Sciencemadness Discussion Board

Make Potassium (from versuchschemie.de)

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Sedit - 5-12-2010 at 18:36

I think Naptha would be a better alternative since it is non reactive, can have high BP and on top of that its overall contents can be found thru MSDS.

Does anyone have spare trace amounts of K or Na to test my theory? I don't and I would have to make some thru electrolysis before I could confirm my theory. I have been thinking about this alot and im not 100% sure that the alcohol does squat other then remove the protective coating which could be done better by amalgamating the Mg prior to use.

Preformed Potassium tert-butoxide should be used IMHO instead of the alcohol since it would not evaporate on contact with the high temperature reaction.

aonomus - 5-12-2010 at 18:37

Quote: Originally posted by Sedit  
Preformed Potassium tert-butoxide should be used IMHO instead of the alcohol since it would not evaporate on contact with the high temperature reaction.


That was my thought too.

Fleaker - 5-12-2010 at 19:15

I reckon the water present in the KOH probably also activates the surface and any in situ KtBuO is made into elemental K by reaction with the magnesium.

So this tells me that perhaps I can make K from KtBu in say decalin or some other solvent I have available in the laboratory. Hmm...

Jor - 5-12-2010 at 20:40

Quote: Originally posted by garage chemist  
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.

Maybe I will attempt the synthesis of t-pentanol tomorrow. I have never done a Grignard before.
How dry exactly does the ether need to be? I heard for making Grignards such as PhMgBr you need very dry ether, but when using aliphatic halides this is not necessary right? Is drying with anhydrous CaCl2 suffiencent, or should I also dry with Na? And is distillation after drying with Na necessary? AFAIK hardly any NaOH is soluble.
I will make EtBr by the method GC mentioned here, and wich was nicely documented by smuv:
http://www.sciencemadness.org/talk/viewthread.php?action=pri...
It is required that EtOH and H2O are removed from the EtBr, Et2O contamination is no problem in this case. Is it satisfactory to dry it over CaCl2, wich absorbs both EtOH and H2O? I'd rather not wash with conc. H2SO4, because it costs quite a lot of this reagent to remove relative small amounts of impurities. But if it's absolutely required, I will.
Finally, can I use Mg-ribbon (cut into small pieces) as the magnesium source?

Just to say, I'd rather have a little lower yields, than spending a lot of time distilling and wasting chemicals to make the used chemicals very dry for optimal yields.

mr.crow - 5-12-2010 at 20:51

Wow this is quite a thread. Too bad I don't have anything to add.

I might be ordering some tert-butanol, but I'm not sure how feasible it would to send it to other members

Magpie - 5-12-2010 at 20:53

I've always heard that the ether must be dry for a Grignard. I believe this as my attempt at school failed, as did those of all my labmates. We suspected wet ether. The night shift used a new can of ether and they were all successful.

I did a Grignard at home and it worked very well. Made phenyl bromide, IIRC. I used 3A mole sieves to dry my ether. As GC recommended I also added a speck of iodine.

[Edited on 6-12-2010 by Magpie]

garage chemist - 5-12-2010 at 22:00

The ether must be predried over CaCl2 for several days and then refluxed over thin slices of sodium for 24 hours, using a reflux condenser with CaCl2 guard tube until there is absolutely no more hydrogen evolution at reflux temperature. Then it is distilled form the sodium in an oven-dried distillation apparatus also carrying a drying tube, and used immediately. The apparatus itself also has to be oven-dried, as glassware from the shelf contains adsorbed moisture on its surface that will make starting of the grignard difficult.
The EtBr is predired with CaCl2, dried by shaking with P2O5 and distilled in an oven-dried apparatus over a fresh portion of P2O5.
Using those precautions, my grignards start reliably and quickly.

Jor - 5-12-2010 at 22:08

Thanks. I understand this will work, but isn't this a bit overkill? I hate to all the distillations and refluxing for 24 hours. I don't have the time for it. Will drying for a few hours over CaCl2 or otherwise refluxing for an hour over Na, and distilling be fine?
I will dry the EtBr over P2O5, it doesn't introduce an extra distillation step anyway.
The acetone, should this be dried? I have a bottle 'for analysis' from Acros.
By the way, i have about 100-150mL ether left. If I dry it (I will need about 50mL for the Grignard), I will dry it all, and distill it all. i will then have to stabilise it again. I have done this before, by adding a pinch of diphenylamine. But how much is actually needed?

Magpie - 5-12-2010 at 22:26

I agree that your glassware should be well dried in an oven. I forgot to mention that.

aonomus - 5-12-2010 at 22:38

@Jor: just be careful how you dry your acetone, drying agents which have proton abstraction ability and are far too basic can form an aldol condensation product which is a pain in the arse to get rid of. Threw me off once and I saw it on another blog as well (http://curlyarrow.blogspot.com/2010/04/anhydrous-solvents-pa...) . CaCl2/NaSO4/MgSO4 work fine for a quick dry.

ziqquratu - 5-12-2010 at 23:56

Assuming your ether isn't soaking wet, just a couple of hours is more than dry enough for a Grignard reaction if you use an initiator (iodine, ultrasound, 1,2-dibromoethane). If you have benzophenone, add a little as an indicator - if it's blue, you're water and oxygen free; anything else, reflux it longer then add a little more benzophenone, repeat until it stays blue.

Probably the easiest at-home method, however, is to use molecular sieves - 24-48 h over 10-20 %w/v of 3 (or probably 4) angstrom sieves should get it as dry as would sodium wire. See the attached article from J Org Chem.

You can dry acetone over sieves as well, but only overnight (any longer and the aldol condensation aonomus mentioned becomes significant).

In addition, if memory serves, there was a J Chem Ed article a couple years ago that suggested that a Grignard reaction could easily be performed in undried ether if you sonicated the reaction. The yields were somewhat lower, but with acetone and bromo(m)ethane, it's probably not a huge issue.

Attachment: Efficient drying of organic solvents using molecular sieves or activated alumina.pdf (765kB)
This file has been downloaded 8574 times

Ephoton - 6-12-2010 at 01:21

what about making the alcohol from MTBE a common additive for fuel

bbartlog - 6-12-2010 at 06:37

Well if you just had some MTBE sitting around, sure. But otherwise you could as easily order tert-butanol as MTBE, neither is OTC.
Theoretically you could also synthesize a suitable alcohol via Barbieri reaction rather than Grignard (for example, 3-ethyl-3-pentanol from 3-pentanone, zinc, and ethyl iodide... and I think then you don't have to worry quite as much about dryness). But again, unless you just happen to have these reagents (and probably even if you do) it seems easier to just order the tert-butanol.

blogfast25 - 6-12-2010 at 07:19

Read the patent. Shellsol should NOT be essential and a close equivalent should work, assuming the whole thing isn't a hoax, of course! In the other examples of the patent other solvents are used (assuming again these examples are real experiments). An inert, non-polar, mostly paraffinic solvent is what seems to be needed.

And other t-alcohols are also mentioned. I'll be trying 2-methyl-2-burtanol (t-amyl alcohol).

Jor - 6-12-2010 at 07:28

Barbieri reaction seems a nice alternative.
Say I want to make t-amyl alcohol from acetone and EtBr. I have the zinc powder as well. But the problem is, a quick search indicates that THF is most often used, wich I don't have.
Are there alternative solvents?
And what would be a common procedure in this case for anyone who has experience with this reaction (acetone + EtBr).

Pok - 6-12-2010 at 07:58

Quote: Originally posted by watson.fawkes  
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.


