Sciencemadness Discussion Board

Make Potassium (from versuchschemie.de)

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len1 - 7-1-2011 at 19:23

You must use at least 20% Mg powder with the turnings as in my post above. The reasons they dont work 100% is that powder is required to dehydrate the KOH and commence initiation, after that turnings, provided they are thin work, and I have got reasonable results with that as I posted above.

I dont know whats all this wishing for a reaction mechanism, I had verified it long ago and posted the steps.

garage chemist - 7-1-2011 at 20:52

@ Nurdrage: How old was the surface of the magnesium filings that you did your successful batches with? Months? Years?
The surface of my 99,8% magnesium powder was about three years old, and it instantly worked without activation.
In contrast to that, the surface of my magnesium scrap turnings was about 12 hours old, and they didn't work.
Whether or not this says something about the importance of a fresh surface on the magnesium turnings relative to the purity of the metal itself will be determined in my next experiment with freshly made 1-3mm turnings from a 99,8% pure magnesium ingot.

Here are some pictures of the magnesium scrap that I've been working with.
Picture 1: The printing plate and the turnings, you can see how I cleaned the first three sections with a wire brush before drilling into them.
Picture 2: A close-up on the turnings that didn't work. Sorry for the blurry picture, but you can clearly see the size distribution of the particles.
Picture 3: The batch itself after 4 hours of reflux with t-BuOH, KOH and Shellsol D70. The turnings have darkened, but only superficially, and are still solid metal throughout.

I poured the remaining scrap Mg turnings onto a brick and lit them in my fume hood. They burned brightly and fiercely, but this is obviously not an indicator of purity.


Kalium-Test_021.jpg - 61kBKalium-Test_022.jpg - 37kBKalium-Test_023.jpg - 44kB

blogfast25 - 8-1-2011 at 11:03

Garage chemist:

The third picture is striking, as that there seems to be no MgO whatsoever, a clear indication of no reaction. Perhaps the small amount of Al in the alloy does indeed inhibit the reactions?
Quote: Originally posted by len1  
You must use at least 20% Mg powder with the turnings as in my post above. The reasons they dont work 100% is that powder is required to dehydrate the KOH and commence initiation, after that turnings, provided they are thin work, and I have got reasonable results with that as I posted above.


Len1: do you believe water in the KOH (say as KOH.H2O) would inhibit initiation? There seems nothing in the proposed reaction mechanism that makes me think so…


[Edited on 8-1-2011 by blogfast25]

garage chemist - 8-1-2011 at 12:17

Results of the second test batch with selfmade "pure" Mg turnings:
During initial heatup there was the familiar vigorous reaction with gas evolution and production of white smoke. After adding the butanol there was some more gas evolution, and during the reflux no butanol crystallized in the condenser.
But the solution never got turbid, there was no MgO produced and after 4 hours, there was no potassium in the solids, neither compact nor distributed. The batch looked the same as the one with impure Mg turnings.

This is a strange result. The initial dehydration of the KOH took place and most of the t-BuOH seemed to be consumed, but the reaction did not proceed.
Did people have success with grignard turnings only, or was there always some powder addition?
I am inclined to blame my magnesium for this result, as this was from an unknown source and not from a regular seller.
Either way, now I know that I can't make potassium with selfmade turnings.

NurdRage - 8-1-2011 at 13:48

I have had success using straight grignard turnings or powders. I have never mixed both.

blogfast25 - 8-1-2011 at 14:30

It's a very strange result indeed, indicating: 'not all magnesium is suitable for this reaction'.

garage chemist: what observation makes you write: "most of the t-BuOH seemed to be consumed"?

MagicJigPipe - 8-1-2011 at 15:59

Quote:
@MagicJigPipe

Maybe you can "clean" your oil by heating it with just magnesium turnings to destroy whatever "goodies" the companies put in there. Then let it settle overnight and decant off the liquid. In the lab i normally destroy stabilizers and other goodies in solvents with sodium.


I will try that on my next run, but do you really think some vitamin E could interfere with the reaction so much as to produce my results? I'm thinking it could be my magnesium due to garage chemist's and woelen's results. What could be in that firestarter Mg? Would heating it in water produce insoluble particles of Al if it's in there or would the Al react as well due to it being alloyed with the Mg? It's time to research how to determine purity of Mg...

EDIT:

I have decided that I will attempt the experiment with a similar firestarter that is "Made in the USA" instead of China. I think this will shed much light onto the nature of the Mg because I have heard that the new Chinese bars are more difficult to light, indicating less pure Mg.

After this, since I have surprisingly made a little bit of money on eBay, I will purchase some decent Mg and Shellsol. I assume the Shellsol could be recycled. Has anyone done so?

[Edited on 1-9-2011 by MagicJigPipe]

len2 - 8-1-2011 at 20:20

Yes I believe water will stop this reaction dead in its tracks. The Mg reacts with t-BuOK only very unwillingly over several hours producing the semi-soluble 'gel-like' (t-BuO)2Mg. It would unite with any water much faster covering the magnesium surface with insoluble Mg/Mg(OH)2 and its game over.

blogfast25 - 9-1-2011 at 05:58

Quote: Originally posted by MagicJigPipe  
I will try that on my next run, but do you really think some vitamin E could interfere with the reaction so much as to produce my results?

[…]

I assume the Shellsol could be recycled. Has anyone done so?

[Edited on 1-9-2011 by MagicJigPipe]


Vitamin E is mainly α-tocopherol, a well known anti-oxidant with hydroxyl functionality (as the name suggests). If it irreversibly links with K that could interfere but even then it would depend on quantities whether that would significantly affect yield or not. I’ve a hard time believing low quantities of the anti-oxidant will interfere.

As regards recycling the solvent, IMHO it should be possible to do so, even WITHOUT ADDING any further catalyst. If the proposed reaction mechanism is correct, the reduction is carried out under total reflux (no evaporation of alcohol) and no chemical degradation of the alcohol occurs then conservation of mass dictates that all the catalyst initially added is conserved in the solvent. The only thing that could ruin it is that the catalyst, by the end of the reaction present as K-alcoholate, settles as a solid with the MgO (from which it may be hard to recover). In the patent for the caesium example the authors claim to have recovered the catalyst by means of toluene/dioxane mix.

==============
Quote: Originally posted by len2  
Yes I believe water will stop this reaction dead in its tracks. The Mg reacts with t-BuOK only very unwillingly over several hours producing the semi-soluble 'gel-like' (t-BuO)2Mg. It would unite with any water much faster covering the magnesium surface with insoluble Mg/Mg(OH)2 and its game over.


Len1, we’re talking about the same water that is reacted away as MgO/Mg(OH)2 + H2 in the initial drying step. I just don’t see how presence of the alcohol could interfere with that reaction.

You state “It would unite with any water much faster covering the magnesium surface with insoluble Mg/Mg(OH)2 and its game over”, yet we know water is a vital part of the reaction mechanism, providing catalyst regeneration, assuming our equations are correct.

What prompts you to state that (RO)2Mg is gel-like? Do we know this with any certainty?

And it reacts “only very unwillingly over several hours”? Not so sure about that either. In my reaction which lasted only 1 ½ hours I got plenty of K and plenty of MgO, I just didn’t get complete coalescence because I didn’t keep refluxing for another 2 hours. Nurdrage’s Tetralin/t-amyl alcohol test was over and done with, coalescence included, in under an hour…



[Edited on 9-1-2011 by blogfast25]

mr.crow - 9-1-2011 at 12:18

tert-Butanol is really annoying. It came in a 1L bottle but was frozen solid. I had to hotbox the bathroom with a safety heater to get it to melt.

What is the proper way to get the bottle to melt? Put it on the hotplate with a hot water bath?

garage chemist - 9-1-2011 at 12:39

Mine is in a plastic bottle. I smash the bottle on the table until there are enough loose crystals in there to take out the required amount with a spatula.;)

mr.crow - 9-1-2011 at 15:22

Quote: Originally posted by garage chemist  
Mine is in a plastic bottle. I smash the bottle on the table until there are enough loose crystals in there to take out the required amount with a spatula.;)


Well thats one way to do it!

I transferred it to a smaller container and when I put it in the freezer it grew some beautiful crystals on top, almost like whiskers.

In the main reagent bottle too, the whiskers went on the inside of the glass from the liquid at the bottom all the way to the top. The little things never cease to amaze me


[Edited on 10-1-2011 by mr.crow]

rrkss - 9-1-2011 at 19:11

Quote: Originally posted by mr.crow  
tert-Butanol is really annoying. It came in a 1L bottle but was frozen solid. I had to hotbox the bathroom with a safety heater to get it to melt.

What is the proper way to get the bottle to melt? Put it on the hotplate with a hot water bath?


In our organic labs at the university, we always kept it in a metal container on top of the oven we used to dry glassware.

===========================================

http://www.kremerpigments.com/shopus/PublishedFiles/70460_SH...

Wondering if this would work instead of shellsol D70, any ideas?

[Edited on 1-10-11 by rrkss]

Cuauhtemoc - 9-1-2011 at 20:04

Well, that's an advantage of living in Brazil

One pot experiment

blogfast25 - 10-1-2011 at 12:48

Here 6.12 g KOH (flakes), 3.12 g Mg (reagent grade), 50 ml Shellsol D70 and 1.02 g 2-methyl-2-butanol (t-amyl alcohol) were combined directly at the start of the run. 1.02 g 2M2B is about 45 % more (in mol) than used by pok (and most of us so far). No catalyst was added separately at any stage of the run.

