Sciencemadness Discussion Board - Make Potassium (from versuchschemie.de)

if one can replicate it then we'e essentially done here. If not, it's tar and feathers for our German friend...

Did I tell you that I am German?

Some Austrians and Swiss would be deeply offended by ignoring their existence

- most of them also speak German.

And: I would take the tar and feathers if none of you will produce that kind of K balls that I produced! - therefore: please do your experiments with attentiveness - because I don't wanna look like a chicken.

blogfast25 - 16-12-2010 at 12:43

No, but versuchschemie.de is German. I jumped to conclusions...

Pok - 16-12-2010 at 13:44

No, but versuchschemie.de is German. I jumped to conclusions...

Your conclusions are correct

. Although I'm not a "nazi" or so

...I hate these people even more than you (probably). I think everytime when I say that I am German I have to apologize for stupid Hitler and people who followed him.

So maybe now you don't call me a German but just pok?

blogfast25 - 16-12-2010 at 13:53

No problems with being German though...

hinz - 16-12-2010 at 13:55

Hi Pok

before this all ends in an endless flamewar, i would suggest that you use the improvements in technology and set up an live video stream (maybe a chat client like IRC with video support) and let some of the skeptical users here look live at your reaction and comment it. (Some user shall sugest an good live chat client with video support as I don´t know much about it)

If you faked it, this probably also could be faked, but it will be much harder (if someone says you should move your hand in front of the camera and nothing happens, it´s an inevitable prove of faking)

Greez

Pok - 16-12-2010 at 14:12

Quote: Originally posted by hinz

Hi Pok

before this all ends in an endless flamewar, i would suggest that you use the improvements in technology and set up an live video stream (maybe a chat client like IRC with video support) and let some of the skeptical users here look live at your reaction and comment it. (Some user shall sugest an good live chat client with video support as I don´t know much about it)

If you faked it, this probably also could be faked, but it will be much harder (if someone says you should move your hand in front of the camera and nothing happens, it´s an inevitable prove of faking)

Greez

This is a very good idea. But I (1) don't have a webcam - just a normal digicam. (2) I'd have to buy a new stopper

- my last one is "dead". I want to repeat the experiment as soon as I have a new stopper :cool:

...I still have some nice short videos of the boiling reaction mix at different times (from beginning to the end where you can already see the large rolling and jumping K balls) - maybe this could help you?

[Edited on 16-12-2010 by Pok]

[Edited on 16-12-2010 by Pok]

videos

Pok - 16-12-2010 at 14:29

the pictures you already know. now some videos of the same (1st ever) procedure.

(1) one of my produced tiny K balls in water (doesn't prove anything but that I have K).

(2) my first procedure ever: 50ml at beginning - unfortunately very dark.

(3) same procedure: at the end - K balls jumping. (dark, too)

I hope you have Windows Media Player.

Now you can be sure: its either Na, K, Cs or Rb (or an alloy of mainly alkali metal +x) that you see in the last video. No "woods metal" (or other mainly heavy metal alloy), no mercury, no gallium - because no transparent liquid is existing with such a high densitity that these heavy metals could form such balls.

Attachment: Film.wmv (1.5MB)
This file has been downloaded 1211 times

Attachment: 50ml - beginning.wmv (1.4MB)
This file has been downloaded 1208 times

Attachment: 50ml - end.wmv (1.5MB)
This file has been downloaded 1202 times

News from versuchschemie.de / versuchsgenie.de:

german: http://www.versuchsgenie.de/forum/viewtopic.php?f=16&t=1...

google translation:
http://translate.google.de/translate?js=n&prev=_t&hl...

[Edited on 17-12-2010 by Pok]

Magpie - 16-12-2010 at 14:34

Quote: Originally posted by Pok

I think everytime when I say that I am German I have to apologize for stupid Hitler and people who followed him.

Pok, you don't have to apologize for the actions of your ancestors any more than the rest of us. My wife's ancestors (Vikings) terrorized the living hell out of my UK/Irish ancestors. I don't look for apologies from her. I'm in the driver's seat now (I think).

len1 - 16-12-2010 at 17:15

You can hardly call one of the few sets of posts with any relevant scientific content in this thread trolling. So either you are angry or you are just have no idea of how science works.

You wont me to repeat this stuff, fine Ill do it now not because of what Pok posted, I dont want to encourage trolls, but as a last good-will gesture, for you, in a university chemical laboratory, using all the chemicals as listed, temperature controlled oven, reflux condenser. This stuff is normally used to do far more complicated reactions. This is about level 0 on degree of difficulty. So if its true, it should work. If it doesnt goodbye to you and your mate.

NurdRage - 16-12-2010 at 18:34

I want to make the add-on that i'm using magnesium that's been stored in a glovebox since the day it was purchased and I've used it to make grignards before without the need for additional surface activation.

To do my potassium experiments, I take the portion i need out of the glovebox and its on air for about 20 minutes before it's tossed into the reaction mixture.

I'm not sure if the magnesium is the "special sauce" for this, but i just thought i'd inform everyone where it came from.

Sedit - 16-12-2010 at 19:19

I must ask due to videos I have seen on the internet that show ignition of Mg+KOH producing elemental K. Isn't this reaction exothermic? When they ignite the mixture it turns into a roaring flame, How come the reaction here does not pick up pace like that of normal KOH+Mg reaction? What would moderate this reaction to make it benign where as solventless reactions go heywire?

Pok - 16-12-2010 at 23:48

Quote: Originally posted by Sedit

I must ask due to videos I have seen on the internet that show ignition of Mg+KOH producing elemental K. Isn't this reaction exothermic? When they ignite the mixture it turns into a roaring flame, How come the reaction here does not pick up pace like that of normal KOH+Mg reaction? What would moderate this reaction to make it benign where as solventless reactions go heywire?

I could image that the slowness of the reaction here is due to the low solubility of t-BuO K in Shellsol as already considered possible here. And: the low temperature!

[Edited on 17-12-2010 by Pok]

woelen - 17-12-2010 at 00:10

[...] You wont me to repeat this stuff, fine Ill do it now not because of what Pok posted, I dont want to encourage trolls, but as a last good-will gesture, for you, in a university chemical laboratory, using all the chemicals as listed, temperature controlled oven, reflux condenser. [...]

That would be very nice and I really do appreciate that effort!

I am (and most likely many others as well) looking forward to read what your results will be. Most of us have to do with less than ideal approximations of a real lab, so when you can do this under precisely controlled conditions and you have positive results, then that would mean a giant step forward in finding out the conditions, necessary for success.

len1 - 17-12-2010 at 06:27

Well I performed the experiment and the results are in.

First I would like to thank Wilco for getting me to do it. However I cant blame myself for thinking the whole thing was trash given it was clear to me some parts of it were, and I did initially think some parts of it were real - but here is what happened.

100ml shellsol D70 were placed into a 250ml conical flask with 19/26 joint, connected to a Liebig condenser, which was attached to a 3-way adaptor at the top. Argon was flushed through one of the attachments (only at the beginning), the second attachment was vented through an oil bubbler, and the third, vertically above the condenser was plugged with a septum.

100ml Shellsol D70 were placed in the 250ml flask, 12.4gms KOH flakes added through a powder funneled, followed by 6.2gms of completely oxide free magnesium shavings (see photo I dont have the mesh).

The setup was placed in an air oven and heated over the course of 60mins to 200C. At about 130C water evolved - not violently as the patent says, whereupon the KOH flakes rose through the magnesium and assumed a twisted shape (as they gave up water). At 210C reflux commneced, at which point 1.2gms t-BuOH mixed with 1.2gms D70 were added in portions of about 0.5gms every 5 minutes with swirling. Vigorous reflux of the t-BuOH (bp about 80C from memory) ensued in the lower part of the Liebig. After the addition, nothing much happened for about 15mins, and I assumed the reaction will follow its previous path. At about this point however the D70 become very cloudy, and it became obvious a reaction was occuring - this was the first difference from last time. After 30mins shiny balls of molten metal - less than 1mm in diameter started appearing. With shaking and over the course of several hours these grew up to 1cm diameter in size. The whole flask and solution was coated in fine MgO sand. (Picture)

I have not the yield yet - I will establish it accurately tomorrow, but judging by the amount of MgO present, and lack of visibility of any magnesium, it appears in the range 50-100%, making this at least a liquid-phase surface layer reaction. The yield precludes solid phase, and I do not believe it to be liquid phase due to my results before.

So why the change from last time? I can not say definitively at this stage, but my best guess is as I stated earlier in the thread - the KOH and Mg must be left alone to come into close contact - as would be the case for a boundary layer reaction, with t-BuOH probably acting in the fashion described earlier in the thread. If this was purely a liquid phase reaction I would have succeeded in forming at least some K last time since there was plenty of fresh Mg surface. However the patent clearly states stir constantly.

