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

Make Potassium (from versuchschemie.de)

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mr.crow - 1-12-2010 at 18:44

Here is something cool I found, not mine of course

http://www.versuchschemie.de/topic,14677,-Synthese+von+Kaliu...

Translation

Jor - 1-12-2010 at 19:38

Funny thing. The minute you posted it I also found out and was reading it

I immediately wanted to do this as well. However, I will make slight modifications, I will use i-PrOH instead of t-BuOH.
Also I don't know where to buy Shell D70 solvent, so i will use some high boiling hydrocarbon such as mineral turpetine.
I will still have to order some magnesium. I have some ribbon, I think i will use that to try it this weekend, but it's not enough to prepare more than a gram or two (I don't want to waste all the ribbon).

UnintentionalChaos - 1-12-2010 at 20:00

I think I'll be trying this with sodium, asap. I lack solid KOH presently...

condennnsa - 1-12-2010 at 20:16

Very nice thread there! and patent.

What is the purpose of the alcohol in this reaction? Does it react with the KOH to make a more soluble K alcoholate in the shellsol solvent, thus increasing the reaction rate?

I also wonder if solid paraffin from candles could be used instead of shellsol d70.

BromicAcid - 1-12-2010 at 20:20

The thread for this is laying around here somewhere. I tried this back in 2005 using KOH and isopropanol in kerosene with promising results but never revisited it.

http://www.bromicacid.com/experiments/october2005.php#a

In the thread on this site however someone had tried to repeat the experiment and failed to have any positive results. I think this dissuaded others from attempting the experiment. It's amazing how similar some of the pictures look on that site, I could have recognized the reaction by sight

Edit: It was Len1 that tried the experiment on this site, most of the discussion is here (although it starts on previous pages):

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

Edit: Condennnsa, I do believe as you indicated that the alcohol acts as a phase transfer catalyst of sorts to help bring the KOH into solution so it can react with the magnesium surface. Isopropanol seemed to work but I think that a more non-polar alcohol such as t-butanol would give superior results as in the patent. Part of my lack of success may have also been due to adding all reagents together then heating rather than a feed of alcohol.

[Edited on 12/2/2010 by BromicAcid]

bbartlog - 1-12-2010 at 20:23

Fascinating, and elegant! Wouldn't have thought it possible (elemental K by chemical means under such mild conditions). I will be interested to see whether isopropanol actually works as well as the tert-butanol...

crazyboy - 1-12-2010 at 21:01

A fascinating thread indeed but if you read the comments, attempts to reproduce the results with gear oil, kerosene and paraffin oil in place of Shellsol D70 were all failures.

MagicJigPipe - 1-12-2010 at 21:19

This is absolutely amazing. And I was still pondering over electrolysis to obtain alkali metals in propylene carbonate (with AlCl3 catalyst). Will isopropanol work for this? I don't see why not. Potassium isopropoxide should form with just a little bit more difficulty, right?

condennnsa - 2-12-2010 at 05:19

Well I'm completely confused. Like BromicAcid said, Len1 tried the exact same procedure in the unconventional sodium thread , following the patent precisely. And had zero success. No reaction.
He said that he thought the patent is bogus.

Yet on the Versuchschemie.de forum we see those beautiful blobs of potassium. Why would the guy on the german forum lie!? i don't get it

[Edited on 2-12-2010 by condennnsa]

mr.crow - 2-12-2010 at 07:26

It seems people here are too happy to substitute different reagents. Maybe you really need tBuOH and everything has to be anhydrous and at the right temperature. Sort of like how Mg can form grignards or remove the last bit of water from anhydrous alcohol.

I don't know why he would lie, it would have to be the nerdiest troll in history. Why would the patent lie? Those are expensive to get.

I didn't know its already been tried here, but now that we have a new thread everyones seen it

condennnsa - 2-12-2010 at 07:34

Quote: Originally posted by mr.crow

It seems people here are too happy to substitute different reagents. Maybe you really need tBuOH and everything has to be anhydrous and at the right temperature.

Well Len1 said that he did use t-BuOH as well as shellsol d70. Check the link provided by bromicacid, on page 9

This got me really excited, now I'm boiling some mg chips with NaOH in paint thinner, refluxing it. It doesn't have nearly the boiling point required, it boils at about 80C, I just want to see if anything happens. I also added a few drops of isopropanol. So far, nothing happened. There seem to be no gas bubbles on the surface of the Mg chips.
The liquid did get slightly darker, which i can't explain...

blogfast25 - 2-12-2010 at 07:48

Wow!

As the first commeter over there wrote: 'Geil, mir fehlen die Worte !' ('Blimey, words fail me!').

I remember the len1 thread and what dissapointment it caused. So does this really work? A heterogenous reduction of KOH with Mg? You'd expect a fair amount of heat to be developed...

blogfast25 - 2-12-2010 at 07:51

Quote: Originally posted by mr.crow

I don't know why he would lie, it would have to be the nerdiest troll in history. Why would the patent lie? Those are expensive to get.

I didn't know its already been tried here, but now that we have a new thread everyones seen it

There are shedloads of non-functioning patents out there: often companies patent something on an educated hunch, prior to actual testing, just to beat potentila competitors to it.

I'd advise anyone trying this to stick rigorously to the recipe, at least at first...

UnintentionalChaos - 2-12-2010 at 07:53

Quote: Originally posted by condennnsa

Quote: Originally posted by mr.crow

It seems people here are too happy to substitute different reagents. Maybe you really need tBuOH and everything has to be anhydrous and at the right temperature.

The temperature is quite important. If potassium (or sodium in your case) is indeed formed, the mix would need to be above the melting point to allow it to ball up.

Arthur Dent - 2-12-2010 at 08:13

Interesting experiment. What would be the simplest apparatus/setup to "reflux" the process?

The idea of boiling kerosene is a bit scary, but the result would definitely be total awesomeness!

Robert

blogfast25 - 2-12-2010 at 08:24

Quote: Originally posted by Arthur Dent

Interesting experiment. What would be the simplest apparatus/setup to "reflux" the process?