Set up:



Well, this experiment was completely successful and uneventful. Hydrogen started evolving from around 100C with break up of KOH, then from about 150C clouding over starts which then continues with a complete change in the solid KOH/Mg mix. After an hour I was pretty certain I was seeing very small globules of K, after about 2 hours, I could vaguely see 3 – 4 globules of 4 – 5 mm size. In this terribly badly focused shot their presence is betrayed by the flash which penetrates the MgO slag somewhat:



I refluxed uneventfully at > 200C until a total run time of 4 hours was achieved. After cooling much potassium was found including about 4 globules of about 1 cm and many more small ones. Yield will be determined tomorrow.

It shows what I suspected: adding the alcohol right from the start does not impede potassium forming. It also provides indirect evidence that the proposed reaction mechanism is correct, as in my view it predicts that the point of alcohol addition is not important.


In addition I carried out another coalescing experiment, this time on a dozen of small K balls in Shellsol, almost entirely free of MgO dross, by holding them at 80 – 100C on water bath, stirring or tapping the round flask occasionally:



Coalescence is really slow, if not negligible. I tried adding a few drops of IPA (causes reaction), a few drops of 2M2B (also causes reaction) and later even some ethanol (very quick flurry of H2) but all to no real avail.

Nicodem - 10-1-2011 at 13:42

Quote: Originally posted by NurdRage  
Quote: Originally posted by Nicodem  
I wish someone would be interested enough in the mechanism of the reaction to do some scientific work on it.


I wish it were that easy. But i'm sure you know that real scientific work requires time, equipment and money. Most of us have one, some of us have two, but none of us have all three.

For someone who claims of not being able of doing much scientific work, you do a pretty good job in doing it. I did not want to sound like down rating your contributions. Sorry if that sentence sounded that way. Actually, your experiments are informative in regard to the mechanism, especially your subjective evaluations of various alcohols and phenol, or testing solvents. Len's indirect evidence of the other reaction product (MgO) at least gave the basic information of the reaction (stoichiometry and end products). Mechanistic studies are like building a puzzle. It takes a lot of well thought experiments, kinetic and thermodynamic measurements, normalized conditions, etc. The most elementary experiment in this situation which is viable also to amateur setting is the one I described earlier in this thread - checking the Mg + t-BuOK reaction. If I had not changed lab recently I could have easily done this myself, but where I work now it is not really possible for the moment.

blogfast25 - 10-1-2011 at 14:26

I also found a link to an interesting *.pdf on metal alkoxides, on an older sciencemadness thread (posted by ‘dann 2’):

http://www.sciencemadness.org/talk/viewthread.php?tid=15203#...

len1 - 10-1-2011 at 22:34

I had verified the reaction mechanism I posted not just based on the overall stoichiometry from the H2 experiment but in stages as well. I have no time to post everything, because I have other work to do as well.

The reaction of K with t-BuOH is well known, Ive done it many times before. Here the key is that t-BuOK is soluble in paraffin. This is easily verified by hydrolyzing the /u liquid /u which shows the presence of KOH and t-BuOH. You can also see it by eye when opening the apparatus to the atmosphere as I described above.

The second stage of the reaction is also reasonably well verified. I have established the solved alkoxide attacks and dissolves Mg turnings - the product can only be a magnesium alkoxide, although a mixed alkoxide with potassium is also possible. You can also observe this by the gradual dissolution of specks of Mg dust stuck at the top of the lask to its walls during reflux.

The third reaction is in essence dictated by the former two and the overall reaction. However you can also see it from the fact that at slow reaction rates the MgO crust effectively replicates the KOH - that is why in Poks original experiment we could not tell the difference between the initial and final byproduct. At higher temperatures some KOH actually solves in the paraffin (in tetralin although I have not tried I believe it is far more soluble also) and the MgO is formed in solution - that explains the fine sand I have been getting at high reaction rates.

I dont believe any water is formed in the reaction. It would immediately react depositing insoluble MgO on Mg surface. In fact it is not priduced in the dehydrating stage as well. KOH does not dehydrate in and of itself at 150C - it just melts, and that explains the reaction temperature of the dehydration stage.

I have not done any work on the initiation stage - becuase its quite difficult. Is K initially formed by action of Mg on KOH, or is a sizable amount of butoxide produced by reaction of KOH with t-BuOH?

[Edited on 11-1-2011 by len1]

blogfast25 - 11-1-2011 at 02:27

Len:

OK. I went back over your reaction equations and the invaluable work you did on the hydrogen, proving essentially that the end by-product is MgO, not Mg(OH)2. But I disagree with parts of your explanation and found (again) that my proposed set of reactions differs in one respect. Some comments on relevant posts of yours:
Quote: Originally posted by len1  
Potassium reacts with tButanol, this I have done many times to give KOBu. Water is generated. Since it is gasesous at the reaction temperature it should vaporise. The tBu actually consumes potassium.


Unconvincing: t-butanol (or 2M2B) is volatile too. We’re in total reflux. The amounts of gaseous (vapour) alcohol and water must be small: there’s not a lot of headspace. But potassium really does react with t-butanol, with 2M2B too.
Quote: Originally posted by len1  


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

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

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

The propagation equations are then

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

summing these gives the overall reaction

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

If the second equation was

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

the overall reaction would be

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

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

[Edited on 20-12-2010 by len1]


Agreed on the gasometer results. But here you rather craftily skip the formation of the t-BuOK, which really by the most probable means is KOH(s) + t-BuOH(solvent) < === > t-BuOK(solvent) + H2O(solvent). This reaction is in all likelihood driven by entropy, as the right hand side is all solution.

As regards the ‘second equation’:

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

I don’t like it much: it’s another liquid/solid reaction. In my scheme there’s no need for it and the proposed catalyst generation reaction is:

Mg(OR)2(solvent) + H2O(solvent) < === > MgO(s) + 2 ROH(solvent)

… driven by MgO’s Heat of Formation, this should be fast and more or less the ‘engine’ of the whole machine. In an aprotic solvent it’s kinetically easily explained and similar to hydrolysis of a Grignard Reagent.
Quote: Originally posted by len1  

I dont believe any water is formed in the reaction. It would immediately react depositing insoluble MgO on Mg surface. In fact it is not priduced in the dehydrating stage as well. KOH does not dehydrate in and of itself at 150C - it just melts, and that explains the reaction temperature of the dehydration stage.

[Edited on 11-1-2011 by len1]


No, not if catalyst regeneration gets to it first, as I suggest: Mg(OR)2(solvent) + H2O(solvent) < === > MgO(s) + 2 ROH(solvent) with a very high equilibrium constant due to high ΔG.

And during the dehydration phase some MgO is deposited on the Mg surgace anyway: that doesn’t seem to stop the overall reaction from proceeding. Not even when you add the alcohol right from the beginning, as I did…

You claim KOH ‘just melts’: I’ve seen no evidence for that whatsoever in my tests.

******************************

The yield of my one pot experiment was 70 % but this is based only on the largest globules and calculated on the anhydrous part of the KOH. There were some fines and lots of 1 – 2 mm balls too, unfortunately embedded in ‘crusty’ type MgO from which they cannot be separated manually.

I think it will be necessary for anyone who wants to make significant amounts of chemical potassium and hates wasting much of the product to invest in a bit of Tetralin or dioxane for the purpose of separating the product from the slag. Alternatively any old organic solvent based on C-14 (the isotope) and deuterium should also do the trick ;-)

What I also noticed that overnight the solvent had turned a bit like a light jelly, that doesn’t bode very well for solvent/catalyst recovery…


[Edited on 11-1-2011 by blogfast25]

watson.fawkes - 11-1-2011 at 04:29

Quote: Originally posted by blogfast25  
No, not if catalyst regeneration gets to it first, as I suggest: Mg(OR)2(solvent) + H2O(solvent) < === > MgO(s) + 2 ROH(solvent) with a very high equilibrium constant due to high ΔG.
Another possibility about where H2O could end up is as an adduct with KOH. There's a lot of affinity there to begin with. Once the KOH is dehydrated, there's no reason to believe that it wouldn't rehydrate given the opportunity. I've got to suspect that adduct formation has must have faster kinetics than any other candidate reaction in the mix.

blogfast25 - 11-1-2011 at 04:42

Quote: Originally posted by watson.fawkes  
Another possibility about where H2O could end up is as an adduct with KOH. There's a lot of affinity there to begin with. Once the KOH is dehydrated, there's no reason to believe that it wouldn't rehydrate given the opportunity. I've got to suspect that adduct formation has must have faster kinetics than any other candidate reaction in the mix.


Hmm... what does a KOH/H2O adduct look like in an aprotic solvent? And isn't such a reaction likely to produce less enthalpy than the formation og MgO?

*****************

Thinking a bit about inert solvents with higher density, there MUST be some. Chlorinated solvents for example: granted, at higher temperature K will want to snatch the halogen but at barely above K’s MP? Take dichloromethane for instance: cheap and cheerful with a density of 1.33! Nice and thin too...:) Except its BP is too low... :(

Or perchloroethylene (C2Cl4): BP = 121C, d = 1.62

Or TCE (Trike)?

[Edited on 11-1-2011 by blogfast25]

Eclectic - 11-1-2011 at 06:35

!!DANGER!!, DANGER, Will Robinson!
Quote: Originally posted by Axt  
The Explosion of Chloroform with Alkali Metals
Tenney L. Davis, John O. McLean
J. Am. Chem. Soc.; 1938; 60(3); 720-722. [attached]

DJF90 - 11-1-2011 at 07:08

Don't mix halogenated solvents with alkali metals, no matter what the temperature is. Its asking for trouble.

watson.fawkes - 11-1-2011 at 08:53

Quote: Originally posted by blogfast25  
Hmm... what does a KOH/H2O adduct look like in an aprotic solvent? And isn't such a reaction likely to produce less enthalpy than the formation og MgO?
My guess would be an O/H3/O(-) trigonal bipyramid "ion", charge-bound to K(+). A bipyramid might hold together purely on the basis of back-to-back dipole attraction, but I'd guess there's also some QM-specific behavior, too. It smells to me like a more general version of hydrogen bonding, although I'm not convinced there's a proper bond in place. A deuterium exchange experiment that measured the rate of KOH + D2O --> KOD + DHO would give an idea of how symmetrically coupled the H protons are in actuality (or even if there's any exchange at all). To be explicit, I think it's completely plausible that there's no need to consider solvation for such an adduct to be stable.