However a lot of what I was saying before is still true. There are large differences between the post of Pok and reality. I said I was expecting the MgO particles to stick to the K like glue - and they clearly do. Sedit wrote he would expect much MgO sand in the reaction product. I agree its everywhere in my reaction, yet in the germans pictures there is no MgO to be seen. Possibly a little trick.

SO is he a troll? Strictly no, although by playing a few games with us he made me believe so. How did he get those large clean balls? Possibly by purifying and inserting back into the mix so that he would have the best looking results. Clearly he will now claim otherwise. Maybe hes one of the patents authors, I dont know.

I will repeat the experiment a few times varying the reflux temperature, amount of t-BuOH and shaking regime, to see if any improvement is possible.

The potassium balls can be coalesced (I think I have about 10-20 will have to check tomorrow) and this is a viable K generation method for the home chemist. It is not for the professional and I do not see it replacing the electrolytic method I have been using at this stage because the reaction fairly slow judging from the rate of H2 evolution (< 1 bubble per sec) and the potassium needs a lot of work to purify.

woelen - 17-12-2010 at 06:50

Len1, many thanks for doing this experiment and sharing the results. And congrats for having (at least to a great extent) success. VERY good!
There may still be issues with Pok's experiment and writeup but this definitely proves that he is not fucking up a lot of people.

This makes me more and more eager to do thus myself. Hopefully tomorrow!

watson.fawkes - 17-12-2010 at 07:13

I have not the yield yet - I will establish it accurately tomorrow, but judging by the amount of MgO present, and lack of visibility of any magnesium, it appears in the range 50-100%, making this at least a liquid-phase surface layer reaction.

Example 1 in the patent computed yield by a workup that includes solvent removal (vacuum distillation or filtration), a density separation of product K from solids with 1,4-dioxane, upon which molten K will float, and suction filtration of molten K.

Personally, I'd like to know the mass balance of the t-BuO moiety in the products and residues, to see where it ends up and if there are side reactions in practice that lead to its destruction.

len1 - 17-12-2010 at 07:33

Yes, here are some more pictures. I did not have time to let the potassium cool to rt and opened the flask to the atmosphere at about 70C. Immediately the fairly clear D70 above the potassium (as most of the MgO had settled on 1hr standing) went cloudy. This had a positive side to it proving theres solved t-BuOK - and there is nowhere else for the t-butanol to go, in the D70. This must be the key to the reaction and its yield, t-BuOK solved in paraffin reacting with Mg to give K and (t-BuO)2Mg + 2K, the former reacting with the water to regenerate t-BuOH. I dont think this is known is scientific circles.

On examining the product I have only about 5-6 balls of potassium, so the work involved in purification is not extensive, and this might be a viable method in the lab as well. Ill know for sure tomorrow when I get to coalescing/purifying and determining the yield.

[Edited on 17-12-2010 by len1]

condennnsa - 17-12-2010 at 07:35

Wonderful work, Len. Such an easy method to synthesize alkali metals, alkali metal production in the home lab is now child's play!

And Pok, you are my hero. This synthesis deserves a pdf in the prepublication and the sciencemadness library!

Ever since this thread started, i've been looking for t-butanol, unfortunately, I still don't have any. I'm dying to try this...

And Pok , please stick around sciencemadness, your posts are much appreciated.

NurdRage - 17-12-2010 at 11:47

I can also confirm i was able to make potassium using candle wax solvent. It requires a much more extensive and dangerous workup to get clean balls of potassium but it can be done.

I also tested with tert-amyl alcohol and had similar success.

I'll be making a video on my candle wax process in the near future. First i'm going to try ways to make it safer.

In my opinion, the D70 method is safest for the amatuer.

Anyway, I'm formally saying potassium can be made from magnesium and KOH along with a tert-alcohol and an aliphatic solvent.

Pok - 17-12-2010 at 12:09

There are large differences between the post of Pok and reality.

This is true in 1 point for your experiments observations: MgO dust instead of MgO crusts. Nothing else.

The other observations by you sound like mine:
- After 30mins shiny balls of molten metal
- less than 1mm in diameter started appearing.
- yield appears in the range 50-100%

SO is he a troll? Strictly no, although by playing a few games with us he made me believe so.

If you consider "answering questions" as "playing games" - yes. then I played games.

How did he get those large clean balls?

My MgO formed crusts instead of dust. Thats the explanation for no MgO dust on K surface.

Possibly by purifying and inserting back into the mix so that he would have the best looking results. Clearly he will now claim otherwise.

Exactly. I claim otherwise! This would mean that I lied. I didn't lie 1 word in this forum. WHY should I do this? If my K was covered with MgO - I would show it - because this kind of K is 100% better than no kind of K. You only seem to believe your own eyes. If you didn't get it: nobody is able to get it??? No.

Maybe hes one of the patents authors, I dont know.

I would be proud to be one of the authors. Unfortunately, THEY are the real heros here. Not me.

Quote: Originally posted by condennnsa

And Pok, you are my hero.

And the patents authors are MY heros

Quote: Originally posted by condennnsa

And Pok , please stick around sciencemadness, your posts are much appreciated.

Unfortunately, I don't really know so much details about chemistry like you all. This really was my one and only "luck luck luck experiment". I made potassium nitrite in another experiment in high yield and purity. But nothing else. I need the information on sciencemadness and versuchschemie.de much more than you will need me

I can also confirm i was able to make potassium using candle wax solvent.

This is a very usefull information for many people! cheaper and easier available. Little or small K is better than no K for home chemists (although a more dangerous K isolation as you say)! And if this method will be shown to be improvable: perfect!

[Edited on 17-12-2010 by Pok]

blogfast25 - 17-12-2010 at 12:56

len1: fantastic, congrats and thanks for taking woelen’s (and others’) advice and finally replicate it. If anyone here could do it, it was always going to be you…

pok: congrats to you as well. Vindicated and all shadows of suspicion lifted. Please go and inform versuchschemie.de, if you haven’t already.

And I’ve got everything ready to go tomorrow or Sunday. Exciting times!

mr.crow - 17-12-2010 at 13:07

Good job, its Christmas come early! My t-BuOH is coming in January, grr

Fleaker - 17-12-2010 at 15:11

Len,

Thank you for doing the experiment! I was more excited reading your post than watching the game-winning goal of a football game! It is easy to be sceptical and sit on one's ass and speculate on the merits of a reaction but you sir, have never been one to sit in the armchair!

I eagerly await the purification and the yield!

Pok,

I think we're all glad you didn't deceive us and that this patent actually turned out to not to be bullshit.

bbartlog - 17-12-2010 at 16:22

len1: excellent experimental work as always. But I feel you disgrace yourself by trying to (still!) accuse Pok of chicanery, after essentially proving him correct. One would think that after reproducing his success so closely, you would chalk up any remaining discrepancies to other minor differences in experimental conditions rather than trying desperately to salvage this 'pok must be a troll' idea. Why not just man up and apologize?

chen - 17-12-2010 at 17:13

holy s*** nurdrage is here!!

and holy s*** it actually worked!!!

this is like christmas comes early I know someone said it but I want to say it again!!

thanks len!! gj to all of you!! tia

len1 - 17-12-2010 at 19:07

Removing the potassium from the conical this morning I experienced some difficulty in that the largest ball could not get through the 26/19 neck (see picture and note the characteristic blue tinge on the potassium, and the flattened shape <1cm height of the largest blob, which was about 1.4-2cm diameter), so a 24/29 would be better.

The potassium weighed in at 5.8gm - so with KOH being 10% water as an average figure - thats a 75% yield based on the KOH. If we recall that 1.5gm t-BuOH was added as catalyst - which ended up consuming ~0.8gm K as t-BuOK, thats an 84% yield of theoretically expected. A very good figure. The yield based on Mg (which is only illustrative since it is used in excess by about *2) is 33%.

Pok its not true that I do not trust anyone, if you read what I wrote I believed your reports at first. Its only after you wrote some other stuff which wasnt true that I thought the whole thing was bunkum. I believe that you got the MgO in crust form - so did I, but I also got sand. At the moment there is no explanation to why, and I will try changing a few things.

Anyway I thank you for showing us that the patent can be made to work. I

[Edited on 18-12-2010 by len1]

aonomus - 17-12-2010 at 19:16

Perhaps you might have luck heating to 70degC in that dish and forcing the droplets to coalesce while you skim off the 'slag'? Making the potassium is one thing, cleaning it up is another... some people have done good work actually *trying* something. Its not like making K is the holy grail of chemistry, but its still pretty neat that it can be done so readily.

len1 - 17-12-2010 at 19:26

Coalescing and cleaning is no problem - I have done that a lot with electrolysis potassium and this is no different. My biggest worry at the moment is that this might be better than my electrolysis method, there are so few balls that coalescing and purifying should be a matter of 1/2 hour. The only advantage to the electrolysis I see now is that its much faster once youve got the setup, but the setup takes far longer.

