Sciencemadness Discussion Board

Sodium Ethoxide and anhydrous EtOH

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Organikum - 9-10-2004 at 03:36

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.

A great procedure!

Mephisto - 9-10-2004 at 06:57

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?

Marvin - 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.

Organikum - 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- This I cannot answer, not very much if the ethoxide/alcohol is withdrawn from top and not at the bottom.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]

vulture - 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.

Mendeleev - 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]

Marvin - 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?

chemoleo - 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]

Oxydro - 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]

Organikum - 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]

Tacho - 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.

I’m sane.
Well, I think I’m sane.
No. I’m sure I’m sane... pretty sure... sometimes though... maybe that mercury... I mean, how can you really, really know... Is there a test in the internet or something? I once did a test for my IQ in the internet, maybe there is a sanity test!
Most people think I’m sane... They tell me so. They hold my head with both hands, look into my eyes and say slowly "Tacho, listen, you...are... sane!"...
Sometimes though...

Rosco Bodine - 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 .

a123x - 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.

trilobite - 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]

sparkgap - 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)

sparky (~_~)

trilobite - 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]

sparkgap - 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. :)

sparky (~_~)

alkoxide exchange

unionised - 8-8-2005 at 11:07

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.

chemoleo - 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.

neutrino - 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.

S.C. Wack - 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.

praseodym - 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.

unionised - 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.

praseodym - 9-8-2005 at 22:33

Do you have Na metal? Because to get NaOEt, you can react Na with EtOH.

bio2 - 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?

unionised - 10-8-2005 at 13:08

Praseodym,
If you had read my post you would know that I don't have sodium, that's why I'm playing this game.
Also, if you read the title of the thread, you would realise that the solubillity of metal alkoxides in water is of rather limited relevance.

bio2 - 10-8-2005 at 14:35

.......solubillity of metal alkoxides in water is of rather limited relevance............

I'll say; the shit decomposes back to the starting alcohol.

With Al methoxide it is not instant but can be observed occuring over a few minutes at test tube scale. Solubility is said for PhMe, benzene etc. but it ain't much. Even at the boiling point of toluene very little of the methoxide dissolves.

sodium ethoxide//metallic sodium

Dr.Muto - 3-3-2006 at 08:15

Thanks 2 all 4 all the comentary on this topic. Not as positive as I would have liked, but much better than I thought it would be. Mostly sounds like a viable pain in the ass. Viable none the less. ......................................................................................................... Any tricks to the electrolysis of Naoh anyone can enlighten me to?

neutrino - 3-3-2006 at 14:02

There are a number of threads about sodium production in technochemistry.

Rosco Bodine - 31-8-2006 at 15:47

Hmmmm , I keep having this thought that *possibly*
simply dissolving NaOH in ethanol and subjecting
to electrolysis should produce sodium ethoxide .
At first , all that would be occuring is elimination of any
water which is present , but after that , the product
should be sodium ethoxide content increasing as NaOH content decreases in the alcohol , with H2 evolving at the cathode and O2 at the anode .

A simple cell design like a jar having a plastic lid with
two parallel carbon rods , one for the cathode and one for the anode , is possibly all that would be required .

The cell resistance could be high so it might lead to needing a cooled cell or reflux condenser to prevent
excessive evaporation of the alcohol . Compared with
the simplicity of the method which Organikum has
posted here , there seems no advantage to an electrolytic
method , though it may be a valid alternative .

Anyway , attached here is the patent mentioned by Organikum at the beginning of the thread .

[Edited on 1-9-2006 by Rosco Bodine]

Attachment: US2796443 Sodium Ethoxide.pdf (279kB)
This file has been downloaded 1493 times


Marvin - 31-8-2006 at 21:01

As much as I hate to be a stickler for scientific accuracy. Hmm, thats a lie, it is my reason d'entre, but lets pretend I do for the moment....

"Compared with the simplicity of the method which Organikum has
posted here"

Which there is no evidence works for alkoxide. The description is heavily simplified from the patent, which is counter current based.

I suspect electrolysis would probably not yeild oxygen, but rather oxidise the ethanol to produce aldehyde, which would then condense in the strong base to produce more water and unhelpful products.

Rosco Bodine - 31-8-2006 at 22:38

Yes it is likely that Organikum simply cannot be trusted :P

However , setting aside the political credibility issue in the interest of science :D , ........

After all *if* it is a simple solubility driven equilibrium reaction as the patent states , the reaction should proceed to the right according to the greater affinity of the solid hydroxide for water . The hydroxide in excess wins the battle for the water . If this holds true , then the counter current scheme is simply a means of continuous production convenient on an industrial scale , as an efficent method , and keeping the apparatus compact in size . But for a laboratory method , the same reaction should be accomplished simply by long stirring of an excess of what NaOH will be dissolved in
ethanol heated to near its boiling point . Over time the
same equilibrium will be reached . The liquid portions
should separate on standing and the upper layer of
ethoxide in alcohol could be pipetted or decanted ,
or the whole of the liquids decanted from the solids
into a separatory funnel . Perhaps some sort of color indicator could be used to distinguish the phases if needed , but I am going to guess that the ethoxide layer will likely have a slightly yellowish cast and the interface
should be visible .

Dow is a huge chemical company , so it seems unlikely that one of their patents would be total bullshit .

The experiment to examine this is as simple as I have described , and I will get around to it myself unless
someone else beats me to it first .

As for the electrolysis oxidizing the alcohol , perhaps
and that may depend upon the anode material alone ,
or it may require a partitioned cell ....I do not know ,
and I would bet good money that you don't know either .

Until experimental results are found , then my theory , hypothesis , guess or speculation , is as good as yours
and so is Organikums .

Azeotropic removal of water from a non-excess of NaOH in ethanol by using a benzene additive to the mixture to create a ternary system of ethanol-benzene-water , which distills away and from which the water can be separated in a Barrett water trap , leads to a solution of sodium ethoxide in the dried ethanol in the distillation flask . See the patent attached . This is simply another way of shifting the same
reaction equilibrium to the right , correct ?

In another thread someone mentioned having experimented with toluene as an alternative to benzene
described in an even older patent .

Attachment: GB377631 Sodium Ethoxide via Azeotroping.pdf (169kB)
This file has been downloaded 1442 times


not_important - 1-9-2006 at 00:36

The azeotropic method is used commercially. Petroleum ether can be used instead of benzene, even though it's a mix you'll still get an azeotrope effect.

The earlier patent depends of the counterflow to drive the reaction to completion. As the solution moves up the column, there's less and less water for the right hand side of the equilibrium. To do the same thing in a beaker or flask, you would need to repeatedly remove the aqueous layer to change the conditions, or dump in a huge amount of hydroxide. The patent explicitly talks about batch mode, where no replacement hydroxide is added. The system is almost like a reactive column, giving products that seem out of equilibrium with the expected result from the ratios of reactants.

An alternative to third component azeotropic distillation is to dry the water-alcohol azeotrope vapour with cornmeal, then condense the dry alcohol for return to the reaction vessel :
http://journeytoforever.org/biofuel_library/ethanol_grits.ht...

I had the same idea as you regarding electrolysis of a hydroxide-alcohol mix some time ago, and tried it. Started with 95% ethanol, the current did drop off as the water was removed and eventually effectively stopped; the power supply I was using couldn't put out much voltage. There were some non-alcohol oders, discolouring of the solution, and gunk on the electrode. Given that alcohols can be oxidised by air/O2 in the presence of strong base, I believe that was what was happening. However it did not appear to be the primary reaction path. The concept may be practical using a low current density and anode with low oxygen overvoltage. Even if not practica on its own, it might work to remove the last of the water from the hydroxide column method.

For real production one could possibly use pervaporation to pull the remaining water and some alcohol out of the alcohol-alcoholate, run that takeoff through a second higher temperature pervaporater to concentrate the water phase, and use hot xylene to azeotrope off the water from the water-hydroxide phase to recover the hydroxide for reuse. Larger scale than a lab needs, but a nice system. Like reactive columns, it's one of those processes that are on the edge of being too big for lab use.

Rosco Bodine - 1-9-2006 at 06:09

*If * the NaOH formed a series of solid crystalline hydrates having higher and higher vapor pressures of water , instead of simply dissolving directly and separating as an aqueous phase descending , then I could better believe in that idea that the reaction is
dependant on a progressing reaction zone which is dependant upon a more and more anhydrous condition higher and higher in the column being supplied from the bottom . But the way I see this reaction is that it is a *simple equilibrium* between saturated aqueous NaOH
and the alcohol solution of sodium ethoxide .

