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Author: Subject: Make Potassium (from versuchschemie.de)
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[*] posted on 4-1-2011 at 15:19


I just turned off the water completely and let convective air cooling do the job. If too much alcohol gets lost, I'll add some more later.

The reaction products have formed a hard porous mass that encloses the K globules and prevents coalescence, so I interrupted the heating and broke it up with a spatula. I'll further look after it in an hour.
Manual agitation seems to be required at some point.




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[*] posted on 4-1-2011 at 16:41


I added 0,2g extra t-BuOH and resumed the water cooling.
No more alcohol is solidifying on the condenser- a sign that most of it has reacted with the K.
The solvent has become very turbid and opaque. There is still no coalescence, and the reaction mass is hard and crunchy, just in smaller pieces.




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[*] posted on 4-1-2011 at 17:39


If someone is really interested in obtaining tetralin birch reduction of naphthalene yields tetralin and 1,2-dihydronaphthalene

But, this might be beside the point if access to Lithium is limited.

I do not have access to this article but the abstract on the first page seems to be relevant.

Chem. Eng. News, 1966, 44 (51), pp 70–72
DOI: 10.1021/cen-v044n051.p070
Publication Date: December 1966
Copyright © 1966 AMERICAN CHEMICAL SOCIETY
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[*] posted on 4-1-2011 at 19:11


Electroreduction of aromatics using magnesium electrodes in aprotic solvents containing alcoholic proton donors

DOI: 10.1016/S0013-4686(03)00259-7

Quote:

Regioselective electroreduction of the aromatics, including methoxybenzenes, by using LiClO4 as a supporting electrolyte in an aprotic solvent (THF) containing alcoholic proton donors such as t-BuOH was successfully achieved to afford the corresponding 1,4-cyclohexadienes regioselectively in high yield. The electrolysis can be performed under constant current conditions at ambient temperature. The effect of electrode materials was remarkable, that is, the use of Mg electrodes gave the best result. Moreover in the presence of t-BuOD instead of t-BuOH, the deutrated 1,4-cyclohexadienes were obtained in high deuterium incorporation at 1- and 4-positions. A direct electron transfer to the aromatics is unlikely, and it is reasonable that the solvated Li(0), which is generated by the cathodic reduction of LiClO4, is intermediately involved in the electron transfer with being assisted by the anodically generated Mg2+ as an electron transfer catalyst.

Author Keywords: Cathodic reduction; Aromatic compounds


If someone could kindly aquire this paper is would eliminate the issue with Lithium avalibility. Abstract link embedded in the title.





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[*] posted on 4-1-2011 at 19:40


Cool stuff!

Here is an idea regarding how to reduce the amount of magnesium needed (and the amount of Mg(II) byproducts formed, maybe even helping purification if the magnesium alcoholate is soluble enough in a purification solvent) by preforming the potassium tert-alcoholate in stoichiometric amount.

Water-immiscible alcohols form an azeotrope with water, so distilling the alcohol with KOH would displace the equilibrium of alcohol + KOH <==> K-alcoholate + H2O to the right. This is an established method of preparing sodium 1-butoxide. If the solubility of the alcohol in water is low enough, it might even be possible to perform the reaction with a Dean-Stark apparatus to return the alcohol back to the distilling flask while tapping off water. Tert-amyl alcohol is soluble in 8 parts of water, so it is not as favourable as it could be but adding toluene might have a positive effect, also in formation of a tertiary azeotrope. Alternatively one could use a higher-boiling tertiary alcohol.

Azeotropes of t-amyl alcohol (bp 102,25C)
t-AmOH + 27,5 weight-% water bp. 87,35C
t-AmOH + 44 weight-% toluene bp. 100,5C
t-AmOH + water + toluene bp. 82C (no composition data)

After distilling the reaction water, one would add the high-boiling hydrocarbon and distill off all the lower boiling components. Some of the alcohol may stay retained in the potassium tert-alcoholate as alcohol of crystallization, but I suppose it would distill when heated enough. At least sodium t-butoxide may be freed of the alcohol by distillation with xylene. The residue would then be ready for reduction with Mg(0). Recovery of the alcohol would depend on separation of potassium metal from the Mg alcoholate, which could be decomposed with water to give Mg(OH)2 and alcohol.
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[*] posted on 4-1-2011 at 20:49
Failure (Negative Result Report)