Such an adduct, which would first form electrostatically, would have a much higher kinetic rate that anything else that requires bond changes. So while enthalpy would eventually drive MgO formation, it doesn't meant that the reaction is an Mg + H2O one.

blogfast25 - 11-1-2011 at 09:19

Quote: Originally posted by DJF90  
Don't mix halogenated solvents with alkali metals, no matter what the temperature is. Its asking for trouble.


Fair enough. It's not so easy to come up with inert organic solvents of d ≥ 0.9 without hetero atoms or called Tetralin, dioxane or THF!

Edit: and in any case DCM's good for separting the floating K from the slag, at room temperature. I've just done it on a large collection of fines, afterwards rinse with kerosene...

[Edited on 11-1-2011 by blogfast25]

Formatik - 11-1-2011 at 19:31

Quote: Originally posted by blogfast25  
Edit: and in any case DCM's good for separting the floating K from the slag, at room temperature. I've just done it on a large collection of fines, afterwards rinse with kerosene...


Dichloromethane and potassium mixture has a shock sensitivity of 0.06 mkg and yields a "relatively strong explosion" when exposed to the shock. Mercury fulminate in the same instance had a shock sensitivity of 0.04 mkg. Alkali metals and halogenated organic solvents are typically shock sensitive systems. Lithium, sodium, potassium, then Na-K alloy with the solvents are most sensitive, sensitivity increasing respectively from left to right. Some mixtures are so sensitive, that one just has to slightly bump them and they detonate strongly.

Source for this: Über Explosionen mit Alkalimetallen. Zeitschrift für Elektrochemie und angewandte physikalische Chemie, 31: 549–551. H. Staudinger.

Btw. I admire everyone's work here and contribution to this new method of potassium manufacture. Keep up the good work.

[Edited on 12-1-2011 by Formatik]

blogfast25 - 12-1-2011 at 08:51

Thanks Formatik for the well wishes and for the information on DCM/K. I wonder what is meant by ‘mkg’… ‘milli kilogram’? That makes no sense. ‘mass kilogram’? That would be Newton.

Here’s what I did. A few small k balls (total < 0.1 g) were mixed with about 1 ml of DCM, the K floated. Shaking the tube nothing happened. I then carefully heated the tube on steam bath and the DCM started to boil and I got reaction with the K. Nothing spectacular: they didn’t dissolve, explode or burst into flames; they just darkened to the point of blackness (C?) But remember that the K balls had previously been stored under kerosene, so the DCM had been diluted somewhat.

Next I made a mixture of about 50/50 v/v DCM/kerosene (they are miscible) and added a mixture of K-balls (< 0.5 g) and MgO reaction slag, the K-balls floated and the MgO sank to the bottom. The K-balls were then skimmed off and dropped into Shellsol D70 and rinsed a few times with small amounts of that solvent to eliminate any DCM. The separation is neat and quick. And uneventful, all at RT.

The then clean K balls were then subjected again to coalescence experiments involving heating and cooling them to above and to below K’s MP to try and get them to merge. I also (re)tried garage chemist’s trick with the IPA. All in all it’s very slow and frustating process which at times had me wishing those glistening balls were Hg and not K!

So while mixing DCM with K in larger amounts is probably really silly, for the purpose for instance of yield determination, using a mixture of DCM and kerosene as a floatation liquid is probably a relatively safe option. C2Cl4 may be even safer because it has a higher density and less of it can be used in the floatation mix to reach d ≥ 0.9 and get K to float comfortably.

Of course organic inert solvents with d > 0.9 due to hetero atoms are to be preferred but tend to be much more expensive (see dioxane or THF).

Next attempt at coalescing will be using octane (well, 99 octane unleaded car juice), hoping that the lower viscosity of that solvent will make a difference to the rate of coalescence. The BP of n-octane is about 125C.

************

Unfortunately, probably due to additives, the 99 % octane petrol reacts slightly but persistently with the potassium, with gas evolution. The light yellow green colour of the petrol gives way to a darker brownish tone. So as a coalescing solvent it can’t be used, at least not without prior distillation.


[Edited on 12-1-2011 by blogfast25]

MagicJigPipe - 12-1-2011 at 19:35

Normally I would expect that unit to be "meters*kilograms". Does it have another name? I don't know. I've never heard of it.

Something to look up though.

EDIT:

From Wikipedia:
Quote:

Impact Sensitivity is expressed in terms of the distance through which a standard weight must be dropped to cause the material to explode.


Mass of standard weight times the distance?

[Edited on 1-13-2011 by MagicJigPipe]

Eclectic - 12-1-2011 at 20:17

Maybe useful solvents in this application:

http://www.novolyte.com/documents/brochures/glymes-and-grign...

Concerning test values

Formatik - 12-1-2011 at 21:28

Staudinger was using a fallhammer method to get those values. The kilogram-meter (kgm): for every meter, 1kg of mass equivalent.

0.06 m was needed. Then 0.06m*100cm/m = 6cm for 1kg. The usual older tests used about 2kg weights. So, equivalently here, 3cm falling distance for 2kg.

Additional notes:

The increasing size of halogen substitution of the solvent generally increases sensitivity of the aforementioned mixtures also, which is why carbon tetrachloride mixtures are among the most sensitive. Pentachloroethane and tetrachloroethane are even much more sensitive as noted in the next paper, so much so, that after some time of mixing (with Na-K alloy) that without any external shock, explosions have occurred. Staudinger also noted in this other paper, Erfahrungen über einige Explosionen. Angewandte Chemie, 35: 657–659: some neutral sulfur products like carbon disulfide, and some oxygen containing compounds like CO2 when mixed with alkali metals and when exposed to shock, detonate also (no values given).

@blogfast25:
Staudinger reported (first paper above) potassium with pentachloroethane or bromoform at first showed little shock sensitivity, but these reacted chemically forming a grey mass, whereby the shock sensitivity of the system enormously increased, so that spontaneous explosion can occur. The shock sensitivity not being due to heat generated, since even after cooling, it remains, but the sensitivity has to do do with the formation of an unstable intermediate. In some cases, spontaneous explosion occurs, but the intermediate can also decompose quietly into KCl and carbon. Likely, a similar less sensitive intermediate formed in your experiment. A visible change usually must not occur, Staudinger also noted sodium or potassium with CHCl3 or CCl4, etc. showed no reaction or explosion, and that it appeared the metal remained unreacted under the liquid (heterogeneous system). I am also not sure if these mixtures respond to warmth, like they do to shock. Also, mercury fulminate is very shock sensitive, but in an open vessel I would think it might take some good agitation before it would come explosion. Similar to mixtures which are similar in sensitivity. In all cases, I would entierly avoid all chlorinated solvents solely out of the hazard of dangerous intermediates.

[Edited on 13-1-2011 by Formatik]

watson.fawkes - 13-1-2011 at 06:16

Quote: Originally posted by Formatik  
Staudinger was using a fallhammer method to get those values. The kilogram-meter (kgm): for every meter, 1kg of mass equivalent.
Did these devices have standard contact area? If not, I don't see how this measure represents a physical standard, as opposed to a device-bound practical one. The invariant of mass*height is a representative of gravitational potential energy, so you multiply by the earth's gravitational acceleration 'g' to get a physical energy. But if you spread that energy out too far, it's fairly clear that the number become meaningless.

blogfast25 - 13-1-2011 at 09:19

@watson and Formatik:

I believe the test procedure is described in the *.pdf on chloroform explosions linked to here:

http://www.sciencemadness.org/talk/viewthread.php?tid=5112#p...

I quote:

“About 0.3 g. of sodium or of potassium or of the liquid
alloy was sealed up in a small glass bulb, 6 to 8 mm. in
diameter, which had a capillary stem 15 to 20 mm. in
length. This was placed in the bottom of a narrow testtube
and held in place by a collar of glass (a section of
glass tubing) which was sintered to the inner wall of the
test-tube. The latter was then drawn down, chloroform
(1 to 2 cc.) was introduced, and the explosive capsule was
sealed. These capsules could be prepared in advance and
could be kept safely as long as desired.
For studying the products of the explosion a […]”

The bit about ‘could be kept safely as long as desired’ really caught my attention.

I’m also a little surprised that so relatively little mention is made about the potential dangers of alkali metals in contact with haloalka(e)nes (or even aryl halides?) in the wider literature.

Since as the value of 0.06 kgm is indeed not a strict scientific unit, it’s quite relative. If a guest at a dinner table accidentally drops a 1 l bottle filled with water from 6 cm above the table, does that constitute quite a shock or not?

I appreciate that higher chorine content of the solvent aggravates the problem but in the case of C2Cl4, mixing 1 litre of a 0.8 density hydrocarbon solvent with only 0.14 litre of the 1.622 density C2Cl4 yields a mixture of d ≈ 0.9. Solid K would already float in such mix, containing only 0.14/1.14 x 100 = 12.2 v% C2Cl4. At 24 v% the density becomes about 1, at 18 v% about 0.95.

It would be interesting to strap an ampoule of potassium in DCM, TCE or C2Cl4 to a heavy object and drop it from a small height (while keeping a safe distance!) to see if Stausinger’s results can be reproduced and explosion occurs.

