Sciencemadness Discussion Board

Make Potassium (from versuchschemie.de)

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blogfast25 - 10-12-2010 at 10:11

@plastics:

You seem to be injecting the t-butanol straight into the solvent, that’s what I’m planning to do too…

Try if you can to take stock of temperature in the reactor. A household digital thermometer up to 200 C with needle like probe (straight through the bung!) will set you back about 10 quid, Was the solvent at least boiling?

What's the internal diameter of your 'refluxer'?

watson.fawkes - 10-12-2010 at 12:45

Quote: Originally posted by plastics  
There is certainly initial generation of gas after gentle warning and I am able to get the tert butanol to reflux successfully (see pictures). After a couple of hours I just end up with a sludge of magnesium turnings and white powder (? MgO ? Mg(OH)2 this is definitely not the KOH flakes I started with)
Could be potassium tert-butoxide. As Nicodem mentioned, its solubility goes up with temperature. I'll second the blogfast25's call to measure the temperature.

When you're done, it would be useful to know how much of the tert-butanol remains free and how much has reacted. A very simple distillation would be adequate, given the large difference in boiling points.

condennnsa - 11-12-2010 at 00:33

@plastics , it is really disappointing to me to see you failed... considering you used the right alcohol, i think you should've gotten something.

Did the remains of the reaction at least bubble in water when you washed it down?

Let's just see how woelen's setup turns out , but i have a feeling he won't succeed.

would've been awesome to see this work..

woelen - 11-12-2010 at 10:39

I also tried the experiment twice. Both experiments failed.

One time I tried the safe way in a sand bath with temperature monitoring. The temperature remained between 200 C and 250 C during the entire experiment. I used a 10 times as small a scale as Pok did.
Below follow pictures of the setup I used. The third picture shows that the test tube is filled with white fumes and the glass is wetted. There is some refluxing of t-butanol. The fourth picture shows a typical temperature readout during the experiment and the final picture shows the result.

Unfortunately the result is no potassium at all. Definitely there is some reaction (evolution of gas and formation of a white solid, other than KOH), but no potassium metal is formed.



setup.jpg - 50kB setup2.jpg - 47kB

reflux.jpg - 36kB temperature.jpg - 26kB

result.jpg - 38kB



[Edited on 11-12-10 by woelen]

woelen - 11-12-2010 at 10:51

So, in the first experiment I had no results. I added the t-butanol, dissolved in some paraffin oil in two steps with 20 minutes in between by letting it run through the thin glass tube (I did not want the air to be replaced inside the test tube). Initially, before heating, I flushed the air in the test tube with butane gas to be sure that no oxygen remained. I capped the glass tube with a thick dash of paper tissue which was wetted. At the start of the experiment a small amount of gas was released through this outlet, lateron, I pushed it firmly on the tube and no exchange of gas seemed to happen anymore.
I have kept the test tube in the hot sand bath for around 1 hour and then I allowed it to cool down slowly (which took another hour). Not the faintest sign of formation of potassium. No shiny globules. Pok states that in 20 minutes, but certainly in 30 minutes after adding the tert-butanol one should clearly see shiny globules of potassium. Nothing at all!


In the second experiment I did something unwise ;)

I went outside, took a test tube and put in fresh reagents and with a clamp, I kept the test tube above (not in!) the flame of a propane torch. I did this outside just in case the test tube would crack and the very hot mineral oil would catch fire. With this setup much higher temperartures are possible. I had a very vigorous reaction in this experiment. A thick white cake formed, sticking at the glass and excess Mg-powder was floating around in the mineral oil. I allowed this situation to last for around 30 minutes after adding the tert-butanol, but again no potassium is formed. Part of the Mg reacts, a lot of gas and white smoke is produced, but no K-metal. The vigorous reaction only lasts for a few tens of seconds, after that I had a clearly visible formation of gas, but it was not violent at all.

The KOH, used in my experiments is in the form of half-spheres, diameter around 4 mm and thickness at most 2 mm. The Mg is in the form of moderately fine powder (according to label it is 65 um to 100 um particle size).


Tomorrow I hope to find the time to try the reaction with Mg/Al instead of Mg. My Mg/Al is a fine powder, sold to me as magnalium for pyro purposes. It, however, also is a fine reductor and that is what I am using it for.


[Edited on 11-12-10 by woelen]

blogfast25 - 11-12-2010 at 10:56

@ woelen:

Can you describe your experiments in a little more detail? Duration? Introduction of t-butanol, how and when? Was the solvent effectively boiling?

ooops, you beat me to it!

[Edited on 11-12-2010 by blogfast25]

woelen - 11-12-2010 at 11:07

Actually, I am somewhat disappointed with this result. I really tried to assure that the conditions are like what Pok had. I used temperature monitoring (I was babysitting the whole thing and when temperature went below 220 C I switched on the hot plate and when it went above 240 C I switched it off again, and I'm sure that the temperature never went below 200 C and never went above 250 C). I have used the ratios of Pok and I swirled the test tube just a few times, just as Pok did with his erlenmeyer. I took away the initial oxygen to be sure that no excess magnesium is oxidized in the process and I have used a thin glass tube of approximately the same length as Pok did for the 'refluxing'.

There must be some 'magical' thing with the ShellSol D70, or the piece of magnesium, used by Pok was special. It was some piece of scrap, which most likely is not pure but contains other allied metals.

blogfast25 - 11-12-2010 at 11:28

@woelen:

I bet you are, so are we all (disappointed).

You’re starting from the assumption that pok’s results are genuine (and not a hoax) and that they are dependent on some ‘hidden parameter’, right? It’s possible. It just doesn’t really ring true with me.

Most likely (again assuming his experiment is genuine) it would be some overlooked form of catalysis.

Next we’ll be holding a collection for a representative of SM to fly to Germany and go watch <i>jackpok’s</i> ‘magic’ in the flesh (I just hope he’s not an illusionist as well as a hoaxer! ;)

woelen - 11-12-2010 at 11:45

Yes, up to now I assume that Pok is honest. It might be that he won the 'jackpok' but if that is the case then it is a job for us (and him) to find out what is the 'hidden luck' behind this. Maybe something in the ShellSol D70, maybe in the magnesium he used or even in some of the other chemicals. It might be that we even need more info from Pok himself. Maybe he should try with another batch of magnesium metal.

Anyway, I have an order outstanding at Kremer right now for 1 liter of ShellSol D70. Hopefully they dispatch at the start of next week and then I migt have the stuff next weekend. Shipping times usually are two to four days from Germany to the Netherlands.

blogfast25 - 11-12-2010 at 13:08

Quote: Originally posted by woelen  
Yes, up to now I assume that Pok is honest. It might be that he won the 'jackpok' but if that is the case then it is a job for us (and him) to find out what is the 'hidden luck' behind this. Maybe something in the ShellSol D70, maybe in the magnesium he used or even in some of the other chemicals. It might be that we even need more info from Pok himself. Maybe he should try with another batch of magnesium metal.

Anyway, I have an order outstanding at Kremer right now for 1 liter of ShellSol D70. Hopefully they dispatch at the start of next week and then I migt have the stuff next weekend. Shipping times usually are two to four days from Germany to the Netherlands.


Oh, I thought you'd run this one with Shellsol. Mine got here today (just a few days from Germany to Ole'Blighty). I'm just waiting for one or two more components... Next WE should be my 'Big Day', LOL...

I DID MAKE SOME POTASSIUM!!!

woelen - 11-12-2010 at 13:48


I have had the whole setup in the sand bath cool down completely and I found it was time to cleanup the mess. I added some water to the paraffin oil with the sticky goop inside which is mostly white, but also contained some remains of unreacted magnesium. I was met with a crackling and roaring noise as soon as water went into the test tube and bubbles of gas were produced. I decided to allow the concentration rise somewhat by pressing my thumb on the open end of the test tube and wanted to light the gas with a cigarette lighter, just testing for hydrogen. Suddently, while I still had my thumb on the open end of test tube, the gas mix above the oil exploded :o and immediately after the explosion there was a fairly strong underpressure. Luckily the test tube did not crack and I did not drop it! My heart skipped a few beats :o

After I recovered ;), I took the test tube (it still made a crackling noise) and swirled it somewhat and suddenly a 5 cm long pinkish flame erupts from the test tube and a WHOOP sound is produced. My heart skipped another beat ...

This kind of behavior can ONLY be explained by assuming that some potassium is formed. It only is a minute quantity and it is not formed in the form of nice globules, but definately some potassium is formed. The potassium reacts with water, giving hydrogen gas and when the liquid is swirled, then potassium at the liquid/air interface can form very hot spots, capable of igniting the hydrogen air mix above it.


The second experiment with the test tube above a flame of a propane torch indeed did not make any potassium. So, high temperature and fast reaction is not favorable. Slow reaction in a sand bath is one requirement for making potassium.


[Edited on 11-12-10 by woelen]

condennnsa - 11-12-2010 at 13:49

Congratulations , woelen!
This is truly a breath of fresh air in this huge thread!

woelen - 11-12-2010 at 13:57

Yes, this result certainly gives hope for better results. The potassium I had must have been very finely dispersed and mixed with the white material and unreacted magnesium, but at least it is proven now that this reaction indeed can make potassium. It might be that the ShellSol D70 allows better coagulation of potassium into larger spheres. The oil I used is quite viscous, the ShellSol is much more like water and that might help.

Let's see what blogfast's results will be when he has all the reactants. I'll try tomorrow with Mg/Al and (hopefully) next weekend with ShellSol D70.

plastics - 11-12-2010 at 14:22

Spurred by woelen, I too added water to the white/grey sludge left after my previous attempt (photos I displayed earlier). The water sank to the bottom and after swirling around I too was greated with a very effervescent reaction which eventually caused the whole mixture to heat up very quickly and boil (picture attached). No actual flames but not what I was expecting

Now to pour some cold water on everything, I have had Shellsol D70 busily refluxing with Mg turnings and tert-butanol for 3 hours and no potassium beads. I will see what happens when everything is coll and I add water.

I too still have some hope but there is still something missing

IMG_4928.jpg - 33kB

IMG_4927.JPG - 40kB

[Edited on 11-12-2010 by plastics]

watson.fawkes - 11-12-2010 at 18:25

Quote: Originally posted by woelen  
I added some water to the paraffin oil with the sticky goop inside which is mostly white, but also contained some remains of unreacted magnesium. I was met with a crackling and roaring noise as soon as water went into the test tube and bubbles of gas were produced.
Congratulations on the result. Very exciting.

Something you said reminded of a post-reaction workup, which isn't in the patent but might be useful during this experimental stage of progress. I thought well of blogfast25's comment about using a viscosity modifier, but had no ideas at the time. Combining these two, I suggest a workup with heptane.
&nbsp;&nbsp;&nbsp;&nbsp;Melting point of potassium = 63.38 &deg;C
&nbsp;&nbsp;&nbsp;&nbsp;Boiling point of heptane = 98.42 &deg;C
The idea is that you might use a viscosity modifier in a workup phase rather than in a reaction phase. Rather than adding water, add heptane, and put the vessel back under reflux at a lower temperature. The heptane, at much lower molecular weight, should dramatically reduce the viscosity, and thus also the energetic balance that would otherwise favor surface energy in potassium. There's also an activation energy for getting two potassium boundaries to merge, and heating under reflux could provide that. Furthermore, the alkoxides should be more soluble in heptane, dissolving off some of the surface contamination from K globules that interfere with them merging.

It's possible that heptane might be useful in the reaction phase, either with no ill effect or with positive effect. If the overall reaction is limited by alkoxide transport, then adding heptane would increase solubility and hence also its mobility. On the other hand, since the heptane would also be in reflux at the reaction, it would also tend to deposit alkoxide near the evaporation surface, leading to stratification if nothing were done to counteract it. So it's worth trying, but for present I can only recommend it in workup, given the large number of process parameter we don't understand yet.

MagicJigPipe - 12-12-2010 at 01:06

Woelen, how difficult was the "gunk" to remove from the test tube? This is merely a question to do with potential cleanup problems.

blogfast25 - 12-12-2010 at 07:20

Yes, that’s exciting indeed! I wondered why woelen hadn’t tested his gunge by now...

Watson’s idea of a work up with a lower viscosity solvent is very reminiscent of example 2 in the patent: sodium from NaOH. There the original solvent Shellsol D70 is removed after reaction and metal recovery carried out with dioxane (1,4-dioxacyclohexane), a low viscosity, low BP (101C) solvent. It's also polar and possibly a better solvent for alkoxides.