NurdRage - 17-12-2010 at 20:00

Noteworthy results from my latest attempt:

http://www.youtube.com/watch?v=DUzsyNLuLyg

The first half of the video is drops of molten potassium floating in hot solvent. The second half is me testing one of the solidified drops.

Magpie - 17-12-2010 at 20:30

The first half of the video is drops of molten potassium floating in hot solvent. The second half is me testing one of the solidified drops.

Very nice! Those moving droplets are hypnotizing.

len1 - 17-12-2010 at 21:02

Even with a large yield from the experiment, the white solid by product from the reaction in D70 (MgO and some unreacted magnesium) contains some initially invisble specs of potassium.

Unlike sodium, potassium pieces of ~0.5gm or more can explode on contact with water and start a fire. Even if there is no reaction on first adding water, there can still be potassium in the mix because water, and D70 protecting the potassium, dont mix. And a fire can start later on.

I quenched the byproduct by adding ethanol dropwise - which reacts with potassium quickly (t-butanol is too slow) and mixes with D70. The ethanol must not be added too quickly - the mixture can burst into flames - as mine just did. Thats a problem because a fire with burning alcohol and potassium can not be extinguished with water. The lab CO2 extinguisher however made short work of the inflamation.

Cuauhtemoc - 17-12-2010 at 21:52

Hi,

Good experiment, I will be trying it soon with different solvents, I don't think I can get Shellsol D70 here anyway. I just need to buy more magnesium, I'm out of it. So it may take until next year.

Oh, by the way, nurdrage, you didn't use d70 did you? If not could you explain a little further your procedure?
What would be candle wax solvent? I don't think I've ever heard about it here.

[Edited on 18-12-2010 by Cuauhtemoc]

aonomus - 17-12-2010 at 21:57

Wow, this thread got stickied real fast! I can't wait to try this out in the new year. If it works, I'll aim to ampoule some potassium metal under argon (or nitrogen if I can't get argon)

NurdRage - 17-12-2010 at 23:20

Quote: Originally posted by Cuauhtemoc

Hi,

Oh, by the way, nurdrage, you didn't use d70 did you? If not could you explain a little further your procedure?
What would be candle wax solvent? I don't think I've ever heard about it here.

[Edited on 18-12-2010 by Cuauhtemoc]

Candle wax solvent is exactly the title. Candle wax.

Basically you find white unscented candles. These are extremely plentiful this time of year around christmas. The absolutely cheapest brand you can find is what you want as it will be just paraffin wax.

Its melted directly into your reaction vessel and you can heat it to 200 celsius (or more) without boiling.

The major problem with candle wax is that you have to work with it while it's hot, around 80 celsius. Working with it too close to it's melting point, around 60 celsius, isn't good because it solidifies quickly around the neck and sides of your reaction vessel. but at 80 celsius the potassium is molten and its an extreme fire hazard. Even at 80 celsius though, when you pour it, a thin film of wax remains in your container.

As len1 pointed out, tiny bits of potassium are still left in the solvent even after you remove the major balls.

During cleanup this is still essentially "devil's C4" because the bits of solidified wax on the various pieces of glassware you use don't react with water or alcohol unless the wax is cracked or deformed to reveal the potassium bits. This however is extremely unpredictable. heating the apparatus to melt the wax off is also nearly suicidal, i had several tiny fires when i did so.

While the candle wax solvent is viable, i think it's too hazardous because it solidifies onto everything you use.

I'm working on a safer approach but for now i strongly recommend NOT to use candle wax unless you have full fire safety protocols and equipment in place.

Cuauhtemoc - 17-12-2010 at 23:43

I see, I was thinking about using Kerosene, which is also mentioned in the patent. Although I need to be careful because I've read that kerosene may easily ignite at such high temperatures.

I agree that candle wax looks very dangerous, so I won't be trying it.

plastics - 17-12-2010 at 23:55

Well done len1 and Nurdrage.

I have only had success like woelen. For me the key seems to be nice fresh oxide-free Mg. This seems to be the common thread with Pok, len1 and Nurdrage.

Anyone have any ideas how to remove surface oxide from Mg turnings. Or like Pok am I going to have to sit there with an Mg block and file?

Again well done

Taoiseach - 18-12-2010 at 03:29

Ok now how do we go about making best use of this groundbreaking discovery?

I suggest we go on and overcome another difficulty for home experimenters, the aquisition of inorganic azides. It should be relatively straight forward to manufacture potassium azide from metallic K. First you need to react it with dry NH3 to form potassium amide. Next, dinitrogen monoxide (laughing gas = whipped cream propellant) is used to convert it into potassium azide. This is a well known procedure with good yield and no special apparatus or exotic chemicals are needed.

len1 - 18-12-2010 at 05:57

The patent says there is vigorous evolution of hydrogen at 130C. There is certainly a vigorous evolution, but did they check what it is? Certainly Mg reacts casually with hot water to give hydrogen, and with steam at 130C the reaction can be assumed to be vigorous, but the reaction is heterogeneous here, as the water can not dissolve in the D70, meaning some water must escape unreacted.

Now thats were an interesting point comes in. If there is relatively little exposed 'fresh' magnesium surface, such as was the case in my shavings experiment, there will be enough water to react with all exposed magnesium surface - and indeed from memory all magnesium was dulled at the end of my experiments three years ago. If the magnesium is in the form of fresh filings - some magnesium will retain its shine - ie it will not be covered in oxide.

This reaction appears to be one of t-BuOK which from the experiment is clearly soluble to an extent in D70, and Mg. It is likely, as in the case of grignard reagents, that the reaction will not occur with dulled magnesium. But unlike the Grignard case, it is not sufficient to have some active Mg surface to get some reaction. You must have sufficient active Mg surface AFTER the reaction with water at 130C, to get any reaction at all. So the surface area appears to be the key.

A new option however also opens up. If the above is correct, and one does not have fine magnesium, one can hold back part of the fresh Mg until 200C has been reached. It appears also advantageous to let any free water escape and not employ a reflux condenser until after the evolution at 130C

[Edited on 18-12-2010 by len1]

blogfast25 - 18-12-2010 at 07:24

len1:

Are we now sure the byproduct is MgO, not Mg(OH)2? Could you verify that: wash with white spirit or such like, dry at low temp., then determine weight less at 500C or so?

We now need to try and understand the mechanism better, in particular why the t-BuOH seems to have a preference to react with the KOH, rather than with the formed K. And the role of water in all this…

I think it’s safe to say that t-butanol is the catalyst and that the reaction depends physically on some short range – boundary layer diffusion of the activated species (T-BuOK).

The yields are fine!

And nurdrage’s experiments seems to clearly indicate the solvent is not that important, within reason…

watson.fawkes - 18-12-2010 at 07:43

The patent says there is vigorous evolution of hydrogen at 130C. There is certainly a vigorous evolution, but did they check what it is? Certainly Mg reacts casually with hot water to give hydrogen, and with steam at 130C the reaction can be assumed to be vigorous, but the reaction is heterogeneous here, as the water can not dissolve in the D70, meaning some water must escape unreacted.

This same issue has vexed me. It's seems clear that both H2 and H2O are products of various reaction pathways, but I've seen nothing about when and how much these products are produced. I would guess that the original patent authors did originally measure it to understand the science, but then didn't disclose it for the patent because they didn't need to. Patents, after all, are commercial recipes; you have to disclose how to make it go but not why it works. It would seem straightforward enough to put a condenser on a separate collection flask and a syringe to measure non-condensed, evolved gas.

Quote:

Now thats were an interesting point comes in. If there is relatively little exposed 'fresh' magnesium surface, such as was the case in my shavings experiment, there will be enough water to react with all exposed magnesium surface [...] You must have sufficient active Mg surface AFTER the reaction with water at 130C, to get any reaction at all. So the surface area appears to be the key.

A new option however also opens up. If the above is correct, and one does not have fine magnesium, one can hold back part of the fresh Mg until 200C has been reached.

I had a similar thought about a second addition of Mg; see below. The sum of the observations you're making here imply that the solvent-based step of the process is divided into two parts: (1) dehydration of KOH with Mg and (2) KOH / Mg substitution with t-BuO moiety. Previously, these had been done in a single setup, but there's no reason to think of them as two distinct steps. For example, inert gas cover seems far more important for the second step than the first. If the goal is to remove water in the first step, then the right thing to do is to use a different condenser set up in that step. It's mechanically more difficult to change setups will a flask full of hot oil, but it hardly seems insurmountable.

