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

The floating does indeed not seem to speed up coalescence all that much: total process times with Nurdrage's IR grade of paraffin were about the same as yours (len1) and woelen's... So maybe it's not worth pursueing as a possible avenue to cut total process time...

[Edited on 27-12-2010 by blogfast25]

(Btw, i've always wondered why people add "..." at the end of their posts if they don't continue their thoughts.)

Floating doesn't speed up reaction time. If speed is your objective then no, paraffin approach is NOT the way to go.

Although my original intention with testing paraffin was to determine what solvents could be used other than D70.

woelen - 27-12-2010 at 07:08

My experiment with the old Shellsol was not a success. The magnesium hardly reacted at all. The Shellsol had become turbid (white flocculent stuff in the Shellsol) over the few days that it was standing and I assume that the butoxide somehow has decomposed. Probably I also had too little of the butoxide dissolved in the Shellsol.

Probably better results can be obtained if runs are done quickly after each other, but given the long time of a single run that is not a really attractive thing, especially in a home-lab setting, where due to safety concerns one must babysit the reaction all the time.

blogfast25 - 27-12-2010 at 08:47

Quote: Originally posted by woelen

My experiment with the old Shellsol was not a success. The magnesium hardly reacted at all. The Shellsol had become turbid (white flocculent stuff in the Shellsol) over the few days that it was standing and I assume that the butoxide somehow has decomposed. Probably I also had too little of the butoxide dissolved in the Shellsol.

I’m not that surprised, the total amount of K t-butoxide (assuming no losses) must be really small.

How about we make some potassium methoxide (or ethoxide), which as I’ve shown above forms easily by dissolving KOH in MeOH. Would the dissolved ethoxide or ethoxide react with magnesium, for instance at BP of the solution? Any formed K would of course react away back to ethoxide and hydrogen…

Would the ethoxide or methoxide be soluble in paraffin based solvents? Doesn’t sound likely…

blogfast25 - 27-12-2010 at 08:56

Quote: Originally posted by len1

I just found this datasheet for Nujol (mentioned by len1):

http://msds.chem.ox.ac.uk/NU/nujol.html

Stated density of 0.752, still below that of K at MP (0.828).

Anyroads, for those in the UK that want to have a go with ‘mineral oil, ‘paraffin oil’, ‘liquid paraffin’, ‘Nujol’, ‘Aboline oil’, ‘Adepsine’ and other stated synonyms, this stuff below is probably your best bet:

http://www.myvetmeds.co.uk/cat/nutritional-supplements-cat/l...

Veterinary ‘liquid paraffin’, 2 l for £6.43…

[Edited on 27-12-2010 by blogfast25]

condennnsa - 27-12-2010 at 09:35

that's a great find blogfast! I hadn't thought vets also use it. 5 bucks for 1 l is great price.

len1 - 27-12-2010 at 17:27

I had yields of about 50% in paraffin oil, which was dependent on the source of paraffin, I used three types - its not an AR. Due to its large excess, just a small amount of unsaturated, or God forbid -OH groups has a large effect on the yield.

That K floats in whats generally termed paraffin oil is beyond the reach of datasheets. I saw it with my own eyes in two of the oild I used just a few days ago . Key here is the density at 200C

[Edited on 28-12-2010 by len1]

blogfast25 - 28-12-2010 at 06:29

Quote: Originally posted by len1

That K floats in whats generally termed paraffin oil is beyond the reach of datasheets. I saw it with my own eyes in two of the oild I used just a few days ago . Key here is the density at 200C

[Edited on 28-12-2010 by len1]

Oh, but I don't dispute it: nurdrage's both videos are unambiguous. I just wonder why I can't get it to float even in molten petroleum gel (Vaseline) at 200 C, that's all.

But I won't really be pursueing the paraffin line anymore.

aonomus - 28-12-2010 at 09:18

A thought that crossed my mind last night, USP grade mineral oil intended for use as a oral laxative and skin treatment might have to be aromatic free in order to be approved for human use. A low smoke or smoke free paraffin oil may still contain some aromatics, but I would suspect the USP test limits are lower.

