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Author: Subject: Make Potassium (from versuchschemie.de)
Sedit
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[*] posted on 3-12-2010 at 12:47


Any suggestions on an over the counter alcohol that could be used? I don't know if anything has tert-butyl alcohol in it but I do want to attempt this with NaOH. I do feel a high boiling alcohol is needed here and IpOH will not work.

[edit]
BTW, what about Ethylene Glycol, anyone think that may work here? Im not really sure how this reaction would take place so im having a little trouble thinking this thru.

[Edited on 3-12-2010 by Sedit]





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[*] posted on 3-12-2010 at 13:10


Quote: Originally posted by Sedit  
Any suggestions on an over the counter alcohol that could be used? I don't know if anything has tert-butyl alcohol in it but I do want to attempt this with NaOH. ]
According to the NIH Household Products database, Vanish Mildew-Stain Remover contains 1-5% t-butanol and Turbo Octane Boost 108 contains an unstated amount of t-butanol, the remainder being methanol.

[Edited on 3-12-2010 by entropy51]
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[*] posted on 3-12-2010 at 13:32


A_bab:

Once more several assumptions for the price of one. I don’t have a stash of potassium and if I did it still wouldn’t prove anything. Pok’s results may well be an elaborate hoax (and he wouldn’t need potassium to build it – for the early stages lead shot would have been more convenient but other methods of falsification would be even better) but I’m not going to comment on likelihood or motives because I know zilch about that and neither does anyone else.

I say ‘elaborate’ but it needn’t be: heat some alkali metal (some others would do it too) to above MP with an inert solvent and some plausible looking grit and shake it up (some t-butanol may even act as a dispersant…): an emulsion of the metal in the solvent/grit will be created. Allow to stand, swirling occasionally, at above the MP of the metal in question and globules of the metal will start appearing. Simples…

Looking at the whole body of evidence (including tests conducted by bromic, len1 and others) I conclude that it’s hard to conclude anything definitive. Thermodynamically speaking the reaction works. Alcoholate may provide a route. If someone six moths ago had told me that a relatively dry mix of KOH, Mg and a paraffinic (in essence) solvent would start fizzing at around 100 C for ½ to 1 hour I’d have declared them mad, yet we have some bona fide witnesses that testify to it. In itself this is worth investigating. And that’s what I’ll do: at least verify whether, with or without ‘catalyst’, and using dry (or slightly moist but of known composition) ingredients hydrogen gas is obtained or not… Len1’s best shots showed no gas but others reported it.

‘dropping a piece or two of potassium in water while at high shool doesn't count here.’: another wild and condescending assumption not worth refuting.

You make an assumption about my Al balls without even having seen them: clairvoyant scientists must be in real demand…

Still, you’re entitled 110 % to your opinion.


[Edited on 3-12-2010 by blogfast25]
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[*] posted on 3-12-2010 at 13:44


Quote: Originally posted by Sedit  
Any suggestions on an over the counter alcohol that could be used? I don't know if anything has tert-butyl alcohol in it but I do want to attempt this with NaOH. I do feel a high boiling alcohol is needed here and IpOH will not work.

[edit]
BTW, what about Ethylene Glycol, anyone think that may work here? Im not really sure how this reaction would take place so im having a little trouble thinking this thru.

[Edited on 3-12-2010 by Sedit]


Sedit, a humble word of advice: if you’re gonna do it do it properly, we’ll soon have a stream of hasty and half-hearted attempts that prove nothing. So go that extra mile and buy a decent t-alcohol. Avoid or at least measure initial moisture and avoid oxygen. Try and prove that any gas produced is hydrogen or not. Otherwise we'll get none the wiser.

Falsification of this experiment is not difficult as I pointed out upstairs.

I’ll be trying this as soon as I feel confident I can meet these conditions but not before…
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[*] posted on 3-12-2010 at 15:04


I agree 100% about doing it right however, I don't honestly feel there is a right way to do this as everyone thats done it right has failed meaning some other forms of experimentation has to be done to come to any conclusion as to what exactly is going on if any. Of course I want to do this with t-Butyl alcohol if nothing else as a control but I also want to extend the possibility of other things working to give myself an idea on how this is suppose to proceed. In all honesty I don't think it will and I believe that a crown ether MIGHT work since it can carry the cations into solution.