As to the Mg, given that the Mg is reacting in two different ways here, there's no reason to assume that the same mechanical form of the Mg reagent is appropriate to both steps. For the second step, contiguous surface seems necessary, or at least greatly desirable. For the first step, however, total surface area would maximize the rate and completion of dehydration. So it would seem that a protocol might recommend (but not require) very fine mesh Mg powder for the first step and require shavings or turnings for the second.

My previous idea was a variation on this, which is to add fresh Mg to the reaction with each addition of tert-butanol. I didn't mention it because there's no readily available equipment to add large particles a little bit at a time under inert atmosphere; it's not like there's a glass screw feeder that operates in a hermetic atmosphere on most (any?) chemist's shelves. The suggestion to add Mg before the first t-butanol seems to meet my concern and has the advantage of being feasible.

blogfast25 - 18-12-2010 at 07:55

watson:

What about imagining a wholly different way of drying the KOH in situ? Some sacrificial K perhaps?

watson.fawkes - 18-12-2010 at 07:58

Are we now sure the byproduct is MgO, not Mg(OH)2? Could you verify that: wash with white spirit or such like, dry at low temp., then determine weight less at 500C or so?

I recommend a fire assay on this one. Weigh the crucible and Mg(OH)2 / MgO contents first, after dissolving out any Mg(t-BuO)2 and drying out any residual solvent. Then calcine the crucible and dehydrate Mg(OH)2 --> MgO + H2O. The mass difference between final and initial state yields the amount of hydroxide in the initial charge.

watson.fawkes - 18-12-2010 at 08:13

What about imagining a wholly different way of drying the KOH in situ? Some sacrificial K perhaps?

That's a good idea. I'm guessing, without having done a total energy computation, that it's more economical to dehydrate with Mg. But it might be feasible to dehydrate with Al as well, although I'm guessing that free Al post-dehydration might poison the reaction. So start with an estimated amount of Al and run to completion. Then add more KOH (with new water) if there's unreacted aluminum still present. Heat that up to dispose of metallic aluminum. Finally add fine Mg to dehydrate. This protocol uses less Mg per unit KOH. I'm not sure it works, though, and I'm not sure it's worth the bother.

It's probably worth doing the experiment to see how much water can be driven off with just heat.

I don't know if molecular sieves would work for this, but they might. I doubt all the equipment required to regenerate the sieves makes their use better than electrolysis, though.

Nicodem - 18-12-2010 at 08:47

I'm glad all the paranoia and the ridiculous (though very entertaining) flame wars are finally settled away by experimental confirmation.
Congratulations to all who finally got their balls! It was fun reading about these adventures.

We now need to try and understand the mechanism better, in particular why the t-BuOH seems to have a preference to react with the KOH, rather than with the formed K. And the role of water in all this…

I don't think it is proper to say that t-BuOH would have a preference to react with KOH rather than K. Contrary to the reaction of K with water, the reaction

KOH(s) + t-BuOH(solution) <=> t-BuOK(solution+KOH phase) + H2O(KOH phase)

is reversible and in such a multiphasic system depends on various effects. For example, due to solubility issues, the t-BuOH can only react with KOH at the phase boundary with the slight solubility of t-BuOK in alkanes and hydration of KOH being the driving force (anyone who ever used solid KOH for deprotonation of weak organic acids of pKa > 15 in aprotic organic solvents would know what effect I talk about - very useful in organic synthesis!).
It appears from the above described experiments that the moisture in KOH somehow reacts with Mg (prior to t-BuOH addition). I can only imagine this happening at the phase contacts between KOH pellets and Mg metal (assuming the evolved gas is H2 at all). The reaction between Mg and t-BuOH should be much slower and I doubt much K(s + Mg phase) forms before all the protic species (t-BuOH and H2O) are consumed or steam distilled away. Thus, my suggestion to the next motivated enough experimenter is to measure the gas phase temperature of the refluxing mixture as this should give an indication of whether K globules start forming before all t-BuOH is converted to metal akoxides or later only.

Quote:

I think it’s safe to say that t-butanol is the catalyst and that the reaction depends physically on some short range – boundary layer diffusion of the activated species (T-BuOK).

It would be more proper to call it a precatalyst and t-BuOK being the catalyst, or more properly phase transfer catalyst. The most likely explanation in my opinion is that there is a

2 t-BuOK(solution) + Mg(s) <=> (t-BuO)2Mg(solution) + 2 K(s)

redox equilibrium in the solution/solid system. If extrapolating redox potentials form aq. chemistry, one would tend to believe the equilibrium is much in favour of the left side, though generalizing among such very different reaction media and species involved is not reliable at all. Experimentally, it might be possible to evaluate this equilibria by measuring the potential of a galvanic cell having Mg and K electrodes immersed in the "electrolyte solution" of either metal akoxide in a alkane solution (the conductivity would be close to zero, but with a voltmeter of very high internal resistance it might actually be feasible to measure something).
Nevertheless, even if the equilibrium constant at 200 °C is only in the range of a couple of negative magnitudes, an efficient removal of (t-BuO)2Mg from the solution should drive the reaction toward the right (forming more K). Among the (un)available possibilities, the most obvious such reaction could occur at the phase boundary of KOH:

(t-BuO)2Mg(solution) + 2 KOH(s) => Mg(OH)2(s) + 2 t-BuOK

This one should be near to irreversible and can be experimentally verified or disproved.

I don't put much faith in such a catalysis mechanism, but if you need a preliminary hypothesis, that's the best I can offer with my limited knowledge of inorganics. Someone from the inorganic field ought to give a better one. Thus the catalytic cycle would be the combination of the above equations:

2 t-BuOK(solution) + Mg(s) <=> (t-BuO)2Mg(solution) + 2 K(s)
(t-BuO)2Mg(solution) + 2 KOH(s) => Mg(OH)2(s) + 2 t-BuOK

Quote:

And nurdrage’s experiments seems to clearly indicate the solvent is not that important, within reason…

I would not call it unimportant, as the solvent apparently needs to fulfil quite some criteria. It is obvious that it needs to be completely inert to KOH, Mg and K at the reaction temperature (+ high bp). So I would rule out most solvents already on this basis. Only fully saturated alkanes seem appropriate, if nothing else, due to potassium. Making assumption based on the above hypothesis it also appears that the solvent must be as non-polar as possible, even though this means lower solubility of the alkoxides. Higher t-alkoxides (t-amyl alcohol and above) might speed up the reaction if the limited solubility of t-butoxides is found to be an issue.

NurdRage - 18-12-2010 at 08:59

Quote: Originally posted by Nicodem

Higher t-alkoxides (t-amyl alcohol and above) might speed up the reaction if the limited solubility of t-butoxides is found to be an issue.

On this point i'd like to add my purely qualitative observation that t-amyl alcohol did seem to give me greater rate. the hydrogen bubbles seem to go twice as fast than my trials with t-butanol.

I only tried t-amyl alcohol once though and thus i can't say definitively that it's better. But from what i did see, i think it's worth exploring longer chain alcohols.

blogfast25 - 18-12-2010 at 11:28

Quote: Originally posted by Nicodem

Higher t-alkoxides (t-amyl alcohol and above) might speed up the reaction if the limited solubility of t-butoxides is found to be an issue.

Nurdy, how do you add your t-alcohol? All at once? Slowly? In stages?

I'VE GOT BALLS!

woelen - 18-12-2010 at 11:40

Right now the ShellSol D70 is happily refluxing for 2 hours already and I now already have potassium balls ranging in size from 1 mm to 3 mm (pictures are made of these, they will be shown in a final write up). I'll let it reflux for another 2 hours, but it seems to work.

I have taken 50 ml of ShellSol D70 and added 7 grams of KOH and 3.5 grams of Mg powder (appr. 100 um size). I put this in a sand bath. When temperature reaches 150 C, then a vigorous boiling starts and white fumes are produced. At this point I opened the apparatus and allowed the white fumes to escape. I left it open till temperature of the sand bath reached 170 C. This must be water vapor, condensing in the cold air.
At 170 C the boiling almost comes to a halt and at that point in time I put on the reflux cooler. Temperature went up further and around 200 C there was slight simmering, at 220 C there was a nice constant simmering. This must be the boiling ShellSol D70.

In the meantime I prepared 0.7 ml of tert-butanol with 7 ml of ShellSol D70. When temperature reached 225 C I added appr. 2 ml of this mix. Five minutes later I added another 2 ml. Ten minutes later I added a third portion of 2 ml and almost 10 minutes after that I added the final amount. So, in total I added 0.7 tert-butanol dissolved in 7 ml of ShellSol D70. Each time when this was added, a vigorous boil started, which quickly subsided, however. I added these small amounts of liquid through the reflux cooler, so that it is not necessary to open it and loosing lots of vapor.