I did some searching and noted that CFR 172.878 and CFR 178.3620(a) are the current specifications for USP mineral oil (afaik, correct me if I'm wrong) which mention only test methods and UV absorbance limits, but not hard assay limits (mg/mL, etc).

Side note: I find the test procedure in CFR 178.3620 amusing, the procedure involves copious amounts of isooctane (toxic) and nitromethane (wut?) on a silica column. I guess it is pretty hard to separate out aromatics from alkanes, so they had to resort to some exotic mobile phases.....

Magpie - 28-12-2010 at 09:42

Is there some reason, other than cost, than no one has suggested silicone oils? They can have a sp gr around 0.93, are non-toxic, non-flammable, and have high boiling points. It may be possible to find a mass consumer product that isn't prohibitively expensive, like a brake fluid, for example.

Dow provides a lot of technical information, eg:

http://msdssearch.dow.com/PublishedLiteratureDOWCOM/dh_0030/...

UnintentionalChaos - 28-12-2010 at 09:44

Quote: Originally posted by Magpie

Is there some reason, other than cost, than no one has suggested silicone oils? They can have a sp gr around 0.93, are non-toxic, non-flammable, and have high boiling points. It may be possible to find a mass consumer product that isn't prohibitively expensive, like a brake fluid, for example.

Dow provides a lot of technical information, eg:

http://msdssearch.dow.com/PublishedLiteratureDOWCOM/dh_0030/...

Reactivity. Polydimethylsiloxanes would certainly react with alkali metals, especially when molten.

Looks like the hydroxide probably would too...the first page is not showing for me. http://pubs.acs.org/doi/abs/10.1021/ac60271a046

[Edited on 12-28-10 by UnintentionalChaos]

blogfast25 - 28-12-2010 at 10:26

Yes, what UC said. Also, poor solubility of alkoxides in siloxanes to be expected...

blogfast25 - 28-12-2010 at 10:28

Quote: Originally posted by aonomus

A thought that crossed my mind last night, USP grade mineral oil intended for use as a oral laxative and skin treatment might have to be aromatic free in order to be approved for human use. A low smoke or smoke free paraffin oil may still contain some aromatics, but I would suspect the USP test limits are lower.

I did some searching and noted that CFR 172.878 and CFR 178.3620(a) are the current specifications for USP mineral oil (afaik, correct me if I'm wrong) which mention only test methods and UV absorbance limits, but not hard assay limits (mg/mL, etc).

Side note: I find the test procedure in CFR 178.3620 amusing, the procedure involves copious amounts of isooctane (toxic) and nitromethane (wut?) on a silica column. I guess it is pretty hard to separate out aromatics from alkanes, so they had to resort to some exotic mobile phases.....

The vet grade paraffin oil suggested above should also be aro-free and cheaper than anything for human consumption.

Waffles SS - 28-12-2010 at 12:51

What is replacement for shell oil?
I think Parraffin is not suitable because has density of >.86

NurdRage - 28-12-2010 at 14:50

Quote: Originally posted by Waffles SS

What is replacement for shell oil?
I think Parraffin is not suitable because has density of >.86

Please read the thread, Len1 and I have already proven paraffin to work. Although yields with "domestic" paraffin sources are ~50%, which lower than with Shelsol D70.

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

@ everyone else

More negative results: (They say all results are "data" after all

)

PEG-3350 as a solvent --> Catches fire! About 3 minutes after melting the reaction vessel spontaneously ignited and the magnesium and PEG burned together yielding a sponge-like mass of carbon with KOH embedded in it. Messy, but no loss of glassware and safety measures including bubblers and back-flow arresters completely contained the flames. Was fun to watch. I wish i filmed it.

Brake Fluid --> since brake fluid (DOT 3) is essentially shorter chain PEG i decided against directly testing this. I think the PEG-3350 results still apply.

PEG-3350 as an activator/catalyst (in paraffin solvent)--> Failure, hydrogen evolution stopped 20 minutes into the run and subsequent testing with water showed no pops or fizzes associated with even minute potassium production.