[edit]

In order to not double post i'll just add a couple more thoughts here. See something about this that bugs me is if this was faked then he really went all out because you can see progression of the K forming as it starts from small little beads and grows meaning he would have had to add it little by little to convince us or he would have had to stir molten K and then what we are seeing is it re-conglomerating as he heats it . I dont feel it should react yet it appears to be working so im interested.

Something to note is that He did not stir the reaction which would make sense that it would just spread the particles apart making them less likely to react where as Len IIRC stirred the entire time.

Also if water hinders this reaction and KOH rarely comes anhydrous would it not be best to heat powdered KOH in the solvent to drive off the water before adding any Mg?

[Edited on 4-12-2010 by Sedit]





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[*] posted on 3-12-2010 at 17:46


A couple of people on the versuchschemie forum did report partial success - one said he saw half millimeter globules of K when cleaning up after what he thought was a failed experiment. I don't think it's a hoax, just maddeningly sensitive to something we haven't identified.

Quote:
would it not be best to heat powdered KOH in the solvent to drive off the water before adding any Mg?


Well, assuming that pok is correct, the initial evolution of hydrogen around 130C is the magnesium reacting with the water (getting rid of it). While other people seem interested in the alcohol and variations thereon, I think a more relevant question is: what is going into solution during the initial warmup (before alcohol is added), and how is it dissolving? So far as I am aware, neither KOH nor magnesium (nor the passivating layer of MgO) should dissolve in paraffin - not even slightly. So there should be no way for the compounds to react at that point in time, unless some other solvent component is allowing one of them to pass into solution. In particular, something must be able to strip the passivation from the magnesium. Even if we imagine that there are exposed surfaces on the Mg turnings, coated only with oil that is then dissolved in the paraffin, so that H2O freed from the KOH can attack them... they should rapidly become inert.
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[*] posted on 3-12-2010 at 20:41


Quote: Originally posted by bbartlog  
I'm interested in watson.fawkes observation about the boiling point of the tert-butanol. In order for it to not gas off, it would have to react fairly rapidly (turning into alkoxide, presumably) after being added.
That's certainly one possibility; I won't deny that. There are others, though.

I made a very blunt assumption when I first wrote about this for simplicity's sake, namely, that the boiling points of the D70 solvent and the tert-butanol were independent in this admixture. That's very possibly not true. They certainly don't form a low-boiling azeotrope (at least not at this concentration, not lower than that of tert-butanol). They might well, though, form a high boiling one. I'm just guessing that the phase diagram of the Shellsol D70 / tert-butanol system has never been published. There may be those here that would like to contribute to the effort without trying the whole synthesis. Measuring the phase diagram would be some Actual Science, regardless of whether the synthesis itself succeeds or fails.

Another basic parameter I haven't seen is the solubility of tert-butanol in D70. Perhaps they're miscible. And as long as I'm suggesting measurements I'm not set up to make, how about vapor pressure vs. temperature diagram for some concentrations of interest? Knowing these basic physical quantities would help greatly in having some idea of just where the reagent is.

Related to this issue is another: Where does the tert-butanol go? The system exhausts gas (H2, presumably) both initially and after adding tert-butanol. Given the low boiling point of tert-butanol, how much of it could be escaping with the H2 before it has a chance to react? Certainly having the bulk of your reagent evaporate on you would cause a synthesis failure. It seems that it might be a good idea to run two condensers at two temperatures: one to reflux D70, and another to capture what might be tert-butanol vapor. What doesn't react or escape ought to be left in the solvent. It would seem that a simple distillation after the fact could measure the amount of residual tert-butanol and show that indeed it did not react. The third measurement should be to extract the alkoxide and measure that.