The D70 only seems to form fog within the reaction vessel which doesn't rise up into the pipe. But the t-butanol indeed condenses in the pipe during the whole time (especially at the beginning of the t-butanol addition...after some time only small amounts condense) and is refluxed by outside cooling with wet toilet paper around the glas pipe. This are just observations. Of course, you can't identify the two solvents in a gas or fog phase by eye. But something liquid always condenses (no water, of course) which really should be the t-butanol. As I kept the glas pipe quite cool by the wet toilet paper, only t-butanol can rise up - the D70 would have been condensed much earlier on its way through the pipe.

watson.fawkes - 6-12-2010 at 08:45

Quote: Originally posted by Pok  
The D70 only seems to form fog within the reaction vessel which doesn't rise up into the pipe. But the t-butanol indeed condenses in the pipe during the whole time (especially at the beginning of the t-butanol addition...after some time only small amounts condense) and is refluxed by outside cooling with wet toilet paper around the glas pipe. This are just observations. Of course, you can't identify the two solvents in a gas or fog phase by eye.
Identification by eye isn't so much the problem, given the large boiling point differences. At risk of putting words in your mouth, it seems like you're reporting that the reflux is fractionating the two volatile reagents. This is indicative that they're not forming a high-boiling azeotrope. I find it interesting that the top part of the Erlenmeyer flask has enough contact area with air to condense the D70.

The other observation you make is that the reflux rate of tert-butanol seems to diminish over time, which means it's being consumed as a reactant, even if an intermediate reactant. This reminds me of the behavior of AlCl3 as a catalyst, which can end up in a complex with the product. I see several possibilities:

watson.fawkes - 6-12-2010 at 08:55

Quote: Originally posted by blogfast25  
Something like a Banbury internal mixer, huh? [...] It don't hurt to dream! :D
You get right on that, will you? kthnx.

This idea did give me another idea this morning for some reactor engineering. Use a column as a reaction with a tube to lead distillate to the bottom of the column. The idea is that you could then add a perforated pressure plate on the reagents. As potassium is generated you could press it out of the solid reagents and cause it to flow upward past the perforations. The plate would separate product from reagent and would further promote intimate contact between the solid reagents.

blogfast25 - 6-12-2010 at 09:58

I was only half joking, Watson. But you do seem to have an awful lot of faith in pok's results. I prefer to be agnostic for the moment. Someone needs to corroborate this first...

Your second idea? Think Soxhlet with a fused glass filter as 'pressure plate'...

ScienceSquirrel - 6-12-2010 at 10:33

I think this paper might be of interest, can anyone get it;

http://pubs.acs.org/doi/abs/10.1021/ja01690a014

I have compared their results with the electrochemical series here and it is a bit contradictory.
http://en.wikipedia.org/wiki/Electrochemical_series
Reaction 1 seems OK and what I would expect but the others seem a bit odd.
In my opinion potassium should displace magnesium from its salts, not the other way round unless something very strange is happening.

[Edited on 6-12-2010 by ScienceSquirrel]

watson.fawkes - 6-12-2010 at 12:32

Quote: Originally posted by blogfast25  
I was only half joking, Watson. But you do seem to have an awful lot of faith in pok's results. I prefer to be agnostic for the moment. Someone needs to corroborate this first...

Your second idea? Think Soxhlet with a fused glass filter as 'pressure plate'...
I'm more interested in developing hypotheses under the assumption that the synthesis works than believing them myself. I feel like I'm just following the logic of what my turn out to be a counterfactual assumption. It's been profitable as a way of developing things to try and ways to think. I don't have any particular investment in being right, as such. It's a way of promoting creative thought about the subject.

As for using a frit as a pressure plate, I'm guessing the surface tension of molten potassium would require too much pressure to be particularly feasible. Maybe one could locate a plate made of very coarse frit.

blogfast25 - 6-12-2010 at 12:40

Quote: Originally posted by ScienceSquirrel  
I think this paper might be of interest, can anyone get it;

http://pubs.acs.org/doi/abs/10.1021/ja01690a014

I have compared their results with the electrochemical series here and it is a bit contradictory.
http://en.wikipedia.org/wiki/Electrochemical_series
Reaction 1 seems OK and what I would expect but the others seem a bit odd.
In my opinion potassium should displace magnesium from its salts, not the other way round unless something very strange is happening.

[Edited on 6-12-2010 by ScienceSquirrel]


Personally I don’t think the paper you cited will throw much light on the question.

We know that 2 KOH + Mg --- > 2 K + MgO + H2O should work thermodynamically, just calculate the heat of reaction (HoR) from the heats of formation. Similarly 2 KOH + Mg --- > 2 K + Mg(OH)2 works thermodynamically even slightly better (as shown above)…

When solids are reacted and others formed, lattice energies (that make up most of the heat of formation of ionic solids) play a very important part in determining ΔG (and thus also ΔH).

Typical ‘confusing’ situations arise when you rely on the potential series alone:

AlCl3 + 3 Na --- > Al + 3 NaCl:

Estimated HoR: - 649 kJ/mol

AlF3 + 3 NaF --- > Al + 3 NaF

Estimated HoR: - 215 kJ/mol

But NaOH + 2/3 Al --- > Na + 1/3 Al2O3 + ½ H2

Estimated HoR: - 132 kJ/mol

So depending on the solid ionics involved, Na can reduce Al salts but Al can also reduce Na salts.

Here we suspect that the reaction is rather between a dissolved K salt (K t-butoxide) and the solid reagent, yielding K and solid Mg(OH)2…


[Edited on 6-12-2010 by blogfast25]

blogfast25 - 6-12-2010 at 13:00

Fair points, Watson...

blogfast25 - 6-12-2010 at 13:22

I’ve been thinking a bit about the problem of metal coalescence in the case of a typical set up (conical or round flask as reactor) because pok mentioned a test where the metal didn’t seem to want to coalesce into larger globules.

A load of small globules store potential energy which is wholly or partly released when the globules coalesce into larger ones. This is reflected in the thermodynamical definition of surface tension s = dW/dS (J/m2) with W energy and S surface area. To increase the surface area (make smaller, not larger globules), dW = s x dS (with dS < 0). To promote the formation of larger globules you’d have to do something that actually increases surface tension, so the opposite of adding a surfactant. What could achieve that?

Per - 6-12-2010 at 14:08

thermodynamic back and force, I tried it with low boiling paraffin and isopropanole.
first I heated up the paraffin with the Mg and KOH, noething happened as expexted, then I added the alcohol, bubbles at the Mg have been formed, most likely H2(not tested) after about 30 minutes the gas evolution stopped nearly completely.

I used a piece of Mg, not just powdered Mg, so I took one peace out of the oil and dropped it into water, noething happened, no K at all and the piece of Mg was eroded quite well.

I tried aluminium foil as well, which has become attacked at the addition af new isopropanole but once again, no K with the water test.

It seems that only the isopropanole was consumed in my experiment, bacause the gas evolution started always after adding the alcohol, but then again the potassium propoxid should has been formed and should have reacted to give the potassium, but there was no K at all.

Sedit - 6-12-2010 at 14:22

Quote: Originally posted by blogfast25  
I’ve been thinking a bit about the problem of metal coalescence in the case of a typical set up (conical or round flask as reactor) because pok mentioned a test where the metal didn’t seem to want to coalesce into larger globules.

A load of small globules store potential energy which is wholly or partly released when the globules coalesce into larger ones. This is reflected in the thermodynamical definition of surface tension s = dW/dS (J/m2) with W energy and S surface area. To increase the surface area (make smaller, not larger globules), dW = s x dS (with dS < 0). To promote the formation of larger globules you’d have to do something that actually increases surface tension, so the opposite of adding a surfactant. What could achieve that?


My only thought is in order to increase surface tension what is needed is a drop in temperature. Perhaps the larger globs are made by swirling the mixture while it is relatively cold as compared to heavy stirring when its very hot.