I’m pretty convinced that woelen’s Mg was far from ideal for this purpose: 65 to 100 micron is very fine compared to pok’s material. It’s possible that very fine K is formed with woelen’s stuff and that these fine droplets have a hard time coalescing into larger ones, isolated as they are by reagents/debris.

My own Mg seems much closer to pok’s: just about sub mm filings (but reagent grade purity).

What would also be interesting to do as a side show is to verify whether or not t-butanol reacts with KOH. For instance take a ration of 1 mol/0.5 mol t-butanol/KOH and heat it in test tube on a steam bath. Does the KOH dissolve? That would be evidence of t-BuO K forming…


[Edited on 12-12-2010 by blogfast25]

woelen - 12-12-2010 at 07:49

@MJP: The gunk can easily be removed. Just let it soak in water and then rinse with some dilute HCl. It then is gone in minutes.


@blogfast: I also realize that my Mg may be too fine for this purpose. I do have filings, but these, on the other hand, are very coarse with a size of 5 mm and a thickness of 1 to 2 mm. I unfortunately have no block from which I can easily file something and crushing my filings to something smaller is very difficult. The big filings are too coarse, the powder is too fine.

blogfast25 - 12-12-2010 at 10:14

Quote: Originally posted by woelen  
@blogfast: I also realize that my Mg may be too fine for this purpose. I do have filings, but these, on the other hand, are very coarse with a size of 5 mm and a thickness of 1 to 2 mm. I unfortunately have no block from which I can easily file something and crushing my filings to something smaller is very difficult. The big filings are too coarse, the powder is too fine.



Between a rock and a hard place...

And sod it, I’m going to have a go tomorrow. I’ve put together my ‘apparatus’ today. My main problem is that I only have 2-methyl-2-butanol, a tertiary alcohol but not t-butanol.

My experiment will be close to pok’s (same scale but with injection of the alcohol straight into the mix, and using an equivalent amount of 2-methyl-2-butanol). I’ve been messing with a sand bath to find a (gas) setting that takes the reactor slowly to 200C and beyond. I’m as ready as I can be with a bit of a plan to investigate any resulting gunge properly.

Below the ‘magical ingredient’, Shellsol D70. It really is remarkably odourless:


woelen - 12-12-2010 at 13:04

@Blogfast25: To answer your question why I didn't do the check with water earlier, the answer is simple. The goop that I obtained did not look promising at all. It is a somewhat slimy white/grey goop with unreacted magnesium in it. No nice globules of K-metal. Hence I did not expect any interesting properties of this material.


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

I also did the experiment with Mg/Al now instead of Mg. Again I used a powder, particle size at the order of magnitude of 50 um, even finer than my magnesium. I don't have anything else, remember this stuff is intented for pyrotechnics purposes such as glitters and flares.

The reaction with Mg/Al is more vigorous than with Mg alone. At times the liquid started foaming up and white smoke erupted from the liquid. Luckily, the reaction never became so violent that foam went into the thin tube. In this experiment, indeed no K-metal is formed at all. After cooling down after a 2 hour period at 200 ... 250 C I added water. No reaction at all. Shaking also did not result in an interesting reaction. The white material just is a mix of Al(OH)3 and Mg(OH)2 or oxides of Al and Mg. Nothing interesting.

For now, I'll skip the experiments with the Hg. It is clear that K-metal is formed, but now we need to find the conditions for formation of easy to separate beads of K-metal.

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

I have a source of very coarse reagent grade Mg powder (250 um particles, more or leass spherical). What do you think of that? Would that be suitable? I did not yet order it, minimum amount is 1 kg at around EUR 25 including the price of shipping. This is a decent price, but what must I do with 1 kg of Mg-metal if I it is not useful for making K-metal?

blogfast25 - 12-12-2010 at 13:37

Quote: Originally posted by woelen  
I have a source of very coarse reagent grade Mg powder (250 um particles, more or leass spherical). What do you think of that? Would that be suitable? I did not yet order it, minimum amount is 1 kg at around EUR 25 including the price of shipping. This is a decent price, but what must I do with 1 kg of Mg-metal if I it is not useful for making K-metal?


Srtange of the magnalium yielded nothing. The coarser Mg may work better, who knows what in this riddle/puzzle?!?

NurdRage - 12-12-2010 at 21:47

Impressive work. I'm amazed this can be done in such a straightforward fashion.

I might try this with ACS grade Tetralin rather than Shellsol D70. I only got t-amyl alcohol at the moment, hopefully it'll be compatible.

watson.fawkes - 13-12-2010 at 03:53

Quote: Originally posted by blogfast25  
Watson’s idea of a work up with a lower viscosity solvent is very reminiscent of example 2 in the patent: sodium from NaOH. There the original solvent Shellsol D70 is removed after reaction and metal recovery carried out with dioxane (1,4-dioxacyclohexane), a low viscosity, low BP (101C) solvent. It's also polar and possibly a better solvent for alkoxides.
I reread the patent. I have said above that workup isn't in the patent, which isn't quite right. They do have a single, one-sentence paragraph in their narrative which says "alkali metal [...] may be separated", but don't stress the workup. The examples do have workup. Ex. 1 and 2 have workup with vacuum distillation and 1,4-dioxane, ex. 4-6 by filtering through a frit, and ex. 3 has no workup.

In both of the solvent workup examples, the initial Shellsol solvent is first filtered off by vacuum distillation (1 torr in the first example, 15 torr / 120 &deg;C in the second). Example 2 does explicitly state "when the sodium has the desired particle size", presumably because tiny particles are merging. On the other hand, example 1 says "the potassium rises to the surface [of the dioxane] and is communited by stirring to form spheres of suitable size". Communited means "made smaller", which happens only after K "rises to the surface". Is there an innuendo here that potassium beads are merging together as it rises? When they say that stirring breaks up the potassium, I infer that vigorous mechanical agitation is sufficient to keep K particles small. So agitation from vigorous boiling itself may initially produce small particles of K. If this is the case, then reflux at a lower temperature, with less agitation, may induce merging. I imagine that centrifuging would cause merging as well.

I suggested heptane because (1) it is readily available (as Bestine rubber cement thinner), (2) it is in the same saturated alkane class as the solvent, and (3) it has a boiling point higher than that of potassium. It should be miscible with the reaction solvent, alleviating the need for vacuum distillation. I doubt it works as fast as vacuum distillation and a more polar solvent, though. I'm now thinking that using heptane at slightly beneath its boiling point, but above the melting point of K, would eliminate agitation from bubbling. Gentle kneading with a glass rod or the like might help coalesce K particles. I admit this is mechanically problematic to knead under inert atmosphere.

Pok - 13-12-2010 at 03:58

I'm glad that you've done it, woelen.

About the Mg size: At the photo you can see the Mg I used. It looks coarse (scale is in cm - so about 200µm broad), but these are mainly rolls of very thin (maybe < 50µm thick) Mg "foil" - an effect of filing. So if you use 20-50µm spherical Mg powder or so I think this should be the best size. I'm pretty sure that only 2 Mg properties will effect the outcome: surface area (depends on particle size) and purity (98%-100%).

@ woelen: did you also see small flashes of light (besides the explosion noise) when you added water to the mix? At least at 0.5mm K balls I always observed light flashes when I cleaned the erlenmeyer with water after the experiment. Maybe in your experiment the K was too much dispersed and too small to produce light flashes.

An advantage of a larger reaction volume (e.g. 50ml) is, that failures like income of some oxygen or loss of tert.butanol can be accepted up to a certain amount. But if the same amount (for example due to an equally large hole in your balloon or elsewhere) affects 5ml reactions the relative effect would be 10 times higher. Larger scale also is easier to handle (for exampe: stepwise addition of t-butanol within 30 min might be difficult if your total t-butanol is 0.6ml). But you obviously showed that even in small scale the experiment gives potassium.

I will repeat the synthesis as well in the next time to see if it still works as in the past. If it works, I will try to make Cesium as described in the patent. Making the needed dry CsOH may even be more difficult (for me) than making Cesium from it :D.



[Edited on 13-12-2010 by Pok]

watson.fawkes - 13-12-2010 at 04:16

Quote: Originally posted by NurdRage  
I might try this with ACS grade Tetralin rather than Shellsol D70. I only got t-amyl alcohol at the moment, hopefully it'll be compatible.
Example 3 in the patent uses preformed t-amyl alkoxide (called "tertiary amylate" there). Given Nicodem's comment above, it should be readily formed from KOH and t-amyl alcohol in alkane solution.

As far as tetralin, the same example 3 uses decalin as a solvent, the completely saturated version of tetralin, itself the half-saturated version of naphthalene. Shellsol D70 is hydrogen-treated to remove aromatics and alkenes; it has little unsaturated content. I have strong suspicion that somehow this is significant, although I have no particular mechanism to explain why. It might act as a hydrogen scavenger, forcing a side reaction. Double bonds might significant change solvation energy, or preferentially solvate some species and reduce its reactivity.

I'm all in favor of exploring the parameter space of this reaction, but at the current stage of maturity, I wouldn't take failure with this solvent as anything like determinative.

On the plus side, the boiling point of tetralin is a bit higher than that of decalin, which will make the reaction faster. Tetralin as a boiling point right around the lower boiling point of Shellsol D70. That of decalin is in the range of Gamsol, which is easily available and inexpensive in the US.

woelen - 13-12-2010 at 04:23

Pok, I did not see flashes of light in the test tube. I had one big flash of fire, but I think that is due to ignition of a hydrogen/air mix by some reacting potassium at the liquid/air interface.

If I read your response, then the magnesium I used must have been quite suitable with its 60 ... 100 um particle size. Using filings for grignard reactions then is not appropriate, these are way too coarse.

I will try the experiment at a larger scale when the ShellSol D70 arrives. I hope that it arrives before next weekend, otherwise the experiment will be one week later. I only have the time for 4 hours babysitting such a reaction in weekends.

Arthur Dent - 13-12-2010 at 04:48

Well this has been encouraging so far! As I mentioned previously, I have a source of t-Butanol that's reasonably inexpensive in Canada, so perhaps I could buy a 500 ml bottle and split it with one or two folks from north of the border? If anyone is interested, PM me and we'll figure things out.

Right now, I have most of the hardware, but i'll have to find an inexpensive source of KOH and something equivalent to the Shellsol (might as well get as close to the original recipe as possible to ensure success). I've been measuring the exact setting on my hotplate to get near the range of temperatures needed for the sandbath. A big thick-walled 500ml bomex round bottom flask will be used instead of an erlenmeyer, will this be ok?

@Pok: Thanks for suggesting this great experiment, this is the most fun thing i've read in a long time!

@NurdRage: Will we see a cool video of this synthesis soon? :)

Robert

len1 - 13-12-2010 at 05:27

Well I am as surprised as anyone by this, since I had gotten it completely out of my mind as a pretty distasteful waste of time. So to see this work in someone elses hands was a bit like seeing Lazarus get out of the grave.

A few thoughts spring in my head.


It cant be that grinding the KOH allowed it to absorb too much water in my experiment because commercial KOH is 10% water anyway. You can get it anhydrous by holding at 450C for extended periods but the experimenter nor the patent mentions nothing on this. It can only be stored anhydrous in hemetically sealed (not ordinarily capped) bottles, and again there is no mention of this.

Perhaps the problem is with the magnesium. But I used several types. One was straight from the drill press and totally shiny. Maybe you need a certain fineness for best results, but SOME potassium should form on the fresh magnesium surface if this is a real reaction - it might not coagulate well - thats another matter. But a reaction with water should ensue - and there was none.

The most likely explanation for all this is that this is not a mixed phase reaction at all and all the D70 is doing is providing an inert cover. In this case I got nothing because I stirred all the time. Letting it all settle to the bottom will get the magnesium in contact with the now solid KOH and react with it solid phase (mediated presumably by addition of the alcohol), here it is better that the KOH be coarse else the Mg has nothing to 'bite' into. In this case youll get better results heating the KOH with Mg solid.

But I still have some niggles.

Potassium reacts with tButanol, this I have done many times to give KOBu. Water is generated. Since it is gasesous at the reaction temperature it should vaporise. The tBu actually consumes potassium.