The idea of boiling kerosene is a bit scary, but the result would definitely be total awesomeness!

Robert

A long vertical tube... Inert material and aircooled if possible...

blogfast25 - 2-12-2010 at 08:28

It gets crazier: the patent he quoted claims production of cesium from CsOH!

bbartlog - 2-12-2010 at 08:36

The post on versuchschemie describes heating the shellsol d70 to reflux (that's 200C), with hydrogen evolved mostly in the temperature range of 100-130C. Also the tert-butanol / solvent mix gets added while the solution is still at 200C. I don't think low-boiling solvents are going to cut it.

Quote:

There are shedloads of non-functioning patents out there

No doubt! And if it were just a patent, I'd consider Bromic and len1's experiments a fairly decisive refutation of the claims. But now we have a new claim of reproduction, complete with pretty pictures. It's either a very elaborate troll, or else there is some detail that we're missing; take your pick. Who knows, maybe some naphthene compound that is intermittently present in the shellsol in small quantities (depending on batch :-) ) acts as a catalyst ...

blogfast25 - 2-12-2010 at 08:58

From what I undertand it's realy more a reduction of potassium alcoholate; that makes it a lot more plausible, assuming the alcoholate is soluble in the main solvent...

Low boiling solvents won't do it, I'm pretty sure of it...

[Edited on 2-12-2010 by blogfast25]

blogfast25 - 2-12-2010 at 09:03

Now is the time to buy a tonne of Shellsol!

Arthur Dent - 2-12-2010 at 09:17

Shellsol D70 has a composition of 60% parrafins and 40% Naphtenes, according to Shell's data sheet. I checked for availability and it's available in canada, I'll have to enquire if it's available in small quantities.

- Robert

[Edited on 2-12-2010 by Arthur Dent]

watson.fawkes - 2-12-2010 at 10:04

Here a recent datasheet for Shellsol D70. Link is in PDF form.

ScienceSquirrel - 2-12-2010 at 10:29

I really do not know what to think about this. His procedure is significantly different to the patent and I know that some patents are quite frankly fiction.
My gut feeling is that the reaction just would not occur.
Potassium reduces magnesium chloride to form Rieke magnesium, see below;
http://en.wikipedia.org/wiki/Rieke_metals

blogfast25 - 2-12-2010 at 10:53

Thermodynamically it makes sense though: NaOH HoF 298 K = - 426 kJ/mol, KOH HoF 298 K = - 425 kJ/mol, MgO HoF 298 K = - 606 kJ/mol:

NaOH (KOH) + Mg - Na (K) + MgO + ½ H2 HoR 298 = -602 + 426 = -176 kJ/mol. Quite a bit of heat... This would be typical of quite a slow reaction which is what we're seeing...

It can be shown that the Rielke reactions take place too with the same logic.

Other example: reduction of AlCl3 with K (or Na) but also reduction of NaOH with Al powder or Mg...

If the t-butyl alcohol somehow solubilises the KOH then that's half the work done...

[Edited on 2-12-2010 by blogfast25]

blogfast25 - 2-12-2010 at 11:10

Possible mechanism:

KOH + t-butanol -> K t-butanolate + H2O

Equilibrium driven to the right by evaporating H2O and by

2 K t-butanolate + Mg - 2 K + Mg t-butanolate

Mg t-butanolate + H2O - MgO + t-butanol

Random - 2-12-2010 at 11:30

maybe naoh cant be reduced by magnesium and koh can

watson.fawkes - 2-12-2010 at 11:31

I can't say I have a lot of understanding about this whole affair, but one thing does strike me immediately. The boiling point of t-butanol is 82 °C and the reflux is being carried out at 200 °C. The partition of the t-butanol is biased toward the vapor phase. The concentration of t-butanol is going be highest in the distillate before it hits the surface of the liquid in the boiling chamber. If it doesn't get appreciable mixing before it evaporates again, it's like having a whole lot less of the reagent. If this is at all relevant (I'm not sure it is), then there are three things I can think of that would help.

1. Ensure that there's adequate excess t-butanol to occupy the gas phase between the boiling liquid and the condenser and the steady-state liquid flow on the condenser. It's like the amount of holdup in an ordinary distillation. It seems possible that the same amount of reactants should they succeed in a small vessel might fail in one too large.

2. Introduce the distillate to the bottom of the boiling flask with a funnel. That will promote greater intimate contact of the t-butanol and the solid reactants before the t-butanol evaporates. This improves the contact geometry of the reactants.

3. Supercool the distillate (that is, below its dew point) before introducing it back into the boiling flask. The t-butanol will take longer to heat up, promoting greater contact time between the reactants.

The only other idea I've got is to try a higher-boiling alcohol, like 2-methyl-2-butanol (t-amyl alcohol). It's boiling point is 102 °C and may have a more favorable partition and better contact times.

blogfast25 - 2-12-2010 at 12:35

Quote: Originally posted by watson.fawkes

1. Ensure that there's adequate excess t-butanol to occupy the gas phase between the boiling liquid and the condenser and the steady-state liquid flow on the condenser. It's like the amount of holdup in an ordinary distillation. It seems possible that the same amount of reactants should they succeed in a small vessel might fail in one too large.

Patent (Example 1: potassium) mentions a '1 litre four necked flask', production of 34 g of K (87 % of theor.), considerably larger than our German friend...

Example 3 (potassium from potassium alcoholate) mentions potassium tertiary amylate, here heated with Mg chips and decalin as solvent... indicating decalin (if easier available) may be a substitute for Shellsol...

In example 5 (cesium from CsOH), the solvent used is undecane (C11H24), BP 196 C, with some t-butanol, indicating the inert solvent isn't really critical appart from min. BP...

[Edited on 2-12-2010 by blogfast25]

bbartlog - 2-12-2010 at 13:39

Thing is though I wouldn't trust their examples. Len1 already made a convincing argument that some of the things they wrote up as if they had measured them were clearly just numbers that they had calculated (and not calculated correctly at that). They could easily have found one reaction that worked and then (based on a poor understanding of what was going on) written up a bunch of related 'examples' that would have no chance of success.