BTW can anyone think of a source of Mg I might have around the house? It doesn't have to be pure at all.

ScienceSquirrel - 6-12-2010 at 14:27

I am not rubbishing the claimed procedure but trying to take a reasonable position as a constructive critic.
It is possible that it may work under the right conditions but they could be quite critical eg D70 as the solvent, t-butanol as the co solvent / reactant and the right grades of potassium hydroxide and magnesium.
I would be delighted to see it work.

a_bab - 6-12-2010 at 15:33

Quote: Originally posted by Sedit  

BTW can anyone think of a source of Mg I might have around the house? It doesn't have to be pure at all.




Mg_Sharpeners.jpg - 123kB

Jor - 6-12-2010 at 16:04

Is this magnesium fine for a Grignard, or is too fine (and thus too reactive):
http://cgi.ebay.de/100g-Magnesium-Pulver-200-m-Mg-/330501433...

If it fine, can an even finer powder be used for Grignards, as fine powder is more useful for producing some funny compounds such as magnesium silicide.

[Edited on 7-12-2010 by Jor]

Sedit - 6-12-2010 at 16:12

Quote: Originally posted by ScienceSquirrel  
I am not rubbishing the claimed procedure but trying to take a reasonable position as a constructive critic.
It is possible that it may work under the right conditions but they could be quite critical eg D70 as the solvent, t-butanol as the co solvent / reactant and the right grades of potassium hydroxide and magnesium.
I would be delighted to see it work.


I have seen various attempts such as lens fail. He used the best and weighed things to a T yet he failed. As have others.

Im speculating that this has nothing at all to do with the proper equipment or process but am more inclined to believe that this is the result of contamination of some sorts such as perhaps Al in the Mg and the possibiliy that the formed Potassium aluminate is what is being reduced with Magnesium to yeild Potassium. Or there is the possibility that there is some sort of chlorocarbon in the solvent forming some sort of grignard which is acting like a catalyst in the reaction.

I know its far fetched but it seems the simpler someone goes with this the better results we see. Now we are sure that K can be generated by the reduction of KOH with Mg but the solvent and low temperature conditions seem to prohibit it. My logic is I say we start to test as many things as we can in hopes of activating the Mg the best way possible as well as trying various alloys. I personally want to try amalgumated Mg in KOH and see what I get out of it.

Thanks, I forgot all about pencil sharpeners. I know them and fireblocks are all I could think of.

Sedit - 6-12-2010 at 18:10

Ok I have read and read and watched and read some more but.... Something really really bugs me here Pok that I hope you can answer quickly so I don't think you had time to think up a lie.


Where is your Mg in this picture
http://www.versuchschemie.de/upload/files3/2272042_4948.jpg

It appears there is alot of K around and what also appears as a bunch of unreacted KOH yet there is a single problem. There is NO PRECIPITATE. There should be Magnesium hydroxide or something of the sorts in there and there is NON.

I want to believe and I want to prove you right but..... Its starting to appear that the Mg we see in the start is nothing more then K that has been heated... stirred... then allowed to cool while still stirring. Then as we heat the solvent the "Mg" reacts with the KOH producing K. In reality it appears the KOH is doing nothing and there is no Mg there at all.


Im not trying to make an enemy here im only stating what I see and I honestly have a good sence of when someones pulling the wool over my eyes..... What would the motivation for such an act be?


You know what it takes to be a chemist since we are all overly critical so you MUST reproduce this experiment in order for it to stand again peer review. And before doing so you must allow us to tell you how to do it. If you can not do that then.... its not worthy of my money and time im willing to invest to prove your cause.

woelen - 6-12-2010 at 23:59

Next weekend I plan to do a few different tests. I have no ShellSol D70, but I do have high grade paraffin oil (colorless viscous liquid, perfectly odorless and completely free of acidic and unsaturated ingredients). I also have the following chemicals:
- KOH, general reagent grade, 85...90% KOH, balance is water
- Mg turnings, reagent grade
- Mg powder, reagent grade
- Al needles, ultra pure 99.99%
- Al powder, different grades, mostly pyro-stuff
- Magnalium powder (50Mg/50Al), pyrotechnic grade
- tert-butanol, synthetic grade (I need to melt this first, now it is a hard lump of ice, MP is 25 C)
- mercury

I could try different things systematically each time at 5 ml volumes of oil (one tenth of what Pok did):

- original experiment, using KOH, Mg turnings, paraffin oil and a drop of tert-butanol
- variation 1: Use Mg powder instead of Mg turnings
- variation 2: Use magnalium powder instead of Mg (influence of presence of Al)
- variation 3: Use magnalium powder with a tiny amount of Hg added for amalgation and destruction of oxide layer
- variation 4: Use Al needles instead of Mg
- variation 5: Use Al needles with a tiny amount of Hg added for amalgation

All experiments I can do in a test tube with a tight seal and a small thin tube as gas outlet for the hydrogen. Heating to 200 C is no problem with this. The use of mercury is a last resort for me due to the toxicity of the resulting waste and I will only use tiny amounts. I only do these if all other experiments fail.

If you have any ideas of how to improve the experiment series, please let me know. I want to try this next weekend. I really want this to be resolved. Making K-metal in this way sounds too good to be true, but I'll give it a try anyway.

Arthur Dent - 7-12-2010 at 04:31

As soon as I have a minute of my time this Friday, i'll visit a few art supply stores in my neighborhood, maybe I'll get lucky and find that Shellsol solvent. I vaguely remember that name from a few years back as an oil paint solvent and brush cleaner. Maybe I'm wrong but it would be great to find that stuff (or something similar to it).

I just found out that "Bestine" rubber cement solvent is 100% pure heptane! Cool!

I found a good source of Tert-Butanol at $36 for 500 ml... Is that a fair price?

Robert

Pok - 7-12-2010 at 04:44

Quote: Originally posted by Sedit  
Ok I have read and read and watched and read some more but.... Something really really bugs me here Pok that I hope you can answer quickly so I don't think you had time to think up a lie.

I'm not 24/7. But if you wish answers within seconds/minutes, I could tell you a date and time so we can do something like chatting. And I couldn't think about lies.

Quote: Originally posted by Sedit  

Where is your Mg in this picture
http://www.versuchschemie.de/upload/files3/2272042_4948.jpg


It's gone (this is a picture of the end of the reaction!).


Quote: Originally posted by Sedit  
It appears there is alot of K around and what also appears as a bunch of unreacted KOH yet there is a single problem. There is NO PRECIPITATE. There should be Magnesium hydroxide or something of the sorts in there and there is NON.


There isn't alot of K around. The only K you can see are more or less large balls (shiny). The rest (gray stuff) is the Mg(OH)2 or MgO (the patent says "MgO" if I remember correctly). The reason why there isn't a precipitate or dust of Mg(OH)2 or oxide is: I put in the non-modified KOH-Chips at the beginning. And I did it without stirring continiously. Stirring would definitely devide the Mg(OH)2 or oxide into dust/powder.


Quote: Originally posted by Sedit  

I want to believe and I want to prove you right but..... Its starting to appear that the Mg we see in the start is nothing more then K that has been heated... stirred... then allowed to cool while still stirring. Then as we heat the solvent the "Mg" reacts with the KOH producing K. In reality it appears the KOH is doing nothing and there is no Mg there at all.


