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Author: Subject: Make Potassium (from versuchschemie.de)
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[*] posted on 8-12-2010 at 08:14


Quote: Originally posted by Sedit  

However to explain why I see a problem look at the volume of Mg in the pictures you just posted at the start of the experiment. Now at the end there is a huge reduction in the amount of metal in the experiment presumably converted over to K. However I do not see any precipitate which there should be a LARGE amount of if logic serves me correctly. Not only just large but given the fineness and fluffyness of inorganics precipitated in a non polar solvent my instincts would tell me that that entire flask should be filled with a white precipitate yet as the reaction proceeds all im seeing is a conglomeration of metal into shinny spheres and a total reduction in reactants volume with no explination as to where they are going. It also appears as though the KOH flakes are untouched as though the metal has melted around them.


Even before pok answered that, I believe that’s very weak criticism. It’s very hard to estimate the amounts the ‘should be there’ just with the naked eye. Pok’s answer is satisfactory to me.

Quote: Originally posted by Pok  
No reproduction of my results? Because nobody has tried it yet (as I described it)!


I think that’s a completely fair point. But as woelen pointed out, the invention must have some degree of ‘robustness to change’ for it to be truly useful. Otherwise the type of butterflies that breeds in your area and how precisely they flap their wings may make any reproduction impossible…

Quote: Originally posted by aonomus  
Here is a suggestion, if the sludge is indistinguishable between potassium metal and magnesium metal (and its oxides/hydroxides) check the mass balance. If you can account for the loss of hydrogen that should indicate reaction progression.



Good idea but hard to put into practice: hydrogen is very light: 1 mole is only 2 gram. Also, according to some of the mechanisms proposed here (2 KOH (+ catalyst) + Mg --- > 2 K + Mg(OH)2 (+ catalyst)) there really should be no more H2 evolution after initial water from the moist KOH has been reacted away with the stated (patent) excess of Mg…

Quote: Originally posted by Eclectic  
Did we decide that k-1 kerosene with the volatiles distilled off and butoxyethanol as the alcohol were not worth trying?


I don’t recall anyone mentioning ‘butoxyethanol’, what does that look like when it’s at home?

Kerosene remains on the menu as far as I’m concerned. But my Shellsol is in the post…

[Edited on 8-12-2010 by blogfast25]
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[*] posted on 8-12-2010 at 08:26


A point I’ve been meaning to raise was raised by not_important in the ‘len1 thread’:

http://www.sciencemadness.org/talk/viewthread.php?tid=2105&a...

According to him, the patent “EXPIRED DUE TO FAILURE TO PAY MAINTENANCE FEE”. That comment was made in 2007, following len1’s diligent attempt at reproduction. It would be wise to recheck the current status of the patent. A dead patent can point to a dud or severe problems with attempts to commercialise the invention. I believe for US patents to stay protective of their invention, the invention must be commercially exploited within x years of patent filing.


[Edited on 8-12-2010 by blogfast25]
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[*] posted on 8-12-2010 at 08:36


Quote: Originally posted by blogfast25  
I believe for US patents to stay protective of their invention, the invention must be commercially exploited within x years of patent filing.
No. All you've got to do is make the renewal filing and pay the renewal fee along with it.
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[*] posted on 8-12-2010 at 08:38


Quote: Originally posted by watson.fawkes  
Quote: Originally posted by blogfast25  
I believe for US patents to stay protective of their invention, the invention must be commercially exploited within x years of patent filing.
No. All you've got to do is make the renewal filing and pay the renewal fee along with it.


Ah. I seem to recall that what I described applies to some European patents. But I'm no expert...
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[*] posted on 8-12-2010 at 08:51


Quote: Originally posted by blogfast25  
A dead patent can point to a dud or severe problems with attempts to commercialise the invention.
It's almost certainly the case that this patent is not an economical method to produce potassium on an industrial scale. Magnesium is produced with electrolysis. Potassium can be produced with electrolysis. If you're going to build big equipment, why would you do electrolysis for Mg and then a chemical step requires both a yield loss and more capital for another plant? I've got to believe it's cheaper just to build a K electrolysis plant directly.