These three pathways, evaporation, solution, and alkoxide, seem to exhaust the possibilities that have been mentioned to date. Maybe we'll find out that there's something not accounted for.
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[*] posted on 3-12-2010 at 21:09


Quote: Originally posted by blogfast25  
Two other experimenters on versuchschemie claim to have seen ample amounts of hydrogen, one saw a small amount of K. Where does the hydrogen come from? Water + Mg? There's not enough water there. Alkanes + Mg? Hardly likely...
This brings up another thing I noted when reading over the Shellsol D70 data sheet. This specific composition is a hydrogen-treated oil; I'm not sure that's in the data sheet but in some other places where I saw it listed. D70 has very low levels of aromatics (< %0.5 m/m). It's listed as 60% paraffins (alkanes) and 40% naphthenes (cycloalkanes). There's no room for left olefins (alkenes). What I'm getting at is that this solvent is close to completely saturated.

I see two ways this could be relevant. The first is that the presence of desaturated hydrocarbons could be inhibitory to the process. That's a hypothesis about why solvent substitution fails.

The second is that the solvent is actually a reagent (somehow) and we're getting alkene formation in a way that doesn't disturb its bulk physical properties very noticeably. Of the many ways the oil industry has developed for olefin determination, there's got to be one chemically compatible with the reagents and products of this reaction.
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[*] posted on 4-12-2010 at 01:24


I gave this a shot with turpentine instead of the oil and iprOH the Mg was half turnings and half powder/granules and the KOH was large chips. Next time I'm going to try both Mg and KOH powder, the KOH formed large chunks of MgO that are surrounding the unreacted KOH, mag stirring wouldn't work either so I guess I'll have to try again. I cannot see any beads of K let though, I can see a small grey lump that doesn't look like the other reactants but it might not be potassium.
Ok, a huge storm just broke out so I finished the reaction, no K was isolated, there wasn't even much of a reaction with water once the turps was drained off, the KOH must have gone somewhere though, otherwise the Mg would be reacting with it and the water.
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[*] posted on 4-12-2010 at 04:23


As explained by Nicodem, our organic expert, i-PrOH is not likely going to work:

quote:
I did not read all about this topic, but just wanted to comment that the use of isopropanol is most likely futile. Metal alkoxides containing alpha-hydrogens are generally not stable up to 200 °C. They tend to decompose via beta-hydride elimination reactions, the rate at which they do so highly depends on the what metal alkoxide they are. I would expect potassium isopropoxide is not stable at the required conditions. The acetone formed in the decomposition is most likely to blame for the brown-red coloured crap formed via self condensation.

So I would not waste your chemicals on this. Just try to find some t-BuOH, it isn't that expensive, at least much less so than potassium metal.
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[*] posted on 4-12-2010 at 05:30


Quote:
Water + Mg? There's not enough water there


It does seem a little strange, but on the other hand KOH does normally contain a fair bit of water. If you assume Pok's 6g of KOH was 8% H2O, that's enough to generate ~600ml of hydrogen gas.
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[*] posted on 4-12-2010 at 05:40


Technical and lab-grade KOH usually contains about 15% water by weight.
So indeed that is where the water comes from.
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[*] posted on 4-12-2010 at 06:56


Up to 15 % water in technical KOH? That sounds mightily high to me, Jor. I’ve now ordered some KOH and some 2-methyl-2-butanol (t-pemtanol) and will measure the water content of the KOH shortly.

Update: I have to say that Jor seems to be right on the water content of most KOH grades, just about every grade sold on eBay is about 90 % only. It’s been a while since I bought any KOH…

It’s very clear from Example 1 (potassium from KOH) that the initial hydrogen comes from moisture reacting away (assuming for argument’s sake that neither the patent nor pok are hoaxes). During that step no t-butanol is present yet.

The actual reactions in the second step could possibly be (with R a t-alkyl group):

Initiation:

KOH + ROH < --- > KOR + H2O

Wherever this equilibrium lies, it may be pulled to the right by:

Propagation:

2 KOR + Mg < --- > 2 K + Mg(OR)2

Which may itself by pulled to the right by:

Termination:

Mg(OR)2 + H2O --- > MgO + 2 ROH

Of the last step at least we can be reasonably sure: Mg alkoxides should be quite prone to hydolysis.

2 KOH + Mg --- > 2 K + MgO + H2O

… would be the overall reaction in that scheme. A calculation using NIST values of HoF and S at 298 K, shows the ΔG = ΔH – TΔS to be about -48.6 kJ/mole (of Mg reacted, @ 298 K), so just barely thermodynamically possible. For higher temperatures this requires a correction that is likely to be small. The low change in Gibbs free energy could explain low reaction rates.