After the addition of the last amount of tert-butanol I left it refluxing for 1.5 hours and in this time the balls of potassium formed.

To be continued....

NurdRage - 18-12-2010 at 11:50

@blogfast25

I added my alcohol all at once through the top of the condensor without dilution. not really possible to dilute the alcohol since my solvent is candle wax.

@woelen

Excellent work! you should make videos and put them on youtube, you'd do great.

blogfast25 - 18-12-2010 at 13:23

Yep! The K is mine too!

I didn’t get ‘Great balls of K!’ but loads and loads of little ones: about 1 mm or less, which react very vigorously with water.

‘Apparatus’: same as pok, same quantities. Nice and glistening 99 + % Mg:

Syringe with long tube of chemical Nylon tube, allows to inject the t-butanol deep into the refluxer (cooled with iced kitchen roll):

The usual gas evolution started quite early and kept going for 15 – 20 mins (I held at about 170C). Then the ritual of t-butanol/Shellsol mixture over about 15 mins, commenced at 190C – 200C. Almost immediately the solvent started to cloud over and this continued basically all throughout out the experiment with sediment forming gradually and the KOH disappearing. I cranked up the heat till the temperature was outside the range of the thermocouple (200C) and kept it there, all the while tending to the refluxer with iced water. Start time 16.45, end of test 19.15. After cooling this is what it looks like, lots of white powder at the bottom, some flakes (unreacted KOH?) also to be seen:

Feeling pretty sure there was K there, I decided on a partial work up by rinsing the metal with fresh Shellsol a few times: the blueish colour of the metal then became very apparent, although this is a particularly bad shot:

Quite a few were captured on a plastic tea sieve and treated with water: the ensuing reaction produced gas, lilac flames and burnt holes in my sieve! Complete dissolution, no residue.

Tomorrow I’ll attempt to work up some more… and run another test.

But there is no doubt to me: this work but needs improvement…

[Edited on 18-12-2010 by blogfast25]

blogfast25 - 18-12-2010 at 13:31

Thanks Nurd, will keep that in mind for next test.

@woelen: fantastic and neat little idea to mix the t-butanol into a larger volume of Shellsol.

All doubts gone now :cool:

. Bizarre way of making potassium and a victory for backyard science!

blogfast25 - 18-12-2010 at 13:35

Thanks also to Nicodem for his contribution above. This is more or less the reaction mechanism I proposed.

[Edited on 18-12-2010 by blogfast25]

cheeseandbaloney - 18-12-2010 at 13:38

I'm more of a browser than a poster, but I just wanted to say that I fucking love you sciencemadness. What a great thread with a interesting discussion. Gotta add this to the list of reactions to undertake. peace!

woelen - 18-12-2010 at 14:05

Yes, this really is a victory. I now have 3 balls of potassium, one has a diameter of just below 1 cm and the other 2 have diameters of well above 1 cm. Precise measurements are hard, due to the grey liquid in which the balls reside. Besides the 3 big balls I have many smaller ones, stuck in grey gunk. Given the density of potassium which is appr. 0.89 grams/ml I estimate to have around 3 grams which can be isolated. The tiny balls of less than 1 mm diameter I will not isolate, simply too much of a hassle.

This night I will let all of it cool down in a sealed flask and tomorrow I'll transfer the balls of potassium to some clean mineral oil.

len1 - 18-12-2010 at 14:10

The difference in what nicodem wrote from the reaction I and blogfast had been proposing is that MgOH2 is formed instead of MgO.

No Im not sure that I got MgO instead of MgOH2. I know the dry hydroxide decomposes at higher temperatures - I wrote that in the post 3 years ago - but my feeling is that the hydroxide wont survive at 200C with t-BuOK around, that stuff is a desicator, and I presume the thermodynamics will favour dehydration, but maybe not the kinetics. The test of course is very simple and Ill do it.

The other thing is that there IS a constant gas evolution in the course of the reaction. I suppose one could maintain its residual water from the KOH reacting with the magnesium. 5.8gm K corresponds to about 1.8L hydrogen, or about 10ml per minute for a 3 hour reaction. I had been observing about that. On the other hand MgOH2 is known to form crusts, I havent heard of MgO doing that. It could be a combination of both. Ill check.

Well done Wilco!

[Edited on 18-12-2010 by len1]

blogfast25 - 18-12-2010 at 14:13

The other thing is that there IS a constant gas evolution in the course of the reaction. I suppose one could maintain its residual water from the KOH reacting with the magnesium. 5.8gm K corresponds to about 1.8L hydrogen, or about 10ml per minute for a 3 hour reaction. I had been observing about that. On the other hand MgOH2 is known to form crusts, I havent heard of MgO doing that. It could be a combination of both. Ill check.

You sure about that? I mean with all that boiling, how do you distinguish between vapour bubbles and other gas?

It really would be worth separating the 'slag' and checking for MgO v. Mg(OH)2, IMHO...

[Edited on 18-12-2010 by blogfast25]

len1 - 18-12-2010 at 14:27

Remember I had a 30cm condenser connected, so that not much condensable matter would have made it past that. And anything that did would have condensed in the bubbler.

Yet a had a steady stream of bubbles throughout the experiment. So there is no doubt hydrogen is evolved. One could mainatin its all from Mg reacting with the KOH water. Or MgO and MgOH2 being formed simultaneously.

This is such an unexpected result, I thinking only testing can show.

blogfast25 - 18-12-2010 at 14:28

I've notified Theodore Gray (from periodictable.com (in)famy). He might want to put this forum on the map via PopSci...

blogfast25 - 18-12-2010 at 14:30

Remember I had a 30cm condenser connected, so that not much condensable matter would have made it past that. And anything that did would have condensed in the bubbler.

Yet a had a steady stream of bubbles throughout the experiment. So there is no doubt hydrogen is evolved. One could mainatin its all from Mg reacting with the KOH water. Or MgO and MgOH2 being formed simultaneously.

This is such an unexpected result, I thinking only testing can show.

Hmmm...my money's on H2 from water + Mg. Intuitively speaking...

Pok - 18-12-2010 at 14:42

Many small K balls could possibly get unified by adding a few drops of isopropanol to the inert fluid in which the molten potassium balls are. Also oxidized potassium can be cleaned by this method.

http://translate.google.de/translate?u=http%3A%2F%2Fwww.vers...

See description of photo 1 - 4 there.

[Edited on 18-12-2010 by Pok]

garage chemist - 18-12-2010 at 14:50

I'm about to place an order for 3L of Shellsol D70, but just had the idea of trying some of the heating oil that our central heating runs on as a solvent. We have like 3000L of this in our basement.
This is a simple crude oil distillate with a boiling range of 150- 390°C to which a marker substance (furfural) and red dye has been added to allow the detection of its diversion towards use as a diesel fuel, which is illegal due to the reduced tax on heating oil and the high tax on all automotive fuels.

I would have to vacuum distill this to get rid of the furfural in the first fraction and obtain a cut with (atmospheric) boiling range of 200-300°C. I could then try this out as a reaction medium. Shellsol D70 boils from 195-245°C.

However, I've read that potassium cannot be stored under aromatics like toluene since it can abstract an alpha hydrogen, forming potassium benzyl which undergoes further reactions, slowly consuming the potassium already at room temperature.
Heating oil contains aromatics, while Shellsol D70 is specified as aromatic-free.
Does anyone here know anything about the true nature and extent of the incompatibility of K with aromatics?

[Edited on 18-12-2010 by garage chemist]

NurdRage - 18-12-2010 at 15:22

Quote: Originally posted by Pok

Many small K balls could possibly get unified by adding a few drops of isopropanol to the inert fluid in which the molten potassium balls are. Also oxidized potassium can be cleaned by this method.

http://translate.google.de/translate?u=http%3A%2F%2Fwww.vers...

See description of photo 1 - 4 there.

[Edited on 18-12-2010 by Pok]

I can confirm this works.

I reccomend though keeping the temperature around 80 celsius. Too high and you risk spontaneous combustion if a tiny piece of potassium floats up and hits some stray moisture.

garage chemist - 18-12-2010 at 15:30

This was my work. I'm Stefan at Versuchschemie.
I made Na/K alloy from the metals, cleaned the oxide off the alloy by adding IPA to the paraffin oil that it was melted in, and put the metal into a vial.
I use this alloy to dry ether. It's very efficient due to its fine dispersion when vigorously shaken.

S.C. Wack - 18-12-2010 at 16:30

Quote: Originally posted by garage chemist

However, I've read that potassium cannot be stored under aromatics like toluene since it can abstract an alpha hydrogen, forming potassium benzyl which undergoes further reactions, slowly consuming the potassium already at room temperature.