@Ecclectic:

Butyl diglyme is an interesting thought, it’s just borderline in density (0.88). I expect it to be very expensive. But it did jog my grey matter towards other hetero atom containing compounds like light ethylene glycol esters or oxalic (or malonic) acid esters: diethyl oxalate has a density of about 1.08, should be rather easy to synthesise but it’s relatively viscous…



But by and large there are other possibilities that don’t involve coalescing the globules that are too small to recover: provided they can be separated more or less completely from slag, there seems no impediment dictated by the proposed mechanism to add these ‘fines’ to the next reaction. Either this initial K will react with the KOH.H2O (forming more KOH) or it will dissolve as KOR, any left overs may even promote coalescence of newly formed K.


[Edited on 13-1-2011 by blogfast25]

Formatik - 13-1-2011 at 22:57

Quote: Originally posted by watson.fawkes  
Did these devices have standard contact area? If not, I don't see how this measure represents a physical standard, as opposed to a device-bound practical one. The invariant of mass*height is a representative of gravitational potential energy, so you multiply by the earth's gravitational acceleration 'g' to get a physical energy. But if you spread that energy out too far, it's fairly clear that the number become meaningless.


The fallhammer test used a rigid device, you can see an example of one picture of it here.
Quote: Originally posted by blogfast25  
I believe the test procedure is described in the *.pdf on chloroform explosions linked to here:

http://www.sciencemadness.org/talk/viewthread.php?tid=5112#p...

The bit about ‘could be kept safely as long as desired’ really caught my attention.


As long as they aren't touching each other, yeah, they are safe. Not something I would be carrying around in my pockets. Tenny Davis also shows the schematic in his COPAE book (in the forum library), on pg. 403.
Quote:
I’m also a little surprised that so relatively little mention is made about the potential dangers of alkali metals in contact with haloalka(e)nes (or even aryl halides?) in the wider literature.


Gmelin's Handbuch mentions it under all entries of the alkali metals concerning organic solvents. Bretherick's Handbook also mentions it a bit.
Quote:
I appreciate that higher chorine content of the solvent aggravates the problem but in the case of C2Cl4, mixing 1 litre of a 0.8 density hydrocarbon solvent with only 0.14 litre of the 1.622 density C2Cl4 yields a mixture of d ≈ 0.9. Solid K would already float in such mix, containing only 0.14/1.14 x 100 = 12.2 v% C2Cl4. At 24 v% the density becomes about 1, at 18 v% about 0.95.


If C2Cl4 were to form the similar type of highly unstable intermediates as penta- and tetrachloroethane, then it's possible it could pose a spontaneous explosion or decomposition risk, depending on the nature of a reaction in a diluent. Bretherick's at the least states potassium and tetrachloroethylene explode when heated together (to ~ 97 C), except when the metal was free of any oxide film.

blogfast25 - 14-1-2011 at 07:19

@Formatik:

Well, for:

CCl4 + 4 Na === > C + 4 NaCl

The theoretical heat of reaction would be + 366 [CCl4] + 4 x (- 411) [NaCl] = - 1,278 kJ/mol, a very high value, thus a very high negative ΔG and a very strong drive for the reaction to proceed. That would be similar for most halocarbons, with the highest values for those with highest degree of halogenation. Like all reactions you have to surpass the kinetic barrier to get them started (concept of ‘activation energy’) but this kinetic barrier could quite literally be overcome with a kinetic shock, if the barrier is quite low. The 97C in the case of C2Cl4 suggests very low activation energy. A shock wave running through the mixture could greatly increase collision speeds between the halocarbons and the metal, equivalent to heating, with initiation as a (n explosive) result.

Formatik, you can now relax :), I won’t be pursuing this anymore, except for with a sense of mischief to try and get a plastic ampoule with some C2Cl4 and a very small amount of K to explode on impact. Thanks for your input.


[Edited on 14-1-2011 by blogfast25]

Formatik - 14-1-2011 at 15:14

Quote: Originally posted by blogfast25  
A shock wave running through the mixture could greatly increase collision speeds between the halocarbons and the metal, equivalent to heating, with initiation as a (n explosive) result.


Well, yes. Gmelin states mixtures of even lesser sensitive lithium mixtures with halogenated hydrocarbons can be initiated with blasting caps.
Quote:
Formatik, you can now relax :), I won’t be pursuing this anymore, except for with a sense of mischief to try and get a plastic ampoule with some C2Cl4 and a very small amount of K to explode on impact. Thanks for your input.


Sounds like good harmless family fun.;) Take care, since it looks like significant amounts of phosgene could also form.

condennnsa - 16-1-2011 at 15:35

I post this to let everyone know that in case of lack of magnesium for this synthesis, I have bought it from Likurg in Poland. 750 g for 11 euro. The owner , Tomasz, is very reliable , he even talks in realtime via googletalk, and my package arrived in only 4 days.
I tested this magnesium for trace aluminum , by reaction with NaOH solution, there was no sign of reaction.

blogfast25 - 17-1-2011 at 09:51

Not wanting to rain on your parade but a test for traces of Al in Mg by dunking it in NaOH is likely to be very insensitive (poor resolution). Unfortunately there are no quick, cheap and colourful tests available to detect Al, at least not AFAIK. Pyro Mg is unlikely to be uncontaminated by Al though: Mg and Al simply aren’t very likely bed fellows because Mg is produced by electrolysis of MgCl2. At those temperatures, AlCl3 (a largely non-ionic compound) is seriously volatile.

You’ll only find Al in Mg metal if someone deliberately put it there for alloying purposes, not as a ‘natural’ contamination…

blogfast25 - 20-1-2011 at 12:43

’One pot’ with Kerosene

6.12 g KOH flakes, 3.11 g Mg (reagent), 1.32 g 2-methyl-butan-2-ol (t-amyl alcohol) were all mixed together with 40 ml of medium kerosene (deodorised lamp oil, flash point min. 61 C, £2.70/l, supplier Caldo Oils via local hardware store). Note the slightly lower amount of solvent and higher amount of alcohol (0.015 mol).

Temperature was rapidly taken up to 200C which was reached at t = 10 min and refluxing was very strong. I lost a little bit of material through the refluxer (glass tube with iced water kitchen roll wrapped around it) but that subsided shortly after. Strong hydrogen evolution and then the solvent started clouding over, as per usual.

Potassium fines started to appear at about t = 1 h. After about 1 ½ h, hydrogen evolution had largely subsided. I measured hydrogen evolution using a test tube gazometer over 15 mins and found it to correspond to a reaction rate of about 8 mmol/h of K (four millimol per hour of H2), according to 2 KOH + 2 Mg === > 2 K + 2 MgO + H2. To put this into perspective, at a rate of 8 mmol/h of K, reducing about 0.1 mol of KOH would take about 12.5 hours! It’s therefore safe to state that at t = 1.5 h the reduction reaction must have been 90 % or more completed. If I had any dioxane I would stop reaction at about t = 2 h and recover the formed K with that solvent. Alas that wasn’t case and I had to continue for another 2 h until (after cooling it looked like this; mostly 3-4 mm globules, no larger ones this time:



During this test coalescence was very visible, mainly because most of the MgO had formed in clumps, not fine sand (which tends t obscure the view a lot).

Conclusions:

1. box standard kerosene works about as well as Shellsol D70
2. ‘One pot’ method confirmed to work as in previous experiment
3. increased level of catalyst didn’t seem to have a dramatic effect on reaction rate

Tomorrow: yield determination.

Picric-A - 21-1-2011 at 06:32

Has anybody tested this method using diesel fuel?

Diesel is composed of 75% saturated hdyrocarbons (primarily parrafins) and 25% aromatics. After reading the whole thread it looks like aromatics are a no no however has anybody actually tested this?

Diesel has almost ideal b.p. and to say it is incredibly cheap and readily available would be an understatement...

blogfast25 - 21-1-2011 at 09:31

Quote: Originally posted by Picric-A  
Has anybody tested this method using diesel fuel?

Diesel is composed of 75% saturated hdyrocarbons (primarily parrafins) and 25% aromatics. After reading the whole thread it looks like aromatics are a no no however has anybody actually tested this?

Diesel has almost ideal b.p. and to say it is incredibly cheap and readily available would be an understatement...


Nurdrage much above used Tetralin ®, that’s highly aromatic. AFAIK no one has tried diesel yet. With a (Wiki stated) density of 0.832 liquid potassium might float in it, which seems to offer an advantage for coalescing the metal.

&&&&&&&&&

My crop for the latest experiment was about 50 %, lower than with Shellsol D70 and definitely due to poorer coalescing: I lost quite a bit of fine metal. Also the MgO was very crusty (almost no ‘sand’) and some potassium got stuck in it.

The crop: 2 g of potassium in a round flask under Shellsol D70; the poor focus doesn’t do it justice, the metal is very clean skinned:


Picric-A - 22-1-2011 at 11:37

Re. storing alkali metals in aromatics, Acros Organics (part of Fisher) supply powdered sodium metal in a toluene suspension... surely they would not supply a highly reactive form of a metal in something it reacts with?!

blogfast25 - 22-1-2011 at 12:51

Toluene is one I want to try and achieve separation of the K metal and the MgO slag by floatation: toluene has a density of 0.8669 (@ RT) (K, d = 0.862 at RT, d = 0.828 at MP). It may even be possible to achieve coalescence by heating the floating K to just above its melting point. Toluene (or xylene) are of course far too low boiling to consider as a reaction medium.

I bought a heavier kerosene (‘outdoor lamp oil’) with a measured density of d = 0.85 at about 10C. I’ll be trying it as a reaction medium tomorrow, hoping to get the produced K to float and possibly coalesce more, like in Nurdrage’s experiments with ‘IR Grade Paraffin oil’ and Tetralin.


[Edited on 22-1-2011 by blogfast25]

mr.crow - 22-1-2011 at 15:41

Great work on the experiments with lamp oil! I need to find some in the dead of Canadian winter.