Before anyone complains that this was a waste of time and would have never worked in theory anyway I'll say my objective is to test theories.

Eclectic - 28-12-2010 at 17:04

Exxon Isopar L seems to be fairly close to Shellsol D70:

http://exxonmobilchemical.ides.com/en-US/ds128997/Isopar%E2%...

http://www.westmarine.com/webapp/wcs/stores/servlet/Product_...

Synth. of t-alcohols thread in ‘Organic chemistry’ section

blogfast25 - 29-12-2010 at 08:04

In agreement with nicodem a thread has now been started in ‘Organic Chemistry’ to generate knowledge on the synthesis of t-alcohols, in particular longer chain ones:

http://www.sciencemadness.org/talk/viewthread.php?tid=15171#...

May it too become sticky soon! :cool:

AWESOME results with tetralin

NurdRage - 29-12-2010 at 14:26

Once again i was back at it verifying "negative theory". I know it seems like a waste of time, but good science uses theory as guides for experimentation, not substitutes.

Tetralin wasn't supposed to work since it was aromatic, but since i had lots of it (for making lithium sand) i thought i'd try it anyway. I figured if it could stand up to molten lithium it should stand up to molten potassium.

Anyway, When I did it I didn't make accurate measurements since this was a qualitative experiment and just kinda eyeballed the reagents since i had done it a dozen times already. I did the usual dehydration step and popped a shot of alcohol. Then i heated the solution to vigorous reflux and walked away for 1 hour while i tended to other things.

Came back and saw:

I wasn't expecting this so i wasn't filming it, needless to ssay i was

Anyway, the reaction is FAST, in just one hour it went to completion and gas evolution stopped. It might even have taken less time since i wasn't monitoring it for an hour. Perhapes the crudely eyeballed amount of t-amyl alcohol used might be the bigger contributor to the rate so i'll try again under more quantitative conditions later.

Since tetralin is so dense (0.97g/mL) floating coalescence took place and i got all the potassium crammed into one ball like with my paraffin runs. Despite the vigorous reflux the ball still maintained cohesion.

So with tetralin we're getting the advantage of HD paraffin of floating coalescence without the disadvantage of longer reaction times.

I'm not certain about the yield yet, since i didn't accurately measure anything when i started. Like i said i'll do a quantitative run later.

Now i know not everyone can get their hands on tetralin but i think the bigger result here is that aromatics can work. So we don't need to go out of our minds to hunt down perfectly aliphatic solvents. Since aromatics tend to be denser we might have an easier time brewing up some floating coalescence capable solvent mixtures.

On a different note, this also highlights the good practice of verifying all assumptions and theories with hard data or citable sources using exact conditions. Occasionally we might uncover diamonds

I guess Woelen needs to update his page

bahamuth - 29-12-2010 at 15:30

I am interested if anyone tried this?

Quote: Originally posted by Jor

Now this would be a very interesting way to make lithium, but it's really a bummer that lithium has such a low molecular weight, making it impossible to make acceptable amounts of lithium.

Also, a little off topic though,may LiOH be collected/synthesized from worn out Li ion laptop batteries?
This may be a cheap source for Li compounds...

Sedit - 29-12-2010 at 16:44

Awsome results Nurdrage, am I correct in assuming that what I see there on the top that looks shinny is Potassium? If so then that looks very lovely.

I wanted to ask and sorry if it was already but has anyone considered the use of Cyclohexanol? I would think, even more so now that the alkoxide of it would be more soluble in paraffin oils and higher boiling point as a short chain alcohol. At 160 degree Celcius boiling point I would think its use would be well suited here.

NurdRage - 29-12-2010 at 17:11

yeah, the shiny metal "island" in the flask is actually the top of a single large ball of potassium.

As for your cyclohexanol question; since it's a secondary alcohol i think it wouldn't work. Isopropanol, another secondary alcohol, was already tried in this thread and that failed miserably.