The reaction column is only supplied from the bottom in order to provide for gravity separation of the reaction products in the reaction vessel , to avoid requirement of a second vessel for performing the separation of the phases , and allow for the process to be operated continuously ......it is a matter of mechanical design for
convenience of manipulation of materials and economy , not for necessity of the reaction conditions . The reaction should proceed to the exact same equilibrium with everything mixed together in a single reaction zone ,
since there is no chemical difference at the surface of the solid hydroxide anywhere in the column , other than the thickness of the adherent * film * layer of saturated
* aqueous * solution on the pellets or granules of sodium hydroxide .

What seems to be the debate here is semantics about
whether the patents column of hydroxide is a counter current reactor in the classical sense ( which it is not ) ,
from my impression of it anyway , it is a simple reactor
separator . If you consider the case where the pellets
of hydroxide used would be the potassium salt , it is
already the case that perhaps 12-15% water is present
in the fresh pellets which would be unreacted material
at the top of the column at the very start , because of
the virtual impossibility of securing the potassium hydroxide as the anhydrous material . The tenacity of
the fused hydroxide for that much water content even
at fusion temperatures prevents its further dehydration
in common manufacture ......so it isn't a matter of having
any requirement for the absolutely anhydrous material to be at the top of the column , because that condition would not be the case anyway . Therefore , is further
evidence that the equilibrium is related to physical
* film chemistry * of the aqueous layer of hydroxide
accumulation on the solid particles of hydroxide , which
grows in thickness and coalesces into droplets which
precipitate , settling out at the bottom .

It is a simple " salting out " driven equilibrium analogous to what would happen happen if 70% isopropanol was
shaken together with an excess of solid sodium chloride ,
which would result in a phase separation , having something 92% or higher isopropanol in a dehydrated
upper phase , where the salt has won the battle for the water and separated as a brine in the lower phase .
This could also be be done in a column like the patent ,
but the same equilibrium would be reached , and the same result on separation .

[Edited on 1-9-2006 by Rosco Bodine]

Rosco Bodine - 1-9-2006 at 12:08

We have here Organikum and Rosco sitting side by side
on the trick board over the dunk tank , mercilessly ridiculing the crowd of novice baseball pitchers .....

Todays " two for the price of one special " :D
Could it be they will stay dry , or is it bath time
for the reaction dynamic duo ?

Okay all you physical chemistry apprentices :P
it's three balls for a dollar , so step right up
and take your best shot :P:D:P:D:P:P:P:P:P

Marvin - 2-9-2006 at 15:57

This is why there is so much shit on the internet. Someone takes a possibly good method they don't completely understand and takes out the parts they don't think are needed to make it easier. I'm not blaming Organikum, its more likley to have been a hive modification anyway, but I will certainly blame him for passing on an incomplete method as 'Tried and true' which he later climbs down from as far as "I thought I read somebody doing it with success" for production of methoxide. There is also a gap in the claims made by the patent for the ethanol/ethoxide varient that suggest a problem, either of loss of ethanol in the effluent at the base of the column, a higher than it would be advisable to brag about level of hydroxide in the product, or both, and this is when the patent method is followed to the letter.

"Yes it is likely that Organikum simply cannot be trusted"

Organikum has said "I want to point out that I posted the method mainly for making anhydrous alcohol, not ethoxide".
This could well be successful.

He has also said "The working principle is solubility, density and salting out - counterflow makes it continous,"
This is WRONG and contradicts the explanation in the patent.

You have said "After all *if* it is a simple solubility driven equilibrium reaction as the patent states"
Show me where.

"The hydroxide in excess wins the battle for the water . If this holds true , then the counter current scheme is simply a means of continuous production"
This does not hold true for any reasonable excess, and the patent explanation is clear on why the counter current system is not just a means of continuous production but integral to the invention. Looking at the crude batch interpretation at the top of the thread, you have a single equilibrium between the water, alcohol, alkoxide and hydroxide, and the hope this will be enough to drive it mostly to alkoxide. Even in the batch method of the patent it is still a counter current method and is not the same. While the drawing suggests a tall vat, it isn't, its a column and this is made very clear in the text.

Lets take 2 mins to think about the properties of, for example freshly fused KOH as a dehydrating agent. Initially it's very powerful, that is to say it has one of the lowest partial pressures of water of anything in common use, but its capacity at this level is very low. As it has to absorb more and more water the partial pressure rises sharply. As it dissolves in its own water this is a fairly high capacity drying agent, but it is not very powerful. The dilemma, is that in drying a substance you get these properties in the wrong order. The initial powerful drying action occurs when the compound being dried is very wet, and a high capacity drying agent is needed to remove the bulk of the water, while the last stages of drying, when the most powerful drying action is needed occurs when the drying agent is saturated with all the water previously soaked up. The question is, can you reverse this order, to achieve an ultimate water content close to the fused KOH point, and yet still use the full capacity it ends up dissolving in the water.

Quoting from the patent "In operation the reaction equilibrium will be progressivly shifted throughout the column'. This is why it works. The hydroxide and alcohol are put through the column in counter current. The equilbrium at the bottom is wet/liquid hydroxide contacting the alcohol as added and ending up mainly as a dried alcohol/hydroxide and a seperate layer of mainly hydroxide and water.

As you move higher up the column, the rising hydroxide/alcohol solution meets progressivly drier falling solid hydroxide, water is removed and the equilibrium between hydroxide in solution and alkoxide is shifted a little bit more towards alkoxide.

At the top you have a solution of hydroxide and alkoxide in anhydrous alcohol contacting solid anhydrous hydroxide freshly as it is added, driving the conversion to alkoxide as far as it can go, before the liquid at the top is removed as the product.

The very powerful drying action of anhydrous hydroxide is being used where it is needed, at the top to shift the equilibrium as far as possible towards alkoxide. As it moves down the column it progressivly absorbs more water and becomes a weaker dehydrating agent. This is not a problem as it is constantly contacting rising alcohol/hydroxide/alkoxide/water equilibriums with less alkoxide and more water that is easier to remove, until at the bottom at its weakest it is in contact with fresh alcohol being added. If the level of water at the point is enough to alow the hydroxide to form a seperate layer it will and the effluent is a solution containing mainly hydroxide and water with hopefully only a small amount of alcohol.

As another example of a counter current method, and the difference in the results by using it, production of liquid air by venturi expansion (Regenerative process). Depending on the degree of compression, with no counter current you end up with 5 or maybe as much as 20 degrees cooling which is produced by the nozzel and when the aperatus has cooled to this temperature it is spat out the exhaust. Add counter current exchange between the incoming compressed air and the exhaust however and you get an exhaust neer inlet temperature, and the core goes down as much as 300 degrees below inlet at which point the air liquifies. Counter current is a way to push things up the entropy slope. 20 degrees without, 300 degrees with. The energy of course is still the same, but the degree of completion is so much higher.

"Dow is a huge chemical company , so it seems unlikely that one of their patents would be total bullshit ."

While that logic is flawed, it certainly helps if you read and understand the patent.

You can taunt the novice pitchers all you like, but while new to baseball some may turn out to be already rather good at cricket.

Rosco Bodine - 2-9-2006 at 21:53

You are basically declaring that the hydroxide itself
has different levels of hydration according to its height
in the column , wetter at the bottom and drier at the top ,
when this simply is not true .

Only the ethanol with its dissolved components is
wetter at the bottom and drier at the top due to time
of contact with the dehydrating agent , the composition of which is constant , except for the thickness of the aqueous film on its surface .....and the thickness of
that film hasn't any bearing whatsoever on its vapor pressure , but only upon the surface tension which
when exceeded by gravity allows that film to coalesce into droplets .

The thinner film of aqueous hydroxide solution on the hydroxide pellets at the top of the column has no greater dehydrating property than the thicker film of hydroxide
solution on the pellets low in the column , which is actually coalescing and sloughing off .

The dehydrating power of the pellets is *constant* regardless of their height in the column , it is only
the ethanol solution of hydroxide itself that is being
gradually dehydrated over time of passage through
the dessicant .

The hydroxide dessicant in this case is not functioning
in the manner of of a dehydrated salt which would
be absorbing water to arrive at a specified hydration
state for " crystal water " being held in a solid . Fused
pellets of hydroxide are not like sponges , not capable
of physically trapping water , but their dessicant property
is due to a surface reaction of deliquescence . The
affinty for water is to form a saturated solution , which
has a uniform vapor pressure and a constant water
content . So great is the affinity for water that the
solid hydroxide will attract even water vapor from gases ,
or salt out the water from solvents to form an aqueous brine which separates as it does in this case .