The situation is different for a small user of K. If they can fit the present synthesis into their existing infrastructure, then it's not worth spending the minimum to build their own electrolysis facility.
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[*] posted on 8-12-2010 at 09:05


Quote: Originally posted by blogfast25  
But as woelen pointed out, the invention must have some degree of ‘robustness to change’ for it to be truly useful. Otherwise the type of butterflies that breeds in your area and how precisely they flap their wings may make any reproduction impossible…


Yes. Butterflies in Brazil shouldn't affect the process. But I never claimed that. I just said: my method will definitely give you potassium, I never said: my method is the only way to give you potassium. So if you use my method you will be successful and afterwards you can try to substitute shellsol against lamp oil, t-butanol against isopropanol, MgAl instead of Mg and so on. I think we should just wait for woelens or others results. ;)
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[*] posted on 8-12-2010 at 09:21


Quote: Originally posted by watson.fawkes  
Quote: Originally posted by blogfast25  
A dead patent can point to a dud or severe problems with attempts to commercialise the invention.
It's almost certainly the case that this patent is not an economical method to produce potassium on an industrial scale. Magnesium is produced with electrolysis. Potassium can be produced with electrolysis. If you're going to build big equipment, why would you do electrolysis for Mg and then a chemical step requires both a yield loss and more capital for another plant? I've got to believe it's cheaper just to build a K electrolysis plant directly.

The situation is different for a small user of K. If they can fit the present synthesis into their existing infrastructure, then it's not worth spending the minimum to build their own electrolysis facility.



I think that rather depends on the scale of operation, as you say. You are of course absolutely right that it’s quite imbecilic to use one electrolysis to avoid another, considering the cost of one will roughly match the other and that you still need a reduction plant to get to the potassium. Not to mention the cost of displacing the MgO/Mg(OH)2 back to halide…

But for much smaller scale producers, avoiding an electrolysis plant and simply buying in cheap Mg in medium size quantities and converting this fairly problem free (and no electron juice to pay for either) into small batches of K could be economical for those who sell the stuff by the kg and not by the tonne… Ultimately ‘Big is beautiful’ of course…

The patent is therefore not something that would have gotten the big 'K-smelters' unduly worried.

[Edited on 8-12-2010 by blogfast25]
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[*] posted on 8-12-2010 at 09:28


Quote: Originally posted by Pok  
I think we should just wait for woelens or others results. ;)


'Yeah but, no but': in the absence of concrete results from SMers speculation will continue: some useful, some not so useful...

[Edited on 8-12-2010 by blogfast25]
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[*] posted on 8-12-2010 at 17:29


CH3CH2CH2CH2OCH2CH2OH

http://en.wikipedia.org/wiki/2-Butoxyethanol


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[*] posted on 8-12-2010 at 19:56


Like I said, I will be trying this later on. Now it seems like instead of the 23rd I will do it by the 20th. What do you guys think about solvents? What about mineral oil (USP)? VM&P Naptha? Kerosene? I was unable to obtain dodecane.

Obviously, I have had 0 success in obtaining Shellsol D70, in fact, it doesn't seem to be a product promoted and/or sold in the U.S. Closest "US" product I found is D60 and the only source I found was in 55 gal. drums.

I'm thinking it might even be more desirable to use a solvent that can be heated to 200*C without boiling. Obviously I can't repeat the experiment exactly due to this, but, if it works then that will be a confirmation. If it doesn't work, that won't necessarily be a refutation.