Eclectic - 29-12-2010 at 17:58

1-ethylcyclohexanol from ethyl bromide and cyclohexanone might be interesting to try.

len2 - 29-12-2010 at 21:37

Very interesting result Nurdrage. I have not thought of tetralin - and we have none in any of the labs - when when I get a chance Ill try it. The reason the reaction goes slow in heavy paraffins is that the alkoxides are not very soluble in them - even the t-alcohol doesnt dissolve all that easily. The aromatic gets around that problem.

I find the structure on the surface of K obtained this way puzzling. My electrolytic potassium never showed it

NurdRage - 30-12-2010 at 04:14

There is no structure, actually the potassium is still molten and it's shiny surface is reflecting the back of my fume hood. So don't read into it too much

As for Tetralin, it's not easy to get so as a solvent i don't think it's viable for the amateur. The closest domestic thing i can think of is naphthalene, which is somewhat easier. Since it's solid, I think a mixture of naphthalene and paraffin might have all the properties we want of wide liquid range and high density.

Problem is the odor.

Nonetheless, with the realization that aromatics are viable again we can expand our solvent search.

condennnsa - 30-12-2010 at 04:22

Nice work, nurdrage. The naphthalene option is very interesting, do you think you could also run a test with paraffin-naphthalene mixture? naphthalene also has a remarkably high density of 1.14 .

I would try it, but still got no alcohol.

There might be problems with naphthalene which could deposit in solid layer on the cool part of the refluxer...

NurdRage - 30-12-2010 at 04:39

Yeah i'm worried about that too. I'm hoping careful choice of a lower boiling paraffin might wash down any solid naphthalene that gets up there.

I'm still looking into it. when the holidays are over i'll be able to acquire more materials to test.

woelen - 30-12-2010 at 05:09

@Nurdrage: I just am reading now and have no time for experimenting, as I am too much occupied by family matters the last week, but your result with the tetralin is great. That really is a breakthrough in this type of reaction. I'll modify my webpage and mention your results.

blogfast25 - 30-12-2010 at 07:38

Yes, great result on the tetralin! Someone might want to try that on the sodium!

[Edited on 30-12-2010 by blogfast25]

blogfast25 - 30-12-2010 at 07:42

Quote: Originally posted by bahamuth

I am interested if anyone tried this?

Quote: Originally posted by Jor

Now this would be a very interesting way to make lithium, but it's really a bummer that lithium has such a low molecular weight, making it impossible to make acceptable amounts of lithium.

Also, a little off topic though,may LiOH be collected/synthesized from worn out Li ion laptop batteries?
This may be a cheap source for Li compounds...

The patent doesn't mention it. Most people will be deterred, perhaps unnecessarily, by the high reduction potential of Li+ (about 3 V)... And Li alkanoates are likely to be even less soluble in alkanes/naphtenics. But it's worth a try. We're probably more interested in making the more commonly used sodium, though...

[Edited on 30-12-2010 by blogfast25]

blogfast25 - 30-12-2010 at 07:52

Quote: Originally posted by len2

The reason the reaction goes slow in heavy paraffins is that the alkoxides are not very soluble in them - even the t-alcohol doesnt dissolve all that easily. The aromatic gets around that problem.

len1: can you substantiate the claimed higher solubility of alkoxides in aromatics? We’re looking into improving that solubility for sodium alkoxides from the alcohol angle (see organics section) but the solvent may prove more important...

blogfast25 - 30-12-2010 at 08:52

4 l reagent grade Tetralin ® from Sigma-Aldrich: £115.8

http://www.sigmaaldrich.com/catalog/ProductDetail.do?lang=en...

Ding dong!

DJF90 - 30-12-2010 at 10:01

4 consecutive posts in under 80 minutes. No wonder this thread is 22 pages long...

NurdRage - 30-12-2010 at 17:03

Quote: Originally posted by DJF90

4 consecutive posts in under 80 minutes. No wonder this thread is 22 pages long...

I agree, the edit button as well as divider lines should be used more often.

Like i am about to do:
===========================================

@ everyone else

More negative results:

Making sodium does not work in tetralin using t-amyl alcohol as the catalyst. So much for that idea.

tnphysics - 30-12-2010 at 18:59

As for napthalene-you might get napthalides.