This desssicant system is based upon the vapor pressure
of the saturated aqueous solution of hydroxide in contact
with undissolved solid hydroxide which maintains the saturated condition of that liquid solution , whether it
is a pool at the bottom of the column , or a film of varying thickness residing on the surface of the solid hydroxide ,
the vapor pressure remaining the same , and the solid hydroxide dissolving as needed to regulate that vapor pressure .

The alcohol phase is simply stripped of more and more water from time of contact with that dessicant system .

Heat is what makes this thing go , like most reactions .
At 70C , the deliquescence is augmented greatly over what
it is at room temperature , in contact with solution water
available in a solvent like alcohol , because the already great solubility of NaOH in water is driven even higher , and proportionally the NaOH has an even greater affinity for
the water , than for the alcohol , as the solubility curves should confirm .

[Edited on 3-9-2006 by Rosco Bodine]

Nicodem - 2-9-2006 at 23:33

There is a lot of confusion regarding this patent, a confusion which was mainly created from wild speculation reasonably originating from the distrust that sodium or potassium ethoxide and their higher alkoxydes can be prepared so simply when in every lab in the world they are either acquired or prepared from sodium and potassium metal.

I have not followed the whole thread, only its beginnings and even that I mostly forgot, so I apologize if I declare something that was thoroughly discussed and is not an issue anymore. My goal is only to bring back some confidence in the patent. Otherwise the beginners will be repulsed from using it and remain without alkoxydes for the rest of their lives. I personally don’t care for myself as I have reagent grade NaOEt, but I do know how it fells being without it when so many reactions call for this very useful reagent.

So I though to only add these observations and claims to the discussion:

[Edited on 3-9-2006 by Nicodem]

Rosco Bodine - 3-9-2006 at 00:26

So what about Rosco's modification ?

The patent you mentioned is attached and bears out what I have been saying .
Thank You :D

Just to be clear , this reaction should proceed towards the right without any counter current scheme
at all , simply by having * solid * NaOH always present
in excess in the reaction mixture maintained for awhile near the bp of the ethanol .

When the hot mixture reaches equilibrium , the alcohol phase will contain predominately the ethoxide as its dissolved sodium compound .

[Edited on 3-9-2006 by Rosco Bodine]

Attachment: US1978647 Acetone Precipitation of Sodium Ethoxide.pdf (201kB)
This file has been downloaded 1456 times


Marvin - 4-9-2006 at 01:56

"After all *if* it is a simple solubility driven equilibrium reaction as the patent states"

I am guessing that since you did not substantiate this in either of the two posts following my request that it is simply untrue. :P

"The thinner film of aqueous hydroxide solution on the hydroxide pellets at the top of the column has no greater dehydrating property than the...."

I would agree that I also do not consider mechanical reasons to be worth arguing for the performance of the column, or kinetic ones. If reaction rate is the issue a home chemist can always wait longer.

I would state that such possible effects as K2O produced by heating the KOH isn't worth thinking about, to all real intents and purposes it doesn't happen.

I would insist that KOH and NaOH as powders have surface absorbtion properties that are very strong, but I concede that these effects are probably not important on the molar scales in the context of this thread and would not by themselves prove my argument.

"their desiccant property is due to a surface reaction of deliquescence . The affinity for water is to form a saturated solution , which has a uniform vapor pressure "

You are stating that the deliquescence is the basis of the dehydrating power, and that while solid KOH (for example) is in contact with the resulting saturated solution a change in the overhead vapour pressure of water would cause more or less KOH to dissolve until equilibrium is again reached. Since the composition of the solid, and that of the solution don't change only their amounts do, the vapour pressure of water must be constant while both exist. I can't fault the logic. The bad news is that the chemistry is wrong. At RTP solid anhydrous KOH will absorb water, neglecting surface kinetic effects it will continue to do so until it is entirely converted into the monohydrate, which is solid. KOH.H2O will continue to absorb water with slightly less enthusiasm until it is converted entirely to the dihydrate which is...... solid. KOH.2H2O will continue to absorb water with slightly less enthusiasm still until entirely converted to KOH.4H2O which is a liquid. I say liquid and not solution, because it really is a molten hydrated salt, if you cool it you get KOH.4H2O as crystals. The behavior of the solution then gets more complicated, but is outside the required scope for this thread and beyond any reference material in my home library anyway. Sodium, rubidium and Caesium hydroxide have at least 1 solid hydrate each at RTP in addition to the solid 'anhydrous' forms. I use the term with care as in very old textbooks the anhydrous XOH are themselves considered to be hydrates of X2O, so to be clear an XOH and at least one XOH.nH2O both solid at RTP. In the case of sodium there are in total about a half dozen individually distinct hydrates mostly liquid at RTP. Returning to KOH and the patent, at the temperature of the column the anhydrous form and monohydrated forms are solid, the dihydrate and tetrahydrate are miscible liquids, being a changing mixture the dehydrating power is continuously variable depending on its composition. So depending on the point in the column you are looking at you have anhydrous/monohydrate desiccant or monohydrate (solid)/dihydrate (liquid eventually, there is no requirement for a sharp change in melting point) or a liquid consisting of varied proportions of the dihydrate and the tetrahydrate, and with a desiccating power that depends on the ratio in the active phase.

"You are basically declaring that the hydroxide itself has different levels of hydration according to its height in the column , wetter at the bottom and drier at the top "

Yes, that is exactly what I'm stating.


Nicodem,

There is no one here that is challenging the claims in the patent, which renders most of your post a cheerleading chant for a nonexistent cause. The question in the thread is on exactly how the process in the patent works, and if the modification proposed at the top of this thread is equivalent. You do also provide a number of important facts.

"Conclusion: Ethanolic solution of NaOH contain a lot of ethoxide ions in relation with the hydroxide ions."

This is certainly true, and a dilute solution of sodium hydroxide in absolute ethanol can be nearly 100% ethoxide. Ethoxide is the stronger base though, and simply adding an acid, ethanol, to a base, sodium hydroxide, cannot result in a solution that is more strongly basic unless something is doing work. These reactions must be pushed up hill. Of course required purity and solvent is strongly dependant on what the ethoxide is going to be used for. A second point, is that with hydroxide and ethoxide (and water) in the same mix, if the hydroxide is kinetically better than the ethoxide, even a small amount may do a disproportionate amount of damage, as the hydroxide is used up more will be created as ethoxide is destroyed by water to maintain the equilibrium. While your conclusion can well be true, on its own it doesn't mean anything.

"The prejudice that sodium ethoxyde and similar alkoxydes are exceedingly difficult to prepare by a hobby chemist ...."
"Conclusion: Producing alkoxydes higher than methoxyde, without resorting to sodium or potassium metal, is no big deal."

Your conclusion does not follow by any logical or chemical means. This is just rhetoric.

"The H2O in NaOH/KOH alcoholic solutions is not mysteriously bound, only its concentration is bound... "

This may be a useful conclusion, had someone actually assumed this. No one has.

"Conclusion: Organikum’s modification can only be considered faulty as a method of preparing truly and absolutely dry ethanol, but as a method of NaOEt production there is no reason to believe it does not work"

I find it interesting that Organikum's method as an absolute alcohol process is flawed, this passed me by completely, much thanks for bringing the Vogel information to our attention. The second part to your conclusion does not follow logically unless two conditions are met, firstly that the distillation is done slowly with a fractionating column, when azeotropic distillation would dehydrate the remaining hydroxide/ethoxide mixture, a simple or rapid distillation could well fail to further conversion. Secondly it relies on the degree of dehydration of the alcohol used to exceed the water produced by the conversion of hydroxide to ethoxide. Otherwise alcohol will be exhausted as azeotrope before all the hydroxide is gone. Since the alcohol Organikum is using is wet, and no directions on distilling are given neither of these conditions are automatically met and there is considerable question as to how successful it is as a complete method. With modifications though, it could well be made to work with more complicated means and equipment. If the amount of dehydration of the wet alcohol below azeotropic by the sodium hydroxide removed exceeds the amount required to dehydrate the sodium hydroxide that remains dissolved in it, then following fractionation substantial conversion to ethoxide should occur. There are a lot of variables though and the conditions are a long way from being automatically satisfied.

In other words there is every reason to believe it does not simply work as stated.

In addition to previously discussed trinary azeotropic methods, sodium alkoxides can be produced using magnesium metal as detailed elsewhere and in Fieser as a method for making very dry alcohol (Magnesium hydroxide being insoluble in the mixture, the ethanol is decanted and distilled off from remaining magnesium ethoxide).