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[*] posted on 8-12-2010 at 21:12


Quote: Originally posted by MagicJigPipe  
What do you guys think about solvents? What about mineral oil (USP)? VM&P Naptha? Kerosene? I was unable to obtain dodecane.
[...]
I'm thinking it might even be more desirable to use a solvent that can be heated to 200*C without boiling.
I outlined possible alternatives for the US in this posting. Given your query, I decided to look at the products called "odorless mineral spirits". Gamblin sells one, branded Gamsol; here's its MSDS. It's the right chemical class, but with a different CAS #. The trouble is that its boiling range is 185 - 206 ° C, barely intersecting that of Shellsol D70 on the low side. If this reaction is kinetically limited overall, this solvent might not get hot enough to work well; perhaps it works but very slowly. On the upside, Utrecht Art, a national chain, stocks it. And the temperature range isn't outrageously different.

Recochem Odorless Mineral Spirits (MSDS by free login to their site) has the right CAS #, but has an even lower boiling range (151 - 205 °C) than Gamsol.

Not everything called "odorless mineral spirits", however, is the same class as Shellsol D70. D70 is hydrogen-treated, removing both aromatics and alkenes. One product that comes up right away in Google, lists "aliphatic hydrocarbons" CAS #8052-41-3, which exclude aromatics but includes alkenes. Presumably this product is not hydrogen treated. Klean-Strip brand (very common in big box hardware stores) lists this CAS # as well. This class of product might work (if alkenes don't poison the reaction and if it's high-boiling enough), but given that Gamsol is readily available, I'd try it first.
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[*] posted on 8-12-2010 at 21:52


One thought that comes to mind, can anyone try fusing Mg shavings with KOH pellets with a blowtorch in a crucible?
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[*] posted on 9-12-2010 at 03:38


Quote: Originally posted by aonomus  
One thought that comes to mind, can anyone try fusing Mg shavings with KOH pellets with a blowtorch in a crucible?


Sounds like a good way to have a small fire with caustic smoke :(
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[*] posted on 9-12-2010 at 06:24


Quote: Originally posted by Eclectic  
CH3CH2CH2CH2OCH2CH2OH

http://en.wikipedia.org/wiki/2-Butoxyethanol




As stated above, there are good reasons to believe only t-alcohols might work.
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[*] posted on 9-12-2010 at 06:31


Quote: Originally posted by aonomus  
One thought that comes to mind, can anyone try fusing Mg shavings with KOH pellets with a blowtorch in a crucible?


This has basically already been tried by various experimenters, also on this forum, but with NaOH instead of KOH. Leads to a mess with a little, largely unrecoverable sodium. Risk of fire or hydrogen explosions is quite considerable. It should 'work' with KOH too (heat of formation of NaOH and KOH are mearly the same).

Mind you don't lose an eye in the process: 2 KOH +Mg + blowtorch --- > 2 K + MgO + H2 + use your imagination

[Edited on 9-12-2010 by blogfast25]
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[*] posted on 9-12-2010 at 06:59


Over at versuchschemie.de things are still dead, last post on 1/12 by ‘tritiumoxyde’ – TOT, German for ‘death’. Prosaically applicable then…

I’ve been paying a little attention to the alkoxide side of things. Alcohols are of course extremely weak acids, methanol being the ‘strongest’ of the mono alcohol series. I seem to recall an acid constant for ethanol as a water Bronsted acid of about E-29. Alkoxides, the conjugated bases, in watery solution are almost entirely hydrolysed. They do form by reaction of various alkali metals with the pure alcohol.

Reaction of (for instance) methanol with anhydrous KOH will yield very little potassium methanoate, probably even less in a suspension of KOH in a paraffinic solvent. If potassium 2-methyl-propa-2-oxyde (K tButO) shows some solubility in the solvent then that might drive towards the formation of it. Dissolved it would remain almost completely undissociated. Reactive with solid Mg?

Longer chain t-alcohols would be even less strong ‘acids’ but probably more soluble in the paraffinic solvent. Again there may be a real optimum there…
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[*] posted on 9-12-2010 at 10:32


Quote: Originally posted by blogfast25  

Longer chain t-alcohols would be even less strong ‘acids’ but probably more soluble in the paraffinic solvent. Again there may be a real optimum there…


Someone up for making some biodiesel out of saturated fats and then running a grignard? :D

Actually, stearic acid, with some palmitic acid content (from what I've heard) is available as a candle-making supply. That would be the best starting point, I think.