I can absolutely understand that you are sceptically. I for myself would react exactly the same way. The Mg at the one photo (Mg in Shellsol + KOH in bag) really is Mg. (BTW: which photo do you mean? On versuchschemie.de I put a collage of photos where the second one shows the beginng of the reaction (after the total t-butanol addition). The shiny pearls ARE already K! The white stuff is KOH, the dark gray stuff is Mg (in reality more shiny). At the 4th photo you can see larger K balls coated by Mg filings. The grayish stuff here is KOH and MgO, the dark gray stuff is a mix of tiny K balls and Mg) If you would put some bought K into the Shellsol and devide it into such fine drops, it would rapidly oxidise on its surface (especcially if you assume, that I heated something like small K balls which united to form larger K balls). I HAD the problem sometimes, that K balls didn't unite.

Quote: Originally posted by Sedit  

Im not trying to make an enemy here im only stating what I see and I honestly have a good sence of when someones pulling the wool over my eyes..... What would the motivation for such an act be?


The motivation for a lie would be recognition. Of course, I want this :D. Nevertheless it's a true experiment which I've done.


Quote: Originally posted by Sedit  

You know what it takes to be a chemist since we are all overly critical so you MUST reproduce this experiment in order for it to stand again peer review. And before doing so you must allow us to tell you how to do it. If you can not do that then.... its not worthy of my money and time im willing to invest to prove your cause.


I know science. I know that reproduction is neccessary. But if I just do it again and again, this wouldn't help you. Some of YOU has to do it (scientifically - which means: in exactly the same way I did it.)

Tell you how to do it? I thought you already have every detail. Look at versuchschemie.de and here. I made it like the patent with modifications that I already told you. You should easily be able to repeat this synthesis.

I hope you know what "exactly" means: 50ml, 25cm-glaspipe with wet toilet paper (moisten it sometimes to prevent dehydration), balloon at the top with tiny hole, zero oxygen input, Mg filings made from 99-100% Mg made by filing with a medium file. and so on It might be that I really forgot one or two details. If this is the case, I would apologize - just ask! (I'll be back in 24 hours)

I can also give you a really detailed description where no questions will be left. But this takes time and only be possible tomorrow or in max. 3 days.


[Edited on 7-12-2010 by Pok]

[Edited on 7-12-2010 by Pok]

metalresearcher - 7-12-2010 at 07:21

Quote: Originally posted by woelen  
but I do have high grade paraffin oil (colorless viscous liquid, perfectly odorless and completely free of acidic and unsaturated ingredients).

Where can I get this oil ? Or can I melt candles which is also paraffin ?


[Edited on 2010-12-7 by metalresearcher]

wallschem - 7-12-2010 at 07:26

I found this in an old Spanish chemistry book that is very similar to what pok is doing to synthesized Potassium. Instead of using Shellsol D70 and Ter-butol in this book they use Paraffin and Toluene at 180 degree Celsius.


DSC05173.JPG - 59kB

ScienceSquirrel - 7-12-2010 at 07:30

It depends where you live but some UK pharmacists sell very high grade liquid paraffin.
It is about the viscosity of motor oil and completely odourless and tasteless as well.
It used to be taken internally for constipation in humans but I think this use is now deprecated but it is still given to horses to get them 'moving', it is still used for eczema.

Cheap as chips on eBay!

http://cgi.ebay.co.uk/Liquid-Paraffin-500-ml-/260694362661?p...

watson.fawkes - 7-12-2010 at 07:41

Quote: Originally posted by woelen  
- original experiment, using KOH, Mg turnings, paraffin oil and a drop of tert-butanol
- variation 1: Use Mg powder instead of Mg turnings
- variation 2: Use magnalium powder instead of Mg (influence of presence of Al)
- variation 3: Use magnalium powder with a tiny amount of Hg added for amalgation and destruction of oxide layer
- variation 4: Use Al needles instead of Mg
- variation 5: Use Al needles with a tiny amount of Hg added for amalgation
[...]
If you have any ideas of how to improve the experiment series, please let me know.
If you're going to try with Hg, do also try a variation with the different forms of Mg itself as opposed to the magnalium alloy. I'm not sure it's worth doing the experiments with the Al needles unless the ones with Mg or magnalium work. Aluminum isn't claimed in the patent; it's only mentioned there as a reducing agent in thermochemical processes.

Using magnalium is interesting because it might give some indication if Al could either hinder or promote the reaction. Consider doing a mixed x% Mg + (100-x)% magnalium series.

I'm not going to go so far as to suggest that you make alloy compositions in these percentages, but I'll mention it. I find it plausible that certain alloying agents might causing lattice strains that change the activation energy of Mg atoms leaving the lattice. That is, the thermodynamics might well be different for Mg alloys than pure Mg. Here's Wikipedia on magnesium alloys. The patent is silent on the purity and detail composition of the Mg used, but I have to imagine that they might have been using some "technical grade" alloy, say with as much as 3% non-Mg. That would be plenty of material to cause lattice strain and change activation energy.

Related to this, the patent does state that the Mg is employed "particularly in the form of turnings", which I read as meaning that was the form they tried that worked best. Machining processes put really large stresses on crystal lattices, and there's mechanical strain remaining in the chips, as a rule. This mechanical strain is an available energy source to drive the reaction. In other words, the energy content in strained Mg is higher than that in non-strained (annealed, as it were). I note that Pok's Mg was in filings, which would have the same kind of strain resulting from a machining process.

In short, it's possible that strained Mg and non-strained Mg should be treated as different reagents in this process.

blogfast25 - 7-12-2010 at 07:44

Quote: Originally posted by a_bab  
Quote: Originally posted by Sedit  

BTW can anyone think of a source of Mg I might have around the house? It doesn't have to be pure at all.



Pencil sharpeners? Why make pencil sharpeners from Mg when the cheaper and safer Al will do?

Sources of Mg include anodic protection blocks, often found on boilers and such like for corrosion protection, and flint blocks for emergency fire making (hikers, mountaineers, explorers – any multi purpose sports store will sell them). And the streets of eBay are paved with sellers of pyrotechnical Mg powder. Look for fairly pure, fairly coarse stuff…

Quote: Originally posted by Sedit  
Where is your Mg in this picture
http://www.versuchschemie.de/upload/files3/2272042_4948.jpg

It appears there is alot of K around and what also appears as a bunch of unreacted KOH yet there is a single problem. There is NO PRECIPITATE. There should be Magnesium hydroxide or something of the sorts in there and there is NON.

I want to believe and I want to prove you right but..... Its starting to appear that the Mg we see in the start is nothing more then K that has been heated... stirred... then allowed to cool while still stirring. Then as we heat the solvent the "Mg" reacts with the KOH producing K. In reality it appears the KOH is doing nothing and there is no Mg there at all.


Im not trying to make an enemy here im only stating what I see and I honestly have a good sence of when someones pulling the wool over my eyes..... What would the motivation for such an act be?


There is precipitate (or 'stuff' as part of the deception), look closer.

The Mg ‘at the start’ really looks like Mg filings, not like suspended K at all. If it is a hoax (I’m 50/50 on that), an emulsion of liquid K and some grit would have been created separately, then allowed to coalesce into larger globules.

As regards ‘motive’, that is entirely beside the point: if it is a hoax we’ll probably never know the motive, nor does it matter.

Quote: Originally posted by Arthur Dent  
As soon as I have a minute of my time this Friday, i'll visit a few art supply stores in my neighborhood, maybe I'll get lucky and find that Shellsol solvent. I vaguely remember that name from a few years back as an oil paint solvent and brush cleaner. Maybe I'm wrong but it would be great to find that stuff (or something similar to it).

I just found out that "Bestine" rubber cement solvent is 100% pure heptane! Cool!

I found a good source of Tert-Butanol at $36 for 500 ml... Is that a fair price?

Robert


Cool for the heptane, unfortunately it is totally UNSUITABLE for this purpose: way too low boiling point…

t-butanol, $36 for 500 ml? Very, very good, go for it and sell some back to those here who can’t get any!!!