The acetone method is new to me, and I find it suspect. Not because the production of sodium ethoxide by precipitation seems so unusual, but because acetone is well known for being incompatible with strong bases.



Rosco,

"At 70C , the deliquescence is augmented greatly over what it is at room temperature , in contact with solution water "

Having shown the mechanism of dehydration to be different to that assumed this no longer makes sense either. While hydroxide will be more soluble in the water the real question is the vapour pressure over a hydrates salt, which generally increases somewhat with increasing temperature. Heat will certainly reach equilibrium faster, but it may turn out to be better to let the system cool after and crystalise more of the hydroxide as a hydrate.

"Just to be clear , this reaction should proceed towards the right without any counter current scheme at all , simply by having * solid * NaOH always present in excess in the reaction mixture maintained for awhile near the bp of the ethanol "

There are no claims in the patent regarding sodium ethoxide, so just the use of sodium hydroxide and ethanol in the full process may produce unacceptable results and stretches the patent as evidence beyond its breaking point. At best had the yield been higher they would have listed this instead. At worst it does nothing useful at all. This is why I've concentrated my examples on KOH, which the patent can be considered fair evidence for success in the column.

I think it is now clear why a counter current is so important, as put forward and explained by not_important and myself.

Modifying the method may be feasible, much larger excess of hydroxide, fractional distillation of the remaining ethanol, but this is something that would have to be done experimentally. Its not about the potential of a method to do well by our limited understanding of the chemistry, it's about evidence and reliability.

Nicodem - 4-9-2006 at 03:31

Quote:
Originally posted by Marvin
Nicodem,

There is no one here that is challenging the claims in the patent, which renders most of your post a cheerleading chant for a nonexistent cause.

But that is exactly what I wanted it to be. :)
I believe there are still a bunch of hobby chemists who question the patent and there are still many others who don’t even realize that producing NaOEt from NaOH is possible. I did not want to intrude in the “Marvin vs. Rosco debate” about the mechanism behind the NaOEt formation by that process, since it is obvious to me that it is a dehydration process where the ethoxyde/hydroxide ratio exponentially raises with the distance up the column. This is already evident from the equilibrium constant equation and the partition theory of extractions. Furthermore, since the increase of the ratio is exponential with an already good starting point, I would say that one or two Organikum’s partitioning of H2O among conc. NaOH(aq) and conc. NaOH/NaOEt(EtOH) gives a relatively satisfactory resulting purity for the NaOEt produced after distilling off the solvent. Its purity should be enough for most uses (meaning >90%).
As for the other details of the current debate, I’m not really interested enough nor am I competent enough, but I see it mostly as another annoyance that may deter experimenters to use this method or its variants. If you two are working together to find out an optimization then that is just great, but unfortunately it does not look like this at all.
Quote:
The acetone method is new to me, and I find it suspect. Not because the production of sodium ethoxide by precipitation seems so unusual, but because acetone is well known for being incompatible with strong bases.

But at least it was verified by me and another Hive member that I currently don’t remember the name. The link to The Hive at the beginning of this thread points to the particular discussion on NaOEt where this and many other useful things were mentioned.
Acetone slowly self condense to diacetone alcohol in the presence of ethoxyde or hydroxide. This is however not very relevant as the exposure of acetone lasts only a couple of minutes, the time it takes to filter the precipitate which must be washed with a volatile aprotic non polar solvent (petroleum ether, ether, toluene etc.) to remove acetone remains and any side products thereof. This is explicitly mentioned in the patent. Higher condensation products from acetone exposed to strong bases only occur after prolonged exposure or heating which is irrelevant to this method.

Rosco Bodine - 4-9-2006 at 09:33

The patent states that the operative requirement for
the amount of excess alkali hydroxide , is a minimum of
0.1 mole to a maximum of 0.4 mole beyond the theoretical
amount required for the alkali which will be bound chemically in the ethoxide . Having less than 10% molar
excess of alkali hydroxide , the equilibrium for the reaction
will drift back towards the left . This small amount of
10% excess beyond the theoretical requirement is very pertinent to my " making the leap " in proposing that
this reaction shouldn't require any column at all , and
should proceed in a similar way as a batch process
although the minimum excess of theory might be tripled
without a column , to keep the time of reaction reasonably short .

Concerning the hydrates of alkali hydroxides which you have mentioned , I had already thought about the possibility of a series of such possible hydrates which are formed from evaporation from hot melts being concentrated , where in *liquid* form there well may
be a stepwise dehydration marking distinct hydrates ,
which have identifiable melting/solidification points as
distinct hydrates . But those distinct stages of hydration
would only come into consideration for the entire mass in
the molten condition , and have no bearing for the dessicant properties of the material after it has cooled and solidified . After solidification , the absorptive properties of the *solid* are virtually gone with regards to any practical reaction rates and the solid does not
behave the same as would say for example a liquid
hydrate of sulfuric acid , which as a liquid is free to
seek different levels of hydration for the mass ,
as if swelling like a sponge . When you have the
hydration status of a melt * locked * in place upon
solidification and cooling which effectively seals the
interior of the solid from exposure to external moisture ,
then its dessicant properties is not absorptive en masse
but is shifted by the physical limitation of the solid to
become a surface reaction manifested by deliquesence ,
forming an aqueous solution as a film wetting the surface
of a structure beneath which does not change , but
is like an ice cube sitting in liquid water so cold that
the ice cube does not melt further nor further freeze
until the properties of the liquid in contact is changed .

When handling solid prilled alkali hydroxides , you may notice that when the " dry " solid granular material
is exposed to the humidity of the air , during weighing operations for example , it does not gradually swell in volume as a still solid material passing through stages of higher and higher yet still solid hydrates and then sudddenly liquify en masse . But what does happen is an almost instant formation of a film of liquid on the surface of the prills which makes them adhere to each other .

Even if it is aknowledged that different stages of dehydration do occur in a melt which is being concentrated at higher and higher temperatures of fusion , this does not necessarily mean that the
reverse sequence of rehydration occurs in like stages
for the cooled and solidified melt exposed to moisture
upon its surface . It is not the same case or mechanism
as would be for a porous silica gel which physically
absorbs water . Nor is it the same case as for a
porous material like drierite , or for a liquid dessicant
like sulfuric acid . A deliquescent material operates
according to a distinctly different process .

Many of the deliquescent nitrates behave exactly the same way , calcium nitrate and magnesium nitrate are
two which come to mind .

Anyway , for the 10% molar excess to be the lower limit
for driving this reaction *completely * to the right , in a vertical columnar reaction zone having a 24:1 height to diameter ratio in ( inches ) for the reactant is a very mild
contrast of conditions , in my opinion , over what would be the effect in a " one pot " batch reaction , compensating for any absence of the reaction zone gradient which may be the benefit of the column simply by using a greater excess of solid alkali hydroxide in the one pot process .

As an integral part of my contemplated one pot variation , what I had fully intended was to add a fair amount of toluene to the ethanol solution over dissolving and excess undissolved solid hydroxide , *during* the boil , lettting the added effect of the azeotropic
dehydration augment the dessicant property of the alkali hydroxide , and shorten the time required as well as
produce an even more complete conversion .
Rosco's good old country recipe for sodium ethoxide :D
would marry the salting out effect of Organikums column reactor with the concurrent azeotropic removal of water applied as the " kicker " to eliminate the entire issue of debate concerning the column .

Anybody got a ternary mixture data chart handy for
toluene /ethanol/ water ?

Basically all you would need to do is place your weighed
quantity of NaOH prills and a stirbar into a stoppered flask with the measured amount of ethanol , bring it
up to a boil until you have a saturated solution over
the excess of undissolved solid NaOH . Allow the mixture to cool down a bit and add your toluene . Return the mixture to a boil and let it sit there boiling away like a whistling teakettle . There would probably be so little
of this added azeotropic removal of water needed that
on a lab scale it would be sacrificial solvent loss here ,
and little point in using a reflux condenser , water trap
and solvent return as would be used for economy with
large batches .

There could be worked out some optimum proportions
which would make this straightforward , and a standard
method .

The only point of uncertainty about this proposed method
is whether or not separation of the phases might have to be done before adding the toluene for the Azeotroping , and
if any fresh sodium hydroxide being added with the toluene
might also be beneficial to the yield from that point .


CRC lists the ternary azeotrope : bp 74.4
ethanol 37
toluene 51
water 12

The boiling point with decreasing water
should gradually rise towards the binary azeotrope: bp76.7
ethanol 68
toluene 32

Absent any toluene ,
The ethanol water binary azeotrope : bp 78.2
ethanol 95.6
water 4.4

And for pure ethanol the bp is 78.5 C .

