


The preparation of the Mg and Al alkoxides is well covered in the literature, but not the preparation of Na alkoxides from them. Basic and double salts also, but AFAIK not by adding some other hydroxide to the alkoxide.

The Mg and Al alcoholates are usually made by refluxing for a long time with the help of Hg salts or I2. The yield from EtOH is usually lower than from IPA, it seems.

Other items of interest:
RU2178402 Carbonate + ROH. Encrypted.
US3479381 Hydroxide + ROH + ordinary molecular sieves.
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[*] posted on 8-8-2005 at 23:03


Mg and Al alkoxides are insoluble in water. They are normally produced by using the metal itself and a dry alcohol.
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[*] posted on 9-8-2005 at 12:36


Drat!:(
Mg(OEt)2 isn't very soluble. I might have to start with Mg +MeOH. Add NaOH to get NaOMe and add excess EtOH then distill off MeOH to get NaOEt.
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[*] posted on 9-8-2005 at 22:33


Do you have Na metal? Because to get NaOEt, you can react Na with EtOH.
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[*] posted on 10-8-2005 at 12:59


.......The Mg and Al alcoholates are usually made by refluxing for a long time with the help of Hg salts or I2......

Procedure I have for Al Ethoxide is 90% yield and uses HgCl2 and says to add a crystal of I2 as catalyst. I found that the iodine wasn't needed as it began reacting almost right away.

Would post it if I could figure out how to paste an image but it's from inorganic preps IIRC, found on the FTP.

The key here I think is the alcohol must be truly anhydrous. RA absolute EtOH was dried with silica gel and it worked nicely.

Vogel says CCl4 is used as catalyst in the Isopropoxide prep and I think chloroform would work also? Has anyone tried CHCl3
for this type reaction?
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