Maybe the tBu helps the potassium globules coalesce. But I have tried it many times, and I can not get potassium in D70 to coalesce this way. Potassium has a surface tension much smaller than sodium. In D70 I can easily break large globules up, but do what you like I can not get them to coalesce without working within a few degrees of its mp.

And then I have niggles about the potassium. Judging by the dimensions given we have well nigh 4 grams on the photo. If its all from that one run (it seems to have come from the one flask where you see it), its a near 100% percent yield for the reaction, which is well nigh impossible. It must have been from several runs.

Then potassium in D70 (I have been making it other ways, but with a 20L bucket of the stuff left around from this failed experiment I have used it to clean potassium) attracts all kinds of crap to its surface, and needs extensive puryfing. Yet here its absolutely clean. Thats even more amazing for me that it grows into such large globules.

Then potassium has weak surface tension and all my globules, larger than about 0.5 are flatenned.

Finally after a few hours in shellsol D70 it acquires first a bluish and finally a purple/gray tinge. Presumbaly the photo was taken straight after formation.

My best judgement is that what we are shown is real and the reason I got nothing is that I stirred and this is a solid phase reaction. (I will check this). The coalescing into large clean globules I can not explain. I have worked with K in D70 many time, I used t-butanol and never got this kind of coalescense. I believe its a fluke, maybe a wonder. I think Woelens results are much more fathomable.

NurdRage - 13-12-2010 at 07:00

Quote: Originally posted by watson.fawkes  

Example 3 in the patent uses preformed t-amyl alkoxide (called "tertiary amylate" there). Given Nicodem's comment above, it should be readily formed from KOH and t-amyl alcohol in alkane solution.

As far as tetralin, the same example 3 uses decalin as a solvent, the completely saturated version of tetralin, itself the half-saturated version of naphthalene. Shellsol D70 is hydrogen-treated to remove aromatics and alkenes; it has little unsaturated content. I have strong suspicion that somehow this is significant, although I have no particular mechanism to explain why. It might act as a hydrogen scavenger, forcing a side reaction. Double bonds might significant change solvation energy, or preferentially solvate some species and reduce its reactivity.

I'm all in favor of exploring the parameter space of this reaction, but at the current stage of maturity, I wouldn't take failure with this solvent as anything like determinative.

On the plus side, the boiling point of tetralin is a bit higher than that of decalin, which will make the reaction faster. Tetralin as a boiling point right around the lower boiling point of Shellsol D70. That of decalin is in the range of Gamsol, which is easily available and inexpensive in the US.


Interesting, i hadn't taken into account the presence of (or lack of) aromatic hydrocarbons. I'll still give it a go on small scale since i don't have any other high boiling solvents but I won't let failure dissappoint me in that case ;) thanks for the heads up.

@Arthur Dent
No video soon, i'd like the synthesis to mature a lot before i make a video on it. Ofcourse i'll cite this thread and its contributors.

blogfast25 - 13-12-2010 at 08:42

Well, my experiment didn’t go to plan and needs improving.

After the initial evolution of gas, presumed but not tested as hydrogen, the cork bung of the ‘refluxer’ kept being pushed out. I now have some right sized rubber bungs in the mail.



The question then was to abandon or soldier on without refluxer and I chose the latter, this time putting the temperature probe straight into the open Erlenmeyer. I started heating at 14.35 and gas started to evolve after about 5 mins. This carried on for about 30 mins: the consistency of the KOH/Mg mix changed substantially, becoming more ‘homogeneous’ presumed due to break down of KOH flakes. Then the refluxer started failing and I carried on without it.



At 15.00 I reached 190C with a nice boil and started gradually adding the t-amyl alcohol (2-methyl-2-butanol, BP 102C) over a period of about 15 mins.

At about 15.15 the reaction mix reached 200C. Toggling the gas mark slightly between to settings I managed to keep the temperature quite constant and close to 200C:



Heating to 200C was maintained till 16.00. I saw no perceptible changes in the white/grey mass and no globules of K appeared.

Failure may be due to:

1. t-alcohol distilling off: no K t-amylate formed
2. oxygen: seems rather unlikely; there was a net outflux of solvent vapour all the time
3. magnesium: pok’s is about my size but it has very clearly considerably more surface area: mine’s basically small irregular little cubes. For a solution/solid type of reaction high surface area could be crucial to achieve some rate of reaction
4. t-amyl instead of t-butyl
5. other.

The stuff is now cooling down and a sample of the cold solid matter will be examined for evidence of elemental K. If none is found I may add a considerable amount of t-amyl alcohol and reboil.

watson.fawkes - 13-12-2010 at 08:57

Quote: Originally posted by woelen  
The reaction with Mg/Al is more vigorous than with Mg alone.
While looking up t-butanol data, I came across this safety card. Under the heading "STORAGE", it says "Fireproof. Separated from strong oxidants, aluminium." My guess is there's something very exothermic going on, which results in the permanent sequestration of the t-Bu moiety or its destruction.

blogfast25 - 13-12-2010 at 08:59

@len1:

Good to have you on this thread!

How about the t-alcohol reacting with the KOH to form t-BuO K, which then, as solute, react with the Mg to form K and (t-BuO)2Mg. Something along the lines of:

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 + 2 H2O --- > Mg(OH)2 + 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 + Mg(OH)2 overall reaction.

… thermodynamically driven by the lattice energy of Mg(OH)2.

Amazing you can’t get the K to coalesce in Shellsol: I thought that that would be the easy way to faking the whole thing; a dispersion of K and some plausible looking grit and just wait for it to coalesce, big juicy K blobs. It puts a hole in my falsification theory…

blogfast25 - 13-12-2010 at 09:00

@Watson:

It basically says the same about 2-methyl-2-butanol...

watson.fawkes - 13-12-2010 at 09:15

Quote: Originally posted by len1  
Potassium reacts with tButanol, this I have done many times to give KOBu. Water is generated. Since it is gasesous at the reaction temperature it should vaporise. The t-Bu actually consumes potassium.
If it were water, there would be an oxygen deficiency. The gas is hydrogen, as listed on wikipedia page for t-BuOH. A footnote on that page points to an OrgSyn prep that begins with preparation of t-BuOK; this prep verifies hydrogen.

blogfast25 - 13-12-2010 at 09:51

My cooled gunge doesn't react with water at all: the magnesium would even be recovable if the quantity justified it. There's not a fizz, a cackle, a bit of effervescence or a spark to been seen. Total failure.

Next attempt with better set up and t-butanol...

blogfast25 - 13-12-2010 at 10:02

We may have to resort to buying some of pok's magnesium (from the stock that He Actually Used) for a kind of Round Robin test... Whatever 'magic', it doesn't seem to reside in the Shellsol because woelen didn't use that...

I really think we need to postulate mechanisms to try and improve on...

At versuchschemie.de they seem to have given up the ghost altogether: zero activity on chemical K...


[Edited on 13-12-2010 by blogfast25]

len1 - 13-12-2010 at 12:04

@blogfast Yes I saw your theory, and that would be a good proposal except that in that case its a liquid phase reaction and my attempt with the milled magnesium should have produced SOMETHING. The only hope I really have left for success, and that this is not a hoax, is that the Mg and KOH must be allowed to come in solid contact with the tBuOH breaking the final thin D70 barrier.

As you say there is nothing special about D70. Just some hydrocarbons that boil at 190C - has a slight smell to it. I would prefer paraffin oil which doesnt smell, but to be sure I went out of my way to get shellsol those years ago.

Potassium will coalesce in D70, in fact its better in coalescing in it than in parafin because its less viscous, but only mechanically and only near its solidification point arount 64C. If you try that at higher T's the K is too visocus and the globules just slip past each other no matter how you try to push them together.

Falsification - that would have to be bloody stupid because this is one of the simplest experiment one can do - especially now that stirring is not required. He says you must file the Mg not drill it - fine I will invest today in a new file (all mine are slightly rusty) Ill use a single neck erlenmeyer with a liebig vertical reflux (you need no more for reflux at 190C) and argon flow at the top, with D70 and tButanol and will have an asnwer at the end of the day one way or another.

Pok - 13-12-2010 at 12:23

Quote: Originally posted by len1  
It cant be that grinding the KOH allowed it to absorb too much water in my experiment because commercial KOH is 10% water anyway.

Yes. But maybe during grinding you raised the water content from 10% to 20% :D.
Quote: Originally posted by len1  

You can get it anhydrous by holding at 450C for extended periods but the experimenter nor the patent mentions nothing on this. It can only be stored anhydrous in hemetically sealed (not ordinarily capped) bottles, and again there is no mention of this.
People normally expect from others to handle hygroscopic chemicals adequately (this includes grinding them hermetically by ball mills for instance). I think they didn't mention to dry the powdered KOH because they either bought dry KOH powder - or they didn't grind it. In any case, they just expect from others to be conscious that KOH is highly hygroscopic and only a dry preparation will ensure a water content like the bought chips.

Quote: Originally posted by len1  
And then I have niggles about the potassium. Judging by the dimensions given we have well nigh 4 grams on the photo. If its all from that one run (it seems to have come from the one flask where you see it), its a near 100% percent yield for the reaction, which is well nigh impossible. It must have been from several runs.

On the photo you can see ca. 3 grams - smaller K balls also were in the mix but not shown on the photo. This is from one run! Its about 70-80% of theory.

Quote: Originally posted by len1  

Then potassium in D70 (I have been making it other ways, but with a 20L bucket of the stuff left around from this failed experiment I have used it to clean potassium) attracts all kinds of crap to its surface, and needs extensive puryfing. Yet here its absolutely clean. Thats even more amazing for me that it grows into such large globules.

In the liquid state my potassium also attracted some unreacted Mg filings but nothing else. And when the Mg was used up nothing was attached to the K.

Quote: Originally posted by len1  

Then potassium has weak surface tension and all my globules, larger than about 0.5 are flatenned.

My largest balls were about 3 cm in diameter. Even these were almost round and not flattened. The density of K (0.856) only is slightly higher than that of the Shellsol (0.79), so even at weak surface tension K will form balls (this is what I experienced). If your K in your Shellsol really flattens due to gravity - are you sure that you have Shellsol :o or potassium :D?
Quote: Originally posted by len1  

Finally after a few hours in shellsol D70 it acquires first a bluish and finally a purple/gray tinge. Presumbaly the photo was taken straight after formation.

Yes. The photo was taken only about some minutes after taking it from the reaction mix into clean shellsol. After some days it becomes bluish and gray.

Quote: Originally posted by blogfast25  

Amazing you can’t get the K to coalesce in Shellsol: I thought that that would be the easy way to faking the whole thing; a dispersion of K and some plausible looking grit and just wait for it to coalesce, big juicy K blobs. It puts a hole in my falsification theory…

To be honest, your theory isn't so much affected by lens statements. A guy on versuchschemie.de once was successful in trying to get smaller K balls coalesce.
http://www.versuchschemie.de/topic,8254,0,-Natrium-Kalium-Le... - compare photo 2 with 4 - he just added some isopropanol to the molten K (under petroleum - which should have comparable properties like Shellsol D70). I personally never tried this. Maybe this is a possibility to get larger K balls from a finely devided K suspension in the reaction mix.
Quote: Originally posted by blogfast25  

We may have to resort to buying some of pok's magnesium (from the stock that He Actually Used)

My magnesium isn't magic ;). I think, if you do it like me you will get my results as well.

I must admit that I really was very lucky since the photos from the 50ml run are those of my first procedure ever!:cool: ...if it wouldn't have worked I would probably have given up and never tried it again.

These modifications just worked at the first time:

- not grinding KOH chips
- no temperature control
- Mg of suitable surface area (size)
- no stirring

[Edited on 13-12-2010 by Pok]

blogfast25 - 13-12-2010 at 13:06

Quote: Originally posted by len1  
@Potassium will coalesce in D70, in fact its better in coalescing in it than in parafin because its less viscous, but only mechanically and only near its solidification point arount 64C. If you try that at higher T's the K is too visocus and the globules just slip past each other no matter how you try to push them together.

Falsification - that would have to be bloody stupid because this is one of the simplest experiment one can do - especially now that stirring is not required.


Are you saying that K becomes more viscous at higher T? Most liquids, Newtonian or not, have a decreasing newtonian viscosity v. increasing temperature function, usually an exponential relationship...

Falsification: I mean of course that it's possible pok couldn't get the patent to work either and for some (immaterial) reason decided to fake the results. With coalescing K at hand there would be no need for extensive photoshopping...