@woelen:

Finally a systematic approach (leading by example, eh? :)) I’ve got stuff in the post but I’m holding fire until someone gets some results either way. I’m a little too agnostic to start experimenting just yet.

@pok:

Thank you for continuing to follow this thread. Nothing else is really required of you: it is now up to US to corroborate (or not) your results and conclude they are genuine or not. Thanks again!


[Edited on 7-12-2010 by blogfast25]

blogfast25 - 7-12-2010 at 08:01

Quote: Originally posted by watson.fawkes  
Related to this, the patent does state that the Mg is employed "particularly in the form of turnings", which I read as meaning that was the form they tried that worked best. Machining processes put really large stresses on crystal lattices, and there's mechanical strain remaining in the chips, as a rule. This mechanical strain is an available energy source to drive the reaction. In other words, the energy content in strained Mg is higher than that in non-strained (annealed, as it were). I note that Pok's Mg was in filings, which would have the same kind of strain resulting from a machining process.

In short, it's possible that strained Mg and non-strained Mg should be treated as different reagents in this process.


Isn’t nearly all size reduced Mg, with the exception perhaps of Rieke metal, formed by stressing the metal? Turnings are achieved by means of various kinds of stresses, filings (as used by pok) mainly shearing forces. Where would you get non-strained but size-reduced Mg? (Apart from Rieke metals). Event ‘atomised’ Al (spherical Al – not sure whether they do this in Mg too) will contain stresses on rapid cooling of the spray…

blogfast25 - 7-12-2010 at 08:05

Quote: Originally posted by wallschem  
I found this in an old Spanish chemistry book that is very similar to what pok is doing to synthesized Potassium. Instead of using Shellsol D70 and Ter-butol in this book they use Paraffin and Toluene at 180 degree Celsius.




Interesting but really this is something we kind of accept: that it is possible to disperse molten K into a hot inert liquid. That would not be a 'solution' though, it is an 'emulsion'. You could even call it a 'potassium latex'. It should segregate quickly though, like oil and water mixtures (think salad dressing of liquid K and high boiling inert solvent! :cool:)

[Edited on 7-12-2010 by blogfast25]

blogfast25 - 7-12-2010 at 08:10

And as I keep saying till I’m blue in the face, for those who can’t get the Shellsol, a good quality kerosene should come very close. The Shellsol is really nothing more than a high grade paint thinner, I refuse to believe it contains some magical ingredient. But it may have the ideal ration of high BP and low viscosity...

ScienceSquirrel - 7-12-2010 at 08:13

That is a recipe for potassium sand, sodium sand is made in a similar way.
The solvent is heated up with the metal in it then vigorously shaken or stirred while cooling so the metal remains in very fine particles.
The high boiling solvent is then removed under dry argon and replaced with hexane or similar.
It is a very reactive reducing and metallating reagent.

watson.fawkes - 7-12-2010 at 08:15

I went over the patent last night, to see if I'd notice new things not mentioned yet in this thread. I noted a number of things.I had this giant hunch when I woke this morning that this reaction has a relatively narrow Goldilocks temperature range; too hot or too cold and it doesn't go. The boiling range of Shellsol D70 may be part of the secret sauce, in that putting it under reflux at its low boiling temperature (200 &deg;C) may put the reaction into its sweet spot.

To all people replicating this, therefore: Please at least measure your temperature. It's too much to ask most of you to use a PID controller, but (again) I'll mention it.

Writing this post up, I realized that the patent contains two clues about how too hot a temperature might diminish the reaction.

Edit: Added quotation for "reaction accelerator"; those are the words in the patent itself.

[Edited on 7-12-2010 by watson.fawkes]

blogfast25 - 7-12-2010 at 08:19

Quote: Originally posted by ScienceSquirrel  
That is a recipe for potassium sand, sodium sand is made in a similar way.
The solvent is heated up with the metal in it then vigorously shaken or stirred while cooling so the metal remains in very fine particles.
The high boiling solvent is then removed under dry argon and replaced with hexane or similar.
It is a very reactive reducing and metallating reagent.


The closest you can get to sodium/potassium powder, eh? Interesting: I've found a company some time ago that sold Li powder, I wonder if it's made in similar fashion...

watson.fawkes - 7-12-2010 at 08:24

Quote: Originally posted by blogfast25  
Isn’t nearly all size reduced Mg, with the exception perhaps of Rieke metal, formed by stressing the metal? Turnings are achieved by means of various kinds of stresses, filings (as used by pok) mainly shearing forces. Where would you get non-strained but size-reduced Mg? (Apart from Rieke metals).
The nature of the stresses from, say, ball milling Mg are different from that of putting a machine tool on it. Without getting too much into the continuum mechanics, it seems that milling creates a strain with one eigenvalue >1 and one <1 with constant volume, i.e. product = 1. By contrast, milling would tend to strain only one axis, compressing it (eigenvalue <1) and leaving the other axis about the same (eigenvalue ~ 1). I don't know enough of the detailed metallurgy of Mg to defend this hypothesis, but I find it plausible that different manipulations leave different amounts of residual stress.

Perhaps much more important, however, is how recently the machining was done. The passage of time causes annealing, fairly slowly with iron alloys, but I believe much more quickly with Mg. The fact that the Mg filings or turnings are fresh may be the significant factor.

blogfast25 - 7-12-2010 at 08:35

Quote: Originally posted by watson.fawkes  
I went over the patent last night, to see if I'd notice new things not mentioned yet in this thread. I noted a number of things.
  • The abstract mentions an alcohol as a reaction accelerator. I'm guessing the authors managed to make this work without it, albeit in some commercially impractical way.
  • The authors mention doing the reaction under increased pressure. I know one member here with an autoclave (hint, hint). From a physical chemistry point of view, if this is relevant, it means that this reaction is limited by activation energy, because that's what higher pressure gives you&mdash;extra activation energy.
  • The patent states that Mg(OH)2 is formed dehydrating KOH, and that MgO is formed at higher temperatures. Example 1, however, categorically states that the residue was MgO. Is conversion really total? It matters because Mg(OH)2 sequesters more oxygen per mole Mg than does MgO.
  • Example 1 says that potassium t-butylate is left as a residue, which they extract as a solute from 1,4-dioxane. They don't actually state that this is where the rest of the 13% K ends up, and they don't mention if this reside contains any KOH.
  • The residue question is even more interesting in Example 6, where they mention the residue contains Mg butylate. This seems like a bona-fide side reaction to be avoided.
I had this giant hunch when I woke this morning that this reaction has a relatively narrow Goldilocks temperature range; too hot or too cold and it doesn't go. The boiling range of Shellsol D70 may be part of the secret sauce, in that putting it under reflux at its low boiling temperature (200 &deg;C) may put the reaction into its sweet spot.

To all people replicating this, therefore: Please at least measure your temperature. It's too much to ask most of you to use a PID controller, but (again) I'll mention it.

Writing this post up, I realized that the patent contains two clues about how too hot a temperature might diminish the reaction.
  • The formation of Mg(OH)2 is more efficient at receiving hydroxyl groups from KOH. Formation of MgO presumably scavenges more of the available Mg, removing its reaction potential with KOH. MgO is promoted by higher temperatures.
  • The formation of Mg t-butylate removes two reagents from availability. I haven't looked it up, but I'm guessing that the energy to form the Mg butylate is higher than that to form the K butylate. In this case the higher heat leads away from the desired reaction pathway.


Your point 1. I think you’re playing with words here. Isn’t ‘reaction accelerator’ a cautious term for ‘catalyst’? There are many reactions that simply do not proceed without the ‘reaction accelerator’.