It may be practical to follow the course of the reaction
simply by watching a thermometer observing for the cessation of gradual elevation in boiling point and the appearance of precipitated solid sodium ethoxide as the solvent is boiled away . At this point the product
is available as its saturated solution in the supernatant ethanol .

[Edited on 4-9-2006 by Rosco Bodine]

hinz - 16-10-2006 at 09:06

I've just tried to make aluminium isopropylate or Al isopropoxide and it works quite nice without CCl4 or CHCl3, only with HgCl2 and some iodine. I cooked it originally just for fun to get the reagent for the Meerwein-Ponndorf reaction, but it works that nice that it would be probalbly also possible to get sodium ethoxide from it, if you exchange isopropanol by ehanol. Like:

Al(OEt)3 + NaOH ==> NaOEt + Al(OH)3

AlOH3 is insolule, so the reaction is quantitive:cool:

For Al(OIpr)3:
I took Al foil and Al turnings put it in a 2l RBF and added 1,3l isopropanol and a few grams of I2. I heated it, but nothing happened, so I added a bit NaOH , I thought it could complex the oxide layer, but nothing happened too, so I took my last bit of HgCl2, made from a small mercury switch, it was something like 0,25-0,5g and added this and the reaction worked quite vigorous. I could place my hand on the top end of the condensor and felt the pressure of the hydrogen evolved.
The isopropanol was technical, 95% or so.

trilobite - 14-11-2007 at 19:33

After reading the thread alcoholic KOH & experiment suggestions I remembered that there were some reasons to revive this topic. Usually it is assumed that ethoxide is a stronger base than hydroxide. Needless to say, properties of the solvent affect the properties of a solute, so one should be careful when making conclusions based on measurements in one solvent only. I still managed to find the reference that says that ethoxide is the weaker base in ethanol, while the opposite is true in water. While the conclusion is based on extrapolation it makes a good argument. I didn't bother to get the article from the library and scan it, but luckily the abstract is like they used to be.

<b>The equilibrium between ethoxide and hydroxide ions in ethanol and in ethanol-water mixtures.</b>
Caldin, E. F.; Long, G.
Journal of the Chemical Society, , 3737-42 (1954).
CAN 49:18704 ISSN 0368-1769

<u>Abstract</u>
Equil. consts. (K') for the reaction, OEt- + H2O = EtOH + OH-, in EtOH and in EtOH-H2O mixts. were detd. by use of a colorimetric method. The indicator was 2,4,6-trinitrotoluene, which ionizes in EtOH in the presence of a strong base to give a purple soln. K' is defined as [OH-]aa/[OEt-]aw, where brackets indicate molar concns. and aw and aa are activities of H2O and alc. with respect to the pure liqs. K' varied between 1.0 and 3.1 for solns. contg. between 12.6 and 47.0% H2O. Extrapolation of the values gave K' = 0.5 for pure EtOH, in good agreement with the results of Hine and Hine (C.A. 47, 1473a) for iso-PrOH. It is concluded that solns. of alkali hydroxides in EtOH contain mostly OEt- ions rather than OH-. Thus, in 0.1M solns. of NaOH in 100 and 99% EtOH the total base present as OEt- is 99.2 and 95.9%, resp. Similar conclusions can be made for MeOH solns. Solns. made by dissolving Na in EtOH contg. a small amt. of H2O also contain mainly OEt- rather than OH- ions; only a small fraction of the H2O is removed by this means. The values of K' furnish an approx. measure of the relative acid dissocn. consts. of H2O and EtOH. In the solvent EtOH, H2O is a somewhat weaker acid than EtOH, whereas in the solvent H2O, EtOH is weaker.


The problem with using toluene azeotropes is, according to literature, that it takes a lot of time and a good column to drive the reaction to completion. For us the heat economy may not be such a big issue as for the industrialists but reliability is. You'd either have to analyze your product or have a really consistent method. Following water evolution isn't going to help much when you are distilling it with ethanol and toluene.

There are some situations where it is desirable to have as little water present as possible. For example, to avoid the use of diethyl malonate one can prepare organic acids by alkylating ethyl acetoacetate in the traditional way. The product is then boiled with NaOH in ethanol (called acid hydrolysis because the product is an acid). Much better yields are achieved if the reaction is performed with EtONa in ethanol, distilling off the ethyl acetate that forms in the reaction as ethanol azeotrope. The outcome pretty much depends on how little water you have in the mixture, and if I recall correctly, the problem is worse with single-alkylated than double-alkylated acetoacetates. To my knowledge the copper-catalyzed conversion of bromobenzenes to methoxybenzenes with MeONa doesn't like water either, but that's a different story as you cannot use EtONa in the same way, and I think you cannot use toluene for making the MeONa either.

Here's an abstract of a study of EtONa production with azeotropic distillation with benzene. I think the interesting parts are their method for estimating the completion of the reaction, also the boiling points and yield.

<b>The application of a fractional distillation method in the preparation of sodium ethoxide from caustic soda.</b>
Walker, T. Kennedy.
Journal of the Society of Chemical Industry, London, 40, 172-3T (1921).
CAN 15:19359 ISSN 0368-4075

<u>Abstract</u>
EtOH, H2O, and C6H6 form a const. boiling ternary mixt., and with a perfect stillhead any sample of moist EtOH could theoretically be dehydrated by distn. with C6H6 (Young, J. Chem. Soc. 81, 707(1902)). W. employed this method to remove the water from the system NaOH + EtOH <=> NaOEt + H2O. The chief difficulty in the sepn. lies in the fact that the b. p. of the ternary mixt. is only a few degrees lower than that of the binary mixt. of EtOH and C6H6 or of the pure substances; complete conversion of the NaOH cannot be expected since the effective proportion of volatile matter toward the end of the process is small. The proportion of EtONa formed was detd. as follows: 10 cc. of liquid were titrated with 0.1 N HCl to det. the total alkali (NaOH + EtONa). Another sample of 10 cc. was boiled with 12.5 cc. AcOEt and 50 cc. EtOH, when the following reaction took place: xNaOH + yEtOH + zH2O + wEtONa + uAcOEt = (x + z) AcONa + (w - z) EtONa + (u - x - z) AcOEt + (y +x + 2z) EtOH. The whole was then titrated with standard benzoic acid in EtOH. A diminution in alkalinity was thus observed as compared with the previous titration against 0.1 N alkali; this diminution in alkali gave (x + z), the max. quantity of H2O that would be obtained by evapg. 10 cc. of alk. fluid from the still to dryness without loss of water. The total alkali gives (x + w), while (w - z) was the minimum quantity of EtONa to be obtained by evapg. to dryness. The number 100 (w - z)/(x + w) is the "mol. percentage yield" of EtONa. Blanks indicated that the method gave results 4% low. The C6H6 used was purified, dried, and redistd. Anhydrous EtOH was prepd. from com. 96% alc, by dehydration with CaO and metallic Ca. The solns. before distn. were made up by mixing weighed quantities of EtOH, Na, and H2O. Preliminary expts., in which the distn. was carried out through a Vigreaux column 40 in. long and 0.75 in. in diam., gave a 25% yield, which by redistn. with fresh alc. and C6H6 was raised to 28.1%. By distn. through a Dufton stillhead (C. A. 13, 917) 1.5 m. long with an annular space 1.5 mm. wide, the yield was increased to 33.8%. About a 57% yield was obtained with a column 6 ft. high and 1.6 in. in diam., filled with 4000 Raschig rings of sheet iron; 40 g. NaOH (23 g. Na and 18 g. H2O), 1012 g. alc., and 600 g. C6H6 were distd. at 1 drop per sec.; 149 g.distd. over up to 66.55 Deg; the distn. was stopped at 68.2 Deg, leaving a flask residue of 785 cc. Analysis of this residue showed a 54.2% yield, while from the quantity of ternary mixt. obtained the value 61.2% was found.


Here are some patents on alkoxides and dehydration of alcohols, some of them are mentioned already in this thread. The dichloromethane-water azeotrope used to dehydrate allyl alcohol is often overlooked as OTC substitute for carbon tetrachloride.

GB304585 BuOH + NaOH -> BuONa
US1681600 ethylene glycol + NaOH -> Na-glycolyloxides
US1712830 EtONa by azeotropic distillation with benzene
US1816843 aka GB334388 EtOH + NaOH in paraffin oil
US1907834 drying of ethanol with Na-glycolyloxides
US1910331 aka GB377631 EtONa by extractive distillation with benzene
US2179059 dehydration of allyl alcohol by H2O-DCM azeotrope
US2278550 miscellaneous alkoxides
GB698282 and DE628023 MeONa by distillation of methanol with NaOH, recirculation of methanol

LSD25 - 17-2-2008 at 23:05

Quote:
Originally posted by Organikum
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.