Jor - 13-12-2010 at 14:20

blogfast25, nice that you have also done the experiment.

I still think that a highly likely cause for the failures is that as soon as the t-alcohol hits the surface if the very hot hydrocarbon, it boils of instantly. Because KOH is insoluble in these kind of solvents (at least I expect it's solubility as nearly zero), no or hardly any salt of t-alcohol can be formed, because the KOH is at the bottom of the vessel and not at the surface.

I think this could be solved by adding the t-alcohol at the surface of the KOH, adding the t-alcohol to the mixture when the reaction mixture is not yet at the boiling point (when it's hot I suspect that the K-salt of the alcohol is formed as well, maybe not as fast, but it doesn't matter as it doesn't boil away). After the K-salt is formed, this is no longer volatile, and you can start to heat up. Also some sodium or potassium salt of t-alcohol could be added to the mixture instead of the alcohol to prevent it from boiling away.

Also, maybe the water present in the KOH reacts with the K-butylate forming the alcohol (wich boils off) and KOH.
If this were true, there would be the following reaction:
H2O + t-BuOK <--> KOH + t-BuOH
As the H2O is strongly bound to the KOH (KOH will only go anhydrous at very high temperatures), t-BuOH is the only volatile component in the system and will thus quite fast be removed from the system (ok, it will be condensed, but always some is lost, eventually you will lose all). I think the only way H2O is removed from the system is by the reaction with any formed K-metal. So even if some K-metal is formed, it will react with the H2O, greatly diminishing yields. Now the question is, will Mg react with water in KOH at those temperatures, wich ofcourse also depends on the contact of the KOH and Mg, wich is likely to be very small (unless you use very fine powdered Mg), having two solids.

Maybe a reason for the succes of Pok is that he used very efficient condensing (wet towels) preventing escape of alcohol, while the other reaction (formation of K, followed by reaction with water) removed all H2O before all t-alcohol was removed from the system. So I recommend very efficient condensing. Or better use dry KOH (so fuse at high temperatures for some time (nickel crucible or maybe steel)), cool down while covered, and immediately transfer the dry KOH to the flask filled with Shellsol D70, wich is dried with some Na. if you want to work really dry, you could transfer the crucible with fused KOH to a clear plastic bag, together with flask containing the dry Shellsoll D70 and a small amount of P4O10 to dry the atmosphere, and close the bag. Then transfer the KOH to the Shellsol. Or you can use a glovebox :D

I am only looking for reasons why it doesn't work most of the time, so some might be wrong, but i think the reason is the H2O in KOH, wich is strongly bound to the KOH, thus shifting the equillibrium to the right side of the above equation.

Woelen, I think your experiment confirms the formation of potassium, but we don't know how much. If you try it again, and you can't extract potassium out of the mixture can you try to determine how much K is in there by covering the mixture with some inert solvent (ligroin) and adding some isopropanol . If it is present hydrogen will be produced in a safe manner, and you can collect the gas formed in a inverted cylinder and calculate the amount of potassium present. If this is a acceptable amount, you can focus on a way to isolate it from mixture. Trying to collect something if you don't know how much there is (maybe it's only a very small amount of K) makes no sense.


[Edited on 13-12-2010 by Jor]

len1 - 14-12-2010 at 00:56

Pok I think you have misunderstood me. That KOH contains 10% water is not a reflection on anyones skill in handling chemicals, its just a fact of this world. I have ground and not ground KOH many times, the water content is always in the 10-12% range so that (if it is not a joke) is not so.

As for potassium, if its perfectly round and not flatenned at 3cm diameter its not pure potassium, which would be the case if KOH was contaminated with even 10% NaOH, changing its properties drastically. The potassium I use is >99% pure (I have IonicC analyzed it) so theres no problems there. Let alone the above statements be also interpreted personally, as forumers have a propensity to do, let me at once state I intend them as facts of life as they are written, and not in some sort of subtextual ego statement.

len1 - 14-12-2010 at 03:59

Sorry blogfast that was a typo. I meant less viscous. I have read about using isopropanol or butanol to coalesce Na and K. But I must say I must have tried it 5-6 times all without success. It produces a brown gunk on the metal surface which must then be removed by more complicated means.

A 70-80% yield from a solid state reaction? - that I do not believe. You can only get this if its a liquid-solid interaction which my previous experiments preclude - because stirring would aid the reaction in that case. It would only hinder it if this is a solid-solid reactio.

And I have been thinking further. What are these 3cm balls of K you talk about? Thats about 14gms - yet you claim you made 3gms? The other times you ran the reaction you say you only got small globules. That means you must have coalesced them, yet it is clear from what you write that you do not know how to do that.

Then why are the K balls perfectly shiny? The reaction produces water too. And fine MgO which should stick to the K surface like glue. Then I have made potassium t-butoxide before, its solubility in hexane is tiny - in D70 even less, I dont see how this can be a solid-liquid reaction. I am tossing up whether to try this. It seems almost a guaranteed waste of 4-5 hours, and Ive been had by this reaction already. Probably the round shiny balls we are seeing are woods metal. All indications to me are that at least a large part of this is bogus. Mind you fairly eloborate bogus, thats what got me to believe in it at first.


[Edited on 14-12-2010 by len1]

Pok - 14-12-2010 at 05:32

Quote: Originally posted by len1  
Pok I think you have misunderstood me. That KOH contains 10% water is not a reflection on anyones skill in handling chemicals, its just a fact of this world. I have ground and not ground KOH many times, the water content is always in the 10-12%

I know this. But if you grind the commonly available 10% water containing KOH in normal air, it will absorb water from the air and the 10% water content will rise up to 10+x%.

Quote: Originally posted by len1  

As for potassium, if its perfectly round and not flatenned at 3cm diameter its not pure potassium, which would be the case if KOH was contaminated with even 10% NaOH, changing its properties drastically.

"Perfectly round"? Who claimed that?? I said that my largest balls were "almost round". You can see it here http://www.versuchschemie.de/upload/files3/64730229_4948.jpg
Not pure potassium?
My KOH is techn. pure. It will contain much much lesser than 5% Na. A probable worst case alloy of 95% K and 5% Na wouldn't affect the density so much that it could form round balls instead of flattened balls. And even if it would effect it: Na is more dense than K: so an alloy would most probably be even more flattened than pure K. My produced K burns with a brillant purple flame without any trace of yellow Na light. It has a melting point of exactly that of literature potassium (+-1°C for measurement reasons). An eutectic alloy of Na an K has a melting point below room temperature! Sodium impurities would strongly decrease the melting point. Sodium also isn't produced as easily as potassium in this method (see examples for sodium in the patent). I'm sure my K contains 0.00% Na :D...OK, maybe 0.01% Na:cool:.
Quote: Originally posted by len1  

A 70-80% yield from a solid state reaction? - that I do not believe.

Its not a solid state reaction.
Quote: Originally posted by len1  

You can only get this if its a liquid-solid interaction

Thats what it is.
Quote: Originally posted by len1  

What are these 3cm balls of K you talk about?

From a 6 times larger procedure (look at thread).
Quote: Originally posted by len1  

Thats about 14gms - yet you claim you made 3gms?

No. Thats 12 gms. And: only an estimation of size!
("about 3cm in diameter" I said).
In my first experiment ever: 50ml procedure with 3g K outcome.
In a 6 times larger experiment: 18g outcome. Maybe only 2.5cm diameter = 7 grams = easily credible.

Quote: Originally posted by len1  

The other times you ran the reaction you say you only got small globules. That means you must have coalesced them, yet it is clear from what you write that you do not know how to do that.

You obviously didn't read all my comments. I did it several time in 50ml scale, several times in 300ml scale. 50ml always worked. 300ml worked at the first times, but later: only small globules which didn't coalesce.
Quote: Originally posted by len1  

Then why are the K balls perfectly shiny?

Nice, uh?:cool:

Quote: Originally posted by len1  

The reaction produces water too.

No. Hydrogen.

Quote: Originally posted by len1  

And fine MgO which should stick to the K surface like glue.

Why like glue? It's not fine. The MgO forms crusts!

Quote: Originally posted by len1  

Then I have made potassium t-butoxide before, its solubility in hexane is tiny - in D70 even less

Thas why the reaction takes so much time!!!!
Quote: Originally posted by len1  
Probably the round shiny balls we are seeing are woods metal.

Yeah. Probably, probably, probalby....probably you made some mistakes in your experiment??? :P

--- I try to answer questions. But if your next question is "Is this really an erlenmeyer in which you did the reaction? I assume its a tea cup" - then I will not answer anymore. For me, this is a waste of time. I wanted to give the people on versuchchemie a new method for potassium synthesis. I was INVITED to come to sciencemadness. I said "OK" - But if you, by all means, just want to believe that I made a hoax - I don't care anymore. You will be able to get my results if you simply simply do what I did. Some of you will do this and then I can go home :) ---

[Edited on 14-12-2010 by Pok]

watson.fawkes - 14-12-2010 at 06:42

Quote: Originally posted by len1  
A 70-80% yield from a solid state reaction? - that I do not believe. You can only get this if its a liquid-solid interaction which my previous experiments preclude - because stirring would aid the reaction in that case. It would only hinder it if this is a solid-solid reactio.
Your previous experiments do not preclude a liquid-solid reaction. They might preclude one with products of indefinite lifetime, but not such an interaction with products of short lifetime. Your earlier results preclude certain simplistic models of the process, but nothing about this process leads me to believe that there is a simplistic explanation. It seems apparent to me that some ordinary, normally silent, assumption is false here, and it's always difficult to question an unstated premise.

This has been bothering me for a while, ever since Nicodem posted about acid-base interactions in this context. It seems clear enough that on the K side, there's formation of t-BuOK in a liquid-solid interaction. The t-BuOK becomes partitioned into different phases, a certain amount as solute and vapor at least, possibly also some non-dissolved t-BuOK, perhaps as an emulsion. Whatever goes on at the Mg surface, it seems to be interacting with at least one of these phases, or even some derivative of t-BuOK formed at the surface; even a vapor phase of t-BuOH isn't inconceivable. In addition, the patent discloses the use of an alkoxide instead of an alcohol. All this seems to diminish the likelihood that solid-solid contact of KOH and Mg is critical.

On the other hand, however, it does seem to be the case that KOH and Mg need to be in intimate contact, but that does not imply, not in any singular way, that solid-solid contact is the reason for this. An alternate hypothesis is that something at the Mg surface produces a substance/state that quickly decomposes in solution, and so it has to have a short solvent-path to a KOH surface before it becomes something other than a reactive species. This would have similar external requirements on geometric configuration for success as solid-solid contact would require, but not require solid-solid contact itself.

I don't have particularly satisfying options for what the short-lived varieties might be. It might be a compound that undergoes a reaction. It might be an adduct that separates. A reaction at the Mg surface seems to require two necessary components, though: (a) Electron transfer from the half-reaction Mg -> Mg(2+) + 2 e(-) and (b) overcoming the lattice energy of Mg in its crystal. Another possibility, one I find unlikely, is that Mg comes off the crystal without ionizing; how would that happen? Degradation of an intermediate with respect to Mg has three possibilities (1) Mg(2+) reduces as a free Mg atom (presumably reactive), (2) Mg(2+) reduces and attaches back to an Mg crystal, (3) Mg(2+) remains bonded to something, even if it wasn't bound to before. In case (2), that would tend to change the surface structure of the Mg; if anybody has a spare electron microscope to take a peek, do speak up.

condennnsa - 14-12-2010 at 06:56

I think woelen proved beyond a doubt that this reaction yields potassium.
Eversince I first read Poks synthesis on versuchschemie, I believed it's real . The reason I had was because he said at the end "I am extremely proud of my so easily and cheaply produced potassium and expect praise!! " I reckon a forger would not say this

blogfast25 - 14-12-2010 at 07:12

Quote: Originally posted by condennnsa  
I think woelen proved beyond a doubt that this reaction yields potassium.
Eversince I first read Poks synthesis on versuchschemie, I believed it's real . The reason I had was because he said at the end "I am extremely proud of my so easily and cheaply produced potassium and expect praise!! " I reckon a forger would not say this


Unfortunately that's exactly what a forger would say, IMHO. You're trying to read pok's mind: can't be done. To maximise the credulity of the already credulous is much easier than you think.

I remain agnostic until my own set of experiments and hopefully someone else's decide either way.