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[*] posted on 9-12-2010 at 11:17


Quote: Originally posted by blogfast25  
Reaction of (for instance) methanol with anhydrous KOH will yield very little potassium methanoate, probably even less in a suspension of KOH in a paraffinic solvent. If potassium 2-methyl-propa-2-oxyde (K tButO) shows some solubility in the solvent then that might drive towards the formation of it.

Dissolving KOH in anhydrous methanol gives a solution of almost entirely methoxide (MeOH in MeOH is a stronger acid than H2O in MeOH). Generally, in most, if not all, organic solvents, hydroxide is a much stronger base than alkoxides, thus the equilibrium favours the alkoxide (the less anion solvating power the solvent has the more basic hydroxide is in respect to alkoxides).
t-BuOK is quite soluble in most organic solvents, particularly polar ones. But even in nonpolar ones it dissolves to some extent. In my experience, it is relatively soluble in boiling toluene or xylene, so I would tend to believe that at 200 °C its solubility in even less polar solvents like paraffin oils would nevertheless be substantial.
Obviously, in such nonpolar solvents as alkanes, t-BuOH is much more acidic than H2O (due to the obvious solvation reasons). In water, though, the roles are inverted and t-BuOH is about a hundred times less acidic than H2O.

I don't know why, but most people have these basic concepts mixed up and this must be at least the third time I have to explain it.

So, to recapitulate, any t-BuOH added in such a threephasic system is going to get deprotonated at the surface of KOH pellets and get back to solution as t-BuOK. At the Mg surface it can react to give off hydrogen and enter in the solution as (t-BuO)2Mg, provided of course that the surface of Mg is activated (which might not be so trivial in such a messy system). In principle, (t-BuO)2Mg can react on the surface of KOH to give Mg(OH)2 and t-BuOK, but this needs experimental proof. If you have t-BuOK in solution and activated Mg surface, then you also have a redox equilibrium Mg(II) + 2K(0) <-> Mg(0) + 2K(I), which, as far as I know (and I don't know much about inorganic chemistry) favours the right side. Though, like in every equilibrium, if you introduce a driving force to remove Mg(II) from the system, you should in principle force it to the left.

But in my opinion it is futile to speculate and build up a hypothesis before doing the experiment. You need hypotheses when there is something you need to optimize or modify, but this looks as simple as it can get. t-BuOH is cheap and easily available at any chemical supplier. Same goes for KOH and grignard quality Mg powder. Paraffin oils with bp at about 200 °C should not be a problem either.




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[*] posted on 9-12-2010 at 12:48


I just ordered one liter of ShellSol D70 at Kremer Pigmente. I hope it is shipped quickly, it is an international order from Germany to the Netherlands.

Next weekend I'll try the small experiments I mentioned in a previous post, I do not expect the ShellSol D70 to be here next weekend, so these experiments will be with the paraffin oil.




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[*] posted on 9-12-2010 at 13:40


Quote: Originally posted by Nicodem  
Dissolving KOH in anhydrous methanol gives a solution of almost entirely methoxide (MeOH in MeOH is a stronger acid than H2O in MeOH). Generally, in most, if not all, organic solvents, hydroxide is a much stronger base than alkoxides, thus the equilibrium favours the alkoxide (the less anion solvating power the solvent has the more basic hydroxide is in respect to alkoxides).
t-BuOK is quite soluble in most organic solvents, particularly polar ones. But even in nonpolar ones it dissolves to some extent. In my experience, it is relatively soluble in boiling toluene or xylene, so I would tend to believe that at 200 °C its solubility in even less polar solvents like paraffin oils would nevertheless be substantial.
Obviously, in such nonpolar solvents as alkanes, t-BuOH is much more acidic than H2O (due to the obvious solvation reasons). In water, though, the roles are inverted and t-BuOH is about a hundred times less acidic than H2O.