UnintentionalChaos - 2-12-2010 at 13:45

Two thoughts:

Preform the potassium tert-butoxide to avoid volatility issues. Either use a small amount of sacrificial potassium in tBuOH or 3A molecular sieves with KOH and tBuOH.

The author lists the source of his magnesium as "Ebay, junkyard - Chip filed down myself." This is (almost surely) not reagent grade...what if it had some aluminum alloyed into it? Perhaps this allows the KOH to etch the surface of the chips giving a much higher surface area. Does anyone have this sort of alloy available for testing?

aonomus - 2-12-2010 at 17:26

One thing that sounds like a possible idea to try is to add in a small amount of a phase transfer catalyst, or even something like a crown ether. Heck, trying something like glycerine (cheap, readily available, potential chelator for K+ in non-polar solvents) comes to mind too.

I don't have any MgO or cheap KOH (just a bottle of sigma reagent grade I'm saving) so I'm holding off for now. I'm also waiting to try to get my hands on a tank of nitrogen gas to create an inert atmosphere to do this experiment right.

condennnsa - 2-12-2010 at 19:42

well I since gave up boiling my paint thinner , mg, naoh, isopropanol slurry. nothing was happening, when watered down, no reaction, but i'm not giving up on this, i'll try with other solvents i have around, I heard that lamp fuel has a boiling point in the range of 200 C. If this whole business has any chance of working , i'm pretty sure it would be due, as bbartlog said, to a catalyst that's already present in the solvent and not so much the tert-butanol.
But we need more folk to have a go at this!

[Edited on 3-12-2010 by condennnsa]

Sedit - 2-12-2010 at 19:55

No it has nothing to do with a catalyst in the solvent, this is expressed in the patent.

froot - 2-12-2010 at 23:11

I'll have a go later.
Watson.fawkes mentioned one concern of mine so I wont mention it again. Potassium melts at 63 deg C and both the t-butanol and isopropyl alc. boil at 82 degC. I'm not sure why the temperature has to be as close to the bp of the solvent apart from optimising the rate of reaction?
The initial stage if the reaction on that forum involves heating the reactants before the butanol is added. I'm assuming this is to rid the contents of any water so 200 degC or where the solvent starts boiling is good. After the evolution of H2 has completed I would let things cool to the bp of the alcohol (80 degC) and do everything to prevent air accessing the system. So what if things proceed slowly.

2 things that don't make sense to me though. According to the reaction equation he provides MgO is formed. I don't see any in his flask. What happens ot the MgO?
Also, I find it unnatural for the potassium to form beads like that. As the reaction proceeds the precipitated K would have to be ultra clean for it to merge like that even at those temperatures without some sort of 'fluxing' condition helping it along. I would expect a gunky mass to accumulate at the bottom of the flask.

bbartlog - 3-12-2010 at 07:08

Quote:

So what if things proceed slowly.

If the reaction rate were to be halved for every 10C lowering of temperature, then you'd be looking at a reaction that took 4000x as long or so. The author on versuchschemie (pok?) suggests that 20-30 minutes is long enough to be able to see whether K globules are forming (though not to run to completion... that takes four hours). So potentially eight to twelve weeks at 80C to see whether you'd be getting anywhere, over a year to finish the reaction. Sometimes slowly really is ... too slow.

Quote:

MgO is formed. I don't see any in his flask.

There are bits of crumbly-looking white/gray stuff in his flask (up until he transfers the K to another container). In the first couple of pictures I assume the bits are KOH, but after that presumably MgO.
The beads, dunno. Surface tension should be high (I don't find the beading up strange), but I wouldn't be surprised if there's a ton of gunk trapped within the beads.

I'm interested in watson.fawkes observation about the boiling point of the tert-butanol. In order for it to not gas off, it would have to react fairly rapidly (turning into alkoxide, presumably) after being added. I suppose that the reason that it is added mixed in with some solvent is to provide a little bit of thermal buffering, giving some more contact time before it evaporates. Even so, it seems strange that it would be able to form a substantial amount of alkoxide in the short time it would be in the solution, especially given that it is just dripped in on top (no thistle tube or anything like that to try to bring it closer to the KOH on the bottom).

There's another thing about the post on versuchschemie that strikes me as odd: after excluding air via the balloon-with-pinhole (nice trick), he must have allowed quite a bit of air back in while slowly adding the catalyst solution with a dropper. Did he repeatedly unstopper the flask (he says the t-butanol was added over half an hour)? If so it certainly suggests that inert atmosphere is not required. But maybe he dripped the stuff in through the existing tube after removing the balloon, which sounds awkward but would I guess have made for a lot less air getting in the flask.

blogfast25 - 3-12-2010 at 08:09

Quote: Originally posted by froot

2 things that don't make sense to me though. According to the reaction equation he provides MgO is formed. I don't see any in his flask. What happens ot the MgO?
Also, I find it unnatural for the potassium to form beads like that. As the reaction proceeds the precipitated K would have to be ultra clean for it to merge like that even at those temperatures without some sort of 'fluxing' condition helping it along. I would expect a gunky mass to accumulate at the bottom of the flask.

No, no, the MgO is there alright.

And this is exactly how I would expect liquid potassium to behave in an inert liquid: with purity that behaviour has little to do. The only thing that somewhat stops the globules from merging into larger ones is surface tension. Think mercury, really. A round bottomed flask instead of a conical flask would give a slight advantage because it would promote contact between the reagents but also slightly push the formed globules together. I think gentle stirring rather than vigorously is probably to be preferred to promote metal coalescence. Keep it moving but not flying...

The potassium is likely to be quite free of impurities as neither the solvent, the alcohol nor the MgO are soluble in it. That’s somewhat similar to the formation of a regulus of molten metal during various Goldschmidt style (thermite) reduction reactions: there too fine globules of molten metal merge into a pool, pushing out any insoluble impurities (slag).