I think if there's no chlorobenzene it's OK. Xylene is often recommended when handling K, like in OS. I wonder why xylene, specifically, instead of only slightly more volatile toluene.

Working with K under any hydrocarbon can really suck, under certain conditions:
http://pubs.acs.org/cen/safety/potassium.html

len1 - 18-12-2010 at 16:30

I have coalesced purified potassium and sodium under toluene and xylene. And although I finally preferred to do it under D70 or paraffin, due to the inflamability of toluene and zylene, there was no visible reaction at up to the boiling points of these liquid (time scale of several days). Over the course of a year there is some reaction in D70, although this maybe be due to some non-parafinic impurities in the latter - it is jot an analytic reagent after all.

I have tried Stefan purifying Na or K under D70 with isopropanol, but could never get reproducible results that way, so I now developped other means.

With regards to the reaction, the key argument seems to be whether t-BuOK dehydrates Mg(OH)2. If it does hydrogen and MgO will be the reaction products, if not there is no gas released by the main reaction at all (contrary to the precise measurement in the patent).

This reaction should also be sensitive to temperature - clearly dehydration will occur spontaneously at about 580C. Just perhaps, at lower temperatures t-BuOK does not dehydrate MgOH2. So that the latter would be the final product. While at higher temperatures it would, and fine MgO sand would form. If this is true it would explain why Pok produced no MgO sand - he certainly could not run his experiment at high reflux due to the nature of his 'condenser'.

NurdRage - 18-12-2010 at 17:15

I just confirmed i can get decent potassium balls (1-3mm) using parafin oil.

Procedure was:
20mL paraffin oil, 5g KOH, 2g Mg heated to 170C, then 1mL of t-amyl alcohol added and condensors and bubblers were put on. Heating was brought up to 220C and continued until unambigous shiny potassium balls were visible. (~3hr)

Next i'm going to try pharmacy mineral oil.

After that i might try gamsol or varsol. Even though the boiling point is lower i find the temperature isn't too critical, just as long as you're willing to wait long enough for it to go completion.

chen - 18-12-2010 at 17:38

paraffin oil as in kerosene?

NurdRage - 18-12-2010 at 17:50

mine was higher density. the bottle reads 0.86g/mL, i think commercial kerosene is lower.

Unusual observation

NurdRage - 18-12-2010 at 18:39

Is the alcohol required?

During one run i hadn't added the alcohol yet but had to cancel the reaction.

So i started to clean up my fume hood and tossed the reaction mixture of paraffin oil, KOH and Mg into a water bath and to my surprise i saw some tiny but distinct flashes of lilac flames.

I had heated the mixture to 170C for ten minutes but *did not yet add alcohol* before i had to stop and cancel the run.

It might be that my parafin oil, KOH, or Mg was contaminated with some t-alcohol. and produced a minute amount of potassium. But if that wasn't the case, then my observation might mean the alcohol is not critical to the reaction.

I won't have time to try again for another few days so if anyone wants to do a test run without any alcohol it would be interesting to see the results.

[Edited on 19-12-2010 by NurdRage]

len1 - 18-12-2010 at 18:41

The other thing of course is that formation Mg(OH)2 in the overall reaction

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

consumes half the magnesium than the reaction claimed in the patent

2Mg+2KOH -> 2MgO + H2 + 2K

which would mean the magnesium is present in over *2 excess in the patent.

Just judging by the amount of Mg in the final product - which was little, it appears the first reaction is either excluded, or forms a minor component. Hence I think the t-BuOK dehydrates the Mg(OH)2 - but will now check.

With regard to nurdrage's result, Im not surprised. Obviously the direct reaction as written can also occur. Provided some fresh Mg surface survived the water evolution, some K will form directly by the overall reaction. Its just that there is bad contact between the Mg and KOH, and my guess is you wont get an 80% yield. The t-BuOH provides and indirect route which gets around the kinetic impediment.

But here is an interesting observation. The KOH liquifies at about 150C in the D70 - at the point where initial reaction with Mg is observed - so this is a liquid-solid, not solid-solid reaction, and so is expected to go a lot better. The liquification is due to the crystalization water in the KOH - which however is not evolved at this temperature (in the dry it starts dehydrating at about 250C). So Mg is needed to dehydrate the liquid KOH, whereupon it is locked inside a dry KOH lattice. Any quality Mg I believe will work in the dehydration stage.

[Edited on 19-12-2010 by len1]

More potassium porn

NurdRage - 18-12-2010 at 20:03

Selected highlights from today's run using paraffin oil

http://www.youtube.com/watch?v=AgfxmCAS4hs

First segment is the reaction mixture timelapsed over an hour. You can see layer separation with the metals rising on top of the oxides/hydroxides.

Second segment is about 3 hours into the run and potassium balls are clearly visible in the mixture.

Last segment is after hydrogen evolution completely ceased. Large potassium spheres (>5mm) are now visible.

I apologize for the poor video quality. My HD camera wasn't available so i had to use my pocket camera.

[Edited on 19-12-2010 by NurdRage]

len1 - 19-12-2010 at 00:13

Well Ive made another run and there are some conclusions.
The gas given out during the reaction was collected in a gasometer and its quantity eliminates Mg(OH)2 as any sizable byproduct of the reaction. It must be MgO.

Using the previous amounts, gas evolution prior to the addition of butanol amounted to 1740ml. Towards the end the evolution slowed down considerably - but never really stopped. The amount is 5% in excess of whats expected from

Mg + H2O -> MgO + H2

with 10% nominal H2O in KOH in the reaction, but thats well within expectations. There is no support for any quantifiable amount of K being evolved prior to the addition of butanol.

Addition of t-Butanol did not immediately change things. However within 10mins the H2 livened up again, and after 1/2hr it was gathering in the gazometer at a sizeable rate. The appearance of the reaction mixture also changed. The D70 - fairly clear to this point became turbid, and fine MgO sand started gathering at the bottom.

This supprots my previous theory. After all water has been reduced, there seems to be a period of initiation, similar to what happens with grignards. This is when the first amounts of t-BuOK form and reach their final concentration. After this it rapidly reduces potassium and oxidizes magnesium. And the reason why my attempt from 3-years ago failed appears to come out of this. With less fresh magnesium at the start, it was all covered with oxide when the t-BuOH was added, and the initiation failed. Holding back some fresh Mg might have helped here.

With the reaction simmiring, stirring, judging by the rate of gas evolution does not seem to have any influence on the reaction rate. And during the initiation phase it seems to hinder it - with no bubbles evolved for about 30s each time the contents were stirred. Its only role I think is to coalesce the K.

The reaction appears complete in 3hrs after commencement of heating - with the last stages only coalescing potassium - after the balls have reached 3-4mm in diameter no new potassium is formed.

The gas evolved during the second stage of the reaction amounted to 2250ml. There were 1.4gm t-BuOH added, contributing 180ml to this. Theoretical is 2.43L by the below reaction

2Mg + 2KOH -> 2MgO + 2K + H2

112->24l

11.3 -> 2.43l

so the yield when I come to weighing the potassium should be 85%. Much as I had before.

Once again I got plenty of MgO sand. The balls are not clean and will need purifying. I would be interested to hear from anyone who succeeded in getting just crust.

And heres another interesting point that seems to flow out of that. Sodium should be more difficult to form by this method. Its butoxide (and hydroxide) are far less soluble in non-polar solvents than the potassium salts.

[Edited on 19-12-2010 by len1]

NurdRage - 19-12-2010 at 00:53

alright, i guess my potassium observations before the addition of alcohol are probably just a cross-contamination issue.

thanks for checking len1.

The gas production in support of the Mg + H2O -> MgO + H2 theory is important though.

As for "crusts" if you look at my latest video (http://www.youtube.com/watch?v=AgfxmCAS4hs ) at the end of the reaction, instead of fine sand for the white MgO, i get "chunkier" solid MgO. Is this what you're getting? or do you think my stuff qualifies as "crusts"?

[Edited on 19-12-2010 by NurdRage]

len1 - 19-12-2010 at 01:37

Hi Nurdrage, I looked at your video, but its hard to tell due to the resolution. Maybe you can tell me. If its crust, then its rock hard, you cant put a stirring rod through it, if its sand, then you can. Your K balls also float - I guess you are using paraffin - near 80C K actually floats in it, but at higher and lowerr Ts it settles. In D70 it is alsways heavier. The rising and falling might actually be benificial, with the lumps getting rid of the MgO that way. Your balls are certainly much cleaner than mine. Maybe paraffin is better - what kind of yield are you getting?

As for the reaction - my gazometer results are accurate to only a few percent. The KOH could easily have 9% or 11% water. SO it does not preclude a small amount of potassium forming before the t-BuOH is added. But it does preclude large amounts.