One way forward would be for pok to make available the two most variable ingredients in Goldilock's Mix: the magnesium and the KOH.

Other than that, this is fast risking to become Backyard Science's own 'Cold Fusion'...

[Edited on 14-12-2010 by blogfast25]

blogfast25 - 14-12-2010 at 07:20

@jor;

Absence of alcohol/too short a time presence of alcohol is a likely cause of failure in my case, I agree. If oxygen had been the cause, I should by rights have found a mix of Mg(OH)2 and K2O/KOH, without or with much less Mg. But it was clear that apart from the initial driving off of water (as H2, probably) not a g-ddamn thing happened. My next one will be hermetically sealed, so to speak...

Pok - 14-12-2010 at 08:19

Quote: Originally posted by blogfast25  
One way forward would be for pok to make available the two most variable ingredients in Goldilock's Mix: the magnesium and the KOH.


Do I understand this correctly: you want me to send you some of my Mg and KOH? If yes: I could do it (send me some money for shipping! :D). But you shouldn't waste your money for shipping costs. I'm absolutely sure any 98-100% Mg of suitable size will do it. And I got my KOH from ebay. Any good technical KOH will do it, I'm sure. The patent doesn't specify the purity or size of KOH / Mg. This indicates that there isn't any hidden and difficult mechanism with other catalysts than t-butanol. I think the mechanism is absolutely simple: somehow the mechanism that you (blogfast25) assumed once.

I really made it some times successfully. And therefore I can't imagine that the experiment depends on a special method or so:

most important:
- avoid oxygen
- add the t-butanol gradually (within 30 min)
- ensure totally reflux of t-butanol
- stirr occasionally, not continiously
- don't grind KOH chips, add them as they are (qickly taking, weighting and putting into shellsol/Mg mix - to avoid water absorption from air)

and:
- use t-butanol (quite pure)
- use shellsol D70 (quite pure)
- use 20-50µm Mg (quite pure)

PS: I'll be back next year or so - or if I see the first pictures of large K balls here :cool:. I think my presence isn't needed anymore. Read the patent, read versuchschemie.de, read here: you will do it ;) - if I forgot about this thread: keep me in mind as the hero who gave you somehow the "gold of the hobby chemists". :D - have a nice chrismas ;)....

[Edited on 14-12-2010 by Pok]

blogfast25 - 14-12-2010 at 08:41

pok:

If no one here, to the best of their abilities, can reproduce your results, then we may have to assume that some ‘hidden parameter’ is lurking in the specific ingredient mix that YOU (and ONLY you used). In that scenario, I would pay for cost of chemicals (KOH + Mg – Shellsol D70 I have and must be presumed same as yours, “good” t-butanol should also be the same as yours) and shipping from Germany to UK, in order to try and lay this controversy to rest for once and for all.

But for now no one has really replicated your conditions accurately, not me either and not woelen. len1 came extremely close to the patent conditions, bar the high speed stirring but didn’t get a semblance of K, not even a trace…

You keep saying that we should be able to reproduce your results easily but right now we don’t even have a plausible mechanism by which this reaction is supposed to proceed. My own postulate is quite flawed. And I don’t believe in a low temperature solid state reaction…

Yes, Merry Xmas to you as well...

[Edited on 14-12-2010 by blogfast25]

len1 - 14-12-2010 at 12:37

Potassium balls that size are not even 'nearly round '.

Im not even going to reply to his post further since this is obviously a troll, and feeding them is what keeps this going but and interesting thought is that the original patent is bogus .. and Pok's bogus. Could it be the same bogus? The expired patent holders are from Germany, and whoever posted this prank on versuchschemie obviously also knows German. In this case his name would be one of the three gentlemen on the patent ..

blogfast25 - 14-12-2010 at 12:51

Quote: Originally posted by len1  
Potassium balls that size are not even 'nearly round '.




In an undisturbed liquid of the same density as liquid potassium they'd HAVE to be.

blogfast25 - 14-12-2010 at 13:01

And did you see this:

Peter, 28/11/2010; 1:36 at:

http://www.versuchschemie.de/hartmut.php?t=14677&postday...

He wrote near the end of his report: “Überraschung: Plötzlich stiegen einige, max. 0,5 mm größe K-Kügelchen auf.”

Transl.: “Surprise: suddenly a few K-globules rose up, max. 0.5 mm.”

Wishful thinking? Susceptibility? Or truth?

len1 - 14-12-2010 at 13:15

Thats an interesting point. However I am not 'theorising' I have them sitting around, and potassium nuggets that large are buttons, no more than 1cm high.

That also follows from theory. When they have exactly the same density as the solvent (as K does at about 80C in paraffin) they ARE perfectly round, but then they do not sit on the bottom but float all around the liquid. The reasomn the nuggets are flattened in D70 is that at its solidification point the density of K rises to about 0.89, that of D70 is about 0.75 - a differential of 0.14 gm/ml. An approximate formula for the height of globules is

h = sqrt(2 * surface tension/(g * density differential))

giving about 0.5 cm, in rough agreement with my results. As you can see 3cm 'nearly round' globules are impossible in D70. So a merry troll Xmas to Pok, and you can see the reason for his desire to quickly disappear from here.

woelen - 14-12-2010 at 13:17

@len1: Please do not say Pok is a troll. Let's keep this constructive ;) My experiment at least shows that K-metal can be made. Now we need to find a method to make it more effective and to have the K-metal clump together into balls. Remember, we are talking about making K-metal, something which is not supposed to be easy at all. So, I am determined to put more effort in this to find out how this can be optimized.

Your remark about a solid-solid reaction is correct, but I agree with Pok that this does not need to be a solid-solid reaction. In my experiment I saw the KOH liquefying when the temperature went through 200 C or so. The Mg touches the somewhat liquiefied KOH (I actually think it is a liquid layer around the solid material, the liquid maybe some K-butoxide and KOH mixed). So, occasional stirring helps, bringing fresh Mg to the KOH, but constant stirring does not help, because then the Mg particles float around in the paraffin oil and do not get in contact with the KOH/K-butoxide.

Another reason why occasional stirring might help is that it brings back t-butanol in solution. In my experiments I noticed formation of droplets, sticking to the glass (you can see that in one of the pictures I made) and when you shake, then these droplets again are brought into the solution again.

I am eagerly waiting for my ShellSol D70 to arrive. I still did not receive it, while the order already was confirmed by Kremer-Pigmente last Friday :( When it arrives, I will exactly copy Pok's experimental setup, using my 65...100 um magnesium powder and then I'll let it reflux for 4 hours or so. If this setup still does not work, then maybe we should consider purchasing some of Pok's Mg-metal, but for the time being I am inclined to believe that any source of decent quality Mg, any source of decent quality KOH and any source of t-butanol should do the job. I indeed think that the size of the magnesium particles is critical and that this is the hard part of this, simply because Pok made the particles himself from larger pieces of Mg-metal.

Sedit - 14-12-2010 at 13:22

This is what I expect the reaction to look like at the end even if it where successful yet poks reaction displayed no precipitate,



It is the results of Peters first experiment you can see what I would expect and that is a flask full of Magnesium hydroxides(oxides?) precipitate. You suggested it was a weak argument but im not so sure im as easy to convince and even less so now that I see an experiment repeating poks that looks the way I would expect it to.

I wish he had taken pictures of the so called formed K that came up with the addition of IpOH at the end so I could come to something more conclusive.

blogfast25 - 14-12-2010 at 13:48

@Sedit:

Hmmm… looks to me that Peter’s precipitate is simply more floccular than pok’s but what does that mean or prove to you? To me not much, IMHO.

The bottom gunge does look quite different from what magnesium powder/sand is supposed to look like but photos are notoriously deceptive in these circumstances.

blogfast25 - 14-12-2010 at 13:56

@woelen:

I wouldn’t have chosen the term ‘troll’ but I really think we cannot a priori exclude the possibility of a hoax by pok. I’m not fussed about possible ‘motives’: unless he confesses these are essentially ‘unknowable’ and thus of no concern to us.

Having said that, I’m proceeding as if I’m sure it’s not. Next run with t-butanol, Shellsol D70 and conditions replicating pok’s as much as possible.

An interesting thought is that if pok’s right and honest about the 70 – 80 % yield then the reacting away of Mg should in itself make coagulation of the K much easier than if you only produce tiny amounts…

Sedit - 14-12-2010 at 14:17

Quote: Originally posted by blogfast25  
@Sedit:

Hmmm… looks to me that Peter’s precipitate is simply more floccular than pok’s but what does that mean or prove to you?


I can't say that it proves anything other then Peter honestly performed the experiment. It does imply to me however that Poks results are not adding up with experimental data.
Quote: Originally posted by blogfast25  

The bottom gunge does look quite different from what magnesium powder/sand is supposed to look like but photos are notoriously deceptive in these circumstances.


Its not so much that Poks precipitate looks off what strikes me as odd is the complete lack there of. The more I look at it the less sense it makes to me. It just looks like metallic sand of some kind(perhaps even the woods metal len was speaking of) just melting down and forming bigger balls around some KOH flakes. I don't see a change in the KOH only a change in the "Mg" thats around it and that bugged me from the start.

I have some faith in the reaction considering Mg+KOH ignited produces Potassium but im having a hard time getting my head around this working the way its claimed. I see little to gain in faking it other then false praise so its warping my mind a bit.

len1 - 14-12-2010 at 16:40

Destructive is when you try to bring another persons argument down with no good reason. And there is plenty of this around here. However when a person is lying to you, its constructuve to put an end to it. Im used to getting crap from students, although unlike here most of them do it honestly - beating about the bush in that instance is only superficialy constructuve. It actually harms people in the long term. Same here.

Further posting a lot of stuff about this before you have been able to reproduce his wonderful balls of K only serves to encourage future trolls - hes having a great time watching you all.

Regarding the liquid-solid reaction, its interesting that a person who knows not the difference between water of crystalization locked inside a dry powder and water picked up from the atmosphere by a deliquesent substance, would suddenly categorically state that this is a solid-liquid reaction, without having shown any inclination towards theory before that statement. He did this of course to get out of a tight situation in what he thought would be the best way he could.

[Edited on 15-12-2010 by len1]

watson.fawkes - 14-12-2010 at 17:54

Quote: Originally posted by len1  
Destructive is when you try to bring another persons argument down with no good reason. And there is plenty of this around here. However when a person is lying to you, its constructuve to put an end to it.
If you have definitive evidence that he's lying, please present it. Nothing I've seen here rises to that status. There are suggestive things that support a hypothesis of lying, but neither are these unambiguous nor do they support only a single hypothesis. I have not yet seen anything like clear and convincing evidence of a lie here.

What we do have here, and most certainly so, is an absence of evidence that he's not lying, given the absence of reproduction of his synthesis report. I shall assume that we here all know the difference between absence of evidence and evidence of absence, so I shan't belabor the point.

condennnsa - 14-12-2010 at 21:14

Quote: Originally posted by Pok  
I'll be back next year or so - or if I see the first pictures of large K balls here :cool:. I think my presence isn't needed anymore. Read the patent, read versuchschemie.de, read here: you will do it ;) - if I forgot about this thread: keep me in mind as the hero who gave you somehow the "gold of the hobby chemists".
Quote: Originally posted by Len1  

and you can see the reason for his desire to quickly disappear from here.



I agree with Len1. Pok, if you are truly genuine, you have the responsibility to stick with us all the way. After all, I don't think we take up much of your time with our questions, so why leave us? While some here may be convinced this is fake, there are others who are not, and woelen's results are encouraging. And yes, If this really turns out to work, you will be a hero here, one further reason to stick around...

DJF90 - 14-12-2010 at 21:28

Whats the point in him sticking around. He's probably fed up of those making accusations, and he's told us everything we need to know to make it work like he did. Until someone replicates the experiment several times under exactly the same conditions he used and still have no result, I'm going to regard this as genuine.

Magpie - 14-12-2010 at 21:44

Next year = 2 weeks of Christmas vacation. Perhaps pok is going sailing in the Caribbean. :D

Pok - 15-12-2010 at 02:56

@ len1
Quote: Originally posted by len1  
The reasomn the nuggets are flattened in D70 is that at its solidification point the density of K rises to about 0.89

Where do you got this info from? english wikipedia? Solid K is 0.856 - 0.862 (different sources). As far as I know, potassium doesn't have a density anomaly like water. In the liquid state it therefore should have an even lower density: density of liquid K < 0.86.