Point 2. Increased pressure gives you higher activation energy? Please explain that for the case of an incompressible reaction medium.

Point 3. As indicated above the reaction 2 KOH + Mg --- > 2 K + Mg(OH)2 seems thermodynamically more favourable than 2 KOH + Mg --- > 2 K + MgO + H2O (somewhat to my surprise). According Wiki, Mg(OH)2 only dehydrates at 332C, as len1 also pointed out. And as he he also correctly pointed out, the patent makes some statements that are patently (no pun intended) the result of calculation (or reason, if you prefer) rather than observation (see the 'hydrogen evolved' issue). Their claim on MgO may not have been the result of analysis...

Goldilocks? It’s possible, then pok just hit the jackpot… (jackpok??) :D



[Edited on 7-12-2010 by blogfast25]

blogfast25 - 7-12-2010 at 08:47

Regards stresses in metal ‘powders’, it could possibly explain some strange discrepancies I’ve jotted down over the years with various Goldschmidt (and analogous) reactions.

In the mean time, somewhat disappointingly, no movement at versuchschemie.de on this issue at all... :(

[Edited on 7-12-2010 by blogfast25]

ScienceSquirrel - 7-12-2010 at 08:52

We used to use it to make sodium cyclopentadienide aka NaCp.
This was used to make other substituted cyclopentadienes and then remetallated and then used to make metal complexes.
The sand can be very fine if it is made properly, you just need a bit of nerve and a fast, powerful stirrer. I have seen it made by wrapping the flask in a towel and shaking vigorously while it cooled, that would probably result in a whole litter of Elf and Safety kittens these days.
A description of the process is here;

http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv7...

watson.fawkes - 7-12-2010 at 09:18

Quote: Originally posted by blogfast25  
Your point 1. I think you’re playing with words here. Isn’t ‘reaction accelerator’ a cautious term for ‘catalyst’? There are many reactions that simply do not proceed without the ‘reaction accelerator’.

Point 2. Increased pressure gives you higher activation energy? Please explain that for the case of an incompressible reaction medium.

Point 3. As indicated above the reaction 2 KOH + Mg --- > 2 K + Mg(OH)2 seems thermodynamically more favourable than 2 KOH + Mg --- > 2 K + MgO + H2O (somewhat to my surprise). According Wiki, Mg(OH)2 only dehydrates at 332C, as len1 also pointed out. And as he he also correctly pointed out, the patent makes some statements that are patently (no pun intended) the result of calculation (or reason, if you prefer) rather than observation (see the 'hydrogen evolved' issue). Their claim on MgO may not have been the result of analysis...
1. The words "reaction accelerator" were in the original; I've edited my original post to make that clearer. My point is that it's possibly that the author may have managed to make this work without a catalyst at all, possibly under pressure.

2. Increased pressure is a source of free energy to overcome activation energy. I should have worded that more carefully. Pressure is just a kind of energy per unit volume. Microscopically, it's from electron clouds not wanting to overlap (both electrostatically and from Pauli exclusion).

3. Mg(OH)2 is thermodynamically favorable simply because its dehydration to MgO is endothermic. Nevertheless, it seems that MgO might be kinetically favorable at higher temperatures. If so, that would make the claim of the authors I discussed realistic. It would also entail that Mg(OH)2 is kinetically favorable at lower temperatures, meaning the process as a whole might not work well if you get it too hot.

The 332 &deg;C temperature is for a solid phase decomposition. That's only one possible pathway and it puts an upper temperature bound on what other pathways would be observable. It doesn't exclude other pathways, though.

I do agree that the examples have a certain amount of idealization in them. The degree of it, however, does not rise to lead me to believe that they were entirely fabricated from whole cloth. My working model of what's going on there is commercial obfuscation rather than patent fraud.

blogfast25 - 7-12-2010 at 10:01

Thanks sciecesquirrel for the link. Live and learn...

Looks like we've really covered quite a lot of theoretical ground here. Someone here (moi!) will be mightily disappointed if all this turns out to be a dud... :o

A vessel filled with an ‘incompressible’ fluid is sealed off with a movable, massless piston of surface area S. We apply a force F to the piston: pressure inside the cylinder now increases by P = F/S. Lets assume the piston also moves by a small amount Δs (on application of the force F), the work done is W = Δs x F. Is this the Free Energy increase of the system, Watson?

[Edited on 7-12-2010 by blogfast25]

condennnsa - 7-12-2010 at 10:07

Well is there anyone here that has all the required reagents to give it a shot?

I ordered the shellsol, i'm working on the t-butyl, but it'll take some time .

woelen - 7-12-2010 at 10:29

Quote: Originally posted by Pok  

I know science. I know that reproduction is neccessary. But if I just do it again and again, this wouldn't help you. Some of YOU has to do it (scientifically - which means: in exactly the same way I did it.)
With this I absolutely agree. Someone else than you should be able to reproduce the experimental outcome.

Quote: Originally posted by Pok  
Tell you how to do it? I thought you already have every detail. Look at versuchschemie.de and here. I made it like the patent with modifications that I already told you. You should easily be able to repeat this synthesis.

I hope you know what "exactly" means: 50ml, 25cm-glaspipe with wet toilet paper (moisten it sometimes to prevent dehydration), balloon at the top with tiny hole, zero oxygen input, Mg filings made from 99-100% Mg made by filing with a medium file. and so on It might be that I really forgot one or two details. If this is the case, I would apologize - just ask! (I'll be back in 24 hours)
With this I cannot agree. If the experimental outcome really depends on how you filed your pieces of magnesium, if it only works with magnesium from exactly that piece of scrap which you obtained from eBay, if it only works with a 25 cm glass pipe and with a volume of 50 ml, then it is not good at all.

You also made modifications (such as using ten times smaller amounts) and you thought out your own setup, etc. Scientific reproduction and exact writeup is important, but it must be within reason.

Does the color of the balloon also affect the outcome. Please specify the color in the form of an RGB color, like #rrggbb :D

Next weekend I'll try the series of experiments I wrote in my previous post. If all that fails, then I'll even consider buying ShellSol D70 (it is not expensive, EUR 7 for 1 liter at Kremer pigments and there is no shipping issue with this). But if even with these things it still doesn't work, then I'm inclined to think it is either fake, or you were extremely lucky and indeed won the jackpok.

woelen - 7-12-2010 at 10:34

Quote: Originally posted by metalresearcher  
Quote: Originally posted by woelen  
but I do have high grade paraffin oil (colorless viscous liquid, perfectly odorless and completely free of acidic and unsaturated ingredients).

Where can I get this oil ? Or can I melt candles which is also paraffin ?


[Edited on 2010-12-7 by metalresearcher]

This oil can be purchased at drugstores. It also is available online in the Netherlands.

http://www.drogisterij.net/diversen-schoonmaak-Chempropack-p...

I am very happy with the quality of this oil. It is clear like water, has no color and no smell and it is really inert. I store Na-metal, K-metal and Li-metal under it and I also use it sometimes in distillations for making seals really airtight. Bromine only discolors very slowly when added to this oil, indicating that it does not contain unsaturated hydrocarbons.

[Edited on 7-12-10 by woelen]

ScienceSquirrel - 7-12-2010 at 10:52

It sounds like it is this stuff, it is used in creams and all sorts of things, so any pharmacy (chemist) should stock it or you could get it from a place that sells massage oils or such like.

http://www.henriettesherbal.com/eclectic/bpc1911/paraffin2.h...

Pok - 7-12-2010 at 10:54

Quote: Originally posted by woelen  
If the experimental outcome really depends on how you filed your pieces of magnesium, if it only works with magnesium from exactly that piece of scrap which you obtained from eBay, if it only works with a 25 cm glass pipe and with a volume of 50 ml, then it is not good at all.