1.58 g of Mg "shavings" from a firestarter and 3.06 g of reagent (though quite old) KOH hemispheres were added to a 125 mL flask containing approx. 50 mL of Kroger brand *mineral oil "laxative". The flask had a rubber stopper with a long glass tube in it. At the top of the tube was a balloon with a few small "needle-holes" poked in it. A wet paper towel was wrapped around the tube near the top. The flask sat in some sand in a 600 mL beaker. The stopper was placed on and the sand was heated (I can't find my damn glass thermometer so I couldn't measure the temp. directly) to a temperature of 200*C over a period of about 30 mins. Then, a hypodermic needle containing 0.6 g of mineral oil and 0.6 g of Reagent t-butanol (the alcohol did slowly dissolve with just a bit of stirring) was thrust through the stopper and injected into the solution over about 25 mins. I noticed "instant" boiling of the alcohol/oil soln. as soon as it hit the main mixute. Once the temp. got up to around 250*C there was sporadic boiling. S

tarting at about 200*C (I'm not quite sure on this exactly and will have to check my notes at home) there was quite a bit of gas evolution and it stayed steady for several hours. The t-butanol did seem to increase the amount of bubbles but it could've just been boiling. The reaction did not seem to proceed at all and after about 3 hours a few black specks were noticed floating around. Then the Mg started to turn black and eventually I started to notice a white "precipitate" on the bottom of the flask. It seems as if the reaction did not occur and the KOH just settled at the bottom and the Mg was oxidized (by H2O/air?) I desperately added quite a bit more t-butanol (unmeasured) and just got refluxing t-butanol in the rxn mixture. I finally stopped heating after 6 hours.

My guess is that there was not enough mixing due to the sporadic boiling. Also, the KOH could've been poorly stored and could thus be mostly the carbonate. It could also be the Mg but I doubt it. I simply shaved off Mg from the block with the tool that was included (and damn that took a long time and a lot of elbow grease; the battery to my drill was dead).

And before anyone says anything: Yes, I know there are different and better ways to set this up.

*The mineral oil said it contained vitamin E as a stabilizer.

The setup:
SETUP.PNG - 216kB SETUPCLOSEUP.png - 173kB

About an hour after adding t-butanol and 1.5 hours into the rxn:
MIDRXN.png - 541kB

The results:
RESULT.PNG - 374kB

Thanks for your time!




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[*] posted on 4-1-2011 at 22:33


Naphthalene will not work in this reaction since it reacts very easily with potassium to form the naphthalinide. I have prepared potassium naphthalinide several times for use in forming exotic organometallic anions. Upon mixing potassium and naphthalene in a glove box and shaking for several minutes a greenish color develops. After adding tetrahydrofuran, the mixture turns to a very dark green color and the potassium soon dissolves, I would assume that no potassium would be obtained if naphthalene was used as a solvent. I am very interested in this work since I would like to have some potassium of my own for syntheses when sodium metal cannot be used. I plan on using my glove box for the isolation procedure though for added safety.



[Edited on 5-1-2011 by benzylchloride1]




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[*] posted on 5-1-2011 at 00:54


MJP, Looking at your setup, you very well may have lost your t-butanol to the atmosphere.

About the t-butanol crystallization problem, how about adding a touch of hexanes?

[Edited on 1-5-2011 by smuv]
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[*] posted on 5-1-2011 at 03:55


I did not have the problem of t-butanol crystallizing. I blew air through the Liebig cooler at a speed of 2 liters per minute, using a small aquarium pump. This provides sufficient cooling not to loose any t-BuOH, while at the same time this does not solidify. My column was warm when touched, but not hot.



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[*] posted on 5-1-2011 at 05:41


MJP while I don't feel that stirring is to good of an idea in this reaction I do believe that during the addition of the alcohol it would be best for one to stir to react the alcohol as rapidly as possible with the KOH. The of the reaction goes thru a different mechanism that im not sure if stirring would be best.




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[*] posted on 5-1-2011 at 07:05


Here's the rest of the report on the first batch:

Last night, I was too tired to look after the reaction after the 4 hours of reflux were over, so I let it run overnight, for a total of 12 hours reflux after addition of the t-BuOH.
The next day, after cooling down, it was still turbid and there was a large amount of brownish powder and crumbly substance under the solvent, with no visible potassium any more.
I emptied the flask into a beaker and sorted through the crud with a spatula and tweezers.
There was a large irregular dirty lump of what first appeared to be byproducts, but upon cutting turned out to be massive potassium metal! The whole lump was softer than sodium and could easily be deformed between two fingers.
It weighs 2,1g and is shown on the first picture, under clean fresh Shellsol. Anyone familiar with potassium will recognize the bluish appearance of the recently cut surface.
I could have cleaned its surface with 1-butanol, but since it rapidly regrows the oxide layer upon storage, this was omitted.
There was another smaller lump of potassium, weighing 0,4g, and several more 3mm pieces that were dropped onto ice outside, making a nice show of flames and sparks.