Xylenes are also available as label remover at home depot, but may have surfactants in it.

Couldn't you just mash your balls together (hehe) in a test tube with mineral oil? Or will this just make it split up even more?

blogfast25 - 23-1-2011 at 06:33

Quote: Originally posted by mr.crow  
Couldn't you just mash your balls together (hehe) in a test tube with mineral oil? Or will this just make it split up even more?


The latter, mr crow. Dispersing K is a lot easier than making it coalesce. But you gave me an idea: mechanically pressing cold, solid K balls into each other may be a possibility, followed by melting. A bit like pressing oil from olives - LOL.

Satan - 23-1-2011 at 08:53

Quote: Originally posted by blogfast25  
Toluene is one I want to try and achieve separation of the K metal and the MgO slag by floatation
In those conditions (K + MgO + toluene) potassium will metalate toluene, giving benzyl potassium.
Ref: http://pubs.acs.org/doi/abs/10.1021/jo01122a005

[Edited on 23-1-2011 by Satan]

blogfast25 - 23-1-2011 at 10:06

Quote: Originally posted by Satan  
Quote: Originally posted by blogfast25  
Toluene is one I want to try and achieve separation of the K metal and the MgO slag by floatation
In those conditions potassium will metalate toluene, giving benzyl potassium.
Ref: http://pubs.acs.org/doi/abs/10.1021/jo01122a005


Satan: you wrote: “those conditions”. Which conditions exactly are you referring to? I can’t access that article. Floatating K at RT using toluene or xylene is probably possible provided the metal is sufficiently clean (solid impurities would increase its density).

I seem to remember that ‘sodium powder’ is produced by rapidly stirring molten Na in toluene, followed by quenching, but don’t quote me on that w/o references.

Edit: Perhaps more relevant to the metalation of aryl and alkyl compounds with organometallic Li/Na/K compounds is this Google book reference here. Methinks that we’re far off the conditions described if toluene, xylene or possible diphenylmethane (higher density) were to be used as K floatating agents (rather than as reaction media or coalescing media).

And here's a Wiki reference to ‘sodium sand’:

In former times, the sodium was provided in the form of "sodium wire" or "sodium sand", a fine dispersion of sodium prepared by melting sodium in refluxing xylene and rapidly stirring, was common.

Again no reference to sodium metalating the xylene.


%$£$%$£$%$£

I didn’t have time to run a complete test with the heavier kerosene, so I did a couple of other things.

The density of the ‘Caldo Oils’ medium kerosene used above was measured and clocked in at 0.81, much lower than the ‘outdoor grade’ (0.85), which will be tested tomorrow.

Then I did another coalescence experiment, this time putting all my K-balls (large, medium, small) in one basket (pardon the pun) under Shellsol D70 and heated them in the usual apparatus to BP of the solvent (somewhat above 200C). Here coalescence was very visible and very swift: in about 30 mins most of the K had fused into a few larger balls. You can see (the solvent is clean) the balls merge, each instance of balls merging being almost instantaneous.

So basically my hunch that coalescing at just above the MP (where the potassium’s viscosity is highest) is plain wrong, it appears the opposite: lowest solvent viscosity and lowest metal viscosity together lead to fastest coalescence. Here’s the bottom of the 100 ml Erlenmeyer during cooling:



5 g of fairly large potassium reguli under light kerosene:




[Edited on 23-1-2011 by blogfast25]

[Edited on 23-1-2011 by blogfast25]

DJF90 - 25-1-2011 at 04:38

Quote:
Sodium oxide, calcium oxide, or magnesium oxide with potassium metal is a
new and unique metalating agent which has made possible the metalation of
toluene in a yield as high as 90 %. Sodium is not effective under these conditions.
In the absence of these oxides, metalation by either potassium or sodium occurs
to a small extent only.

blogfast25 - 25-1-2011 at 08:01

Well, that more or less clears it up then, although it remains strange that Nurdrage’s best result was obtained with Tetralin.

What are you quoting from, DJF?

DJF90 - 25-1-2011 at 08:06

The article you said you couldn't get. Thats the summary.

blogfast25 - 25-1-2011 at 08:38

For some reason in the case of ACS publications my browser (IE) doesn’t show this first page they now offer instead of an abstract, it’s quite frustrating: I can only read the title, authors and biblio references…

DJF90 - 25-1-2011 at 12:36

No I mean I looked at the paper and copied the point of interest (which was best summarised by the summary, funnily enough).

"baby oil" confirmed working

NurdRage - 25-1-2011 at 19:15

"Baby oil" grade paraffin has been confirmed to work in a standard reaction with t-amyl alcohol.

Trick is to use "hypoallergenic" variety that is pure liquid paraffin without additives, stabilizers or perfumes.

Density was too low for "floating coalescence" but modest "bubble agitation coalescence" was present.

Speed was fairly pedestrian at ~4hrs.

blogfast25 - 26-1-2011 at 09:57

Nurdrage:

Have you tried the one pot method yet?

I've been off ill the last two days but hope to test a heavy kerosene soon.

mr.crow - 26-1-2011 at 10:10

So baby oil is able to obtain the 200 degree temperature required

Why would you put that on your baby?

blogfast25 - 26-1-2011 at 12:09

Quote: Originally posted by mr.crow  
So baby oil is able to obtain the 200 degree temperature required

Why would you put that on your baby?


A bit of a silly remark, Mr Crow. Baby oil is almost exclusively paraffins (i.e. alkanes) which by their very nature are quite chemically inert, hence the resistance also to high temperatures. The chemical inertness is however an asset to anything you’d want to rub into a baby’s (or an adult’s for that matter) skin. Simples.

Paraffins have also been used as a laxative and for frying chips (aka 'fries').

MrHomeScientist - 26-1-2011 at 13:28

I've followed this thread for a while, and it's been extremely interesting. It's a real testament to the skill and dedication of the members of this site, with so many different people researching different aspects of this reaction.

Could we compile a single list (maybe another sticky thread) with all the different solvents, alcohols, etc. that have been found to work? It's time consuming trying to sift through 27 pages of replies :) I really want to try this myself, and I'd like to know what works before I try to hunt down the more exotic reagents like the shellsol and t-butanol. Apologies if this has already been done and I missed it.

blogfast25 - 26-1-2011 at 14:26

It’s not really necessary to compile such a list. Solvents that work need basically to be inert to liquid potassium at 200C and have a sufficiently high BP (range) to allow running them at that temperature. For the most part that means saturated hydrocarbons mixtures: paraffin oil, kerosene (medium or heavy), Shellsol (a ‘branded’ kerosene, essentially), paraffinic baby oil, blends thereof and similar will all do. No-gos are things like silicone oil, glycols, halogenated hydrocarbons and such like because they react with K like the clappers.

T-butanol isn’t ‘exotic’: I’m sure you can buy it unrestricted in the US from many sites. If not try t-amyl alcohol, it’s just as effective. And if you come across perchance of any other tertiary alcohols, ring the ‘Alarm bell’: some here are waiting to test such alcohols…

MrHomeScientist - 26-1-2011 at 15:42

Thanks for the info. I guess it all sounds pretty exotic to me, since I never deal with organic chemistry. I'll see what I can find. I still need a good hot plate for it though, so that might need to wait depending on cost.

mr.crow - 27-1-2011 at 18:03

I tend to make lots of silly remarks.

So I bought some of the baby oil in question, says it contains only "paraffinium liquidum" and no fragrances or aloe. I'm wondering how it is different from the regular "heavy mineral oil" sold as a laxative. I hope no one tries to fry chips with it!

So that means I have all the chemicals and equipment. Hopefully the home made Mg powder/turnings works. I can't do it outside, too cold, and probably isn't safe enough to do indoors.

Nurdrage has a list of solvents. t-butanol and 2-methyl-2-butanol both work fine, but the latter is more convenient. Mg particle size is important too, needs to be a powder.

How about the KOH, are 85%+ pellets fine?

blogfast25 - 28-1-2011 at 08:27

Quote: Originally posted by mr.crow  
I tend to make lots of silly remarks.

So I bought some of the baby oil in question, says it contains only "paraffinium liquidum" and no fragrances or aloe. I'm wondering how it is different from the regular "heavy mineral oil" sold as a laxative. I hope no one tries to fry chips with it!

So that means I have all the chemicals and equipment. Hopefully the home made Mg powder/turnings works. I can't do it outside, too cold, and probably isn't safe enough to do indoors.

Nurdrage has a list of solvents. t-butanol and 2-methyl-2-butanol both work fine, but the latter is more convenient. Mg particle size is important too, needs to be a powder.

How about the KOH, are 85%+ pellets fine?


Heavy mineral oil? An elastic term. Suppliers rarely provide density. Baby oil would work as a laxative and in a chip fryer.

Magnesium: chips, powder, it’s mostly all good. Rarely has someone gotten a poor result due to Mg alone. A decent, unadulterated pyro grade should do fine.

KOH: pellets are fine.

Unless you have a dedicated chem area inside, outside is best, some sheltered area I guess. Remember that temperature is ‘the great doubler’ here, you need to maintain 200+C for at least a couple of hours before you even get small globules which might not even be visible because of the formed MgO. Carrying on for 4 to 4 ½ hours is best: may the K-balls be yours!

Concentration dependence of alcohol

NurdRage - 28-1-2011 at 08:44

I tried the reaction again (baby oil + t amyl alcohol method) and this time i minimized my use of solvent to *just* cover the solids but used the same amount of alcohol. Proportionally by volume the amount of alcohol was probably >20% (stochiometry is still the same though at 10% of Mg).

I also used the one-pot approach put for by blogfast.

Reaction didn't work. the mixture got cloudy and dehydrated but failed the potassium+water test. i did this twice to confirm.