I don't know why, but most people have these basic concepts mixed up and this must be at least the third time I have to explain it.


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[*] posted on 9-12-2010 at 14:31


Hehe, good to see people caught onto the fact that blowtorch + H2 = boom. I must say that was probably the only semi-baited posting I've ever made here.

I'm working on getting some tBuOH and I'll give it a shot next week, most likely with a parafin oil. I'm tempted to see what happens when silicon oil is used, and whether the KOtBu generated in-situ can decompose the silicon oil. Though its expensive and probably harder to work up, it would give higher temperatures than otherwise possible with Shellsol D70.
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[*] posted on 9-12-2010 at 16:24


I'm thinking I will use mineral oil. Despite having more aromatics (I'm sure) I think it should be sufficient. It's boiling point is, on average, 300*C and it consists of mostly alkanes and cyclic parrafins according to Wikipedia. I know that could be "not true". No one has said anything about that specifically, so what do you think? I don't really have $50+ to spend on a bottle of heavy oil that may or may not work (and that is really worth much less than that to me).

I already have some mineral oil sitting around... (not odorless mineral spirits)




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[*] posted on 9-12-2010 at 18:21


Quote: Originally posted by MagicJigPipe  
I'm thinking I will use mineral oil. Despite having more aromatics (I'm sure) I think it should be sufficient. It's boiling point is, on average, 300*C and it consists of mostly alkanes and cyclic parrafins according to Wikipedia. I know that could be "not true". No one has said anything about that specifically, so what do you think?
Well, recall Nicodem's point about the temperature-dependent solubility of the butoxide in non-polar solvents. If it's less soluble in heavier alkanes, it's going to slow down the reaction. With a boiling point of 300 &deg;C, it's going to be mostly heavier alkanes. This might or might not be a rate-limiting step. There are a lot of parameters here at play, and I doubt we all here will understand what's most important unless we start with a working synthesis and work outward from there.

Having said that, you might first see if your mineral oil fractionates. There might be enough of a lower boiling fraction that you're more closely matching the boiling range of Shellsol. Or, call your local art supply store and see if they have Gamsol. It's about $11 for a pint jar.
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[*] posted on 10-12-2010 at 06:20


Quote: Originally posted by MagicJigPipe  
I'm thinking I will use mineral oil. Despite having more aromatics (I'm sure) I think it should be sufficient. It's boiling point is, on average, 300*C and it consists of mostly alkanes and cyclic parrafins according to Wikipedia. I know that could be "not true". No one has said anything about that specifically, so what do you think? I don't really have $50+ to spend on a bottle of heavy oil that may or may not work (and that is really worth much less than that to me).

I already have some mineral oil sitting around... (not odorless mineral spirits)


How about blending it with some decent kerosene, to get a wider boilling point range?

I received my 2-methyl-2-butanol and my KOH today but my Shellsol (driectly from Kremer) may be stuck in the atrocious weather we're seeing over Europe...

[Edited on 10-12-2010 by blogfast25]
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[*] posted on 10-12-2010 at 08:47


Like many on this thread I am intrigued and fascinated by Pok's work

My head says that this synthesis just cannot be possible but my heart is enthralled by the site of those beads of potassium metal

I have tried a couple of 'quick and dirty' goes at the synthesis myself using liquid paraffin without success. There is certainly initial generation of gas after gentle warning and I am able to get the tert butanol to reflux successfully (see pictures). After a couple of hours I just end up with a sludge of magnesium turnings and white powder (? MgO ? Mg(OH)2 this is definitely not the KOH flakes I started with)

I have now received some shellsol D70 from Kremer pigments and will try again over the w/e. For anyone interested I ordered the solvent from Kremer and it arrived in the UK today (Friday) - not bad service

IMG_4903.JPG - 31kBIMG_4902.JPG - 19kB
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