Yup, and if someone who has access to 'Nature Journal' could get this letter and if possible to follow the research trail it starts, we may even have some proof to support Organikum's suggestion:

Caldin & Long, Equilibrium between Ethoxide and Hydroxide Ions in Ethanol, Letters to Nature, Nature 172, 583-584 (26 September 1953) | doi:10.1038/172583b0: http://www.nature.com/nature/journal/v172/n4378/pdf/172583b0...

It is not that I doubt Org, it is simply that I would like to know how this works and what sort of yield one could expect.

Reference Information

solo - 18-2-2008 at 00:20

The Equilibrium between Ethoxide and Hydroxide Ions in Ethanol
and in Ethanol- Water Mixtures.

E. F. CALDIN and G. LONG.
J. Chem. Soc, 1954. 4, pp.3737


Abstract
The equilibrium constant for the reaction
OEt- + H, OSEt OH + OH-
in ethanol-water at 25" has been determined. In mixtures containing a few
per cent. of water, and in pure ethanol, the equilibrium favours ethoxide ion.
It is deduced that solutions made by dissolving alkali hydroxides in ethanol
will contain mainly ethoxide ion rather than hydroxide. For instance, in a
0-lM-solution of sodium hydroxide in 99% ethanol, 96% of the total base is
ethoxide. Solutions made by dissolving sodium in ethanol or aqueous
ethanol are also considered. The relative acid dissociation constants of
ethanol and water, in (a) ethanol, and (b) water, are estimated; in each
solvent they differ by less than a power of 10.


Note: the Nature citation is also included here for comparison,

http://mihd.net/nkluf2

Attachment: Equilibrium between Ethoxide and Hydroxide Ions in Ethanol.pdf (1.2MB)
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LSD25 - 19-2-2008 at 13:00

That first article kicks shit out of the Nature letter.

Thanks Solo;)

Perhaps, given the importance of Sauron to this forum, it is time to shift this stuff over to WD? I mean, I ain't allowed to irritate the fuck and much of my contribution irritates it. QED, the majority of my contributions are not welcome on this site.

PS This fits in well with the suggestion made elsewhere too, don't it?

[Edited on 24-2-2008 by LSD25]

Siddy - 15-3-2008 at 20:35

Ok, so practically when 1/3rd of a sep funnel is filled with NaOH anhydrous pellet and then the rest of the fulled is filled with 95% EtOH (with the remaining 5% MeOH and Water), the HIGH majority of NaOH does not dissolve.

Is this correct? Or should the NaOH all dissolve?

Has anyone tested to see if the 3/4 that supposedly cotains EtOH, Na(+) and (-)OEt is actually water free?

The_Davster - 15-3-2008 at 20:39

Quote:
Originally posted by Siddy


Has anyone tested to see if the 3/4 that supposedly cotains EtOH, Na(+) and (-)OEt is actually water free?


Any water will react with ethoxide producing more hydroxide.
It is impossible for it to contain water.;)

Siddy - 16-3-2008 at 23:22

Quote:
Originally posted by The_Davster


Any water will react with ethoxide producing more hydroxide.
It is impossible for it to contain water.;)


Can that be used to test if it is water free?
Ie, if water is added to the proposed EtOH and Ethoxide, will a notable reaction take place? Will the pH lower?

Are there any other methods for testing? I thought of adding CaCl2 chips but they will dissolve in alcohol just as well as water...

Nicodem - 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!).

LSD25 - 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]

grind - 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.

LSD25 - 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?

LSD25 - 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)
This file has been downloaded 2208 times


Use sodium oxide

jarynth - 4-10-2008 at 13:51

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.

no1uno - 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.

not_important - 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.


ethylate

ballzofsteel - 31-3-2009 at 22:24

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.

unome - 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)?

Sciocrat - 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]

Anders Hoveland - 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]

flavaflav - 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.

Sciocrat - 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.

un0me2 - 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)
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Attachment: Irabien.Viguri.Ortiz.Thermal.Dehydration.of.Calcium.Hydroxide.2.Surface.Area.Evolution.pdf (1.4MB)
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[Edited on 7-8-2010 by un0me2]

un0me2 - 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)
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Al Dehyde - 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?

roamingnome - 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...


aliced25 - 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.

mario840 - 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]

497 - 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?

mario840 - 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

497 - 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.

aliced25 - 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;)

Waffles SS - 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]

blogfast25 - 6-12-2013 at 13:36

Well, I haven't read this whole thread but I do remember there was a bit of commotion in the organics section re. a patent for the preparation of simple Na or K alkoxides, by dissolving the relevant hydroxide into the relevant alcohol and precipitating the alkoxide using pure acetone as anti-solvent:

http://www.google.com/patents/US1978647

Some in the organics section claim this works well for Na/K meth/ethoxides. Worth digging a little deeper for those interested, I think...

[Edited on 6-12-2013 by blogfast25]

madcedar - 6-12-2013 at 18:56

There is a valuable post by Nicodem near the start of this very thread. Here is the link. Rosco Bodine has provided the patent you mentioned in the very next post after Nicodem's. It turns out the sodium ethoxide is really easy to make with nothing more than sodium hydroxide, ethanol and acetone.

[Edited on 7-12-2013 by madcedar]

blogfast25 - 7-12-2013 at 06:54

Quote: Originally posted by madcedar  
There is a valuable post by Nicodem near the start of this very thread. Here is the link. [Edited on 7-12-2013 by madcedar]


That closes the circle, thanks!

AJKOER - 7-12-2013 at 10:00

Quote: Originally posted by aliced25  
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;)


I believe Iodine may accelerate the reaction with Aluminum to remove water from the Ethanol before adding NaOH. My take on a possible reaction chain:

2 Al + 3 I2 --> 2 AlI3

but, in the presence of water, a hydrolysis occurs:

2 AlI3 + 6 H2O <--> 2 Al(OH)3 + 6 HI (see http://en.wikipedia.org/wiki/Aluminium_iodide )

removing some water. Also, with the newly created HI:

2 Al + 6 HI --> 2 AlI3 + 3 H2 (g) (see Wiki same link)

or, for all three reactions the net reaction is:

4 Al + 6 H2O + 3 I2 ----> 2 Al(OH)3 (s) + 3 H2 (g) + 2 AlI3

and, upon adding 6 more H2O to consume the formed 2 AlI3, per the first equation above:

4 Al + 12 H2O + 3 I2 ---> 4 Al(OH)3 (s) + 3 H2 (g) + 6 HI

and adding 2 more Al to remove the 6 HI, again per the 3rd equation above:

6 Al + 12 H2O + 3 I2 ---> 4 Al(OH)3 (s) + 6 H2 (g) + 2 AlI3

Note, the process can be repeated, but the quantities '3 I2' and '2 AlI3' will remain constant while the amounts of Al, H2O and H2 increase. As such, the Iodine may accelerate the reaction with Aluminum although not technically acting as a catalyst (only a small amount is required to initiate the reaction regardless of the amount of water, and only that initial Iodine is transformed). This analysis can also suggest the amount of Aluminum to employ to remove any water presence from the alcohol. It is best not to have an excess of Aluminum metal, as upon heating, even a nearly dry Ethanol will start to attack Al forming the alkoxide (see, for example, http://adsabs.harvard.edu/abs/1979JChPh..71.1537E ). Also, note that when all the water is consumed, the chain breaks and the initial number of moles of Iodine added is now 2/3 as many moles of AlI3.

I recall reading about the use of HgCl2 to promote the reaction as well.
-------------------------------------

Now, your idea of using Aluminum and NaOH together would react as follows:

2 NaOH + 2 Al + 6 H2O --> 2 NaAl(OH)4 + 3 H2(g)

successfully removing water from the alcohol, but also introducing Sodium aluminate into the reaction.


[Edited on 7-12-2013 by AJKOER]

S.C. Wack - 7-12-2013 at 10:52

Quote: Originally posted by madcedar  
It turns out the sodium ethoxide is really easy to make with nothing more than sodium hydroxide, ethanol and acetone.


But don't count your yields before they're weighed.

blogfast25 - 7-12-2013 at 10:53

AJ:

It's not that simple.

Your second reaction, the hydrolysis, is an equilibrium reaction:

2 AlI3 + 6 H2O < --> 2 Al(OH)3 + 6 HI

Of AlI3 a hexahydrate can be prepared, indicating AlI3 and water can coexist to a degree. And the lower the concentration of AlI3, the less it hydrolyses.