The K pieces were recovered from the byproducts by manual sorting based on appearance (the K pieces were dark colored on the outside, contrary to the white/grey Mg oxide byproduct). On the second picture, the reaction mixture minus the first lump of K is shown.

So I conclude that Pok's method has worked for me and made 2,5g of useable potassium from 6g KOH and 3g Mg powder in 50ml of Shellsol D70.
Deviations from the procedure were that I added 0,2g more t-BuOH after losing some to evaporation, and boiling for 12h instead of 4h.
The most important observed difference was that I did not get any shiny potassium balls, but large lumps with oxidised surface.

Kalium-Test_1.jpg - 30kBKalium-Test_2.jpg - 37kB




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[*] posted on 5-1-2011 at 07:47


garage chemist:

2.7 g potassium is a yield of 71 % (based on 10 % water in the KOH). And one has to account also for the K that’s bound up as K alkoxide (as the Mg has reacted completely away). Not bad at all!

I think you and magic may also have been using a little too much head space over the reacting mix: that with intensive cooling would promote t-butanol to crystallise out in the upper part of the reactor.

What someone really importantly should do is recover all post reaction solvent: as it should contain all the catalyst and is bone dry, it should theoretically be directly reusable, no fresh catalyst needed!

And another important test would be to, say, double the amount of t-butanol/t-amyl to check effect on reaction time...

[Edited on 5-1-2011 by blogfast25]
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[*] posted on 5-1-2011 at 07:58


Quote: Originally posted by Sedit  
MJP while I don't feel that stirring is to good of an idea in this reaction I do believe that during the addition of the alcohol it would be best for one to stir to react the alcohol as rapidly as possible with the KOH. The of the reaction goes thru a different mechanism that im not sure if stirring would be best.


Sedit, if you've ever seen with your own eyes what happens when you add the aclcohol to the hot solvent, you wouldn't be writing that. The moment the alocohol/solvent pre-mix starts hitting the hot solvent, very intensive refluxing starts and this continues mostly during addition of the alcohol. Vigorous boiling of the solvent/alcohol mixture means that stirring is a complete waste of time.

I'm contemplating using much longer chain alcohols like 2-methyl-2-octanol, they have much higher boiling points and there some gentle swirling may be advisable. But with Shellsol D70 and gentle boil is achieved eventually, making stirring unnecessary.


[Edited on 5-1-2011 by blogfast25]
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[*] posted on 5-1-2011 at 08:20


With the next batch, I will try to use selfmade Mg turnings from scrap Mg using a drill press.
I will also make myself some 2-methyl-2-hexanol via n-butylmagnesium bromide and acetone. Like Nicodem, I think that this alcohol may be one of the best to use- relatively easy to make, suitably high boiling point (141°C) and not too long-chained.
The longer the carbon chain of the alcohol, the higher is its molar mass, and the more you will need of it to achieve a given molar concentration in the solvent. With the "margarine tert-alcohols" proposed in the tert-alcohol thread, it may become unattractive since you will probably need quite a large amount of your precious grignard-made alcohols.




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[*] posted on 5-1-2011 at 09:10


Quote: Originally posted by blogfast25  


Sedit, if you've ever seen with your own eyes what happens when you add the aclcohol to the hot solvent, you wouldn't be writing that. The moment the alocohol/solvent pre-mix starts hitting the hot solvent, very intensive refluxing starts and this continues mostly during addition of the alcohol. Vigorous boiling of the slovent/alcohol mixture means that stirring is a complete waste of time.

I'm contemplating using much longer chain alcohols like 2-methyl-2-octanol, they have much higher boiling points and there some gentle swirling may be advisable. But with Shellsol D70 and gentle boil is achieved eventually, making stirring unnecessary.

[Edited on 5-1-2011 by blogfast25]


Really, if I ever seen it with my own eyes I wouldn't be writting this? Have you hacked my brain and seen what my eyes have seen?

Nope not a chance else you would know my reason for writting it was because the reflux is mostly superficial and the surface rapidly boils with no complete mixing of the reactants. You have a bunch of settled solids with a heavy boiling at the top. Its akin to what causes bumping only in this case the lower boiling substance is on the top and not the bottom.

