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Author: Subject: Sodium Ethoxide and anhydrous EtOH
Nicodem
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[*] posted on 17-3-2008 at 01:33


Siddy, try reading the whole thread. I'm sure you will better understand that you are dealing with equilibriums. The ethanolic phase can, and surely does, contain some water (as long as there is some hydroxide anions, there is H<sub>2</sub>O as well since OH<sup>-</sup> + EtOH <=> H<sub>2</sub>O + EtO<sup>-</sup>;). Hence, if you want a very pure ethoxide ethanolic solution, you need a continuous equilibrium shifting (like is for example done in the title patent where a counterflow method is used). But for most applications, a single or double equilibration will give you sufficiently pure ethanolic sodium ethoxide solution (it all depends on what you'll be using it for).
If you are planning to prepare solid NaOEt, then you can remove traces of water by adding some toluene, distill off to half volume and continue by drying under vacuum (always keep in mind its pyrophoric properties!).
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[*] posted on 29-3-2008 at 01:39


Provided anhydrous ethanol were used and the resultant NaOEt was sufficiently clear of NaOH & water (which would otherwise cause saponification, yes?), could this be used for the Claisen condensation? The pKa of Ethanol is approx 16, while that of the ester is around 22, so if one made up 1 litre of a 1N NaOH/EtOH solution (given around 90% of the base is present as OEt), would this be sufficient?

I note that the orgsyn procedure uses a solution strength of around 3.7N so I am unsure - surely telling me ain't spoonfeeeding here?

PS The answer is provided in one of the patents cited earlier - what chance using some other solvent which forms an azetrope in order to pull the water out of this reaction? Would 4A sieves work?

[Edited on 29-3-2008 by LSD25]




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[*] posted on 29-3-2008 at 21:00


Quote:
The pKa of Ethanol is approx 16, while that of the ester is around 22, so if one made up 1 litre of a 1N NaOH/EtOH solution (given around 90% of the base is present as OEt), would this be sufficient?

I don´t think that this works. Yields will suffer dramatically and side reactions like saponification will occur to a great extent.
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[*] posted on 29-3-2008 at 22:37


Why not just throw excess ethyl acetate at the problem? I mean, saponification should really only affect the ester, not the enolate? By increasing the relative amount of the ester, surely one would increase the probability of product forming? I mean the relative ratio of the reagents in the a-phenylacetoacetonitrile (http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=CV2...) synthesis is 1:5 (PhCH2CN:OEt). A whole lot of the ethoxide is just basing out the reaction, it is not actually contributing directly. The excess NaOH should really saponify (for the most part) excess EtOAc to give additional ethanol and NaOAc shouldn't it? Thus taking no further part in the reaction?

I am thinking on it because a lot of biodiesel research is looking into the transesterification of glycerides with alcoholates, using alkali hydroxides as the catalyst - generating the OC(x)H(y) in situ. If this works with that, I'd be interested to know why it wouldn't work here (provided always that the reaction is kept dry - maybe mix some Na2O/CaO in with the NaOH to pull the water out of the reaction).

Hang on, thinking here, what would happen to the equilibrium of the reaction if the NaOH/EtOH<=>NaOEt/H2O was made up first (with CaO - which should pull out the water and be transformed to fucking near insoluble CaOH) - with its reported 90% OEt, then the serious excess of EtOAc was added, allowed to sit for a while then the PhCH2CN were added? Would the saponification of the minimum amount of EtOAc remove the remaining NaOH from the equation by forming the NaOAc? How quickly would the ethylacetoacetate form? Would sufficient EtOAc remain to react with the PhCH2CN?




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[*] posted on 2-6-2008 at 18:08


OK, just found this (I condensed two articles into one download), the first details the use of cation exchange membranes (as used in water treatment - although a decent route to the same would be nice) and electrolysis to produce NaOMe from MeOH & NaOAc, excess CH3COOH is given off as ethane & CO2, so more is added as the reaction progresses. The second details the use of the same reactants and the same system to produce Acetoacetic ester. Wonder if there is a way around those bloody membranes?