In this scheme ROH is not consumed and thus would play the part of a real catalyst. But pok reports more hydrogen gas during the second phase (and this scheme generates no free H): is part of the formed K reacting with the water? Part of the Mg?

I refuse to rule out a complete hoax at this point: if perhaps one of the best experimenters here – len1 – could get neither hydrogen nor potassium then that counts for a lot. It might just be a case of waiting for pok to state: ‘Aha, gotcha, you s*ckers!!’

There is that possibility of an ‘unstable experiment’, one that works only in very specific conditions but that makes it even less likely that pok, with his very rudimentary set up, struck gold (I mean potassium)…

One thing that keeps nagging me are the Rieke metals: the situation there is quite similar to this patent/pok. Even for the synthesis of the highly reactive Mg the reduction reactions with alkali metals are thermodynamically favourable, yet there seems no way the thermodynamic hindrances can be overcome at these temp. And yet it works…

[Edited on 4-12-2010 by blogfast25]
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[*] posted on 4-12-2010 at 07:37


Where can I get 2-methyl-2-butanol ?
On ebay I cannot find it.
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[*] posted on 4-12-2010 at 07:54


I couldn’t find t-butanol on eBay either, so I found this:

http://www.purechemicals.net/buy-2-methyl-2-butanol-2m2b-42-...

In GBP, might not ship elsewhere, not sure…

t-butanol should not be hard to find in US, try Chemsavers?

[Edited on 4-12-2010 by blogfast25]
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[*] posted on 4-12-2010 at 09:44


A detailed reading of len1’s and pok’s experiments, literally looking at the texts side-by-side, does seem to reveal two main differences. In len1’s last experiment, the one where he fully discloses the experimental set up:

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

… he uses a “special fast overhead motor”, i.e. for high speed stirring. This is not mentioned in the patent and pok only uses occasional swirling of the conical flask throughout the whole experiment (assuming it is one). Does high speed stirring in a heterogeneous mix really promote contact between reagents or does it possibly achieve the opposite effect? It’s certainly not conducive to any liquid K coalescing from small drops into larger ones, IMHO…

Secondly, the introduction of the t-butanol after the alleged initial hydrogen surge (alleged by patent and pok): neither pok, patent or len1 much elaborate how that is done. Pok mentions a syringe but we don’t get to see it. From len1’s photo I tend to conclude the alcohol has been added at the top of the refluxer by means of a separation funnel, with the argon switched on. But he writes high up: “The heat output was upped so the reagents were at a vigorous boil at 200C”. With a BP of 82C, this way introduced the t-butanol might not even actually make it into the reactor flask, if the Shellsol is ‘boiling vigorously at 200C’. As by far the most volatile component of the mix, it would tend to gather in the hotter, Shellsol vapour phase… Similar to the point made by Watson… This would of course also be eventually be true of t-butanol injected into the liquid Shellsol phase but at least it would have some time to react with the KOH and any K t-butoxide (K t-BuO) formed would not be volatile anymore.


[Edited on 4-12-2010 by blogfast25]
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[*] posted on 5-12-2010 at 02:54
its me


Hi,

I am the guy who made the synthesis on versuchschemie.de.
I didn't read all the stuff here. If you have questions, just ask.
- http://www.freepatentsonline.com/4725311.html -

I did the original procedure exactly as it was desribed (this seems to be important!) - exeptions here:
Things that I did in another way than in the original procedure:

- I did a 10th volume procedure (e.g. 50ml Shellol instead of 500ml and so on).

- I did it in a sand bath on a gas cooking plate

- I didn't work with argon atmosphere, but with a stopper, glaspipe, balloon (with a tiny hole at the tip) - so escaping hydrogen is allowed, influx of oxygen mostly prevented.

- the glaspipe (25 cm long!) was surrounded by wet toilet paper (which cooled the pipe so that pipe acted like a reflux condenser)

- used a syringe to add the tert.butanol through the hole of the balloon, letting the tert.butanol flow through the pipe into the reaction container

- I didn't stir continiously. Stirring only: every 3 minutes for 10 seconds at butanol addition --- and every 30 minutes for 10 seconds afterwards (before butanol addition: every 10 minutes stirring for 10 seconds) - by taking the reaction vessel with gloves and carefully swinging around in circles.