Would it be so damn hard to just stir the reaction on the addition of alcohol since it would ensure more alcohol reacts with the KOH in a decent time frame?





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[*] posted on 5-1-2011 at 09:35


Quote: Originally posted by Sedit  
Really, if I ever seen it with my own eyes I wouldn't be writting this? Have you hacked my brain and seen what my eyes have seen?

Nope not a chance else you would know my reason for writting it was because the reflux is mostly superficial and the surface rapidly boils with no complete mixing of the reactants. You have a bunch of settled solids with a heavy boiling at the top. Its akin to what causes bumping only in this case the lower boiling substance is on the top and not the bottom.

Would it be so damn hard to just stir the reaction on the addition of alcohol since it would ensure more alcohol reacts with the KOH in a decent time frame?


Ok, calm down already.

Firstly it’s understood that len1’s experiment of years back failed, probably due to too fast stirring. pok’s experiment involved occasional swirling of the flask and that is what most here have been doing successfully, including myself.

Secondly, refluxing isn’t ‘superficial’. Assuming you have total reflux and not a huge amount of head space, most of the alcohol is in the liquid phase, in homogeneous solution. If that wasn’t the case you wouldn’t get reaction at all. The ENTIRE solution boils, not just the surface, as can very clearly be seen during the reaction and as logic dictates. On top of that you’ve got hydrogen generation and heat generation (quite a bit too: 2 KOH + 2 Mg === > 2 K + 2 MgO + H2 is highly exothermic mainly due to the formation of the high lattice energy MgO). And the alcohol is added as a premix of alcohol and solvent. Combine all this and you have a heterogeneous solid-liquid system but with a highly homogeneous liquid phase.



[Edited on 5-1-2011 by blogfast25]
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[*] posted on 5-1-2011 at 09:46


Quote: Originally posted by garage chemist  
With the next batch, I will try to use selfmade Mg turnings from scrap Mg using a drill press.
I will also make myself some 2-methyl-2-hexanol via n-butylmagnesium bromide and acetone. Like Nicodem, I think that this alcohol may be one of the best to use- relatively easy to make, suitably high boiling point (141°C) and not too long-chained.
The longer the carbon chain of the alcohol, the higher is its molar mass, and the more you will need of it to achieve a given molar concentration in the solvent. With the "margarine tert-alcohols" proposed in the tert-alcohol thread, it may become unattractive since you will probably need quite a large amount of your precious grignard-made alcohols.


Personally I’m hesitating between 2-methyl-2-hexanol and 2-methyl-2-octanol, the former being cheaper to synth., the latter being even higher boiling (>180C) and the alkoxide possibly more soluble in C12-C15 based solvents (for both K, Na). The higher boiling point of 141C will still lead to considerable refluxing unless you add it at low temperature. But refluxing isn't a problem anyway, it's part of the solution (no pun intended). 2-methyl-2-hexanol is only 2 C longer than t-amyl alcohol, would that make a decisive difference? n-hexyl bromide, BTW, is surprisingly reasonably priced by Sigma-Aldrich...

As regards ‘The longer the carbon chain of the alcohol, the higher is its molar mass, and the more you will need of it to achieve a given molar concentration in the solvent’, that’s been dealt with in the organics thread. I don’t think it’s an attractive proposition to make longer chain 2-methyl-2-alkanols if it turns out they’re not considerably more active (than t-butanol or 2-methyl-2-butanol), thus requiring lower molarities for same or better reactivity. In particular higher solubility of the K alkoxide should speed up the first two steps of the (proposed) reaction mechanism. Personally I'm convinced it's possible to bring the whole procedure down to under an hour, as Nurdrage with his Tetralin/t-amyl alcohol qualitative experiment already showed.

But the main purpose of longer chain t-alkanols must remain IMHO the higher solubility of the corresponding sodium alkoxides, opening up the way to Na production.


[Edited on 5-1-2011 by blogfast25]
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[*] posted on 5-1-2011 at 13:30


I cleaned up my potassium pieces by melting them in dioxane.
This works marvellously, much better than my old method with IPA in kerosene.
In hot dioxane, the molten potassium floats and the impurities either sink to the bottom (MgO and Mg) or dissolve (KOH), leaving absolutely clean shiny potassium spheres.
However, on weighing the cleaned potassium again, I found that the large 2,1g piece now only weighs 1,5g. There is a lot of brown crud in the dioxane.
So my yield is not as high as initially assumed, and the potassium directly from the Shellsol D70 method contains a lot of adhering or suspended impurities and should be cleaned up with hot dioxane.