The patent does recommend passing the molten metal over a suitable filtering media. Purity could be checked superficially by reacting the metal with water and slightly acidifying the obtained KOH solution: any remaining occlusions in the metal should show up as turbidity of the solution…

What I did find especially strange about the patent is where in Example 1 (Potassium from KOH) they use KOH with about 9 % water and then compensate the amount of Mg used exactly to stoichiometry (acc. Mg + H2O = MgO + H2O). Deliberate? In Example 2 (Sodium from NaOH) there’s no mention of water and the alcohol is added straight from the beginning…

I will be trying this relatively shortly with ‘deodorised kerosene’ and 2-methyl-2-butanol (t-pentanol). If it does work there are some areas where improvement could be envisaged to turn this into a slightly more productive method:

1. higher boiling solvent to cut down on reaction times
2. other, perhaps longer chain t-alcohols
3. lower ratio of reaction media to reagents: our German friend simply divided all quantities in Example 1 by 10. Had he used only 25 ml of ‘solvent’, would it have made much difference? An idea of the solubility of the K t-butanoate in Shellsol might be enlightening there…

[Edited on 3-12-2010 by blogfast25]

blogfast25 - 3-12-2010 at 08:22

Quote: Originally posted by bbartlog

There's another thing about the post on versuchschemie that strikes me as odd: after excluding air via the balloon-with-pinhole (nice trick), he must have allowed quite a bit of air back in while slowly adding the catalyst solution with a dropper. Did he repeatedly unstopper the flask (he says the t-butanol was added over half an hour)? If so it certainly suggests that inert atmosphere is not required. But maybe he dripped the stuff in through the existing tube after removing the balloon, which sounds awkward but would I guess have made for a lot less air getting in the flask.

No, he states he used a syringe (‘Spritze’) although he doesn’t show it. Presumably the syringe goes straight through the rubber stopper or is inserted between the stopper and the flask neck. He does recommend drop funnel ('oder besser: einem Tropftrichter in 2-Hals-Kolben').

A two holed bung with refluxer and drop funnel in place would be the simplest solution…

[Edited on 3-12-2010 by blogfast25]

blogfast25 - 3-12-2010 at 08:47

In the mean time there have been failures to reproduce the method too, for instance:

Echelot on 28/11, 15.36, who got nothing:

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

One guy also warns against using molten potassium in conjunction with Teflon coated stirrer bars;

(CF2)n + 2 n K == n C + 2 n KF is potentially nasty!

a_bab - 3-12-2010 at 09:14

OK, if I were to bet my money I'd go for "bogus".

The author simply had some potassium to fake the pics with, because it IS potassium. Time will show I was right and pretty much all the above discussions are a waste of time and Internet bills.

Len also said something about feeling "ashamed" for the wasted time, chems and credibility he credited this project with once the failure was so obvious. WiZ should hopefully jump in with some stories such as "gold from seawater".

This case clearly belongs to my collection of such holly grails:
- phorone
- Hodge's easy WP (albeit he clearly made it as a joke)

The only reaction I've been impressed with all these past 10 years is the Calgon reduction for WP making.

-----------------------------------------------------------
"if it sounds too good to be true, it probably is"

a_bab out.

condennnsa - 3-12-2010 at 09:21

I've been asking myself, eversince I saw Pok's results, is if this really does work, how come the alkali metal producers did not implement it? It should certainly be way cheaper than any electrolytic process.

All the reagents are dirt cheap, and the alcohol non-consumable... something's fishy here.

a_bab, it is possible that he faked it , but what the heck would be his motive? Just making a mess with solvents, adding rubble and potassium to it, melting it , coming up with calculations and costs, just to post it on some forum??? that'd be crazy

[Edited on 3-12-2010 by condennnsa]

froot - 3-12-2010 at 09:29

Ok so I began my attempt this afternoon.
For the solvent I used technical oil which was a judgement call on my part and for the catalyst I used isopropyl alcohol.
I could not locate any magnesium today so I used a camping firestarter which is a block of Mg (supposedly) with a piece of flint. I used a drill press to produce the shavings.

I used the same setup as described.

I do not have the means to read temperatures above 110 degC but I heated the flask to well above 100 degC. As described I got the evolution of gas but did not test for hydrogen. This evolution of gas slowed up and I turned the heat off and let it cool to 80 degC.

I prepared a syringe with isopropyl alcohol and some of the technical oil as described. I expected the 2 to be miscible with each other but this was not the case.

As it was cooling I added a drop from the syringe and waited if the alcohol fizzed into vapour from the heat. When it was cool enough I added the lot, swirled it, and left it for a few minutes for it to get used to what I did to it. It turned a tea colour. I then turned up the heat again and continued swirling it and the tea colour darkened. No 'beads' appeared.

This is it so far and I'll post again later.

condennnsa - 3-12-2010 at 09:42

I found another post by Pok detailing this procedure , before he started a dedicated thread, this post has more pictures and some additional information:

here is the translation:
http://translate.google.com/translate?hl=en&sl=de&u=...

he says:

" (?) The dropwise addition of tert-butanol must be slow (! Exactly as directed), it evaporates much more abruptly when it hits the hot Shellsol mix, but again condenses and forms K-alkoxide can react with Mg (? ). "

the info and pics he provided makes it really credible...

[Edited on 3-12-2010 by condennnsa]

[Edited on 3-12-2010 by condennnsa]

blogfast25 - 3-12-2010 at 09:48

Quote: Originally posted by a_bab

[big snip]because it IS potassium. Time will show I was right and pretty much all the above discussions are a waste of time and Internet bills.

So you want to rebutt the guy's post on the basis of the photos, photos which tell you 'it IS potassium'... And you don't see the contradiction there? How do you know from the photo 'it IS potassium'? For all you know they're aluminium balls: I have one just like it...

We all do well to be sceptical about this but I believe it's worth trying to repeat the experiment.

Two other experimenters on versuchschemie claim to have seen ample amounts of hydrogen, one saw a small amount of K. Where does the hydrogen come from? Water + Mg? There's not enough water there. Alkanes + Mg? Hardly likely...

No, this is definitely worth revisiting, IMHO...