DJF90 - 19-12-2010 at 02:06

Your balls are certainly much cleaner than mine.

Lol.

On a more serious note, in all the excitement it looks as if some serial doubleposting has gone un-noticed. How hard is it to actually use the edit button... geez.

But excellent work with the potassium. Its nice to see that this has come to a nice conclusion, unlike many other patented procedures.

NurdRage - 19-12-2010 at 02:08

I destroyed the mixture with alcohol before i could really test its physical properties so i don't know for certain if it's rock hard. i'll check again in a few days when i have time to run another reaction.

As for the yield, i didn't do quantitative measurements but by eye it seems to be better than 50%. There were several large clean balls in the final product.

condennnsa - 19-12-2010 at 02:22

Is the alcohol required?

During one run i hadn't added the alcohol yet but had to cancel the reaction.

So i started to clean up my fume hood and tossed the reaction mixture of paraffin oil, KOH and Mg into a water bath and to my surprise i saw some tiny but distinct flashes of lilac flames.

I had heated the mixture to 170C for ten minutes but *did not yet add alcohol* before i had to stop and cancel the run.

It might be that my parafin oil, KOH, or Mg was contaminated with some t-alcohol. and produced a minute amount of potassium. But if that wasn't the case, then my observation might mean the alcohol is not critical to the reaction.

[Edited on 19-12-2010 by NurdRage]

Interesting. But how could there be alcohol contamination considering your reagents were fresly added?

The patent only says that the alcohol is a reaction accelerator.

If it's true that it works without the alcohol, what would the process look like?

I can't believe that we're in late 2010 and this amazingly easy synthesis of alkali metals is so unknown ...

This a truly glorious outcome for this thread!

woelen - 19-12-2010 at 02:59

In my experiment, the liquid become turbid more and more and at the end, it was dark grey and you only could look into the liquid for a few mm.

A crust was formed at the bottom and the balls of potassium were on top of this.

Today is the day of working up all of this and processing the pics and vids I made. Next week I hope to make a real web page about this, for the time being just the pics and vids.

len1 - 19-12-2010 at 03:25

Above about 230C I found patassium can metallate some paraffins and lead to decomposition - so perhaps the temperature was a tad high.

As for the reaction without butanol - its just staight according to the overall formula. Possibly KOH has some residual solubility in D70 (I observed what could be some precipitation on heating and cooling KOH in D70 alone. In any case the reaction does not go very far.

Its a good method for producing K for the home chemist. But its not really a blight on the 21st century that its not being used elsewhere. Its quite wasteful of magnesium - in itself an expensive metal, and is very slow. My electrolysis gear produces about 80gms K in the same time that this takes to make 5.8 in similar sized aparatus. What I find most interesting about it is the reaction.

[Edited on 19-12-2010 by len1]

woelen - 19-12-2010 at 05:53

There is, however, one very important difference between this reaction and the electrolysis you performed. This reaction can be carried out with only very simple apparatus, while the electrolysis requires a much more elaborate setup. This reaction makes obtaining elemental potassium much more feasible for a broader range of amateurs.

So, in terms of efficiency, both in used chemicals and required energy, the electrolysis is much better, but in terms of feasibility for a broad range of amateur chemists, this reaction is better.

blogfast25 - 19-12-2010 at 06:48

But here is an interesting observation. The KOH liquifies at about 150C in the D70 - at the point where initial reaction with Mg is observed - so this is a liquid-solid, not solid-solid reaction, and so is expected to go a lot better. The liquification is due to the crystalization water in the KOH - which however is not evolved at this temperature (in the dry it starts dehydrating at about 250C). So Mg is needed to dehydrate the liquid KOH, whereupon it is locked inside a dry KOH lattice. Any quality Mg I believe will work in the dehydration stage.

[Edited on 19-12-2010 by len1]

Well, in my test hydrogen started to evolve way before that: 50 - 70 C. Will try and confirm that today.

Interesting points about the amount of H2 and MgO.

What about using octane (readily available as lead free high octane car juice) as a work up fluid: very low viscosity... Won't boil on steam bath but you sure do need a decent consenser...

woelen - 19-12-2010 at 08:38

This is my writeup of the reaction, with pictures and a few videos:

http://woelen.homescience.net/science/chem/exps/synthesis_K/...

Have fun reading it. It might be changed in the next few days, but I now already want to share the pics and videos. Forgive me if there are still some typos or other errors.

DJF90 - 19-12-2010 at 09:04

Very nice woelen, excellent write-up as per usual.

Jor - 19-12-2010 at 09:12

I really love this synthesis. I'm also gonna try it in January.
I ordered 250g of magnesium fillings for Grignard from Carl Roth. Are these fillings suitable? What did you use woelen? Did you attempt to make the Mg metal oxide-free?
My hotplate will go to 250C, but i think it won't be able to reach a temperature of 220C if I am using a sand bath, so I will put the erlenmeyer straight on the hotplate, and put the hotplate inside a tray.
Woelen, can you try the same reaction using 'lampenolie' ? As far as i know , this is also hardly contains aromatics and has a high boiling point. If it works I won't have to buy the Shellsol (I don't have a lot of money at the moment). i can't try it myself as I first need to obtain some t-butanol (i don't want to buy a liter, so I'm looking for a source where I can obtain about 50-100mL.
Finnally, is the glass etched when performing this reaction. Although the solvent hardly solvated the KOH, I can imagine that the very hot solid KOH etches the surface of the erlenmeyer?

Has anyone tried this with the preparation of other reactive metals? Metals like barium and strontium would be interesting, but they cannot be isolated I think, because their melting point is simply too high.
Sodium does not seem interesting when you can obtain it, as it is very cheap (about 15 cents per gram for me).
Now this would be a very interesting way to make lithium, but it's really a bummer that lithium has such a low molecular weight, making it impossible to make acceptable amounts of lithium.
But who has either RbOH or CsOH? It would be awesome to make some of the beautiful golden metal! I think CsOH could be made from CsCl (I have 100g of this) by either decomposition of the insoluble CsMnO4 and dissolving in water (woelen told this in an earlier topic) or heating CsCl with conc. H2SO4, diluting and neutraling, and adding a solution of barium hydroxide. However I have no idea how to get the CsOH in a solid form from the solution without damaging glass or any plastic container you use. i think you need a TEFLON vessel.

[Edited on 19-12-2010 by Jor]

[Edited on 19-12-2010 by Jor]

blogfast25 - 19-12-2010 at 09:45

My second run still used the old Shellsol but I made two small modifications:

1. 1 ml t-butanol in 1 ml Shellsol, so slightly higher amount of alcohol.
2. addition of alcohol/Shellsol all at once at the end of ‘step 1’.

Well, well well. That was kind of interesting. I simply couldn’t really add all the t-butanol at once because of very strong reflux, with whiffs escaping from the top of the cooled refluxer tube! I started adding at about 180C, about 30 mins into the run and with difficulty and almost violent refluxing managed to add it all in about 5 mins. I’m sure I lost some alcohol, some water or both through the refluxer top. I very much did not see that yesterday. It really looks (but it may be perception pure and simple) that reaction takes place the moment the alcohol hits the solvent, with such a much invigorated boil but no discernable temperature increase.

Secondly I can confirm that I’m seeing bubbles from about 70C onwards. Low boilers? Hydrogen?

Thirdly the K-balls were a bit larger but still no good coalescence was obtained… I’ll rinse them later on. Total duration of test 1 ½ hours.

At the end during cooling:

Tomorrow a test with ‘medium Kerosene’, FP 61C, ‘Lamp oil’, or ‘ROLF = Reduced Odour Liquid Fuel’…

I also tried to coalesce yesterday’s batch using a steam bath and White Spirits. It didn’t really work although the balls are a bit more visible. The solvent got coloured (test duration also 1 ½ hours) amber-red:

Tomorrow I’ll also try Stefan’s (GarageChemist) tip of mineral oil with IPA.

blogfast25 - 19-12-2010 at 10:02

@woelen:

Yes, very nice write up. Looks like I may just be too impatient: I’ve never gone past 2 hours and you’re talking about 3 ½ hours! Patience is a virtue and good things come to those who wait... Did you calculate yield based on the megaballs only? What value?

Looks like we’ve all agreed on a mechanism now and yes, it’s rather wasteful of magnesium…

Your phased addition of the t-butanol is probably the most practical for small scale tests, I’ll be using it in a test with ‘lamp oil’ tomorrow… One could make up larger quantities of the alcohol mix as a ‘masterbatch’ to be used on several occasions.

I like the idea of the air cooled Liebig a lot, much less messy than water...