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[*] posted on 5-1-2011 at 13:55


0.2 g t-butanol accounts for about 0.1 g of K (bound as alkoxide), so that doesn’t explain the lower yield.

But I think if you repeat the experiment, this time avoiding some of the butanol freezing out near the refluxer, you’ll find your results to be close to all of ours.

And since as you have dioxane you could try and help coalescence by replacing the Shellsol (after all hydrogen has stopped evolving and cooling) with dioxane and remelting the fine K…
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[*] posted on 5-1-2011 at 15:19


Quote: Originally posted by garage chemist  
I cleaned up my potassium pieces by melting them in dioxane.
This works marvellously, much better than my old method with IPA in kerosene.
In hot dioxane, the molten potassium floats and the impurities either sink to the bottom (MgO and Mg) or dissolve (KOH), leaving absolutely clean shiny potassium spheres.
However, on weighing the cleaned potassium again, I found that the large 2,1g piece now only weighs 1,5g. There is a lot of brown crud in the dioxane.
So my yield is not as high as initially assumed, and the potassium directly from the Shellsol D70 method contains a lot of adhering or suspended impurities and should be cleaned up with hot dioxane.


If you have some pictures I would like to see. Did you prepare the dioxane yourself from condensation of ethylene glycol or is it purchased?
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[*] posted on 5-1-2011 at 16:14


The dioxane was purchased. This is a quite expensive and difficult to obtain chemical, but it really pays off to get some if you're trying to make clean potassium.
The K looked even better than the large ball of liquid potassium in tetralin that NurdRage obtained. It was like floating mercury.
I didn't make any pictures, but I will make some the next time I purify a batch of potassium.

Dioxane can be prepared at home from ethylene glycol and sulfuric acid. A friend did this a few years ago and still remembers vividly how much work it was to get a pure product from the crude distillate. All kinds of impurities, aldehydes, unsaturated products, and so on are formed as byproducts in the synthesis.
Dioxane is also very difficult to obtain anhydrous. If you don't have any sodium or potassium to reflux it with, you'll never make an anhydrous product.
With the commercial product, specified at max. 0,1% water, there was clearly visible hydrogen evolution at the molten K- but this did not impair the process, since KOH seems to be soluble in hot dioxane.




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[*] posted on 6-1-2011 at 02:41


Because some of us here will now be looking at other catalysts, I’ve rewritten the proposed reaction mechanism presented on woelen’s page in a format that represents what we believe happens in a clearer manner (I think). Just four reactions are involved, arranged here into two SETS, where ‘(sol)’ stands for ‘dissolved in inert, aprotic, apolar solvent’ and ROH is either t-butanol or 2-methyl 2-butanol.


SET 1:

1.1. Alkoxide formation:

2 KOH(s) + 2 ROH(sol) < == > 2 KOR(sol) + 2 H2O(sol)

1.2. (= 2.2.) Redox reaction:

2 KOR(sol) + Mg(s) < == > 2 K(l) + Mg(OR)2(sol)

1.3. (= 2.3.) Hydrolysis:

Mg(OR)2(sol) + H2O(sol) < == > MgO(s) + 2 ROH(sol)

Sub Overall 1:

2 KOH(s) + Mg(s) < == > 2 K(l) + MgO(s) + H2O(sol)

===================

SET 2:

2.1. Hydrogen generation:

K(l) + 2 ROH(sol) < == > 2 KOR(sol) + H2(g)

2.2. (= 1.2.) Redox reaction:

2 KOR(sol) + Mg(s) < == > 2 K(l) + Mg(OR)2(sol)

2.3. (= 1.3.) Hydrolysis:

Mg(OR)2(sol) + H2O(sol) < == > MgO(s) + 2 ROH(sol)

Sub Overall 2:

Mg(s) + H2O(sol) < == > MgO(s) + H2(g)

Overall = Sub Overall 1 + Sub Overall 2:

2 KOH(s) + 2 Mg(s) < == > 2 K(l) + 2 MgO(s) + H2(g)

===================

This scheme relies on the same reactions woelen’s does but eliminates one step I’ve always found highly contentious (I quote from woelen's scheme):

(t-BuO)2Mg + KOH → MgO(s) + t-BuOK + t-BuOH

I just don’t see this happening. My scheme doesn’t have to call on it while still respecting catalyst conservation and overall stoichiometry.