For instance, what would happen if the non-aqueous solution contained dry ethanol, dry ethyl acetate and dry sodium acetate?

Any ideas floating about?

[Edited on 2-6-2008 by LSD25]

Attachment: Electrodialysis.in.nonaqmedia.for.sodiummethoxide.pdf (749kB)
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[*] posted on 4-10-2008 at 13:51
Use sodium oxide


Relying on the alkoxide-hydroxide equilibrium will inevitably lead to contamination and meager yields unless you can afford professional equipment and carefully controlled conditions.

My suggestion is the solvolysis of sodium oxide by means of diethyl ether. There's no issue of eliminating water to drive the equilibrium. Instead, the problem is that ether, unlike ethanol and especially water, has no acidic character (or at any rate a negligible acid dissociation constant). This might be partially compensated by the strong basic character of the sodium oxide, but I think most importantly the reaction is catalyzed by the small amounts of ethanol present.

Sodium oxide can be obtained by roasting the carbonate. Even if this step doesn't go to completion, the only macroscopic impurity we'd get is the carbonate, notoriously insoluble in alcohol, let alone in ether. This precipitate can be collected and measured, which allows you to compute the ethoxide yield.

Sodium oxide can also be made (though in somewhat impure form) by heating the nitrate/nitrite or the sulfite well below the melting point of the carbonate.
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[*] posted on 27-3-2009 at 04:50


Is there any reason we can't use other metalllic oxides in the mix to dehydrate the mix (removing the water as it is formed)? I'm thinking maybe Calcium Oxide (CaO + H2O => Ca(OH)2)? Mix some calcium oxide with the sodium hydroxide, use more calcium oxide in order to dry the ethanol being supplied to the reaction.

That would mean nothing more than heating the solution, then filtration, acetone precipitation, filtration then drying/vacuum to get the sodium ethoxide? (this is predicated on the presumption that anhydrous ethanol will react preferentially with sodium hydroxide rather than calcium hydroxide when both are present, thus calcium hydroxide will be mixed with the insolubles filtered from the solution at the start? The use of calcium oxide to dry alcohol suggests (to me at least) that the formation of calcium ethoxide(s) are slow/non-existant to the extent that they are ignored.

However, the efficacy of calcium oxide in drying ethanol is known, the only real question I would suggest remains, would be how best to do this

Mix finely divided calcium oxide with freshly dried (fused/crushed) NaOH, put it in a vessel, pass in dried ethanol (passed through H2SO4, then over CaO) stir well, then rapidly disconnect everything and pass the solution through the finest filter possible (too slow and you can start again), the clear solution can then be precipitated with acetone, then fillter to collect the crude sodium ethoxide. Weigh it, then make up the required solution? If you are doing this at home, you probably should use it immediately (or asap), I wouldn't even try storing it.
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[*] posted on 27-3-2009 at 06:32


Quote: Originally posted by no1uno  
... pass in dried ethanol (passed through H2SO4, then over CaO) ...


You don't dry alcohols with H2SO4 - ester formation EtO(HO)SO2 & (EtO)2SO2, dehydration to ethers and alkenes.

Outside of that it could work, although for EtOH azeotropic removal of water does the job.

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[*] posted on 31-3-2009 at 22:24
ethylate


There is a patent somewhere which suggests refluxing the mix under a column packed with CaO to remove water.

This may have been posted already.Ill remove if need be.