Quote: Originally posted by len1  

that of D70 is about 0.75

http://kremer-pigmente.de/shopint/PublishedFiles/70470shd.pd... - page 5 "dichte" (=density) - 0.792 at 20°C - lower at 15°C (0.787) - at 100°C maybe even higher ? (I don't know): density of shellsol at 100°C > 0.792.
Quote: Originally posted by len1  

- a differential of 0.14 gm/ml.

the real differential therefore should be around 0.07gm/ml or so.
Quote: Originally posted by len1  

As you can see 3cm 'nearly round' globules are impossible in D70.

As you can see: they are possible!
Quote: Originally posted by len1  

So a merry troll Xmas to Pok, and you can see the reason for his desire to quickly disappear from here.

Quickly? I've answered ANY question here. Even strange ones. Merry troll Xmas to you as well, len :P.

@ woelen
Quote: Originally posted by woelen  

If this setup still does not work, then maybe we should consider purchasing some of Pok's Mg-metal

If you want it: please give me a U2U message as I won't find any request here in the thread after some days. But as I said: you won't need my Mg.

@ Sedit
Quote: Originally posted by Sedit  

Its not so much that Poks precipitate looks off what strikes me as odd is the complete lack there of. [...] The more I look at it [...] It just looks like [...] I don't see [...]

Your scepticism seems only to be based on my photos. And I already explained the reason for lacking of a dust-like percipitate. Do you really think I would fake the hole synthesis so perfectly and forget about such a simple thing??? If you want to believe in a fake, you will see "woods metal" and "unchanged KOH". If you are a scientist: repeat and value afterwards!
Quote: Originally posted by Sedit  

I don't see a change in the KOH

At the beginning it is very white and voluminous at the end the so called "KOH" is gray MgO (due to impurities I think)! If I look at it, I can follow your argument. But it is not true. Another point (I'll answer it befor the question may arise): look at versuchschemie: The first pictures show the erlenmeyer deep in the sand (50ml mark) - at the last pictures the erlenmeyer stands on the sand. The hight of the ingredients doesn't seem to have changed - but if you now know this background maybe this point is explained now.

@ len1
Quote: Originally posted by len1  

hes having a great time watching you all.

No. I'd rather see my experiment reproduced. You shouldn't be so aggressive only because I did the experiment successfully where you made mistakes. Everyone makes mistakes. Even me :D (remember my not willing to coalesce tiny K globules).

Quote: Originally posted by len1  

a person who knows not the difference between water of crystalization locked inside a dry powder and water picked up from the atmosphere by a deliquesent substance

You really must be joking now :o. I told you at the first time: I know that techn. KOH contains water. I also said: thats why hydrogen is evolved before t-butanol is added. I said "But maybe during grinding you raised the water content from 10% to 20%."
Quote: Originally posted by len1  

categorically state that this is a solid-liquid reaction, without having shown any inclination towards theory

Look at versuchschemie.de. I already referred to this site here. I said (translate with google if you want)
"Erklärung:

Magnesium reduziert Kaliumhydroxid zu Kalium. Tert.Butanol dient als eine Art Katalysator, wobei mit KOH Kalium-Alkoholat gebildet wird, welches mit Mg reagieren kann und den Alkohol wieder freigibt (schätze ich).

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

Die anfängliche Wasserstoffbildung vor tert.Butanol-Zusatz ist durch Restwasser aus dem KOH zu erklären:

H2O + Mg -> MgO + H2"

K-alcoholate (solved in Shellsol) - behaves like a liquid and reacts with Mg (solid).
Quote: Originally posted by len1  

He did this of course to get out of a tight situation in what he thought would be the best way he could.

"Of course". You can read my mind! (OMG!)

@condennnsa
Quote: Originally posted by condennnsa  

Pok, if you are truly genuine, you have the responsibility to stick with us all the way

Sorry. But I don't have the responsibility. I have a real life as well. and: (1) Nobody is asking me anything anymore (2) Nobody has come close to my procedure. - I'm not interested in the exact mechanism behind the reaction. So I don't want to read everything in this thread (which also deals with the mechanism). If you wanna ask me, please put an "@Pok" infront of your question (if there are still questions left which only I can answer - this shouldn't be the case if you adhere to the patent where absolutely everything is described in the neccessary extent). Otherwise I will not answer, because I don't wanna search the word "pok" in the hole text. I will answer them tomorrow. This will be the last time as I don't wanna go into the internet and everyday reading tons of suspicions without anyone having done it my way and adhering to my advices (see in this thread).

@Magpie
Quote: Originally posted by Magpie  

Next year = 2 weeks of Christmas vacation. Perhaps pok is going sailing in the Caribbean. :D

No. I'm just bugged by questions I already answered :). Or by statements which don't deserve an answer (e.g. "woods metal"). Caribbean would be nice - but Xmas without snow? :D.

len2 - 15-12-2010 at 03:08

Here's my advice:

Learn to read english sentences

check the density of potassium

en.wikipedia.org/potassium

learn about the behaviour of density for most liquids with temperature

stop trolling, and leave these people alone.

Pok - 15-12-2010 at 03:25

@len2
Quote: Originally posted by len2  
Here's my advice:
do you talk to me?
Quote: Originally posted by len2  

Learn to read english sentences

I'll try it. :D And you please learn german ...and hindu :D.
The only sentence I can imagine what youre aiming at is len1's statement that I don't know the difference between "water of cristallization locked inside a dry powder" and a "wet" compound. I know that KOH can contain water of crystallization. But it doesn't matter in which form the water is in the KOH. KOH * H2O also absorbs water from the air and forms KOH * 2H2O or even KOH * 4H2O. The amount of H2O molecules will increase in bought KOH (containing a little KOH*H2O) if you don't handle it as fast as you can (not grinding) while keeping it in air (e.g. for weighting). It can take up to 4 molecules of water (wikipedia says) and will still look dry. THAT (more or less) may have happened during grinding in len1's case.

If I remember correctly, len1 said that both the unchanged and the ground KOH contained 10-12% water. I really have doubts about that. You can't grind KOH in air without raising its water content because air contains water and KOH is highly hygroscopic. A question to len1 would be: (how) did you measure the water content? and how long did you let the KOH remain in air (how long did you grind)?
Quote: Originally posted by len2  

check the density of potassium

en.wikipedia.org/potassium


It says:
"Liquid density at m.p.: 0.828 g·cm−3" - my 0.07 difference compared to shellsol D70 therefore should be lowered to 0.03 or so :cool: - I therefore should thank you for your advice :D
Quote: Originally posted by len2  

learn about the behaviour of density for most liquids with temperature

Yes. "most liquids" - I also gave you the pdf of Shellsol D70 with 2 temperatures and increasing density from 15°C to 20°C. From this point, I asked "maybe?" and "I don't know"
that density may rise with rising temperature (although even I wouldn't understand it).

Quote: Originally posted by len2  

stop trolling, and leave these people alone.

stop being aggressive without arguments. this is poor. :P

[Edited on 15-12-2010 by Pok]

[Edited on 15-12-2010 by Pok]

Pok - 15-12-2010 at 04:09

@ len1: I forgot about this:
Quote: Originally posted by len1  
An approximate formula for the height of globules is

h = sqrt(2 * surface tension/(g * density differential))

giving about 0.5 cm, in rough agreement with my results.


I must admit that I don't know this equation (as I'm not a chemist I don't have to know it :D...but its interesting...you always learn:)) and therefore I don't know whether it is correct. It looks plausible to me. But: where do you got the value (which you didn't show us) of surface tension of K? And: isn't it also dependent on the sourrounding medium? (I only remember water tension which shows differences if its pure or when soap is added) - where do you got this value from?

I rearranged the equation to get your value. If your "g" is 9.81, it should be: 0,0171675 N/m. -> where do you got this from?
After (hopefully correct) recalculating it (0.036 (more exactly than 0.04) instead of 0.14 density difference), it would still only be 0.99 cm height.

possible reasons (because my balls in ca. 3 cm dimensions were nearly round):

- surface tension value isn't correct
- equation isn't correct (or another factor is needed to include the medium properties (shellsol) on surface tension)
- K and Shellsol have even much lower difference in density*
- something I don't know

*the difference in density of K and Shellsol would have to be 0.005 instead of 0.036 to let the potassium height rise up to 2.6cm (if surface tension and equation are correct) - a value quite imaginable in my case. I'm quite sure about the K and Shellsol densities. But the difference of 0.005 and 0.036 only is 0.031 - maybe the densities change in a different way at high temperatures (esp. that of the Shellsol - maybe decreasing in a much lesser extent (or even increasing?) than that of K)- and/or the surface tension or equation may not be correct.

-> I suspect both: (1) different density change at rising temperature and (2) surface tension value is not correct (my calculation based on a rounded figure (0.5cm) based on a probable estimation (?) of surface tension by you).

BTW: you talked about "0.5 cm, in rough agreement with my results" - and I also think you said that you've observed liquid K in Shellsol D70 - I showed you my "balls" - would you like to show me your "balls" now? :D :D :D

[Edited on 15-12-2010 by Pok]

len2 - 15-12-2010 at 06:24

I see from the general gist of what you are saying and your mistakes that you have no idea what potassium does near its melting point. You have never worked with it have you? Thank you and goodbye

Pok - 15-12-2010 at 06:57

@len2
Quote: Originally posted by len2  
I see from the general gist of what you are saying and your mistakes

the general things? I can't give you a contra argument if you remain unclear. Give me an example! I think you don't have any. That's why you have to remain unclear. Mistakes? I think I was very lucky NOT to make mistakes in my procedure.
Quote: Originally posted by len2  

that you have no idea what potassium does near its melting point.

It either solidifies or liquefies. What do YOU think it should do? Emitting light? Do you know what potassium is? Its a metal. Have you ever seen it? I did. Have you ever made it? I did.

Quote: Originally posted by len2  

You have never worked with it have you?

What a nice question. Everyone shall think that YOU of course did work with it. Am sure you never did. :o I really think that YOU have never SEEN it. :D
Quote: Originally posted by len2  

Thank you and goodbye

You don't have to thank me. I didn't do it for you :D.

blogfast25 - 15-12-2010 at 07:08

Quote: Originally posted by len2  
Here's my advice:

Learn to read english sentences

check the density of potassium

en.wikipedia.org/potassium

learn about the behaviour of density for most liquids with temperature

stop trolling, and leave these people alone.


Assuming len2 = len1 then len1 is now showing serious mean spiritedness for which at times he’s been known in the past. Anger is not always a good adviser… He seems now to have descended into conspiracy theory… His argument abiout densities, surface tension and height of K balls seems thrown together from a far distance and doesn't convince me...

Pok, you are more than entitled to leave and have no responsibility to stay, whether or not your results are genuine. You’ve gone out of your way to accommodate us, even if that still doesn’t prove it’s genuine: only full replication can prove that.

It’s up to us to try and replicate your results, something which, all said and done, almost no one here has really seriously tried with all four cylinders firing. We have woelen’s experiment as with promising but incomplete results and that’s about it. I wil try full replication soon...

And if we really believe you’re hoaxing us then that really is up to us to prove it…

Those who want to stay do so of their own volition. But gentlemen, this is supposed to be science, not a chat room for hormonally charged teenagers!

And pok's English is fine, a damn sight better I bet than most here people's German.


[Edited on 15-12-2010 by blogfast25]

woelen - 15-12-2010 at 07:55

@len2: Please stop flaming! I now have seen enough of it and I am getting tired of it. Whether Pok's results are real or fake is not up to you as long as you don't have any real evidence. All the equations and numbers you are bringing with you do not do justice to Pok's work.

From now on I want to see decent scientific discussion and no more flaming. Further flaming will be removed. Only scientific arguments against or in favor of Pok's experiments are accepted from now on. I really do not want this interesting thread to see derailing into an ordinary flame war.

The fact that no one could reproduce Pok's results for the time being is not an indication of trolling or faking. As Pok has written, no one did the experiment as he did. I am in the process of doing so (waiting for my ShellSol D70) and I think in the next week or so more people will try that when their chemicals arrive. Let's see what all those attempts bring us and then we have new results to base our judgements on.