===================

Some might still wonder how come the potassium isn’t attacked by water and I believe it can be explained as follows. Firstly we must realise that we’re in an aprotic, apolar medium. For the potassium – water reaction we normally have:

2 H2O(l) < == > H3O+(aq) + OH-(aq) and K(s) + H3O+(aq) == > K+(aq) + ½ H2(g) + H2O(l), the latter which proceeds very quickly due to K’s small first ionisation energy, K+ solvation enthalpy and escape of the hydrogen. This of course then drives the dissociation of water to the right. But in an apolar, aprotic solvent, water isn’t dissociated.

Secondly we have to assume that Mg(OR)2(sol) + H2O(sol) < == > MgO(s) + 2 ROH(sol) does proceed very quickly with the very high lattice energy of the MgO being the ΔG driver here. Kinetically the reaction somewhat resembles hydrolysis of a Grignard Reagent: although the water isn’t dissociated, water molecules still have a partial positive charge δ+ on their hydrogen ends and a partial negative charge δ- on the oxygen end. Similarly the Mg(OR)2 moiety may not be dissociated but contains a partial positive charge δ+ on the Mg part and partial negative charges δ- on the oxygen ends of the alkoxide part. This makes rearranging H2O + Mg(OR)2 into MgO + 2 ROH very likely and very fast, much like the hydrolysis of Grignard Reagents.

In any case we don’t have much choice but to assume that reaction water from 1.1. regenerates the catalyst, by reacting with Mg(OR)2 rather than reacting with K, as otherwise no K could be formed AT ALL.

I believe Mg(OR)2(sol) + H2O(sol) < == > MgO(s) + 2 ROH(sol) are the ‘four cylinders’ which drive much of the whole edifice, at least thermodynamically speaking.

I also see no good reason why the preliminary drying step has to be carried out without the catalyst present. As long as a small excess of Mg is accounted for as H2O + Mg === > MgO + H2 the catalyst doesn’t seem to interfere or be interfered with.


[Edited on 6-1-2011 by blogfast25]
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[*] posted on 6-1-2011 at 10:02


Very nice, blogfast. I was really starting to wonder what was driving this remarkable synthesis. When you first look at it, it doesn't seem possible that a less active metal (Mg) will act to reduce a more active metal (K+). Then I assumed that the generation of H2 and the solid MgO were drivers and that the kinetics were favorable enough even at the relatively low temperature.

I was going to calculate the delta G for the overall reaction when I saw your post. Have you done this yet?




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[*] posted on 6-1-2011 at 10:46


Quote: Originally posted by Magpie  
Very nice, blogfast. I was really starting to wonder what was driving this remarkable synthesis. When you first look at it, it doesn't seem possible that a less active metal (Mg) will act to reduce a more active metal (K+). Then I assumed that the generation of H2 and the solid MgO were drivers and that the kinetics were favorable enough even at the relatively low temperature.

I was going to calculate the delta G for the overall reaction when I saw your post. Have you done this yet?


Thanks Magpie… Kinetic barriers are lowered by the catalyst, that's its definition, basically...

ΔH @ 298 K = + 428 kJ/mol – 601 kJ/mol = - 173 kJ/mol K, quite respectable… :)

[Edited on 6-1-2011 by blogfast25]
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[*] posted on 6-1-2011 at 12:28


Quote: Originally posted by blogfast25  

I also see no good reason why the preliminary drying step has to be carried out without the catalyst present. As long as a small excess of Mg is accounted for as H2O + Mg === > MgO + H2 the catalyst doesn’t seem to interfere or be interfered with.


Because the temperature at which drying takes place is well above the boiling point of the alcohols.

The alkoxides don't form until the system is dry so by the time it reaches that temperature most of the alcohol will be in the vapor phase. When drying starts the fast hydrogen production pushes the alcohol vapor out of the condensor and is gone.

The condensor won't recapture all of it since the vapor pressure is obviously less than 100% (since it's being carried by the hydrogen).

Only when the alkoxide is produced will it become non-volatile.

If we can start working with alcohols that have boiling points higher than the solvent then we might be able to do everything in one pot.

Alternatively i suppose you could simply load up with LOTS of alcohol and hope enough is retained to form the non-volatile alkoxide

But that's just my theory, experimentally i find performance better when i add the alcohol later. but i haven't performed a rigorous test of the theory yet.

Just like i tested aromatics and it worked, if you want to go ahead and test your alcohol theory i'd love to hear the results. You might find out that my "vapor loss" theory is a negligible contributor.

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