Removal of hydroxide ion from alkoxide ion solutions

EXAMPLE 1

Removal of Sodium Hydroxide from Sodium Methoxide in Methanol with Methyl
Acetate

A 2791 grams (g) sample of 25 percent solution of sodium methoxide in methanol was analyzed and found to contain 0.36±0.05 percent sodium hydroxide (10.1 g, 0.25 mol). To this was added 20.5 g (0.28 mol) of methyl acetate. The reagents were mixed and the mixture was allowed to stand at ambient temperature overnight. The solution was reanalyzed and found to contain 0.05 percent sodium hydroxide. Analyses were done by Karl Fischer titration using an automatic coulometric titrator to generate the Karl Fischer reagent and conduct the titration. Benzoic acid was added to the cell electrolyte before titration as a buffering agent.

Similarly, to an 18.1 kilogram sample of a 25 percent solution of sodium methoxide in methanol found to contain 0.42±0.02 percent sodium hydroxide (76.0 g, 1.90 mol) was added 141 g (1.90 mol) of methyl acetate. The reagents were mixed and the mixture was allowed to stand by ambient temperature overnight. The solution was reanalyzed and found to contain 0.04 percent sodium hydroxide.
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[*] posted on 23-1-2010 at 04:03


Nicodem, you have stated that you have used this reaction to obtain NaOEt (via Acetone precipitation IIRC)... What sort of yields did you get (I can't find it in the hive files I can access) and what percentage solution of NaOH/EtOH did you use? Also, how did you remove the excess NaOH from the NaOEt (if it in fact existed, which I would have to presume it would)?
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[*] posted on 5-7-2010 at 12:29


This experiment has been based on the patent (US2796443) mentioned in this post. Since I need to acquire some sodium ethoxide, I decided to give this method a chance, because I would very much like to avoid ordering pure Na, which is quite expensive where I live.

Following the ROH + MOH <-> ROM + HOH equation, and having in mind that the patent states that a surplus of 0.1 to 0.4 (in my case ~0.12) moles of NaOH should be added, in addition to the stoichiometric amount, I added about 250 ml of 96% ethanol (4.11 moles) to 200 grams of NaOH (5.00 moles).

When the ethanol was added to the sodium hydroxide, a quite amount of heat was evolved relatively fast. A moment later, the cylindrical beaker was placed in a hot water bath in order to achieve and mantain the temperature around 70°C. When the solution reached the mentioned temperature, I left the reaction to carry on for half an hour.


(click for a larger pic)

The result can be seen on the image above. As stated in the patent, there is indeed a dense fraction that can be seen reaching a bit above the layer of NaOH (picture to the right). This should be composed mainly of water and sodium hydroxide. The solution above the mantioned fraction should contain mostly ethanol and sodium ethoxide.

I didn't expect these two fractions to be that clearly separated, but it certainly did help in the separation of the two, which was done by slowly extracting the alkoxide layer with a pipette, soon after the picture was taken.

The obtained solution has just a slight yellowish color, although you can see that the upper layer of the NaOH is clearly yellow colored, while at the bottom, the NaOH has a clean white color which is also conforming to the patent which claims that the alkoxide rises upwards during the process.

I'm not sure what would be the best way to extract the ethoxide from the solution. I was thinking about distilling the solution until the ethoxide starts to crystallize and then cool the solution to the lowest temperature that I can achieve over a longer period of time (~ -26°C ), in order to lower its solubility. Then the ethoxide would be separated by simple filtration?

[Edited on 5-7-2010 by Sciocrat]




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[*] posted on 5-7-2010 at 23:22


This is really nice. I am surprised that making NaOEt is so easy.
Now I can do that DADNE synthesis I have been wanting to do.

I am going to try CaO and isopropyl alcohol. I am not sure if adding Na2CO3 to this would help.
I have mixed Na2CO3 and CaO in water to make NaOH, since with CaCO3 precipitates out.



[Edited on 6-7-2010 by Anders Hoveland]
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[*] posted on 7-7-2010 at 20:49


Why is the second solution black? Try adding 10ml of this yellow solution to 20mls of ethyl acetate, it should form an immobile mess of sodium acetate. This EtONa is heavily contaminated with NaOH which can only be removed by azeotropic distillation pushing the equilibrium in favor of the ethoxide or by using molecular sieves to do the same thing.
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[*] posted on 7-7-2010 at 23:14


The solution on the right had the same color as the one on the left, it looks dark because of the bright background and because the photo was taken witn no flash (so that the dense layer could be clearly seen).