In alkaline conditions, as 'aliced25' intended, it get even more complicated, because of the formation of iodide and iodate: 3 I2 + 6 OH- === > 5 I- + IO3(-) + 3 H2O and because in any case no free free H3O+ can exist. H3O+ is what causes your perceived reaction of HI with Al; not like you write it but in reality:

HI + H2O === > H3O+ + I-
Al + 3 H3O+ === > Al3+ + 3 H2O + 3/2 H2

In alkaline conditions, depending on [OH-] either mainly Al(OH)3 or mainly aluminate will form.

These are nonetheless silly ways to try an eliminate water.

[Edited on 7-12-2013 by blogfast25]

AJKOER - 7-12-2013 at 11:25

Thanks Blogfast for the review.

I have corrected the hydrolysis reaction, but I believe my chain is still valid assuming the HI formed is not too dilute to attack the Aluminum effectively per the reaction:

2 Al + 6 HI --> 2 AlI3 + 3 H2 (g)

which would then push the hydrolysis reaction by consuming HI.

Well, at least, this is all my opinion as I have not performed this dehydration reaction myself. If effective, its relative merit would be accessibility (everyone has Al foil).

I also have made my comments clearer by noting that this dehydration step is prior to the addition of NaOH. The amount of Iodine is small so subsequent reactions with NaOH should not be of concern. I am not recommending an excess of Al, however, due to the introduction of NaAl(OH)4, which is soluble in alcohol (per Wikipedia http://en.wikipedia.org/wiki/Sodium_aluminate).

[Edited on 7-12-2013 by AJKOER]

blogfast25 - 7-12-2013 at 12:34

Quote: Originally posted by AJKOER  
Thanks Blogfast for the review.

I have corrected the hydrolysis reaction, but I believe my chain is still valid assuming [...]


You'll need to test this hypothesis empirically first. Prepare some iodine water and add a (not too large) excess of Al foil to it. If it converts almost completely Al(OH)3 you're right.

Personally I don't think it will happen or that it will be a very slow boat to China.

blogfast25 - 8-12-2013 at 06:01

Also, you really don't need iodine for it. Any old simple Al compound hydrolyses to some extent (assuming the anion is the conjugated base of a strong acid, just to keep things simple). That's because the [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> ion reacts acidic:

[Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> + H<sub>2</sub>O < === > [Al(OH)(H<sub>2</sub>O)<sub>5</sub>]<sup>2+</sup> + H<sub>3</sub>O<sup>+</sup>

Subsequent steps of deprotonation would lead to Al(OH)3.

Solutions of said simple Al ionic compounds react acidic and would slowly dissolve Al metal, resulting in Al(OH)3 precipitation. Again, 'slowly' is the operative word here: the pH of these solutions is in the range 3 - 4 only.

Sedit - 10-12-2013 at 23:52

I know its derailed but shouldn't Aluminum be able to replace Phosphorus in the reduction of Benzylic alcohols using HI as long as there is enough Al present?

Sedit - 11-12-2013 at 17:31

I placed a small amount of I2 into a test tube with Aluminum shavings and H2O. At first there was a vigorous reaction which I quenched by placing it on an ice bath to ensure Iodine was not lost. Faint bubbles of Hydrogen where evolved from the Aluminum which slowly came to a stop. Heating brings these back so the hydrolysis reaction discussed above takes place at a very slow rate if heat is not applied.

As expected the Iodine color was replaced by a clear solution with a faint brownish tinge presumably HI solution. The Aluminum has been very slightly etched after 24 hours so this is a painfully slow process if there is no heat applied.

Organikum - 12-12-2013 at 02:23

Quote: Originally posted by Sedit  
I know its derailed but shouldn't Aluminum be able to replace Phosphorus in the reduction of Benzylic alcohols using HI as long as there is enough Al present?

Not completely but for parts yes.
There is an old german reference where they substitute P with Al and other metals, IIRC tin gave best results but I am not really sure on the tin.
But it was translated by me and posted at the HIVE as a good way to reduce the amounts of P needed, but it was of course immediately shot down by WixardX who produced some completely nonsensical sermon on poisonous tin complexes which mighr form and kill everybody around. Something like this. Well the man also said that reductive aminations cannot be done in iron vessels for Hg reacting with the iron forming complexes (there must be some obsession on this, or more probably he just doesnt know what this is).

Anyways it is doable, Zn is said to be a possible metal to use by anonymous sources doing unspeakable things to benzylic alcohols over and over again. Horrible. Imagine, if they are not careful they might end up with illegal drugs! Not harmless high explosives but drugs! Drugs kill!

The article is sadly gone with the rest and only memories remain in my devastated brain (drugs, you know...) and most probably a copy on lugh´s harddrive which he may or may not be able and willing to dig up.

/ORG

Waffles SS - 13-1-2014 at 08:39

I tried to make sodium ethoxide by reaction of anhydrous ethanol with Sodium hydroxide.

I distilled mixture for bring out the Ethanol that make azetrope with water and added more anhydrous ethanol for pushing this reaction to right side.

NaOH + EtOH <==>NaOEt + H2O

I used 40gr(1 mole) NaOH and 92gr EtOH(2mole) also added 400 cc excess Ethanol for making azetrope with water but it seems no Ethoxide formed.

I checked density of distilled ethanol and that was 0.793(near dry ethanol)and also i added calcium oxide to it but temp didnt increase.
somebody has experience on this reaction?


DSC00071.JPG - 58kB
Setup

DSC00072.JPG - 74kB
Ethanol + Sodium hydroxide

DSC00073.JPG - 92kB
Color of mixture changed after some heating

DSC00074.JPG - 76kB
Small solid start to float on mixture

DSC00075.JPG - 76kB
more solid

DSC00076.JPG - 76kB
after adding all Ethanol

DSC00077.JPG - 75kB
when all of ethanol distilled



[Edited on 13-1-2014 by Waffles SS]

blogfast25 - 13-1-2014 at 13:19

Waffles:

Have you tried the procedure that uses acetone as an anti-solvent to precipitate the NaOEt? See top of this page for patent and links. That appears to work well for short chain alkoxides.

Nicodem - 14-1-2014 at 10:18

Quote: Originally posted by Waffles SS  
I distilled mixture for bring out the Ethanol that make azetrope with water and added more anhydrous ethanol for pushing this reaction to right side.

The difference of bp between the azeotrope and pure ethanol is less than half Kelvin. There is no way you could quantitatively remove water with a distillation employing such an insignificant difference, not even if you used a distillation column. You would need to add toluene to make a ternary azeotrope with some significant bp difference and still you would need to apply a distillation column.

Waffles SS - 14-1-2014 at 22:07

Quote: Originally posted by Nicodem  
Quote: Originally posted by Waffles SS  
I distilled mixture for bring out the Ethanol that make azetrope with water and added more anhydrous ethanol for pushing this reaction to right side.

The difference of bp between the azeotrope and pure ethanol is less than half Kelvin. There is no way you could quantitatively remove water with a distillation employing such an insignificant difference, not even if you used a distillation column. You would need to add toluene to make a ternary azeotrope with some significant bp difference and still you would need to apply a distillation column.


How much toluene is needed?can you send picture of your suggested system?just i add distillation column?

Quote: Originally posted by blogfast25  
Waffles:

Have you tried the procedure that uses acetone as an anti-solvent to precipitate the NaOEt? See top of this page for patent and links. That appears to work well for short chain alkoxides.


I tried it before.
I reflux 1mole KOH with 2 mole EtOH for 3 hours and then i added anhydrous acetone but i got no precipitate


Metacelsus - 15-1-2014 at 08:25

You don't need to reflux; just make sure all the KOH dissolves. I would suggest making a saturated solution.

clearly_not_atara - 3-4-2019 at 12:01

When I discovered this thread, I, like everyone else, thought it was a huge breakthrough in the preparation of NaOEt. After all, NaOH is much easier to obtain and cheaper than Na metal. However, the method has not been successfully applied very often, and the use of Na is still preferred.

I think this is all related to the difficulty of working with anhydrous NaOH. The dry powder is very hygroscopic and easily dispersed; it will burn your eyes and skin; and it must be dried at red heat. Anhydrous pellet NaOH, as used in the patent, is virtually impossible to prepare without specialized equipment.