Then I performed the same reaction and added more solvent, so the proportion of alcohol was probably less than 10% by volume. This time the reaction proceeded forward and i had potassium formation.

So i believe this shows a concentration dependence in terms of alcohol. Too high a concentration somehow inhibits the reaction.

So when troubleshooting your reaction consider that a sufficient amount of solvent is needed.

I only really tested with the one pot method. I'm not sure if the same issue occurs with the two step method.

blogfast25 - 28-1-2011 at 13:11

Very interesting Nurdrage. Like you, I’ve been cutting down on solvent somewhat, using 40 ml instead 50 for instance, with no adverse effects. But I didn’t go as far as you did. Your results seem to show that there’s a lower limit of solvent/reagents ratio though.

I’m a little surprised at that result: there’s really nothing in the (proposed) reaction mechanism that would seem to imply too little solvent would be detrimental. And where I’ve increased alcohol content (again fairly modestly) I saw no significant influence on overall reaction time.

At the same time it remains difficult to make definite (theoretical) pronouncements with regards to the influence of the initial concentrations of one or more reagents on the overall reaction rate, without actual equilibrium constants. But even if we had them, calculations would involve a system of partial differential equations with 8 variables: time, KOH, ROH, KOR, Mg(OR)2, H2O, K and MgO…

aonomus - 31-1-2011 at 00:49

Just a thought that crossed my mind, as absurd as it may sound, could some t-BuOH loss be accounted for as isobutene gas? It would take severely forcing conditions but perhaps at 200+degC, with t-BuOK as a base and both potassium and magnesium metal being present and able to reduce the eliminated water to hydrogen gas it could happen.

It sounds absurd, but perhaps this could account for why increased contact time with the KOH/Mg, especially during the induction phase could cause reduced yield/increased t-BuOH requirements.

mr.crow - 1-2-2011 at 08:19

Just so everyone can weigh in on this,

Adding the catalyst at the beginning produces more decomposition side reactions due to longer contact time. This increases water and decreases catalyst concentration.

Isobutene sounds like the product. The easiest test would be to bubble the gas through water/solvent with bromine to test for unsaturation.

Has anyone analyzed the purity of the K?

blogfast25 - 1-2-2011 at 08:51

Aonomus and mr crow:

Sigh…

The catalyst is very stable even at these temperatures and conditions. Adding the catalyst at the beginning makes not a blinding bit of difference, as evidenced by mine and Nurdrage's experiments.

Get experimenting, stop faffing about.


[Edited on 1-2-2011 by blogfast25]

NurdRage - 1-2-2011 at 09:14

I'm still a little undecided on the right time to add the catalyst.

Sometimes my one-pot experiments fail but then work when i inject more alcohol later. I tend to have more consistent results with timed addition. But this is based on a limited set of experiments. i need to run a dozen times more before i can iron out any patterns with the one-pot approach.

I will say this though: one-pot does work and does give potassium. but i am unsure if it is as good. Nonetheless it is good enough that it should seriously be considered when making potassium. Especially where the inconvenience of injecting alcohol into an assembled apparatus (like a single neck flask) must be dealt with.

blogfast25 - 1-2-2011 at 09:48

Nurdrage:

I haven't had the time I wished I had, following political events around the world. My one test with one pot showed not a blinding bit of difference, in any aspect. I followed that one eyeballs glued to the flask, as it were.

If the proposed mechanism is correct, there nothing in the theory that should make anyone suspect that adding the catalyst at the start is a no-no. I believe that to believe otherwise is merely 'chemical superstition'...

NurdRage - 1-2-2011 at 10:18

I will not base an entire theoretical construction on your "one test with one pot".

Consistent, reproducible results are far more valuable.

in addition, just because a theory says no one should suspect something doesn't mean we should ignore it without question. It was my blatant questioning of the aliphatic solvent theory that lead me to discover 1,2,3,4-tetrahydronaphthalene (tetralin) to be viable.

I have performed about 6 experiments with the one pot approach, with so far inconsistent results (in terms of time, quality and yield). Quite frankly i trust my 6 vs. your 1 experiment.

I'll make it clear, i acknowledge you *may* be right, and i have not yet decisively proven there is a difference. But without a great deal more data, experiments and verification by others under controlled conditions, for now I outright reject your assertion that there is "not a blinding bit of difference."

blogfast25 - 2-2-2011 at 14:24

Quote: Originally posted by NurdRage  


in addition, just because a theory says no one should suspect something doesn't mean we should ignore it without question. It was my blatant questioning of the aliphatic solvent theory that lead me to discover 1,2,3,4-tetrahydronaphthalene (tetralin) to be viable.

I have performed about 6 experiments with the one pot approach, with so far inconsistent results (in terms of time, quality and yield). Quite frankly i trust my 6 vs. your 1 experiment.

I'll make it clear, i acknowledge you *may* be right, and i have not yet decisively proven there is a difference. But without a great deal more data, experiments and verification by others under controlled conditions, for now I outright reject your assertion that there is "not a blinding bit of difference."


Your reasoning is deeply flawed. When everyone here was still deeply skeptical about pok’s results, it was generally (and correctly) assumed if one person could reproduce his results then we knew pok was speaking the truth (and he was). It’s the same here: my experiment showed that one pot made not a bit of difference.

Your statement about theory is a straw man: I never said such a thing. When theory and experimental results are in line, theory is vindicated and vice versa.

I don’t believe you have carried out 6 one pot experiments without changing any other variables, just like I don’t believe you have carried out 6 not one pot experiments without changing any other variables.

NurdRage - 2-2-2011 at 19:52

No it is not the same.

When testing the potassium the line of reasoning could be simplified to "Is X possible? if not, it is impossible"

All we needed was one successful replication to prove it was possible, we didn't need to prove it was possible all the time or even if it was exactly replicable. Just that it was possible.

That's a fairly straightforward proof.

Now your statement with "not a blinding bit of difference" is a far taller order.

Simplified its goes something like "Is A not different than B for all cases?". This goes to "Is A = B for all cases?"

Where the cases are yield/rate/purity for all experiments and all reactions conditions provided two experiments are performed identically (A being one-pot and B being staggered addition).

This is NOT something you can prove with just one experiment, because that's only one case. You need to thoroughly test more cases before making reasonable determinations.

That is what i'm currently trying to do, so i think its premature to say "A=B for all cases" is true. You may be vindicated in the future, but i don't think one experiment can say that yet.

I'll settle for "A=B for most cases" as its not very practical to test all cases.

A more narrowly defined, but still very important logic statement is "Can one-pot addition work?" This simplifies to "Is X possible?"

For this you just need one experiment, you've performed one and were successful (lucky bastard! :)), I performed some and was successful in fewer. So yes, we have proven one pot is possible. A billion failures still doesn't prove its wrong, just one success proves it right. Same thing with the original belief in Pok's work.

Is there no difference? I don't know, and i don't think you know either.... or at least haven't proven it to a reasonably acceptable degree.

If the situation were reverse, like we started with one pot and then tried staggered addition later. I'd still argue that there may be differences and a great deal more work was needed to prove otherwise.

I think a better order "is there a practical difference?" This requires very few replicate measurements, and can strongly be supported with just one experiment. Although i'd personally prefer a few more. much more variability can be tolerated.

But i think "not a blinding bit of difference" is too strong a statement to make and believe it vindicated by just one experiment.


As for this: "I don’t believe you have carried out 6 one pot experiments without changing any other variables, just like I don’t believe you have carried out 6 not one pot experiments without changing any other variables."

alright, if you're questioning my honesty or experimental approach, i question your honesty and experimental approach as well. I do not believe your one experiment actually showed "not a bit of difference."

Sounds really broad... Are yield, rate and purity all exactly the same? how did you characterize your results?

in my 6 experiments two of them produced no product at all after 4 hours. One went very fast and produced small balls within 2 hours, finally coalescing after 4 (in total). Yield was extremely variable, 0% to 60%

On the other hand once i got the staggered addition parameters worked out i could make potassium in several replicate experiments with about 40-60% yield every time.

So by my experiments there is "a bit" of difference.

But then again, i may be a liar, or an idiot. Your one experiment clearly proves without the need for replication that one pot will always yield exactly the same results as any other one-pot experiment or staggered addition experiment for that matter.

Replicate experiments are only for liars and/or idiots like myself.

You'd be an awesome drug developer. An old drug works but then you make a new drug and from one test with one patient you find "not a blinding bit of difference." Therefore all patients must react exactly the same way to a new drug as the old drug, no need for clinical trials.

So please ignore my results with tetralin. Clearly past experiments with pure aliphatic solvents working and aromatic containing solvents not working vindicate the aliphatic solvent only theory. I never should have questioned that and tried tetralin.

I officially retract my assertions that tetralin works.

condennnsa - 2-2-2011 at 22:49

Hey Nurdrage, this is only the internet, you tend to give different phrases used by others too much 'hidden' meaning...
I don't think Blogfast meant anything derogatory... lets not argue and pickle each other on this wonderful thread...

I can't believe that this synthesis hasn't been researched by some team of chemists, properly funded by a university, so that they could determine at least part of the variables and how they affect the outcome. The silence in the literature about this is staggering.

NurdRage - 3-2-2011 at 00:13

My theories and ideas can be attacked, ridiculed and all the manner of the criticisms applied. That's good science, a truly useful theory will withstand any attack and be strengthened by future experiments. If i turn out to be a complete moron, its about time.

But if my performance and observations are doubted or rejected, bringing into question my honesty or ability, then there is really nothing i can contribute. I have no means of actually proving i really did my experiments and produced my results or observations. Unlike theories and mechanism, another person cannot verify actual performance... well they can, but i doubt anyone will come over here to watch me stare at a flask for 5 hours. I know taking a person's word on faith is generally a bad idea, but without such faith in the contributors of the thread progress would be extremely slow.