Quote: Originally posted by flavaflav  
This EtONa is heavily contaminated with NaOH which can only be removed by azeotropic distillation pushing the equilibrium in favor of the ethoxide or by using molecular sieves to do the same thing.


This was exactly what I did. About 400 ml of the solution that was obtained, was distilled down to about 60 ml of yellow colored solution. Small particles were visible in the solution already at high temperatures. Upon cooling, even more of this yellowish substance precitipated.

Unfortunately, at the moment, I don't have any more of this solution, but will be making more soon, and will try the reaction that you mentioned.
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[*] posted on 6-8-2010 at 22:16


Ok, so all you did was add commercial prilled NaOH to EtOH? Where has all the crap come from? Was there anything organic left in the EtOH? Or is the crud from the commercial NaOH?

I think in order to get anything worthwhile out of that you'll have to work out a way to do the reaction in such a way as to ensure the two major layers are a LOT cleaner than that.

As to the precipitation of the ethoxide - Nicodem has stated (upthread) that acetone will precipitate the NaOEt from solution (ie. it reducing solubility leading to precipitation.

As to the use of Sodium/Potassium/Lithium/Calcium Oxide so as to change the equilibrium, check out the following papers. Working out a workable MW route to the dehydrated hydroxides should be of paramount concern.



[Edited on 7-8-2010 by un0me2]

Attachment: Irabien.Viguri.Ortiz.Thermal.Dehydration.of.Calcium.Hydroxide.1.Kinetic.Model.and.Parameters.pdf (1.8MB)
This file has been downloaded 2255 times

Attachment: Irabien.Viguri.Ortiz.Thermal.Dehydration.of.Calcium.Hydroxide.2.Surface.Area.Evolution.pdf (1.4MB)
This file has been downloaded 1373 times

[Edited on 7-8-2010 by un0me2]




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[*] posted on 6-8-2010 at 23:37


Bugger, I have to double post in order to add the other file (the one dealing with the preparation of sodium/lithium oxides by heating them with CaO).


[Edited on 7-8-2010 by un0me2]

Attachment: Maiti.Baerns.Dehydration.of.Sodium.Hydroxide.and.Lithium.Hydroxide.Dispersed.over.Calcium.Oxide.Catalysts.for.the.Oxidat (866kB)
This file has been downloaded 1325 times





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[*] posted on 26-2-2011 at 11:15


will lithium metal suffice as a substitute for sodium in this reaction? i havent been able to find too much information on lithium ethoxide, is there any reason why sodium is so well documented?
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[*] posted on 5-3-2011 at 14:19


I just arrived at the same inquiry.

Since I dont have a IR machine i was pondering the possible products of a potential reaction forming schiff base in ethanol. I added some lithium to the yellowed solution of mainly ethanol. The lithium did react and bubble. It was very anhydrous sitting on top of 3A molecular seive beads. So did Li ethoxide form and for my purposes did it react with my potential schiff base.

but now i don't have any more lithium handy to put into neat ethanol. darn...

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[*] posted on 10-6-2011 at 22:07


What about using triethyl borate to react with the ethoxide, removing it from the equilibrium as the Metal Tetraethoxyborate (which when heated gives the Metal Ethoxide and the triethyl borate back)? The solubility side of it would be interesting, but the triethyl borate would give the sodium borate (whichever one) without producing any water. The azeotrope could still be removed (along with all the liquid), while I'm sure the tetraethyl borate ester could be separated from the borate salt using a solvent.