[Edited on 3-12-2010 by blogfast25]

blogfast25 - 3-12-2010 at 09:59

Quote: Originally posted by froot

Ok so I began my attempt this afternoon.
For the solvent I used technical oil which was a judgement call on my part and for the catalyst I used isopropyl alcohol.
I could not locate any magnesium today so I used a camping firestarter which is a block of Mg (supposedly) with a piece of flint. I used a drill press to produce the shavings.

Your experimental conditions are rather far removed from pok’s. No temperature reading is a disadvantage, if reduction does take place, the rate is very strongly dependent on temperature.

But you did get ‘gas’, assumed hydrogen. How long did that last for?

Nicodem - 3-12-2010 at 09:59

I did not read all about this topic, but just wanted to comment that the use of isopropanol is most likely futile. Metal alkoxides containing alpha-hydrogens are generally not stable up to 200 °C. They tend to decompose via beta-hydride elimination reactions, the rate at which they do so highly depends on the what metal alkoxide they are. I would expect potassium isopropoxide is not stable at the required conditions. The acetone formed in the decomposition is most likely to blame for the brown-red coloured crap formed via self condensation. On the other hand, t-BuOK is commonly used as a stable alkoxide at fairly high temperatures, so it might actually be long living at 200 °C as well. Also, tert-butoxides should be much more soluble in paraffins, which is another reason to stick to tert-butanol or higher homologues. The formation of t-BuOK should be fairly rapid at the given conditions, especially if Mg is finely divided and/or activated, so I don't think most of t-BuOH escapes, particularly if stirring is used or heating up is slow.

As to how this redox works, I have no good idea. I don't see how a "2 t-BuOK + Mg <=> (t-BuO)2Mg + 2 K" reaction could ever be favoured on the right side. Though at least the other part of the cycle, "(t-BuO)2Mg + KOH <=> 2 t-BuOK + Mg(OH)2" makes sense direction-wise. Quite weird. Anyhow, I would guess the most important factor is the activation of the Mg metalic surface and the quality&mesh of the metal.

a_bab - 3-12-2010 at 10:08

"blogfast25: So you want to rebutt the guy's post on the basis of the photos, photos which tell you 'it IS potassium'... And you don't see the contradiction there? How do you know from the photo 'it IS potassium'? For all you know they're aluminium balls: I have one just like it..."

This clearly shows you never had the opportunity to play with potassium or at least clean potassium balls, and notice their unique bluish hue as they start to oxidize. Few of us did, but those who did will understand what I mean. Their clean aspect in the pic is obvious because they melted and lost their cover, explaining the greyish color in the flask.

froot - 3-12-2010 at 10:15

Blogfast the bubbling lasted for about half an hour.
Thanks for the input Nicodem, that explains my tea then, the solvent cleared up after a while and a brown coagulate started appearing which confirms what you said.

blogfast25 - 3-12-2010 at 10:33

Quote: Originally posted by a_bab

"blogfast25: So you want to rebutt the guy's post on the basis of the photos, photos which tell you 'it IS potassium'... And you don't see the contradiction there? How do you know from the photo 'it IS potassium'? For all you know they're aluminium balls: I have one just like it..."

This clearly shows you never had the opportunity to play with potassium or at least clean potassium balls, and notice their unique bluish hue as they start to oxidize. Few of us did, but those who did will understand what I mean. Their clean aspect in the pic is obvious because they melted and lost their cover, explaining the greyish color in the flask.

Camera's lie: a lot. Had I posted a photo of my Al ball covered in a light paraffin oil and claimed it was potassium you'd have gladly believed me.

And I have played with potassium, thanks very much. You assume too much...

blogfast25 - 3-12-2010 at 10:37

Quote: Originally posted by froot

Blogfast the bubbling lasted for about half an hour.
Thanks for the input Nicodem, that explains my tea then, the solvent cleared up after a while and a brown coagulate started appearing which confirms what you said.

You see, purely intuitively, that seems a long time for just any water (from the oil and the KOH) reacting off...

a_bab - 3-12-2010 at 11:53

Blogfast, I am not assuming too much. Show me your stash of potassium and I believe you. BTW, dropping a piece or two of potassium in water while at high shool doesn't count here. I can, though, show you several pics with K, Na and Al in the same bottle under paraffin oil so you can see the difference. I'm aware about the camera faults, but K does have a bluish tinge

I would have not believed you with an aluminium ball. Don't forget the balls are slightly irregular, just like the real stuff. You obviously don't know what you're babbling about period.

Sedit - 3-12-2010 at 12:47

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

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

[Edited on 3-12-2010 by Sedit]

entropy51 - 3-12-2010 at 13:10

Quote: Originally posted by Sedit

Any suggestions on an over the counter alcohol that could be used? I don't know if anything has tert-butyl alcohol in it but I do want to attempt this with NaOH. ]

According to the NIH Household Products database, Vanish Mildew-Stain Remover contains 1-5% t-butanol and Turbo Octane Boost 108 contains an unstated amount of t-butanol, the remainder being methanol.

[Edited on 3-12-2010 by entropy51]

blogfast25 - 3-12-2010 at 13:32

A_bab:

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

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

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

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

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

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

[Edited on 3-12-2010 by blogfast25]

blogfast25 - 3-12-2010 at 13:44

Quote: Originally posted by Sedit

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

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

[Edited on 3-12-2010 by Sedit]

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

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

I’ll be trying this as soon as I feel confident I can meet these conditions but not before…

Sedit - 3-12-2010 at 15:04

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

[edit]

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

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

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

[Edited on 4-12-2010 by Sedit]

bbartlog - 3-12-2010 at 17:46

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

Quote:

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

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

watson.fawkes - 3-12-2010 at 20:41

Quote: Originally posted by bbartlog

I'm interested in watson.fawkes observation about the boiling point of the tert-butanol. In order for it to not gas off, it would have to react fairly rapidly (turning into alkoxide, presumably) after being added.

That's certainly one possibility; I won't deny that. There are others, though.

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

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

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

These three pathways, evaporation, solution, and alkoxide, seem to exhaust the possibilities that have been mentioned to date. Maybe we'll find out that there's something not accounted for.

watson.fawkes - 3-12-2010 at 21:09

Quote: Originally posted by blogfast25

Two other experimenters on versuchschemie claim to have seen ample amounts of hydrogen, one saw a small amount of K. Where does the hydrogen come from? Water + Mg? There's not enough water there. Alkanes + Mg? Hardly likely...