Therefore, it is desirable to replicate the method without using anhydrous NaOH. I was inspired by the reaction:

Ca(OH)2 + Na2CO3 >> CaCO3 + 2 NaOH

Unlike NaOH, Ca(OH)2 is not hygroscopic and is much less caustic. Na2CO3 is still very hygroscopic, but it can be dried at lower temperatures and does not saponify human skin. In order to prevent caking, I suggest the following protocol:

A suspension of one equivalent powdered Ca(OH)2 in freshly dried (eg MgSO4) ethanol is mixed with a suspension of 1.5 equivalents anhydrous Na2CO3 in dried ethanol and stirred for several hours in a sealed jar or under nitrogen (NaOEt reacts with O2/CO2), filtered, and the filtrate is treated with acetone to yield, hopefully, a precipitate of NaOEt.

SWIM - 3-4-2019 at 17:43

Quote: Originally posted by clearly_not_atara  
When I discovered this thread, I, like everyone else, thought it was a huge breakthrough in the preparation of NaOEt. After all, NaOH is much easier to obtain and cheaper than Na metal. However, the method has not been successfully applied very often, and the use of Na is still preferred.

I think this is all related to the difficulty of working with anhydrous NaOH. The dry powder is very hygroscopic and easily dispersed; it will burn your eyes and skin; and it must be dried at red heat. Anhydrous pellet NaOH, as used in the patent, is virtually impossible to prepare without specialized equipment.

Therefore, it is desirable to replicate the method without using anhydrous NaOH. I was inspired by the reaction:

Ca(OH)2 + Na2CO3 >> CaCO3 + 2 NaOH

Unlike NaOH, Ca(OH)2 is not hygroscopic and is much less caustic. Na2CO3 is still very hygroscopic, but it can be dried at lower temperatures and does not saponify human skin. In order to prevent caking, I suggest the following protocol:

A suspension of one equivalent powdered Ca(OH)2 in freshly dried (eg MgSO4) ethanol is mixed with a suspension of 1.5 equivalents anhydrous Na2CO3 in dried ethanol and stirred for several hours in a sealed jar or under nitrogen (NaOEt reacts with O2/CO2), filtered, and the filtrate is treated with acetone to yield, hopefully, a precipitate of NaOEt.


Stirred for several hours?
That sounds awfully fast for such low solubility.

Fyndium - 18-3-2021 at 13:24

Quote: Originally posted by Organikum  
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.


I used some NaOH to distill my wiper fluid ethanol, which was +80C to start with, and it appeared to me that it came over pretty much anhydrous. Apparently, NaOH/Ethoxide is able to hold a lot of water behind.

Supposedly, pre-drying the ethanol with MgSO4 would reduce the need.

Last time I stripped some fluid, I measured it was about 90% with ABV meter. I did this without NaOH, as I thought it might not be necessary and to save some expensive NaOH. It seems that if I want less watery EtOH, I'll need to use it anyway.

artemov - 18-3-2021 at 19:54

Quote: Originally posted by Fyndium  

I used some NaOH to distill my wiper fluid ethanol, which was +80C to start with, and it appeared to me that it came over pretty much anhydrous. Apparently, NaOH/Ethoxide is able to hold a lot of water behind.

Supposedly, pre-drying the ethanol with MgSO4 would reduce the need.


Does distilling ethanol with NaOH damage boro glassware?
I need some anhydrous ethanol from ethanol contaminated with some water and ethyl acetate, thinking of using NaOH to get rid of the latter two, but I dun want to damage my precious glassware :P

Cheers.

Texium - 18-3-2021 at 21:02

Quote: Originally posted by artemov  
Does distilling ethanol with NaOH damage boro glassware?
I need some anhydrous ethanol from ethanol contaminated with some water and ethyl acetate, thinking of using NaOH to get rid of the latter two, but I dun want to damage my precious glassware :P
Unfortunately it can. I have a flask that I distilled ethanol out of over NaOH, and it has a ring around the circumference of the inside, where I guess some of the NaOH was deposited and etched the glass. It's harmless, but it certainly isn't aesthetically pleasing... my only 2 liter round bottom too.

Fyndium - 20-3-2021 at 11:52

I have distilled ethanol many times with NaOH and never had etched glassware. I have used silicone oil and CaCl2 baths to heat them, though - direct heating methods can cause hotspots which will make NaOH eat glass at the end of the distillation, and when liquid is splashed to the walls. With baths, maximum surface temperature can never exceed the temp of the bath, which is around 100-120C with CaCl2 and 130-150C with oil, due to lower thermal conductivity.

I got a frozen joint though, which caused me a loss of one 3n 1L rbf and a stillhead. Thou rest in peace in the graveyard of experimentalism and amateur chemistry.

But, to the topic. I couldn't wait for my lab and did some rudimentary tests with my own boka-fractionated, polished and diluted 40ABV vodka and NaOH.

I first added a small amount of vodka, and added NaOH pellets and stirred, until separate layers were formed.

funnel1.jpg - 86kB

Then I added more, between 200-250mL, and added more NaOH and stirred, and the liquid heated up a lot, but not too much to handle the funnel. I kept adding NaOH little by little, until layers were separated. They formed bubbles and the boundary closed in just as in any biphasic extraction. The activated carbon residue used to polish the distillate can be clearly seen at the boundary.

funnel2.jpg - 91kB

The extracted upper layer was at first attempted to drain into a tube for ABV meter, but it had a pinhole, which caused a little leak. Actual measuring cylinder was dug through the rubble, and the ABV measurement and volumetric measurement were conducted. The ABV reported pretty much exactly 0ABV, but I hardly believe the alcohol layer would have been the bottom layer, as the upper layer had a very burning alcoholic odor, but the ABV meter is very sensitive and the most likely explanation is dissolved NaOH, ethoxide and other stuff. The total collected volume was 86.5mL as seen, which checks in the ballpark with the input alcohol, at 40ABV by rough calculation and dilution.

The amount of NaOH required is a lot, but not hefty, and obviously as this was 40ABV, the required amount would be inherently larger than when separating azeotropic product.

It should be distilled and remeasured to see if the resulting product is closer to anhydrous ethanol, but at least the very clear layer separation caught my curiosity if ethanol can be salted out from water with NaOH to make actual close-anhydrous ethanol.

Btw, the OP's funnel concept works for the part that it does form a layer close to the edge of the NaOH pellet mass - but the NaOH will completely clog the funnel and form a cone-shaped conglomerate that is very hard to break down and will prevent draining any liquid, even when poked with piano wire from the drainhead. Instead, just adding ethanol to funnel and adding NaOH pellets and shaking until layer separation occurs could be a more viable option. I have currently a batch of azeo EtOH stirring over MgSO4/CaSO4, and I'll probably strip a sample to see if any desiccation has been occured, and if not, I will do a full-scale treat on the batch and see what the resulting product will be. It is likely I'll do it anyway, because I'm running out of ethanol and simply treating the wiper fluid directly with mass of NaOH will both destroy MEK, detergents and additives and keep the residual water, so pretty decent ethanol should be able to be stripped directly off it.

For the record, I have stripped several batches of ethanol from this wiper fluid product. It consists of azeotropic ethanol, denaturants, detergents and additives to enhance the welfare of automobiles, aka crap that NaOH eats for breakfast. I double distilled this stuff, first directly and then with NaOH, although I only used 4% per weight mostly to kill MEK, I remember getting readings over 100ABV with the warm condensate, so apparently most of the water was also left behind. Using more, up to 10% per weight could be even more efficient, but in the long run it would be nice to find the spot at when the returns do not increase anymore and the rest is wasted as ethoxide.

[Edited on 20-3-2021 by Fyndium]

artemov - 20-3-2021 at 22:16

Quote: Originally posted by Texium (zts16)  
It's harmless, but it certainly isn't aesthetically pleasing


It's harmless right? I only have 3 rbfs, so I use them for everything, including boiling and concentrating sulfuric acid :P

artemov - 20-3-2021 at 22:23

Quote: Originally posted by Fyndium  

Btw, the OP's funnel concept works for the part that it does form a layer close to the edge of the NaOH pellet mass - but the NaOH will completely clog the funnel and form a cone-shaped conglomerate that is very hard to break down and will prevent draining any liquid, even when poked with piano wire from the drainhead. Instead, just adding ethanol to funnel and adding NaOH pellets and shaking until layer separation occurs could be a more viable option.
[Edited on 20-3-2021 by Fyndium]


I need some sodium ethoxide in ethanol for my barbituric acid synthesis, so I am thinking of trying OP's/your method:
In a small beaker add 1/3 by volume NaOH, then fill almost to the brim with near anhydrous ethanol, stir, seal, wait for a day or two, then slowly syringe off the top layer.

The top layer is mainly sodium ethoxide in ethanol I hope?

[Edited on 21-3-2021 by artemov]

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