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[*] posted on 28-6-2011 at 11:53


I made today sodium methoxide. Pure Na metal (25 g) slowly dissolved in 350 - 400 ml methanol , to even 5% water (ethanol 95%) catch a fire is almost impossible (almost ;)). The first 3-4 portion of sodium didn't heat a lot of whole mixture but still i put baker in cold water (cooled very good) , at about 15 g sodium and more the whole mixture turned a slightly yellow and at the end just more yellowish colour , liquid had much more density (look like sirup). I put the baker on heating mantle and start to boil off solvent , after evaporate about 100 ml the white mass start to floating on top of surface liquid, after evaporating the whole solution there was in a baker hard rock mass, i quick turn these "rocks" into dust, and i got about 50 g of pure, anhydrous, white sodium methoxide. Very nice synthesis but some kind of time consuming.

PS. "dust" mask and googles and of course gloves in key safety gear in this synthesis , dyust of Ch3ONa is very unpleasant

[Edited on 28-6-2011 by mario840]
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[*] posted on 28-6-2011 at 17:35


If there was 5% H2O in your ethanol, then your methoxide is going to contain quite a bit of hydroxide.. Use anhydrous ethanol.

So all this talk about CaO has reminded me of an idea I had a while ago.
Couldn't you get around the whole problem of dehydrating/decomposing salts to acquire oxides by reacting the metal directly with methanol? Of course Ca metal isn't too easy to get, but Mg is. And the reaction of Mg with MeOH is extremely easy. If the MeOH is halfway dry the reaction starts on its own. At least it did for me.
Couldn't you just mix an H2O contaminated NaOEt solution in EtOH with excess Mg(OMe)2? Its a fact that it will quantitatively react to form Mg(OH)2. I would guess it the reaction would go to further completion much faster than using CaO since Mg(OMe)2 is slightly soluble in EtOH. A small amount of Mg(OMe)2 may be left in solution, but that won't matter for most applications.
I suppose you could add the calculated amount of Mg(OMe)2 directly to a solution of NaOH in an alcohol.. But dealing with a large amount of Mg(OH)2 precipitate may be problematic. I'd azeotrope the majority of the H2O out of the NaOH alcohol solution then mop up the remainder that's tough to get out by azeotroping with Mg(OMe)2.
Does this sound reasonable? Am I missing something?
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[*] posted on 28-6-2011 at 22:36


I know but read more carefully, i wrote that alcohols contaminated with 5% water are still safe to throw sodium into it without risk catch a fire, of course anhydrous ethanol is expensive compare to 95%, but i dissoled 25 g (massive amount) sodium for long time 1,5 hours or so , in some point sollution became more dense and sodium just floating around and react very slowly, but when we add fresh methanol reaction starts again
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[*] posted on 1-7-2011 at 01:09


But that 5% water didn't just disappear.. It doesn't matter how much excess Na you throw in, you will still end up with a mixture of NaOH and NaOEt.. Depending on what you're doing with it, this may not be a big problem, I don't know.. But there are definitely some reactions where the NaOH present would cause poor or no yields.. It's not that tough the get the 5% H2O out before you add the sodium, I would recommend doing so.
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[*] posted on 6-3-2012 at 03:59


Here's an idea from left-field

Add some aluminium -

I'm thinking that the NaOH will react with any oxide/hydroxide to give the sodium aluminate IN the presence of water, while any water will react with any clean aluminium to give the aluminium hydroxide, while the ethanol would react with clean aluminium to give the aluminium hydroxide via metathesis of the formed alkoxide.

Just a thought, there's presumably any number of problems with it, just thinking and typing while I'm looking up something else in another tab;)




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[*] posted on 6-12-2013 at 11:04


I read all of this topic and didnt find out :does any one has success in making ethoxide (pure) by this method?

I want to make Potassium ethoxide but i didnt find complete instruction for all steps.
Has anyone tried this method successfully?
What is best ratio?what will happen for excess KOH?how we can separate KOH from KOEt?i have any chance for making pure Potassium ethoxide by this method?


[Edited on 6-12-2013 by Waffles SS]
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