Theories can be torn apart left right and center, but data needs a minimal level of trust.

I don't think Blogfast meant anything derogatory, but he does doubt i actually performed my experiments as i describe/imply. If my data is not to be trusted, i no longer see the value in posting it.

blogfast25 - 3-2-2011 at 07:59

Quote: Originally posted by NurdRage  
I don't think Blogfast meant anything derogatory, but he does doubt i actually performed my experiments as i describe/imply. If my data is not to be trusted, i no longer see the value in posting it.



No, I don’t actually (and I didn't say that).

Simple question: did you run 6 experiments with one pot in IDENTICAL CONDITIONS? If not you didn’t strictly speaking replicate. If yes then show us the data: you’ve always been completely transparent before (and I’m not claiming you’re not now either).

And a more specific minor question: did you replicate the Tetralin experiment? There was minor doubt about the experimental conditions as I recall…

***********

As I didn’t see you comment above the one I’m responding to, I can only say ‘calm down’ to that one.

You resorted to literalism there: when a reaction proceeds in exactly the same fashion as any other, the term ‘no difference’ applies. You’re reading too much into what I wrote. But I'll tone it down: 'One pot is possible' is simple statement of fact nad more accurate than my previous statement. Agreed?

*******

I’m not too enamoured with some of your approaches to theory either. Considerably higher up you wrote ‘Experiment > Theory’. That’s quatch.

One of the reasons why I found the one pot method result significant is that although it isn’t PROOF that the theory works but that it certainly counts as EVIDENCE that it does.

Another piece of evidence that I’ve been meaning to generate is to recycle the solvent (presumably with catalyst – partly or wholly as KOR) and run a reaction with that (w/o adding “virgin” catalyst). If the alcohol truly is a catalyst it MUST be recyclable, at least to some degree (losses and degradation need to be taken into account). Some very practical problems have stopped be so far from doing it…

[Edited on 3-2-2011 by blogfast25]

[Edited on 3-2-2011 by blogfast25]

condennnsa - 4-2-2011 at 07:57

Well I finally got my Merck t-butanol today!
To my surprise it came in a glass bottle.
Isn't there the risk as it keeps freezing and melting to crack the glass?
Right now it's frozen but my room can exceed 26 C.
I can also constantly hear the frozen alcohol make tiny cracks, so I guess it's still in the process of freezing.
Should I move it to plastic or is it fine? How did you guys get your butanol in?

mr.crow - 4-2-2011 at 08:13

Quote: Originally posted by condennnsa  
Well I finally got my Merck t-butanol today!
To my surprise it came in a glass bottle.
Isn't there the risk as it keeps freezing and melting to crack the glass?
Right now it's frozen but my room can exceed 26 C.
I can also constantly hear the frozen alcohol make tiny cracks, so I guess it's still in the process of freezing.
Should I move it to plastic or is it fine? How did you guys get your butanol in?


Mine is in a 1L glass bottle. It shrinks when it freezes so there is no risk of breaking. Most liquids do this, except water.

When it freezes do you get beautiful needle like crystals forming all over?

condennnsa - 4-2-2011 at 08:18

Yes it really is beautiful crystals! And a very strong interesting smell!
Kind of reminds me of those cheap minty candies, which is worrisome.
I didn't know it shrinks when freezing, that is reassuring. I guess i don't have to worry.

blogfast25 - 4-2-2011 at 09:05

I second mr crow: no risk of cracking, it's not water based.

condennnsa - 4-2-2011 at 10:23

mr crow, are you sure about the t-butanol shrinking on freezing?
because mine has been for 2 hours over a heater and the still frozen piece is obviously floating in the melted alcohol.
though very little of it protrudes from the surface.
it could also be the water in my alcohol messing things up, as the label lists water<0.1%

mr.crow - 4-2-2011 at 11:51

So its like an iceberg. I couldn't find any evidence online, but my experience is freezing isn't a problem. If it does expand it will be a lot less than what water does and the pressure will go upwards towards the air gap above the liquid.

blogfast25 - 4-2-2011 at 12:17

Quote: Originally posted by condennnsa  
mr crow, are you sure about the t-butanol shrinking on freezing?
because mine has been for 2 hours over a heater and the still frozen piece is obviously floating in the melted alcohol.
though very little of it protrudes from the surface.
it could also be the water in my alcohol messing things up, as the label lists water<0.1%


Is the frozen piece solid t-butanol or does it contain air pockets?

condennnsa - 4-2-2011 at 12:32

you're right blogfast, upon close inspection i see that frozen piece has lots of air bubbles.

[Edited on 4-2-2011 by condennnsa]

blogfast25 - 4-2-2011 at 13:46

Mine did too, that's what made me think of it. Cases where the solid is less dense that the corresponding liquid are really quite rare. Ice does it and w/o that property there may not be life on Earth as we know it...

You guys now have enough t-butanol to make a tonne of K! Proverbially speaking at least: about 0.1 mol catalyst per mol K, work it out!

[Edited on 4-2-2011 by blogfast25]

condennnsa - 4-2-2011 at 15:13

Now that i melted it all and let it freeze again, the frozen chunks all sink to the bottom, so mr crow is right.
I guess t-butanol can dissolve gases quite well.

About the tonne of K, Blogfast, after all, isn't the solvent together with the already dissolved alcohol supposed to be recyclable? Has any of you guys tried to have a second run in the same solvent, after getting rid of the MgO crust/sand? I guess it should work. Then one could get away with 10ml of alcohol for years.




[Edited on 4-2-2011 by condennnsa]

blogfast25 - 5-2-2011 at 09:50

Quote: Originally posted by condennnsa  
About the tonne of K, Blogfast, after all, isn't the solvent together with the already dissolved alcohol supposed to be recyclable? Has any of you guys tried to have a second run in the same solvent, after getting rid of the MgO crust/sand? I guess it should work. Then one could get away with 10ml of alcohol for years.




[Edited on 4-2-2011 by condennnsa]


Yes, as I’ve been arguing for some time now: if t-butanol (or 2-methyl-butan-2-ol) is a true catalyst then it should be recyclable. And if experimental proof of that would be obtained then it would be further evidence that the proposed reaction mechanism is indeed correct.

In reality it’s not that simple. Theory tells us that the catalyst at the end of the reaction should be mainly (if not solely) in the form of potassium t-butoxide). At RT this may not be very soluble in the solvent and may separate out. My solvent after cooling is usually quite ‘jelly like’, this may be due to separation. Then again that may not be the case, I just don’t know at this point.

The second problem is effective separation of the solvent (+ K t-butoxide) from the potassium metal and MgO. The post-reaction solvent is difficult to filter with conventional filtration means (at least with ‘garden shed’ filtering media!) But complete separation may not even be necessary: there’s nothing in the proposed mechanism that would indicate that small amounts of fine K or MgO would inhibit the reaction.

My advice to both of you is to not blindly throw the solvent down the drain after you’ve recovered the potassium metal, but to set it aside, well stoppered, for further experimentation…

condennnsa - 6-2-2011 at 03:49

Well I had my first try at this.
I added 50 ml vet grade paraffin oil to a 200ml beaker, 6.1 g KOH, and 3.1 g Mg shavings.
On top of the beaker I added a glass jar filled with ice water, to act like a primitive refluxer, as I don't have a flask so small to fit my 29/32 ground joint condenser.

I put the beaker on the hot plate, and in about a minute, maybe less!, there was a very violent reaction, like woelen describes in his write-up. Lots of splattering, and white fumes of presumably paraffin oil mist.
So there was clearly a very exothermic reaction taking place. the whole beaker heated up very fast.
I attributed this violent reaction to the magnesium reacting with the water in the KOH.
Also during this time, the 'liquefaction' of the KOH flakes took place, as described by others, particularly Len1.

After this violent reaction was over I could not believe what I was seeing in the beaker! Tiny, sub millimeter , spheres of metal! they were moving across the solvent, and i looked very carefully, they were not gas bubbles of any kind!, they were shiny, and did not go up! I was sure thess were potassium metal balls.

Now note that until this time I had not yet added the alcohol.

I continued to heat the beaker, the temperature stabilised at 240C, the loss of solvent was negligible, and it was not boiling. The water jar on top was doing its job quite well.
All this time gas bubbles continued to emerge from all solids, presumably H2.

But the tiny metal balls did not continue to appear, nor did they grow in size.

I now added 0.6g t-butanol to 6ml paraffin oil, mixed it in till it was homogeneous, and put this to the beaker.
On contact, there was lots of boiling, but most of it condensed back and dripped back in. I don't think i lost a large percentage of the t-butanol due to this boiling, as there was almost no smell of it.
The solution became cloudy, as reported by others,
but everything went wrong from there.
The metal balls disappeared, and I continued to heat the beaker at 240C for an hour, during which time the H2 evolution continued to drop...
After about 1.5 hours of heating there is still no metal balls, and the gas evolution was almost non existent.

tnphysics - 6-2-2011 at 05:39

Will Al react in this way if finely divided? It is cheaper so if it does it is to be preferred, but the reaction would need to be driven by reduction of the water produced to H2. It may not react.

blogfast25 - 6-2-2011 at 08:18

Condensa:

Very weird stuff going on there…

Some remarks/questions:

Do you know the density of your solvent? Measure it by weighing a known volume (in a measuring cylinder e.g.).

240 C seems a very high temperature for no boiling to occur. Are you sure the bubbles are H2 and not simply vapour bubbles from the solvent? These would form preferably on the solids (acting like boiling stones).

Are you sure you saw potassium PRIOR to alcohol addition?