- I didn't measure the temperature. I estimated it by eye (you only have to distinct boiling from not boiling)

- tiny potassium globules begin to form at about 20 min after beginning to add tert.butanol. - in the worst case: after 30 min after ending tert.butanol addition.

- If you stir as I did it: the potassium will flow together and form larger globules by boiling turmoil. If you stir continiously (as described in the patent) you might only get many small instead of few large potassium globules.

- let the reaction mixture cool down to room temperature and grab the potassium globes with a pincett, rapidly putting them into clean shellsol D70 (or another inert fluid).

The potassium seems to be very pure since you can see a crystallized surface! Magnesium doesn't dissolve in potassium at low temperatures, so you can be sure, not to have an alloy of K-Mg.

Important: I did this experiment in 300ml-Shellsol dimensions as well and it worked equally well. BUT: my last 2 procedures only yielded many many tiny K-globules which are useless I think. They didn't flow together and didn't form larger globes (I absolutely don't know why!)

I never saw potassium before, so for me it was very exciting to examine its properties (soft like hard butter, super pink flame in contact with water and so on)

one picture of the 300ml-procedure which is not on versuchschemie.de(http://www.versuchschemie.de/topic,14677,0,-Synthese+von+Kal...) - it's not a good picture as I already put every good pictures on versuchschemie.de :cool:reaction vessel on aluminium foil with still liquid potassium globes:

potassium.JPG - 83kB

larger (original size and quality) pictures of potassium (balls and cut potassium) - vertically (its too big):

potassium 2.JPG - 64kB

potassium 3.JPG - 40kB

By the way: Neither the authors of the patent, nor me are bamboozlers (I don't know the exact english word) - but I might think this, too. If I were in your position :D.

I would advice you first to do the experiment exactly in the way I described it with much attentiveness and caution. If this works you can try variations with isopropanol or lamp oil or other stuff. If it does not work you can call me a liar (but: I'm not).

@froot: your reaction vessel with the glas pipe is an open system: oxygen will influx easily and oxidise you potassium immetiately. Your Mg also seems to be too coarse. My Mg was finer (I did it with a file by muscle work)

left: Mg in Shellsol D70. right: KOH chips in plastic bag.
mg.JPG - 25kB

If you have any questions: ask.


[Edited on 5-12-2010 by Pok]

[Edited on 5-12-2010 by Pok]
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[*] posted on 5-12-2010 at 03:38


truly thankyou I shall try this.

I did not think it was possible and for you not to be a chemist makes it even better.

why note make the T butanol from MEK reduction if we can not

buy the alcohol.



[Edited on 5-12-2010 by Ephoton]




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[*] posted on 5-12-2010 at 05:13


Reducing MEK will give 2-butanol, not t-butanol.
I think a Grignard reaction of acetone and MeMgBr will work (or use EtMgBr for 2-methyl-2-butanol (EtBr is much easier to handle and dry).
Or Grignard reaction between ethyl acetate and MeMgBr.

If you want to to this reaction you have the Mg anyway, so you can just as well do the Grignard. You just need dry ether, MeBr (wich is very easy to make from H2SO4+MeOH+KBr, you just need to dry it), and a inert atmosphere (however I think this can just be the refluxing solvent vapours). And you ofcourse only need quite small amounts of t-butanol for usable quantities of potassium.

If I ever try this I will buy the t-butanol.

[Edited on 5-12-2010 by Jor]
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[*] posted on 5-12-2010 at 05:28


Thank you very much Pok for joining us here on scimadness.

Your presence here is really encouraging !

This is one of the most amazing syntheses ever posted here, in my opinion.

Pok, what is your opinion on Len1's attempt at this?
Couple of things that come in my mind, is that he used much coarser magnesium, and that he stirred the mix continuously. It seems that mg powder might be more suitable for this procedure.

I can't wait to see if we will be able to replicate your results! I'll have a go myself as soon as i can get ahold of some t-butanol!
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[*] posted on 5-12-2010 at 05:50


Yes pok, thanks very much for dropping in at sciencemadness!