This brings up another thing I noted when reading over the Shellsol D70 data sheet. This specific composition is a hydrogen-treated oil; I'm not sure that's in the data sheet but in some other places where I saw it listed. D70 has very low levels of aromatics (< %0.5 m/m). It's listed as 60% paraffins (alkanes) and 40% naphthenes (cycloalkanes). There's no room for left olefins (alkenes). What I'm getting at is that this solvent is close to completely saturated.

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

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

spong - 4-12-2010 at 01:24

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

Jor - 4-12-2010 at 04:23

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

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

So I would not waste your chemicals on this. Just try to find some t-BuOH, it isn't that expensive, at least much less so than potassium metal.

bbartlog - 4-12-2010 at 05:30

Quote:

Water + Mg? There's not enough water there

It does seem a little strange, but on the other hand KOH does normally contain a fair bit of water. If you assume Pok's 6g of KOH was 8% H2O, that's enough to generate ~600ml of hydrogen gas.

Jor - 4-12-2010 at 05:40

Technical and lab-grade KOH usually contains about 15% water by weight.
So indeed that is where the water comes from.

blogfast25 - 4-12-2010 at 06:56

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

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

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

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

Initiation:

KOH + ROH < --- > KOR + H2O

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

Propagation:

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

Which may itself by pulled to the right by:

Termination:

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

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

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

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

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

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

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

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

[Edited on 4-12-2010 by blogfast25]

metalresearcher - 4-12-2010 at 07:37

Where can I get 2-methyl-2-butanol ?
On ebay I cannot find it.

blogfast25 - 4-12-2010 at 07:54

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

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

In GBP, might not ship elsewhere, not sure…

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

[Edited on 4-12-2010 by blogfast25]

blogfast25 - 4-12-2010 at 09:44

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

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

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

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

[Edited on 4-12-2010 by blogfast25]

its me

Pok - 5-12-2010 at 02:54

Hi,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

reaction vessel on aluminium foil with still liquid potassium globes:

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

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

.

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

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

left: Mg in Shellsol D70. right: KOH chips in plastic bag.

If you have any questions: ask.

[Edited on 5-12-2010 by Pok]

[Edited on 5-12-2010 by Pok]

Ephoton - 5-12-2010 at 03:38

truly thankyou I shall try this.

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

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

buy the alcohol.

[Edited on 5-12-2010 by Ephoton]

Jor - 5-12-2010 at 05:13

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

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

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

[Edited on 5-12-2010 by Jor]

condennnsa - 5-12-2010 at 05:28

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

Your presence here is really encouraging !

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

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

I can't wait to see if we will be able to replicate your results! I'll have a go myself as soon as i can get ahold of some t-butanol!

blogfast25 - 5-12-2010 at 05:50

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

One very obvious question is where did you by the Shellsol solvent? It's not what we call here an 'over the counter' solvent, it's rather specialist.

condennnsa - 5-12-2010 at 06:00

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

Art supply stores usually carry lots of Kremer products, even if they don't stock it, they should be able to order it for you from Kremer. This is my best bet in eastern europe, as I couldn't find ShellSol anywhere.

blogfast25 - 5-12-2010 at 06:07

I believe that where I wrote:

Quote: Originally posted by blogfast25

Termination:

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

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

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

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

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

and

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

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

[Edited on 5-12-2010 by blogfast25]

Pok - 5-12-2010 at 06:47

Quote: Originally posted by condennnsa

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

I only read a little bit of his attempt:

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

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

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

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

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

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

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

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

Quote: Originally posted by blogfast25

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

As condennnsa said, "Kremer Pigmente" offers Shellsol D70. This is my source.

blogfast25 - 5-12-2010 at 06:59

Thanks condenssa and pok for the Shellsol info...

UnintentionalChaos - 5-12-2010 at 08:21

Quote: Originally posted by Pok

Quote: Originally posted by condennnsa

Quote: Originally posted by blogfast25

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

As condennnsa said, "Kremer Pigmente" offers Shellsol D70. This is my source.

Pok- How sure are you that the printing plates are high purity magnesium? Have you successfully tried this procedure utilizing reagent magnesium?

Additional thoughts for others-

If this turns out to be an issue of reagent contact, perhaps we should try using the NaOH/KOH eutectic mixture. If that works, we can go back to the drawing board with plain KOH.

aonomus - 5-12-2010 at 08:22

Something that comes to mind is that the magnesium is passivated to a degree. Grignards usually need a little kick to get started (crushing with a glass rod, some iodine crystals, sonication). Perhaps either creating the magnesium chips under oil (ie: drill press on a Mg block under mineral oil) or transferring Mg chips in and adding a pinch of iodine prior to addition of KOH might activate the surface more. Since the solvent is non-polar I'd expect any chemistry to really only happen right at the contact point between a piece of Mg and a pellet of KOH.

Another thought is that while probably few people have access to the equipment (or have never used it), mechanical stirring might provide better results due to the above reasoning due slight grinding action, and also just because the paddle can help push potassium globules together.

Once I can find any sand for a sand bath, I'll have to give this a go on small scale with many varying conditions before scaleup.

At least for myself I want to try the following conditions

Reagents: Mg prepared under oil, Mg with iodine surface activation, Mg with mechanical crushing next to KOH
'catalyst': (calling the tBuOH this since we can't seem to confirm its mechanism yet) use KOtBu charged directly in during the reaction, ethylene glycol, glycerine
Solvents: hexanes, distilled mineral oil

UnintentionalChaos - 5-12-2010 at 08:25

Quote: Originally posted by aonomus

At least for myself I want to try the following conditions

Reagents: Mg prepared under oil, Mg with iodine surface activation, Mg with mechanical crushing next to KOH
'catalyst': (calling the tBuOH this since we can't seem to confirm its mechanism yet) use KOtBu charged directly in during the reaction, ethylene glycol, glycerine
Solvents: hexanes, distilled mineral oil

Didn't it get posted upthread that alkoxides with alpha hydrogens are unstable? That would certainly be the case for glycerin and ethylene glycol.

aonomus - 5-12-2010 at 08:35

@UnintentionalChaos: Hm, good point, I must have missed it.

blogfast25 - 5-12-2010 at 08:56

The question of good reagents contact (between Mg and KOH) would become less pressing with the mechanism I proposed above: a K t-alkoxide might be reasonably soluble in the solvent (Shellsol or equivalent) and easier to reduce than KOH itself. The t-alcohol would then play the classic role of an activation energy reducer, i.e. a catalyst...