You also made modifications (such as using ten times smaller amounts) and you thought out your own setup, etc. Scientific reproduction and exact writeup is important, but it must be within reason.

Does the color of the balloon also affect the outcome. Please specify the color in the form of an RGB color, like #rrggbb :D


The experiment won't depend on so many things. But if you want to exclude sources of errors you should follow my description. Why my filing description? Because I saw the difference in Mg particle size between my and lens Mg filings. Why 25cm pipe? Too short: t-butanol can't condense totally and will be lost. Too long wouldn't be a problem. Why 50 ml? The relation of volume of air above your reaction vessel and reaction volume in very small dimension might be so large, that enough air is available to oxidise the K. I also don't think that you can come to such a point. Smaller reaction volume should be ok (5 ml, too). Why balloon? Because someone did it without ANY inert atmosphere (open pipe) - this is just one possible way to exclude air. Maybe just one thing: the purity of your t-butanol (I used > 99,3%) and Shellsol D70 can be of crucial importance. BTW: my ballons were black. This may influence the absorbtion of sunlight and so the reflux of t-butanol if it gets to warm by the sun :D.

But if you want to call me a liar you have to do it in my way. Why trying so many possibilities? Just try 1: my. If it fails: i lied.

condennnsa - 7-12-2010 at 11:33

i see on wikipedia, and also other sources (Ullmann's encyclopedia of industial chem for example) that

"tert-Butanol is used as an octane booster for gasoline, as an oxygenate gasoline additive,"

You won't get a lot of butanol in gasoline, but then a little is all is needed, right?

If this really proves itself to be a reliable procedure, it might be a source for folk who can't get any.

watson.fawkes - 7-12-2010 at 11:45

Quote: Originally posted by blogfast25  
A vessel filled with an ‘incompressible’ fluid is sealed off with a movable, massless piston of surface area S. We apply a force F to the piston: pressure inside the cylinder now increases by P = F/S. Lets assume the piston also moves by a small amount Δs (on application of the force F), the work done is W = Δs x F. Is this the Free Energy increase of the system, Watson?
Not exactly. You can have a change in energy with no displacement change in an incompressible fluid. Incompressible is an idealization, but it doesn't affect the math. One way of approaching this with a "constitutive relation", a proposition that pressure in material X is a function (only) of its density and internal energy: p = p<sub>X</sub>( &rho;, e). For most materials, such functions are monotonic in internal energy; when e increases, so does p.

You can see this use of pressure as an energy density in Euler's equations, in the divergence normal form equation for conservation of energy. More elementary, but also perhaps useful, is Hyperphysics on pressure. I will freely admit that I didn't find any specifically targeted references for this topic.

In more general equations of motion, which might include reactions, the internal energy can be interconverted between chemical species.

Pok - 7-12-2010 at 12:40

some more pictures of the 300ml experiment.


beginning at about 100-180°C - massive evolution of H2. Al-foil prevents heat loss. Toilet paper at stopper prevents cooling water from above toilet paper to come in contact with hot parts. Toilet paper around pipe (here longer than 25 cm) wetted for cooling/reflux. balloon at the end (fixed with cellotape) with an invisible small hole.


T-Butanol not added yet. Picture of massive H2 evolution. Shiny: Mg filings. lumps: KOH.


At t-butanol addition.


After t-butanol addition. Thick fogs of Shellsol and/or t-butanol above the mix. First K bullets visible.

You can remember these pictures if you want to reproduce the experiment and compare them with the things you will see in your experiment.

blogfast25 - 7-12-2010 at 13:47

@ Watson:

Yep, pressure as energy density. Fluid mechanics (after 30 year of hibernation) is slowly waking up in that bunch of neurons up here.

At the molecular/atomic level, repulsion between orbitals...

[Edited on 7-12-2010 by blogfast25]

blogfast25 - 7-12-2010 at 13:50

@ pok:

I small a rat: you never told us the black balloon was shiny or matt! And the toilet paper: recycled or embossed? :P

And still radio silence from our other German friends, so much for 'Deutsche grundlichkeit'... ;-)

[Edited on 7-12-2010 by blogfast25]

Pok - 7-12-2010 at 15:19

Quote: Originally posted by blogfast25  
@ pok:

I small a rat: you never told us the black balloon was shiny or matt! And the toilet paper: recycled or embossed? :P


Oh sorry. The toilet paper was recycled as I don't want to kill trees for my ass :(. If you use normal toilet paper the synthesis will not work: Recycled toilet paper can take up more water. With normal toilet paper you will not gain potassium but gold. And that's really not the goal here. :D

Quote: Originally posted by blogfast25  

And still radio silence from our other German friends, so much for 'Deutsche grundlichkeit'... ;-)


Germans aren't grundlich. I had to make the chaotic background black in the picture here http://www.versuchschemie.de/upload/files3/67979825_4948.jpg , because I'm so very un-grundlich.:D

MagicJigPipe - 7-12-2010 at 15:27

I will try this around the 23rd of December if no one has had success by then. I have all the materials required but I lack one thing right now: the time. Due to some plans that I have that is the soonest I can try (not to mention finals).

Even if someone has tried it by then I will probably still attempt it. I have decided to use VM&P Naptha the first time. If that fails, I will attempt to use use dodecane. KOH and t-butanol will both be reagent grade.

I'm looking forward to it!

mr.crow - 7-12-2010 at 16:41

Thank you Pok, you are awesome

Sedit - 7-12-2010 at 17:48

Sorry pok im not trying to offend in anyway but I do see a serious flaw that I can not explain and its something thats bugging me. I would love to reproduce the experiment but with the holidays ahead I just can not.

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




Like I said im not trying to make an enemy here Im just curious as to why there does not appear to be any MgOH precipitate of any kind. Perhaps the tert-butanol is making them soluble for all I know but as the reaction is outlined something appears amiss.

ScienceSquirrel - 7-12-2010 at 18:09

We are mad scientists and it would be nice if we could express our opinions in a way suited to the Common Room of the University Of Mad Science ( Chemistry School ).
I have my doubts about this reaction; I cannot explaln the results published by Pok and no one has managed to reproduce his results to date.
But that does not mean to say that he is wrong or a liar and I think he deserves a respectful hearing.
Science should be a creative activity carried out in a spirit of constructive criticism, I hope that we can try to be true to that ideal.

PS: I reserve my right to spear the odd troll! :D


Sedit - 7-12-2010 at 18:47

Quote: Originally posted by ScienceSquirrel  
We are mad scientists and it would be nice if we could express our opinions in a way suited to the Common Room of the University Of Mad Science

But that does not mean to say that he is wrong or a liar and I think he deserves a respectful hearing.

Science should be a creative activity carried out in a spirit of constructive criticism, I hope that we can try to be true to that ideal.


This is my intent completely. Parts of the reactions progress confuse me and do not appear to proceed the way they should so I ask Pok and all for that matter, Where is the Hydroxide or oxide precipitate that should be created in abundance. I am at a lose as to what motivation Pok would have to lie so im left to believe this really happened. The question is why can no one reproduce it and where is the missing mass.

Pok - 7-12-2010 at 18:53

Quote: Originally posted by Sedit  

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

Like I said im not trying to make an enemy here Im just curious as to why there does not appear to be any MgOH precipitate of any kind. Perhaps the tert-butanol is making them soluble for all I know but as the reaction is outlined something appears amiss.


