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Author: Subject: Sodium Ethoxide and anhydrous EtOH
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[*] posted on 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]




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[*] posted on 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]
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[*] posted on 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!




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[*] posted on 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]
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[*] posted on 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.




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[*] posted on 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]




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[*] posted on 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]
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[*] posted on 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.




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[*] posted on 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.




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[*] posted on 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?




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[*] posted on 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.





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[*] posted on 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




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[*] posted on 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]
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[*] posted on 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.




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[*] posted on 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.
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[*] posted on 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

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[*] posted on 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.



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[*] posted on 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.




[Edited on 04-20-1969 by clearly_not_atara]
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[*] posted on 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.





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