[Edited on 5-12-2010 by blogfast25]

blogfast25 - 5-12-2010 at 09:41

The third photo posted by pok here (true size):

http://www.sciencemadness.org/talk/files.php?pid=194182&...

… does show, no matter how the metal was obtained, that it includes small pockets of impurities. Not a problem that re-melting, or better filtering, couldn’t solve though…

1281371269 - 5-12-2010 at 12:08

Would this work with NaOH?

NaOH would presumably be easier to work with - fewer problems relating to absorbed water

On the other hand the higher melting point of the Na might be an issue?

blogfast25 - 5-12-2010 at 12:24

Quote: Originally posted by Mossydie

Would this work with NaOH?

NaOH would presumably be easier to work with - fewer problems relating to absorbed water

On the other hand the higher melting point of the Na might be an issue?

Please read the patent (first page of this thread)...

Pok - 5-12-2010 at 12:26

@ blogfast25:

I also think that the tert.butanol works like a catalyst (see also description on versuchschemie.de).

About my 3rd picture: this aren't impurities. This are just some oxide coated smaller K globules wich got flattened by pressure of my pincette and attached to the clean cut surface of the massive K block. The K really is very pure. Just believe me

Pok - 5-12-2010 at 12:40

Quote: Originally posted by UnintentionalChaos

Pok- How sure are you that the printing plates are high purity magnesium? Have you successfully tried this procedure utilizing reagent magnesium?

The seller (once upon a time on ebay) told me about the Mg printing plates: "99% pure Mg" if I remember correctly. They were mainly used to print signs for wine bottles.

No, I didn't try it with reagent magnesium. But as the authors of the patent didn't specify the purity (I think), I just think it should be quite pure Mg. 99,9% Mg should work equally good or better.

watson.fawkes - 5-12-2010 at 13:13

Quote: Originally posted by aonomus

Something that comes to mind is that the magnesium is passivated to a degree. Grignards usually need a little kick to get started (crushing with a glass rod, some iodine crystals, sonication). Perhaps either creating the magnesium chips under oil (ie: drill press on a Mg block under mineral oil) or transferring Mg chips in and adding a pinch of iodine prior to addition of KOH might activate the surface more. Since the solvent is non-polar I'd expect any chemistry to really only happen right at the contact point between a piece of Mg and a pellet of KOH.

This hypothesis about a contact-point reaction is interesting. If true, there are a number of things to draw from it and ways to improve the reaction:

It would account for there being a very slow reaction in a well-agitated mixture, since the particles would be suspended and only occasionally in contact with each other.
Doing the reaction in a vessel that increases particle contact area should improve the rate, such as a pear-shaped flask. A column might work if it helps to compact the reactants.
Instead of stirring, kneading the reactants with a masher of some form. A glass rod with a high-tech rubber boot might be feasible. Something like a pair of muller wheels might work, in a configuration as is used to reprocess foundry sand or grind black powder. It's certainly not an off-the-shelf item, but it's easier than making a ball mill that operates at 200 °C under reflux.
Create a more intimate contact by preprocessing the reactants. Fuse KOH and make shavings of Mg right into the molten KOH. Pour out in a layer to cool and then break up into pieces.

All this is contingent upon point-contact being significant, but it's also a way of testing the hypothesis.

watson.fawkes - 5-12-2010 at 13:17

Quote: Originally posted by Pok

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

Did you notice, perchance, if there were two condensation limit heights in the tube?—one for D70 and the other for tert-butanol. In the configuration you're using, you might have had both. Any observation you've got here would inform how the tert-butanol is behaving in its vapor phase.

Mister Junk Pile - 5-12-2010 at 13:23

I wish I had time to try this. Although I don't think I have a source for Shellsol. Has anyone located it?

blogfast25 - 5-12-2010 at 13:31

@ Watson:

Something like a Banbury internal mixer, huh?

http://composite.about.com/library/glossary/b/bldef-b532.htm

Feed a paste of the ingredients, mix for required time and out comes a paste of potassium and by-products, ready for filtering!

It don't hurt to dream!

blogfast25 - 5-12-2010 at 13:41

Quote: Originally posted by Pok

About my 3rd picture: this aren't impurities. This are just some oxide coated smaller K globules wich got flattened by pressure of my pincette and attached to the clean cut surface of the massive K block. The K really is very pure. Just believe me

I contended much higher up that a clever hoaxer could achieve quite easily these very convincing photos by starting from a highly agitated emulsion of molten K (or Na) in kerosene (or similar) with some convincing looking grit thrown in. I'm not calling you a liar though, because I can't prove anything either way and because I'd like to believe you.

Not long now before a small army or SM experimenters gets to work! :cool:

I will follow the progress at versuchschemie.de too...

blogfast25 - 5-12-2010 at 13:42

Quote: Originally posted by Mister Junk Pile

I wish I had time to try this. Although I don't think I have a source for Shellsol. Has anyone located it?

Look a bit higher up. These people will possibly not know what hit them!

BromicAcid - 5-12-2010 at 15:59

It's the t-butanol, t-amyl alcohol, or other tert alcohol that would be the hangup for me. Can't find anything OTC aside from a few % mixed with other alcohols as entropy51 pointed out (and t-butanol seems to azeotrope with water and methanol so purification would be difficult).

U.S. sources of Shellsol D70

watson.fawkes - 5-12-2010 at 16:26

In the U.S., we don't have the same chain of German-based art supply stores. I've been looking for alternate sources. The CAS number is 64742-47-8. As I learned, it doesn't describe a single product, but a set of them parametrically by distillation parameters IBP (initial boiling point) and DP (dry point).