One very obvious question is where did you by the Shellsol solvent? It's not what we call here an 'over the counter' solvent, it's rather specialist.
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[*] posted on 5-12-2010 at 06:00


Blogfast, in europe it can be found at Kremer Pigments:
http://kremer-pigmente.de/shopint/index.php?cat=030102&produ...

Art supply stores usually carry lots of Kremer products, even if they don't stock it, they should be able to order it for you from Kremer. This is my best bet in eastern europe, as I couldn't find ShellSol anywhere.
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[*] posted on 5-12-2010 at 06:07


I believe that where I wrote:

Quote: Originally posted by blogfast25  
Termination:

Mg(OR)2 + H2O --- > MgO + 2 ROH

Of the last step at least we can be reasonably sure: Mg alkoxides should be quite prone to hydolysis.

2 KOH + Mg --- > 2 K + MgO + H2O



... that even assuming for argument's sake there is such a mechanism as I decribed that the termination step and the overall reaction should be written as:

Mg(OR)2 + 2 H2O --- > Mg(OH)2 + 2 ROH

and

2 KOH + Mg --- > 2 K + Mg(OH)2

For the overall reaction
ΔG = ΔH – TΔS = -75 kJ/mole (of Mg reacted, @ 298 K), so even slightly more thermodynamically favourable than before. I find it hard to believe that actual MgO would form, not Mg(OH)2 in these conditions.

[Edited on 5-12-2010 by blogfast25]
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[*] posted on 5-12-2010 at 06:47


Quote: Originally posted by condennnsa  

Pok, what is your opinion on Len1's attempt at this?
Couple of things that come in my mind, is that he used much coarser magnesium, and that he stirred the mix continuously. It seems that mg powder might be more suitable for this procedure.


I only read a little bit of his attempt:

1. He powdered the KOH. This is not necessary as I used commercially available KOH chips with success (although the chips will clump together at a certain temperature). Maybe, during powdering the KOH absorbs water from the air due to its extremely high hygroscopicity. This could lead to so much water in the reaction that all the Mg is used up to decompose the absorbed water and nothing is left to react with the KOH.
I personally opened the new KOH-bag very fast and weighted the KOH as fast as possible trying to avoid large water absorption from the air.

2. The Mg really looks very coarse. The best way to obtain a suitable size of Mg particles is (as I experienced) to file them off with a normal medium file (not fine, not extremely coarse) by hand - not with a (drill) machine or so. I used Mg of about 99% purity (old printing plates). As the authors of the patent don't give more details, I think Mg of 98-100% purity is OK. I think TOO fine Mg powder also wouldn't be the best way, because it could react TOO fast (especially at the beginning where only water from the KOH is decomposed by the Mg and yields large amounts of hydrogen in a very short time!)

3. I don't think that stirring continiously is really bad here (the patent really sais "continiously stirring"!). But it wasn't necessary in my case. It might yield finely divided K-globules but they definitely should be visible. If you use a teflon coated stirrer, the teflon might react with the potassium as someone on versuchschemie.de seems to have experienced.

4. Exclusion of air (which possibly wasn't ensured at his attempt) AND evolution of hydrogen must be ensured. The cheapest and easiest way for me was a valve-like method (balloon with tiny hole). If you don't exclude the air really tightly potassium (which is swirled around in the mix by turbulences) at the solvents surface will react with the tiniest amounts of oxygen and will be lost.

5. ensuring reflux is very important as the tiny neccessary amount of tert.butanol will otherwise be lost very fast.

6. len1 himself also took into consideration the possibility of impure agents. This could be a problem, of course.

Maybe my modifications just prevented some possible sources of trouble. Just compare my description with his one (also on the original page versuchsschemie.de where you can read it in detail).

As I said, I've done the experíment several times with success. And even when there was "no success" I gained at least many tiny K-globules. So I don't think that you can make many mistakes if you just do it the way i did it.

Quote: Originally posted by blogfast25  


One very obvious question is where did you by the Shellsol solvent?


As condennnsa said, "Kremer Pigmente" offers Shellsol D70. This is my source.
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blogfast25
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[*] posted on 5-12-2010 at 06:59


Thanks condenssa and pok for the Shellsol info...
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