Let experiments say the final words and not one's adrenalin or hormones :)


plastics - 15-12-2010 at 07:56

Quote: Originally posted by len1  
Destructive is when you try to bring another persons argument down with no good reason. And there is plenty of this around here. However when a person is lying to you, its constructuve to put an end to it. Im used to getting crap from students, although unlike here most of them do it honestly - beating about the bush in that instance is only superficialy constructuve. It actually harms people in the long term. Same here.

Further posting a lot of stuff about this before you have been able to reproduce his wonderful balls of K only serves to encourage future trolls - hes having a great time watching you all.

Regarding the liquid-solid reaction, its interesting that a person who knows not the difference between water of crystalization locked inside a dry powder and water picked up from the atmosphere by a deliquesent substance, would suddenly categorically state that this is a solid-liquid reaction, without having shown any inclination towards theory before that statement. He did this of course to get out of a tight situation in what he thought would be the best way he could.

[Edited on 15-12-2010 by len1]


Yes in my limited time on the board len1, len2, lenx whatever is prone to increasingly vitriolic outbursts when his tail feathers are up. Witness his attack when his elegant electrolytic preparation of sodium was replicated by me using a coffee tin and piece of wood:

https://www.sciencemadness.org/whisper/viewthread.php?tid=97...

Anyhow I think it is time to put away the flames and get the test tubes out. There appear to be less than a handful of practical attempts to replicate Pok's work and page after page of non-productive waffle

Rant over

Fleaker - 15-12-2010 at 08:15

@Len, chill out! If this is utter bunk, it'll be ferreted out. I doubt that it's entirely hogwash. You are a superb experimental scientist, and one here that actually does experiments when he says he will and quite well. While I find this patent dubious if you had no success, I cannot now rule it out without having tried it myself (after seeing some reports). Sadly, due to silly socio-political reasons and prevailing zeitgeist, I no longer experiment at home. So trust me when I say would be testing this out as well; I miss this very much! I could have very easily tested this remarkable low T route to a remarkable element.


So far, Woelen has shown some validity to it. I presume he's worked with potassium before and knows the difference between potassium generated fire, and hydrogen produced from water/magnesium/KOH (which may produce KOH-tainted H2) that may light from the heat of the reaction. That was my chief issue with Woelen's experiment--I feel that a mixture of KOH and magnesium in an organic solvent will always generate H2 upon addition of water and that the characteristic spectral colour of K might be seen as bubbles of K+ rich water are present in the mist produced as the metal dissolves. If, however, it [H2] spontaneously flames, that is a strong indication that potassium is present. I think potassium is produced--the thermodynamic data supports this.

The only questions I have are "how efficient, and how easily separated and purified?"



blogfast25 - 15-12-2010 at 09:10

What woelen said… Totally seconded. Even if all this peters out we’ll still have had some fun, IMHO…

I conducted a simple side experiment to try and show reactivity of 2-methyl-2-butanol (2M2B) (as a substitute for t-butanol, in the post) with KOH, NaOH, Mg ‘powder’ and Al.

About 4 ml of 2M2B was subjected to 1 g KOH (tube 1), 2.6 g NaOH (tube 2), 0.2 g Mg (tube 3) and 0.3 g Al (tube 4):



All were subjected with regular 'shaking' to 30 mins of steambath, average temperature about 97C, just below 2M2B’s BP of 102C:



Here are the tubes after wiping dry and cooling:



First off, tube 5 (previously not shown) which is KOH with methylated spirits (EtOH with small amount of MeOH) after boiling. KOH dissolves effortlessly and completely in this solvent to a slightly yellow solution of presumably KOEt, KOMe, EtOH + MeOH and some water. As stated higher up by Nicodem.

Tube 1: KOH and 2M2B started discolouring right way and this continued right throughout the test. But I can’t say whether or not significant amounts of KOH reacted with the 2M2B. 'Something happened'...

Tube 2: similar to tube 1 but less intense discolouration.

Tubes 3 and 4: no reaction or visible change with either the Mg or Al.

It would be tempting to see if 1 and 2 contain some resp. K or Na 2-methyl-propa-2-oxyde but that's not so easy to do. But if any reaction took place it is certainly not comparable in scale to the dissolution of KOH in ethanol/methanol.


[Edited on 15-12-2010 by blogfast25]

len1 - 15-12-2010 at 16:32

Wilco I must say Im surprised at you. Im not sure where you see flaming in my short posts. Just because its not obvious to you the guy is trolling does not mean it is not clear to someone who has done this reaction and has made potassium, and would be harder to hoodwink. In such a situation I would be asking why it is that its fake, rather than acting as if I know it all. I have also used quite a bit of scientific argument which did not get an adequate reply, but it appears no one here, who cared to express an opinion, understood it. Some have even been expressing admiration 'whether he's genuine or not' suggesting they actually want to be conned. So the thought enters my head why have I been wasting my time?

Fleaker, I chose not to do the experiment even though it would only take me a few hours because it became obvious to me the guy is faking - he has no idea of the properties of potassium which somone who made a substantial amount of it would know. The so called-successes can be explained - heating KOH with Mg solid can give some K (I have got it that way in a vacuum) but in these conditions you might be lucky to get a few milligrams.

Unlike the previous posts this last post contains no science because Im just wasting my time. It appears the future of this forum is more Pok's rather than people like myself.
Of course I was angry, for the waste of my time reading this, replying to him and getting ready to do the experiment. But this is a personal thing, if others get joy out of it, thats their right.

[Edited on 16-12-2010 by len1]

Fleaker - 15-12-2010 at 17:03

@blogfast25:
2-methyl-2-butanol is seemingly more reactive than t-butyl alcohol. I don't think that discoloration is as readily seen with t-butanol; I've made KtBuO before, and I don't remember seeing any color to it, but I made it from the metal and the alcohol. Truthfully, I don't remember their being a yellow color to any simple alkoxide in my experience (i.e. NaOMe, NaOEt, NaOiPr) as long as air is kept from them. Of course, heating in the presence of air probably leads to polymers and various degradation products. My base bath at work is KOH-saturated isopropanol and is a sullen gray.

@Len1 :
KOH and Mg producing K is well known; my erstwhile partner has had success with it in a nickel plated steel pipe still if I remember rightly. To be candid, one thing I'm curious about is whether Mg can be substituted for calcium in reducing CsCl or RbCl.

I honestly hope Pok isn't faking as Wilco's results are encouraging.

len1 - 15-12-2010 at 18:47

Wilco and I really thought (not in jest, seriously) more of you than this classic

/quote
All the equations and numbers you are bringing with you do not do justice to Pok's work. /quote

All I can say is that if you dont uinderstand the equations, enjoy his work.

MagicJigPipe - 15-12-2010 at 20:21

Quote:
Pok, if you are truly genuine, you have the responsibility to stick with us all the way


I must take issue with this. Pok doesn't have the responsibility to do shit but post his experiment in the best way that he could. And he didn't even have to do that. I hope he isn't lying for two reasons now. The first is obvious but the second is so that I will stop seeing this bickering (not referring to the above quote).

Time shall tell. Let us keep an open mind until more evidence has been collected contrary to Pok's "results".

I will be trying this as soon as I get back this weekend. Probably Sunday. It probably won't work and, if it doesn't, that will be one more nail in the coffin. However, the coffin will not be completely closed until someone does his experiment exactly (which I will be unable to do due to my inability to gain access to Shellsol D70).

By the way,

My message to Kremer Pigment's in the USA:
Quote:
Hi, I noticed that, in Europe, a product can be had from your company called "Shellsol D70". I was wondering if I, in the United States, would be able to order some from you?

Thank you for your prompt reply!


Their reply:
Quote:
Dear Customer,

This product can be ordered for you, but it would have to come in one of our HAZMAT shipments due to its flammability. Unfortunately, we only receive these shipments three times a year, so there would be a wait time of approximately 3-4 months for this item. The only solvents we carry with regularity are Shellsol T, ethyl alcohol, acetone, and ethyl acetate.

Best Regards,

Becca Pollak
Kremer Pigments Inc. NYC

Phone (212) 219-2394
Fax (212) 219-2395
1 (800) 995-5501

USA Store Location:

247 West 29th Street
New York, NY 10001
(Btw. 7th & 8th Aves.)
Open:
Monday - Saturday 11:00am - 6:30pm[

Sedit - 15-12-2010 at 20:54

Quote: Originally posted by MagicJigPipe  
I must take issue with this. Pok doesn't have the responsibility to do shit but post his experiment in the best way that he could.


This is nonsense and I say this to all those making simular remarks. It is the scientist goal to prove the experiment at hand not the peer reviewers. Yes the peer reviewers are there to ensure a non corrupt system but in the end its up to the "inventor" to justify his claims or take a hike.


PS: for instants would it kill the fellow to show it reacting in water? or even physically prove he even has access to t-BuOH? Yeh its all beating around the bush type shit while countless efforts here and over at the german forum fail time and time again. It will never be up to poks standard in the end and I have reached the conmclusion this issx just due to him making excuses for why there synthesis didn't work....... its over folks you have been had. Prove me wrong and Ill eat my words.

I will leave this thread honestly unlike pok, and not return until someone I trust has big balls of K.

[Edited on 16-12-2010 by Sedit]

woelen - 15-12-2010 at 23:51

My ShellSol D70 arrived yesterday. Next weekend I'll try the experiment, using the same amounts as Pok did. I will use a 100 ml erlenmeyer in a sand bath, keeping temperatures between 200 and 240 C and I'll use a dripping funnel to insert the tert-butanol (dissolved in ShellSol D70) slowly over a period of 30 minutes.

@Len1: What was irking me in your posts (assuming len2 is the same person as len1, correct me if I am wrong) is the anger which shines through them. An experiment may fail and that may be frustrating, but it is not fair to put that anger on the person who made the write-up of the experiment. You even may express your doubts (and to be honest, I also feel some doubts), but using words like 'trolling' and so on is not good manner.
Btw, generally I do appreciate your contributions very much, you have had beautiful results and you shared them with us. So, if you have the chemicals exactly like Pok has and you have the apparatus, then why not give this thing a final shot even if this means risking 4 or 5 hours of your time and a few bucks worth of chemicals? That's why I decided to purchase the ShellSol D70. I just want to settle this for myself and the only way I can do that is by means of experiments. The more people try to replicate the experiment as close as possible to what Pok did, the better the arguments will be either against or in favor of Pok's claims. Again I just want to say: let experimental outcome decide.

Rosco Bodine - 16-12-2010 at 00:22

Has anyone tried making sodium potassium alloy from the mixed hydroxides NaOH and KOH ? It may be easier to completely dehydrate a melt of the mixed hydroxides 50-50 mole %, mp 171C, by steam distilling away the water, before adding the Mg, to facilitate the start of the reaction, and the fluidity of the resulting Na-K alloy would possibly make isolation of the product easier.

[Edited on 16-12-2010 by Rosco Bodine]

NurdRage - 16-12-2010 at 00:50

Woelen, when your run reacted with water, did it look something like this:

http://www.youtube.com/watch?v=NWbod97BjFc

The camera didn't render the lilac color as well as i'd like. i think also my KOH had significant sodium contamination. Nonetheless i'm going to further explore this reaction.

My solvent in this particular run was white candle wax.

woelen - 16-12-2010 at 01:08

Yes, the flash of fire I had looked quite a lot like what you show in the video. A flash of pinkish light and a WHOOSH sound associated with it. The first time, when I had my finger on the test tube there was a real bang and that was quite scary. The second time, when I swirled the test tube, there was the flash and whoosh sound.

NurdRage - 16-12-2010 at 01:14

Yeah, I definitely heard that distinctive woosh sound.

good to know i'm on the right track. i'll see if i can get larger quantities.

The amount of grey gunk produced is staggering, i honestly wonder if its possible to get nice balls of potassium.

Thanks for your help.

len1 - 16-12-2010 at 06:44
Quote:

A warning is not sufficient in such cases. Immediate banning is the only thing which works with trolls and this definitely is a troll


Who wrote this?

[Edited on 16-12-2010 by len1]

dissolution of K in t-BuOH

watson.fawkes - 16-12-2010 at 06:47

In the original patent, tert-butanol is described as a "reaction promoter". I was chided that this is just a synonym for catalyst, but I'm not sure the word "catalyst" applies here, because tert-butanol reacts with the intended product to form the alkoxide. That's not ordinary catalyst behavior. In particular, alkoxide formation from the metal is a competing reaction.