Not exactly a match, but easily the most common, this CAS is about half of what's in WD-40.

There's a product called "Protocol SafeClear Xylene Substitute" with the same CAS number. It's available here, and if you want to spend more money, at Fischer.

Citgo has five different product on this page, with the same CAS number. All of these have a much greater proportion of cycloalkanes than Shellsol D70 does. The one that most closely matches the distillation parameters is the highest-boiling of them; it's called "170 Solvent 66/3".

Equivalent products: Exxsol D80 (from ExxonMobil), Mosspar H (from Petro S.A.), Tarksol SC Plus (from Tarksol, and boils a bit hotter).

In my area, the yellow pages category that holds all the petroleum product distributors is "Oils - Lubricating". These products are all made by oil companies, and finding their distributors is the likely way of getting them at reasonable price.

I also suspect that certain products labeled "odorless mineral spirits" at art stores might be adequate.

watson.fawkes - 5-12-2010 at 16:44

Quote: Originally posted by BromicAcid

It's the t-butanol, t-amyl alcohol, or other tert alcohol that would be the hangup for me.

BLD Science carries 99% grade for about $25 / 250 ml. They sell to small folk. They've got other quantities and grades. This was the only vendor on Google Shopping.

Edit: I thought most would want 99% 250 ml and not the bigger and purer one I posted first. But for whatever reason, there's a $25 hazmat fee in the computer on the small package and not on the larger 99.5% 1 L for $60. Perhaps they'll drop the hazmat if you point it out. Perhaps you'll have extra to share with friend.

[Edited on 6-12-2010 by watson.fawkes]

aonomus - 5-12-2010 at 16:51

@watson.fawkes One thought I had about the contact-reaction hypothesis of mine was that there was some video of 'unconventional sodium' (or potassium?) made by lighting an alkali metal hydroxide and magnesium mixture a while back. I only vaugely recall something about this, mainly because of the magnesium fire and its poor product of low purity.

The process development side of me wants to make a way to make this reproducible in good yield, with readily available reagents... once I get a damned sand bath I can start experimenting...

Another thought is to take an aliquot of the reaction mixture (hot) and evaporate to dryness under inert atmosphere to take a look at what is in the reaction mixture. If KOtBu is being constantly turned over by reflux, then the levels of KOtBu should be fairly constant (bar loss of tBuOH upon sampling).

[Edited on 6-12-2010 by aonomus]

garage chemist - 5-12-2010 at 16:52

T- butanol is easily made by the grignard synthesis from MeMgBr and acetone.
PainKilla has outlined how he successfully made MeBr from NaBr, H2SO4 and MeOH and used it to generate a substantial amount of 1,4-dimethoxybenzene from a solution of hydroquinone in aqueous NaOH.
For turning the MeBr into the grignard, it would have to be purified and dried by a H2SO4 washing bottle and CaCl2 drying tube and condensed in a receiver cooled by ice and salt, or simply dissolved into chilled absolute ether and this solution slowly added to Mg and a little iodine crystal.
For this grignard synthesis, the reflux condenser would have to be colder than 0°C (chilled water/glycol or alcohol as the coolant) to condense the evaporating MeBr, as the formation of the grignard reagent is very exothermic.

I have prepared tert-pentanol from EtMgBr and acetone, and find that this synthesis is excellent to practice the generation and use of a grignard reagent at home.
I mix all of the EtBr with the amount of abs. ether that gives a 2,5 mol EtBr/L solution and add some of it to the stochiometric amount of reagent grade Mg to which a small I2 crystal has been added, so that the Mg is completely covered by the solution.
It is carefully heated by a heatgun or small flame until the solution gets turbid and it starts to reflux on its own (reflux condenser with ice water and drying tube). The rest of the EtBr solution is added at a rate that maintains a gentle reflux of the ether. After all is added, the solution is refluxed for 30-60 minutes so that nearly all of the Mg has reacted (I like to use an excess, but it's not really necessary).
The addition of acetone to this solution is highly exothermic, with fizzing and sputtering. It is done drop by drop under magnetic stirring.
Then work up as usual, with cold dilute aqueous HCl (the tertiary alcohol doesn't dehydrate under these conditions, but be sure to give the ether solution a final thorough wash with aqueous Na2CO3 before distillation, as HCl traces would otherwise cause elimination to 2-methyl-2-butene).

This synthesis of potassium is most remarkable. Regrettably I can't try it at this time, due to lack of Shellsol solvent (what is this solvent made up of anyway?). I do have the t-butanol and 99,8% Mg powder, and the KOH which generally is of 85% purity with the remainder being water.
The t-butanol may not be replaceable in this synthesis, but I hope that the Shellsol is.

watson.fawkes - 5-12-2010 at 16:58

Quote: Originally posted by aonomus

once I get a damned sand bath I can start experimenting...

As a good Canadian, make a Rona run and pick up a bag of backyard playground sand.

Magpie - 5-12-2010 at 17:04

Or, go to your neigborhood pet shop. They should have a fine selection of clean sand.

Sedit - 5-12-2010 at 17:05

OK this is getting tiresome in that I have lost 3 post recently that managed to vanish from the board all together. The last being a post I JUST made here asking if it was possible that the molten K was acting as a catalyst of sorts dissolving trace amounts of Mg and allowing the reaction to proceed quicker. This would mean that until even trace amounts of K where made nothing at all would happen. Yet when the ball got rolling we would be able to generate large amounts of Molten K threw the process.

I would desire to amalgamate the Mg as a last resort to see if the higher reactivity gives faster better results. I had alot more written on the subject but I don't feel like typing it again. Moderators would you please help me in finding out why my post keep vanishing off the forum...... Is someone deleting them?

Fleaker - 5-12-2010 at 17:15

Such a cheap, low temperature route to potassium metal needs to be confirmed. I think that diesel could potentially be substituted for Shellsol since they have similar boiling points and low aromaticity.

This is a supremely interesting thread!

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