Your refluxer idea is probably OK but does it really provide 100 % reflux with your solvent at 240 C? Do you have ice + water in the refluxer at all times? Also, 50 ml of solvent (plus reagents) in a 200 ml beaker leaves a lot of head space: the catalyst (its BP is only 82 C!) at this temperature has a tendency to gather in the vapour phase of the apparatus, even IF refluxing is constant, leaving perhaps little of it in actual solution.

Did you check the post-reaction sludge for potassium?

I would suggest to:

• Try again with at least 100 ml of solvent (double also all reagents and catalyst quantities). This reduces amount of head space.
• Add catalyst right from start.
• Heat a little slower, ending up at about 200 - 220 C.

Tnphysics:

No one has tried Al powder as far as I know. Be the first!

condennnsa - 6-2-2011 at 09:08

Blogfast, i measured the density of my paraffin oil with graduate cylinder and scale, several times. It is 0.83-0.835 at room temperature.
However, during heating i could clearly see that The solvent level has risen noticeably, my guess with about 10-15%, judging by the lines on the beaker.

Yes, I'm sure i saw metal balls before alcohol addition, just after the initial vigorous reaction began to calm down.

I even ran another test with the same oil, 3.1 KOH, 1.5 g Mg. Again the vigorous reaction, and just after that I could see maybe 40 or 50 tiny sub millimeter metal spheres swimming around. I then heated it for 30 minutes more, during time they disappeared. In this test I did not add any t butanol.

My refluxer was in no way making a GOOD seal. but judging by how much the solvent level dropped during one hour, I can say that the losses were very low.
And no, ice water was only at the start, i did not replenish it. But even toward the end the water in the jar was at maybe 30-35C, it was still condensing the paraffin quite well.

I don't know what to make of all this, i'm rather disappointed.
Blogfast, i'm reluctant to have larger batches.
I was thinking of doing it small until i find the one that works best. I want to waste as little of my chemicals as I can.

I'm also going to try it by adding the catalyst from the start, but wasn't it Len1 who came to the conclusion that part of his failure with this reaction three years ago was because he added the alcohol at the start? He said earlier in the thread I think that there should be no alcohol during the 'dehydration' stage..

blogfast25 - 6-2-2011 at 09:43

Quote: Originally posted by condennnsa  

I'm also going to try it by adding the catalyst from the start, but wasn't it Len1 who came to the conclusion that part of his failure with this reaction three years ago was because he added the alcohol at the start? He said earlier in the thread I think that there should be no alcohol during the 'dehydration' stage..


That’s not how I recall it: Len1’s initial failure was attributed to too vigorous stirring and possibly the grade of Mg. In this thread (so, much later!) he did claim ”that there should be no alcohol during the 'dehydration' stage..” but gave no substantiation for that claim whatsoever.

My tests and Nurdrage’s show ‘one pot’ is possible. Nothing in the proposed reaction mechanism says it isn’t either. It has the distinct advantage of avoiding the very strong reflux of t-butanol when you add it to the hot solvent (very much suited to your method of refluxing). It also avoids introducing oxygen into the system…

Your solvent is borderline density for floatation and it expands during heating. But that shouldn’t stop reaction anyway. And a ‘Vet grade’ should be very pure.

Very strange you get K even without catalyst, we’d expect that to happen only at much higher temperatures (> 500 C).

condennnsa - 7-2-2011 at 02:25

Ok, so I ran another test to see if the balls I get right at the start are indeed potassium, this time with a different oil, I used pharmacy paraffin oil.

1 g KOH and 0.5 g mg shavings, in about 12 ml oil.
Reaction proceeded in exactly the same manner, after the vigorous initial bubbling began to calm down, there were lots , maybe a hundred very tiny metal balls swimming around, and this time with a lot of persistance i managed to get one out with a knife.
Upon adding it to water, it reacted just like potassium.
This was exciting for me, as it was the first time in my life to see sodium or potassium with water.

Any one of you guys, who has either paraffin oil, should get these balls right at the start.

Unfortunately, I still haven't had success with the t-butanol reviving this reaction.
I am yet to have a go with a one pot setup.

Every single time, after i get these metal balls, in about 10 minutes they disappear, and no matter if I add the alcohol, the hydrogen evolution slowly dies down, and the magnesium chips get a dark grey/brown luster.

This dark grey luster got me thinking that it might be a silicon contamination which passivates the surface of the chips.

My results seem to be very similar to garage chemist's trials, for which he blamed the magnesium.

blogfast25 - 7-2-2011 at 08:54

What grade of magnesium are you using? Most powdered grades of Mg, AFAIK, are unalloyed. The production process kind of dictates that (electrolysis of molten MgCl2). Try dissolving a small amount in dilute HCl (or any other fairly pure acid - even distilled vinegar will do): you should get a clear sloution (Si will not dissolve unless present as Mg2Si).

Still very strange you're getting potassium w/o catalyst?!?!

condennnsa - 7-2-2011 at 09:25

I have 2 types of magnesium, 63 micron and Mg shavings, both from likurg.pl.

All the trials I had so far were with the shavings.
The only test I did for shavings was by dissolving in 30% H2SO4, I dissolved about 40 grams and the solution was completely clear.

Blogfast, believe me, I am as surprised as you, but if you have doubts, try it yourself, I bet it will work. just heat under paraffin oil KOH and Mg, and in good light I guarantee you will see the tiny balls.

blogfast25 - 7-2-2011 at 10:19

I'm not doubting your word. The only time I didn't get K was when my refluxer broke and I lost all the catalyst through evaporation.

Also, paraffin/kerosene/Shellsol, it matters little, except for Tetralin (much faster acc. Nurdrage).

You'll have to try the 63 micron powder...

Per - 15-2-2011 at 14:01

I've now got the result that Mg from a pencil sharpener works pretty fine while my phlegmatised 99% Mg powder with which I made my first attempts is not working...

metalresearcher - 15-2-2011 at 23:56

Quote: Originally posted by condennnsa  
Well I had my first try at this.
I added 50 ml vet grade paraffin oil to a 200ml beaker, 6.1 g KOH, and 3.1 g Mg shavings.
On top of the beaker I added a glass jar filled with ice water, to act like a primitive refluxer, as I don't have a flask so small to fit my 29/32 ground joint condenser.


Interesting, I'll give it a try. Can vegetable oil or molten candle wax be used as well ? They have a higher boiling point and are readily available.

And what about the fire hazard (above 200oC virtually every oil catches fire) ?


[Edited on 2011-2-16 by metalresearcher]

blogfast25 - 16-2-2011 at 08:30

metal researcher:

No vegetable oil: it contains bound oxygen: -O- + 2 K ===> K2O + much heat!

Candle wax was first used by Nurdrage. Avoid candles that contain stearic acid. Candle wax solidifies at RT, a great drawback for cleaning up the resulting post-reaction mess.

Best cheap solvent, IMHO, is unadulterated heavy kerosene (lamp oil, very OTC).

Fire hazard: use sand bath heating. Make sure you're refluxer works properly.

metalresearcher - 16-2-2011 at 11:46

I tried with a small 100ml beaker in a pan with sand heated by a bunsen burner. On top of it I put a wide crucible to prevent much coming out in case of a vigorous reaction. I put a thermocouple into it but it did no go further than 198oC the lamp oil boiled.
I used 0.93g KOH flakes and 0.31g Mg shavings and immersed it into 30ml lamp oil.

I doubt whether any K metal is formed. I saw some bubbling but that is probably boiling /evaporating oil.

Should I use another (heavier) oil ?

IMG_4464.JPG - 42kB IMG_4463.JPG - 46kB IMG_4467.JPG - 53kB

[Edited on 2011-2-16 by metalresearcher]

condennnsa - 16-2-2011 at 12:02

yes, the alkali metal will react to release hydrogen with the vegetable oil to give the corresponding linoleic/oleic acid salts, (soaps) . http://www.youtube.com/watch?v=5GahIcgDKPs

This reaction is draining the life out of me, I must have had 12 runs till now and I could never get it to work. I tried pyro mg shavings, 63um powder, homemade shavings from sacrificial anode for boilers, even magnesium filings from Mg pencil sharpeners. The only potassium I saw was the one at the beginning which disappears in 2 or 3 minutes, then nothing.
At this point , I guess that good refluxing is critical for this reaction, and I can only point to this as the culprit of my failures. I'll have to buy some smaller volume glassware to fit my condenser and keep trying.
But then i'm still unsure, as Nurdrage if I remember had success with the higher boiling paraffin wax without refluxing at all, where he just used an uncovered beaker, in this video http://www.youtube.com/watch?v=DUzsyNLuLyg.

metalresearcher - 16-2-2011 at 13:33

UPDATE:
I emptied the beaker of my previous experiment from amp oil and found the KOH was sticked together in the beaker. So I added some water and saw a few sparks appearing (so there was a very little K metal formed.
Anyway, I put the Mg turnings (with still some lamp oil on it which does not affect the experiment) + fresh KOH in a steel tube of 2.5cm diameter and 5cm long one end squeezed and welded together. I put a small steel plate on the top of the open end of the tube. During heating the lamp oil remains evaporated and burned and then the KOH reacted with the Mg metal showing purplish flames of buring K metal. Lots of smoke (probably K2O) appeared.

Here a video:

<iframe sandbox title="YouTube video player" width="480" height="390" src="http://www.youtube.com/embed/udYQQ75S3So" frameborder="0" allowfullscreen></iframe>

[Edited on 2011-2-16 by metalresearcher]

peach - 16-2-2011 at 21:28
Quote:
Poppers is a slang term for various alkyl nitrites inhaled for recreational purposes, particularly amyl nitrite, butyl nitrite, isopropyl nitrite and isobutyl nitrite.


Quote:
Isoamyl nitrite decomposes in the presence of base to give nitrite salts and the isoamyl alcohol:

C5H11ONO + NaOH → C5H11OH + NaNO2

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