K + t-BuOH --> t-BuOK + 1/2 H2

This reaction is on the wikipedia page for tert-butanol. There's commentary on this reaction by tracing references in that article. In this OrgSyn prep for β-carbethoxy-γ,γ-diphenylvinylacetic acid, tert-butanol is used to dispose of potassium crusts and unreacted potassium. In this OrgSyn prep for ketene acetal, the first step is the preparation of the tert-butyl alkoxide of a mole of potassium. This reaction for 39 g of K takes eight hours at reflux, not exactly a speedy reaction. The temperature of this reaction is at the boiling point of tert-butanol, much lower than the Shellsol D70 boiling point.

The overall reaction for this process has been described as a displacement reaction from K to Mg. I now believe that this presentation of the process inaccurately omits the effect of tert-butanol on the product. In particular, since the Mg alkoxide is also formed, we should probably consider the total (unbalanced) reaction as follows. I've left off the alkane solvent, although I'm not fully convinced that it doesn't participate.

KOH + Mg + t-BuOH --> K + t-BuOK + Mg(OH)2 + MgO + (t-BuO)2Mg + H2O + H2

In particular, consider the effect of treating K tert-butoxide as an ordinary product. Here's a balanced idealization of the reaction:

( 2 + x ) KOH + Mg + ( x ) t-BuOH --> 2 K + Mg(OH)2 + ( x ) t-BuOK

The easiest consequence of this reaction is that if there's too much tert-butanol present, it's going to diminish yields. The patent already reports that product alkoxides are a side product; see examples 1 and 6 in particular. On the other hand, this need not be a problem in practice. Leftover KOH and t-BuOK can simply be used as reactants for a subsequent batch. Presumably it's readily easy to do a solvent separation of these from Mg(OH)2 and MgO.

adsorption of K onto Mg surface

watson.fawkes - 16-12-2010 at 07:14

This morning, I had an insight about the possible significance of the phrase "particularly in the form of turnings" in the patent description. Let me assume, for the sake of argument, that the final step in formation of K is the following reaction:

2 t-BuOK + Mg --> 2 K + (t-BuO)2Mg

My hypothesis is that this K product remains adsorbed onto the surface of the Mg. As more K is produced, it will agglomerate on the Mg, but the globule so formed can't be much bigger than the Mg particle it's attached to. So the smaller the Mg particles, the smaller the K globules produced, and thus also the harder for them to merge. One of the common assumptions about the form of Mg in this process is that increased surface area would enhance reaction rates. That's likely true insofar as the chemical reaction goes. It seems to me that one of the tacit assumptions here has been that this chemical reaction is the rate-limiting step.

It's not true, though, about the physical transformation of small K globules into larger ones. In this case, it's the characteristic contiguous surface area of the Mg that matters. A useful way to think about this is that contiguous Mg surface acts as a catalyst for the merging of K globules. If you've got micron-sized Mg particles, I can easily see how its entire surface might become passivated by a layer of K that has not yet merged. On the other hand, a chip or shaving has enough surface area that product K will bead up on the surface, not only increasing its characteristic globule size, but also exposing new Mg surface. So perhaps the agglomeration of K is the rate limiting step.

Assuming this hypothesis is true, there should be threshold of some minimum effective contiguous Mg area. It's essentially a tradeoff between K-Mg adhesion and K-K cohesion. Above a certain contiguous Mg surface area, the K will just bead up and extra contiguity won't be any faster. It also means trying the reaction with ordinary Mg strip may be worth trying.

Pok - 16-12-2010 at 07:27

@Sedit
Quote: Originally posted by Sedit  
It is the scientist goal to prove the experiment at hand not the peer reviewers.

its up to the "inventor" to justify his claims or take a hike.

Ok. I'm trying to do this the whole time. And I'm planning do repeat the synthesis within the next time. Tell me what you want to see (photos, videos, whatever) and I will remember this. But I think others will be faster here.
Quote: Originally posted by Sedit  

would it kill the fellow to show it reacting in water?

Correct me if I'm wrong - Do you want me to show the K reacting with water?: I did the experiments 1 year ago but made a video of a tiny amount of K reacting with water. If this is required (and possible to add a video here) I can do it. I also can make a new video of some K I still have. But I could have used bought K, so this wouldn't prove my claims.
Quote: Originally posted by Sedit  

or even physically prove he even has access to t-BuOH?

I could give you a picture of the t-BuOH in the original container - its a crystalline solid at room temperature but very easily supercools (at least in the purity I used) and can thus remain liquid. I also could send you an amount of 1 gram t-BuOH or so. Although this also wouldn't prove anything.

@ watson.fawkes
If my experiences are welcome (although considerred sceptically) I will give some comments about this - otherwise (e.g. not believing me) just ignore me ;).
Quote: Originally posted by watson.fawkes  
If you've got micron-sized Mg particles, I can easily see how its entire surface might become passivated by a layer of K that has not yet merged. On the other hand, a chip or shaving has enough surface area that product K will bead up on the surface, not only increasing its characteristic globule size, but also exposing new Mg surface


I don't know about micron-sized Mg as I didn't use it. I could imagine this. The point with Mg chips or shavings seems very probable to me: I don't think that K builds a layer on the Mg (chips or shavings). After about 20 mins of t-BuOH addition the first K was visible in realy tiny beads. They indeed were attached to the Mg shavings but didn't cover their entire surface. It looked like water drops on a leaf. K and Mg here seem to hehave like a polar liquid attached to an unpolar solid. On becomming bigger (and due to turbulences by boiling) it will more likely coalesce to form bigger balls (some with a surface coated with mg filings) than covering the Mg.

PS: for all wanting to try the synthesis. It might be that too high temperature favors production of tiny balls not wanting to coalesce. I remember that in my "unsuccessful" procedures (with temperature control) I raised the temperature to 230°C or so. Although the patent says "up to the boiling point" (you shouldn't be able to make it "too" warm) it might be the reason for tiny balls - but even here you should get max. 2-3mm K balls.

[Edited on 16-12-2010 by Pok]

blogfast25 - 16-12-2010 at 07:39

Quote: Originally posted by NurdRage  
Yeah, I definitely heard that distinctive woosh sound.

good to know i'm on the right track. i'll see if i can get larger quantities.

The amount of grey gunk produced is staggering, i honestly wonder if its possible to get nice balls of potassium.

Thanks for your help.


Can you please describe your experiment in much greater detail?

woelen - 16-12-2010 at 08:06

Quote: Originally posted by len1  

Quote:

A warning is not sufficient in such cases. Immediate banning is the only thing which works with trolls and this definitely is a troll


Who wrote this?

I fully stand behind these words, a real troll must be banned straight away and the guy from that old 2008 thread definitely was a troll. But who's trolling now in this potassium-thread :mad:?
Instead of spending your time on searching and digging up old and off-topic info, you'd better spend your valuable time on replicating Pok's experiment ;)

[Edited on 16-12-10 by woelen]

watson.fawkes - 16-12-2010 at 08:09

Quote: Originally posted by Pok  
The point with Mg chips or shavings seems very probable to me: I don't think that K builds a layer on the Mg (chips or shavings). After about 20 mins of t-BuOH addition the first K was visible in realy tiny beads. They indeed were attached to the Mg shavings but didn't cover their entire surface. It looked like water drops on a leaf.
Thank you for this observation. It's exactly the sort of thing I'd expect if we have this sequence: reaction -> adsorption -> agglomeration. My guess is that the characteristic minimum contiguous area I'm thinking about is at about the diameter of these first beads you saw or a bit above that (say double). To your best estimate, what was the smallest diameter of the beads of K that you saw? I would surmise that they were somewhere in the 0.1 - 1.0 mm range, but I'd prefer your first-hand account. Thanks in advance.

condennnsa - 16-12-2010 at 08:12

Quote: Originally posted by NurdRage  

My solvent in this particular run was white candle wax.

Yeah, I definitely heard that distinctive woosh sound.

good to know i'm on the right track. i'll see if i can get larger quantities.

The amount of grey gunk produced is staggering, i honestly wonder if its possible to get nice balls of potassium.

Thanks for your help.


Nurdrage, was the potassium in the video produced by this process? If so, it's great work!! Please share some details of your setup!

NurdRage - 16-12-2010 at 08:13

Quote: Originally posted by blogfast25  

Can you please describe your experiment in much greater detail?


25mL of white candle wax (essentially high boiling paraffin) was melted into a 100mL round bottom flask. 5.6g of tech grade KOH (85%) was added along with 3g of magnesium powder (99%). The flask was fitted with a liebig condensor and a bubbler to prevent air backflow. The flask was heated on an oil bath. Evolution of a non-condensing gas (presumably hydrogen) was observed when the oil bath reached 110 celsius. When the oil bath reached 150 celsius 1mL of t-butanol (99% anhydrous sigma aldrich) was injected into the top of the liebig condensor after carefully removing the bubbler. Before the liquid could reach the flask the bubbler was replaced. Vigrous refluxing occured as the t-butanol hit the reaction mixture. A very small, but constant, evolution of gas was noticed in the bubbler. Oil bath temperature was maintained at 220 celsius for 2 hours. It was noticed at several intervals that the t-butanol had solidified in the condensor. cooling water was shut-off to allow the t-butanol to melt and return to the reaction. Water circulation was restored after all the t-butanol melted.

After 2 hours heating was discontinued and the bath was allowed to cool to 130 celsius. The round bottom flask was disconnected from the condensor and its grey-pasty contents was dumped directly into 10mL of water in a beaker. Beaker was heated on the hotplate to remelt all paraffins. a stir bar was added to aid in mixing.

Video in the link before.

A lot of unreacted magnesium powder was directly observed in the bottom of the beaker when stirring was shut off.

NurdRage - 16-12-2010 at 08:18

In hindsight, i should have added the stirbar first and stirred the mixture while it was heating. candle wax doesn't boil at 220 celsius so there was no self-stirring.

Pok - 16-12-2010 at 08:38

Quote: Originally posted by watson.fawkes  
To your best estimate, what was the smallest diameter of the beads of K that you saw? I would surmise that they were somewhere in the 0.1 - 1.0 mm range, but I'd prefer your first-hand account. Thanks in advance.

They were so tiny that I hardly could distinguish between twingling Mg shavings and shiny K beads (I already mentioned on versuchschemie.de that you may mistake Mg with K at the beginning because both are so shiny). I can only say for sure: < or = 0.5mm. But I can't exlude that the smallest ones were about 0.01mm (??) or even smaller (??). I looked at the mix for the hole time (from beginning of tert.BuOH) but only the 0.5mm DEFINITLY was K. Maybe this just was the limit for my eyes. Much less than 0.5mm would be difficult to see even without twingling of the Mg.

[Edited on 16-12-2010 by Pok]

blogfast25 - 16-12-2010 at 08:46

@Nurdrage:

The 'explosions' (small and larger) to be seen in the video of your gunk are of course quite, shall we say, 'titillating'? Something happened in your reaction mix that's not easily explained or explained away...

Could be microdrops of K reaching oxygen, could be something else causing some lilac (with all that KOH around)...

Quite intrigueing... thanks also for the write up. You might want to repeat all with a less viscous 'solvent': higher viscosity impedes metal coalescence, assuming K metal did form...


[Edited on 16-12-2010 by blogfast25]

blogfast25 - 16-12-2010 at 08:53

Quote: Originally posted by NurdRage  
In hindsight, i should have added the stirbar first and stirred the mixture while it was heating. candle wax doesn't boil at 220 celsius so there was no self-stirring.


If you've followed the discussion, then beware (maybe!) of too vigorous stirring but a simmering boil is probably to be preferred.

NurdRage - 16-12-2010 at 09:00

I've been following.

Which is why i didn't do any stirring at first. But i failed to take into account that candle wax doesn't boil near the temperatures we want. Without that self-agitation i'll need to put in some external agitation.

Anyway, i'm looking into various solvents. While i can order anything, i want to stick to the solvents most amateurs have easiest access to. Candle wax is probably the easiest highest boiling solvent, although very far from ideal.

i'm working on it ;)

blogfast25 - 16-12-2010 at 09:41

Sure. We need quite few taking part in this: if one can replicate it then we'e essentially done here. If not, it's tar and feathers for our German friend...

Eclectic - 16-12-2010 at 10:09

K-1 kerosene or low sulfur diesel fuel with the volatiles fractionated off would be the most available for anyone with distillation equipment. It's just a broader boiling range cut of basically the same stuff as shellsol or varsol.
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