Ok. It's easy to explain. The Mg(OH)2 or MgO which you are missing is in the form of a quite hard (but porous) light gray crusts (see here: http://www.versuchschemie.de/upload/files3/2272042_4948.jpg ) - the shiny balls are K, all the rest (lumps of gray stuff) is the Mg(OH)2 or MgO!! You think: so much Mg at the beginning can't be decreased to such small amounts of hydroxide or oxide? The reason is: The Mg filings are highly irregular (due to filing), highly voluminous. What you can see are really only some grams of Mg. The product (hydroxide/oxide) is highly compact. So its volume seems to have decreased unimaginable. - For imagination: If you burn voluminous Mg filings and crush the product into powder, the powder will have a much lesser volume. You will (probably) only get a dust-like precipitate of Mg products if you stirr continiously. I simply didn't. I hope this is OK for understanding.

PS: Its 4 am here and I have to write an exam today so see you in a few hours (I'm not going away :D)...

[Edited on 8-12-2010 by Pok]

Pok - 7-12-2010 at 19:01

Quote: Originally posted by ScienceSquirrel  
I cannot explaln the results published by Pok and no one has managed to reproduce his results to date.


No reproduction of my results? Because nobody has tried it yet (as I described it)!

ScienceSquirrel - 7-12-2010 at 19:15

Quote: Originally posted by Pok  
Quote: Originally posted by ScienceSquirrel  
I cannot explaln the results published by Pok and no one has managed to reproduce his results to date.


No reproduction of my results? Because nobody has tried it yet (as I described it)!


Some people are intending to try it exactly as you have described it.
The best way to prove your point is to hep them to succeed.

bbartlog - 7-12-2010 at 19:48

Quote:
no one has managed to reproduce his results to date.


You mean, besides the guys that filed the patent? And the other guy on versuchschemie who saw some half-millimeter globules of potassium when he tried to reproduce the experiment? I think at this point the skepticism is overblown. Really, you're basically saying that someone filed a patent that was not just in error or overbroad, but a complete fabrication, *and* then someone else created an elaborate hoax based on it, just for kicks. We already know that the patent has some issues (presenting calculated amounts as measured ones, suggesting that isopropanol could be used as well as tert-butanol), but that is hardly the same thing as assuming that the authors just made the whole thing up.

aonomus - 7-12-2010 at 20:26

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

If anyone has a scale with good resolution, they should be able to see if there is any change in mass based simply on the release of hydrogen (which is an indicator of reaction completeness). In an ideal world there would be no solvent loss, but its worth a shot. Maybe better accuracy would be had by filtering the whole mass and washing with hexanes to clean off solvent residues.

The other thought would be to measure the release of hydrogen from the reaction based on volume and calculate backwards to see how many mmol of KOH reacted.

Arthur Dent - 8-12-2010 at 04:50

Quote: Originally posted by blogfast25  

Cool for the heptane, unfortunately it is totally UNSUITABLE for this purpose: way too low boiling point…

t-butanol, $36 for 500 ml? Very, very good, go for it and sell some back to those here who can’t get any!!!


Yeah I know. I just mentioned it because I discovered this inexpensive heptane source as I was reading the MSDS of various "over the counter" solvents to find an equivalent to Shellsol...

As for the t-butanol, if a member of this forum succesfully reproduces the experiment, i'll buy a bottle and split in 2 or 3 among members residing in Canada if I find people that will be willing to share the costs (I think shipping/hazmat fees to the US would be outrageously expensive, so I'd rather keep this north of the 42nd.).

Anyway, I believe that this experiment requires only minute amounts of t-butanol, a syringe's worth?

Robert

woelen - 8-12-2010 at 04:57

Quote: Originally posted by Pok  

The experiment won't depend on so many things. But if you want to exclude sources of errors you should follow my description. Why my filing description? Because I saw the difference in Mg particle size between my and lens Mg filings. Why 25cm pipe? Too short: t-butanol can't condense totally and will be lost. Too long wouldn't be a problem. Why 50 ml? The relation of volume of air above your reaction vessel and reaction volume in very small dimension might be so large, that enough air is available to oxidise the K. I also don't think that you can come to such a point. Smaller reaction volume should be ok (5 ml, too). Why balloon? Because someone did it without ANY inert atmosphere (open pipe) - this is just one possible way to exclude air. Maybe just one thing: the purity of your t-butanol (I used > 99,3%) and Shellsol D70 can be of crucial importance.
Yes, these sound reasonable. No air may come into the reaction vessel, the initial amount of oxygen must be minimal (this can also be achieved by flushing the air away with nitrogen, or any other inert gas, or even hydrogen). I have reagent grade tert-butanol, so that should be no issue.

The key may even be that the tube is short, in the start there is some tert-butanol, lateron it is gone. This is a good point and I'll take this into account in my experiments next weekend (one variation with long tube and one variation with short tube).

Eclectic - 8-12-2010 at 06:58

Did we decide that k-1 kerosene with the volatiles distilled off and butoxyethanol as the alcohol were not worth trying?

blogfast25 - 8-12-2010 at 08:14

Quote: Originally posted by Sedit  

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


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

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


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

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



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

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


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

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

[Edited on 8-12-2010 by blogfast25]

blogfast25 - 8-12-2010 at 08:26

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

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

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


[Edited on 8-12-2010 by blogfast25]

watson.fawkes - 8-12-2010 at 08:36

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

blogfast25 - 8-12-2010 at 08:38

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


Ah. I seem to recall that what I described applies to some European patents. But I'm no expert...

watson.fawkes - 8-12-2010 at 08:51

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

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

Pok - 8-12-2010 at 09:05

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


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

blogfast25 - 8-12-2010 at 09:21

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

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



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

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

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

[Edited on 8-12-2010 by blogfast25]

blogfast25 - 8-12-2010 at 09:28

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


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

[Edited on 8-12-2010 by blogfast25]

Eclectic - 8-12-2010 at 17:29

CH3CH2CH2CH2OCH2CH2OH

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



MagicJigPipe - 8-12-2010 at 19:56

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

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

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


watson.fawkes - 8-12-2010 at 21:12

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

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

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

aonomus - 8-12-2010 at 21:52

One thought that comes to mind, can anyone try fusing Mg shavings with KOH pellets with a blowtorch in a crucible?

ScienceSquirrel - 9-12-2010 at 03:38

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


Sounds like a good way to have a small fire with caustic smoke :(

blogfast25 - 9-12-2010 at 06:24

Quote: Originally posted by Eclectic  
CH3CH2CH2CH2OCH2CH2OH

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




As stated above, there are good reasons to believe only t-alcohols might work.

blogfast25 - 9-12-2010 at 06:31

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


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

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

[Edited on 9-12-2010 by blogfast25]

blogfast25 - 9-12-2010 at 06:59

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

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

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

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

UnintentionalChaos - 9-12-2010 at 10:32

Quote: Originally posted by blogfast25  

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


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

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

Nicodem - 9-12-2010 at 11:17

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

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

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

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

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

woelen - 9-12-2010 at 12:48

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

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

blogfast25 - 9-12-2010 at 13:40

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

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


Repetition is the greatest didactical tool, it is said. That was very clear, never to be forgotten (I hasten to add...:))

aonomus - 9-12-2010 at 14:31

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

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

MagicJigPipe - 9-12-2010 at 16:24

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

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

watson.fawkes - 9-12-2010 at 18:21

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

Having said that, you might first see if your mineral oil fractionates. There might be enough of a lower boiling fraction that you're more closely matching the boiling range of Shellsol. Or, call your local art supply store and see if they have Gamsol. It's about $11 for a pint jar.

blogfast25 - 10-12-2010 at 06:20

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

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


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

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

[Edited on 10-12-2010 by blogfast25]

plastics - 10-12-2010 at 08:47

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

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

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

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

IMG_4903.JPG - 31kBIMG_4902.JPG - 19kB

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