[Edited on 19-12-2010 by blogfast25]

woelen - 19-12-2010 at 12:09

@Jor: The Mg I used is pyro-grade stuff from Keten (a Polish supplier of pyrotechnical chemicals). I purchased 250 grams of this one year ago. It is 100 um size and I think it is of good quality (free of oxygen). It has a light gray metallic luster and not the dark luster of oxidized stuff. Its purity of course is somewhat uncertain, remember, it is just pyro-grade stuff, not the reagent grade you have ordered. If your grignard-grade Mg is fine enough, then it should work.

I am inclined to think that the aliphatic solvent to be used is relatively important. Two things are important:
1) Boiling range (must be a range, around 220 C or so). This allows gentle simmering of the mix. This simmering causes some gentle self-stirring, enough to make the K-globules coalesce, but not so much that there is too little contact between the reactants.
2) Aromatic and unsaturated content must be low. The solvent should be a mix of alkanes and cycloalkanes. Alkenes and aromatics are counter-productive.
I believe that any solvent which satisfies (1) and (2) does the job. I am afraid that lamp oil is less suitable. It has a higher boiling range and is more viscous.

------------------------------------------------------------

I now am thinking of how to improve the efficiency of the process. The remaining clear solvent contains a lot of K-butoxide left in solution. If you pour some of this along a glass, then within seconds the glass becomes covered by a gel-like layer. Maybe this can be used in a fresh batch of Mg with less KOH added. Hence, less losses due to water content and less Mg needed. So, it might be that if one batch is used and the solvent from that batch is used for a second batch that less Mg and less KOH and hardly any t-buOH are needed to obtain the same amount of K-metal. I hope to find some time in the period between Christmas and New-Year's eve to do some research on that.

@blogfast25: I indeed have the feeling that you did not wait long enough. The globules I see in your picture are about the same size as the ones I had after two hours. So, if you keep on simmering/heating, then you may obtain larger balls.

blogfast25 - 19-12-2010 at 12:51

Quote: Originally posted by woelen

@blogfast25: I indeed have the feeling that you did not wait long enough. The globules I see in your picture are about the same size as the ones I had after two hours. So, if you keep on simmering/heating, then you may obtain larger balls.

Yeah, I'm pretty sure too. But I'm thinking of maybe to cut the reaction time down to, say 1 1/2 hours, and develop a work up that can coalesce the crude, formed metal really quickly, less than 1/2 hour or so... I rinsed the latest batch with kerosene and it's all potassium, once the byproduct has been washed away.

[Edited on 19-12-2010 by blogfast25]

a_bab - 19-12-2010 at 12:58

Hmm, this whole thread is really a Christmas gift. I said it won't work but I was proven otherwise, which I'm glad about. At least I was right about the presence of the real potassium in Pok's pics.

It also means it would work for Cs and Rb with a carefull choice of the solvent.

blogfast25 - 19-12-2010 at 13:26

Quote: Originally posted by a_bab

Hmm, this whole thread is really a Christmas gift. I said it won't work but I was proven otherwise, which I'm glad about. At least I was right about the presence of the real potassium in Pok's pics.

It also means it would work for Cs and Rb with a carefull choice of the solvent.

Well, yeah. As long as you’re not going to attempt to break the laws of this universe like you proposed doing here:

http://www.sciencemadness.org/talk/viewthread.php?tid=15061&...

a_bab - 19-12-2010 at 13:36

Don't worry, I am not.

It would still work for Ca though. I believe in thermodynamics too.

condennnsa - 19-12-2010 at 13:46

Quote: Originally posted by woelen

@Jor: The Mg I used is pyro-grade stuff from Keten (a Polish supplier of pyrotechnical chemicals). I purchased 250 grams of this one year ago. It is 100 um size and I think it is of good quality (free of oxygen).

I'm also considering buying some magnesium from either Keten, or Czort. Czort has slightly better prices. Woelen, did you have any trouble with Keten with the order?, I've read on UK pyrotechnics that many people were unhappy with his lack of replies to emails and slow deliveries...

[Edited on 19-12-2010 by condennnsa]

blogfast25 - 19-12-2010 at 14:46

Quote: Originally posted by condennnsa

Quote: Originally posted by woelen

You can buy a wide selection of different Mg powders from eBayers. I’m no longer convinced the grade is as critical as we once believed, in the Dark Ages of three weeks ago. About 4 people have managed it, all with differently sourced materials. There may be slight differences in how one grade reacts vis-à-vis another but anything that’s not too oxidised will probably do the trick… before we establish which grade is ‘optimal’ any of real magnesium should be a decent choice. Keten is slow, I can honestly confirm that…

[Edited on 19-12-2010 by blogfast25]

len1 - 19-12-2010 at 15:02

Nice write up Wilco.

A couple of suggestions.

1) My experiment heating KOH in D70 with no magnesium shows that it does not dehydrate in and of itself. Mg is needed to perform this feat. Therefore I no longer believe any water is given of by the reaction, and the white turbid fumes and liquid apparent on the surface of the liebig at 150C are just D70 spray thrown up by the H2 evolution.

2) The gasometer H2 experiment shows that the hydrogen evolution is much too great for Mg(OH)2 to tbe the final product. The second formula therefore looks like this

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

with the regenerated alcohol reacting with some of the potassium already formed. This I believe is what helps coalesce the potassium.

The propagation equations are then

2t-BuOK + Mg -> (t-BuO)2Mg + 2K
(t-BuO)2Mg + KOH -> MgO + t-BuOK + t-BuOH
K + t-BuOH -> t-BuOK + 1/2H2

summing these gives the overall reaction

Mg + KOH -> MgO + K + 1/2H2. (1)

If the second equation was

(t-BuO)2Mg + 2KOH -> Mg(OH)2 + 2t-BuOK

the overall reaction would be

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

meaning no hydrogen is evolved (and using up 1/2 less magnesium for the same amount of K generated), contrary to experiment, which shows hydrogen to be evolved at a rate close to (1) above.

[Edited on 20-12-2010 by len1]

ScienceSquirrel - 19-12-2010 at 15:06

I did not believe that it would work, I could not explain Pok's results but they seemed unlikely to me.
On the basis of the above results it definitely does work and it is a nicer procedure on a small scale than electrolysis of potassium hydroxide.
Congratulations to Pok and the successful experimenters.

Sedit - 20-12-2010 at 00:00

Please all read this thread Here . For further understanding where im comming from.

I got ahold of a sacraficial Mg anode from an old water heater. I have confirmed it contains Mg although I can comment on the purity since I only got it to ignite a couple times and that was only after blasting it with an acetalyne torch. To make it stanger it was normally only the slag I could get to ignite and not the molten metal this is quite possible due to to much heat being carried away while in its moltan stat. This all seems off topic but bear with me.

See I had some clean metal and I have some "slag" and I have a semipure halfway inbetween pure and "slag".

Well after dropping the three samples into H2O as expected hydrogen was released, BUT, expecting the pure metal to be the most reactive went against intuition. The dirtiest slag that I had reacted right away producing H2 in abundence. The lumpy, shiny yet not quite pure was second in hydrogen production.... However....

The cleanest forms I produced and then cut into strips(some cut even finer to represent shavings) was by far the lest reactive of all, taking sometimes between 10-15minutes before the reactivity caught up with that of the dirty metal.

Prehaps the repeated failiars of this Potassium experiment where not due to shitty Mg or improper reaction conditions but had more to do with time frame and reactivity/purity of the Mg used. Higher purity may not mean better results in this reaction or atlest force an extended reaction time since initiation took much longer then the dirty Mg did.

I can only assume that the shitty slaggy Mg skimmed off the top was so much more reactive due to the larger surface area of active Mg laced into a matrix of oxide and the semi dirty was pretty reactive due to the fact that it also had higher surface area to react with.

Im left with a hypothesis here that Magnesium ribbon is almost doomed to failiar without extended reaction time or prior sanding of the Mg surface to allow larger reaction area.

I think that If I decide to perform this reaction using the Mg(possible alloyed Mg or what ever it is I have) I will first melt down the Mg and then slowly drip it into a cold hi boiling point solvent to make a high surface area simular to mossy zinc if possible. This coupled with Finely ground KOH(NaOH in the experiment I plan on performing because my KOH is a solution at the moment) should allow the fine particals of Alkalihydroxide to settle into the pores giving intimant contant between the two reagents at work.

Hope this helps some one its a work in process and just a few observations I just noticed while attempting to clean up Magnesium for scientific use.

GL
~Sedit

woelen - 20-12-2010 at 01:05

@Len1: Thanks for your feedback. This evening I'll modify the webpage. Good that you did the experiment of collecting the gas, so we have strong experimental backup of your proposed reaction equations.

@condennnsa: I also found Keten to be slow. The quality of their products, however, is good. I'll have a look at the Czort site, I did not know that.

Sedit - 20-12-2010 at 01:33