What about heavier alcohols?

As for making Na and Li, I would try making them by using K to displace from the chlorides. They are

Anyone tried Al lately?

mr.crow - 30-12-2010 at 19:31

Nurd, you are a chemistry hero. Of course having a real lab helps!

Tetralin is a trademark of Sigma Aldrich, you can find it other places as Tetrahydronaphthalene. One of those annoying peroxide formers >: ( Why is this? The benzylic carbon is extra reactive?

blogfast25 - 31-12-2010 at 07:32

Quote: Originally posted by DJF90

4 consecutive posts in under 80 minutes. No wonder this thread is 22 pages long...

I don't think making multiple comments makes much difference. Within reason people should be free to make as many comments as they see fit/need.

Re. the sodium, how long did you 'cook' for Nurdy?

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

Quote: Originally posted by tnphysics

What about heavier alcohols?

As for making Na and Li, I would try making them by using K to displace from the chlorides. They are

Anyone tried Al lately?

Read the tread. There's now a 'longer chain t-alcohols' thread opened for that purpose in the ORGANICS SECTION.

Displacing Li and K with Na doesn't work thermodynamically, as explained at length elsewhere. Search...

EDIT(by woelen): Merged double posts

[Edited on 1-1-11 by woelen]

Arthur Dent - 31-12-2010 at 11:48

This morning, as I was doing some shopping at the drugstore, I saw a bottle of lamp oil on a shelf that featured as its principal ingredient "99% Pure liquid Paraffin" stating further that it's odorless and smokeless. I can attest that it is totally odorless. I'll try to do a "dry run" just to see what the boiling point of that stuff is.

It's distributed by Lembex Import and the product is called "DecoFlam". Website has little technical data on the product and I doubt i'll be able to track down a MSDS, but this "oil" looks promising.

http://www.alibaba.com/product/lembex-11540374-10973924/Pure...

Robert

[Edited on 31-12-2010 by Arthur Dent]

[Edited on 31-12-2010 by Arthur Dent]

Eclectic - 31-12-2010 at 12:35

I got the Ritchie Compass filling fluid URLed in my last post above, and it is in fact Isopar L, and totally odorless.

http://www.westmarine.com/webapp/wcs/stores/servlet/Product_...

http://exxonmobilchemical.ides.com/en-US/ds128997/Isopar%E2%...

[Edited on 12-31-2010 by Eclectic]

smuv - 31-12-2010 at 13:35

The good thing about aromatic solvents, is that while they are non-polar, they have a great degree of polarizability. This allows aromatics to be better solvents for polar compounds than similar non-aromatic solvents. For example, methanol is miscible with benzene, however not with cyclohexane (though it is very soluble in cyclohexane).

Take home message: alkoxides are probably more soluble in aromatics than parafins (or non-aromatic olefins).

starch - 31-12-2010 at 18:03

wow guys this thread is awsome

Excellent discusion, Excellent experimentals, and Excellent results

not something im going try ne time soon but still im in awe :cool:

well done to all those with shiney balls of K

cant wait to read more

take care mad scientists

len1 - 31-12-2010 at 22:53

I meant of course structure of surface of solid potassium obtained this way, not molten potassium (??)

Smuv wrote very nicely about aromatics, I couldnt have done it better myself. Except to add that I expect solubility in long straight chain paraffins to be less than in shorter chains, because of very different molecular shapes

[Edited on 1-1-2011 by len1]

blogfast25 - 1-1-2011 at 09:01

len1: with regards to the aromatics, can we conclude that higher solubility caused (or at least contributed) to the better result with Tetralin? That would stillnot explain the poor result for Tetralin/Na of course...

Perhaps we should be looking at t-alcohols with aromatic functionality, 'smuv' is making some interesting noises along those lines in the t-alcohols thread ('organics')...

woelen - 1-1-2011 at 14:50

Merged double posts and tidied up a little.

tnphysics - 1-1-2011 at 17:08

Quote: Originally posted by DJF90

4 consecutive posts in under 80 minutes. No wonder this thread is 22 pages long...

Quote: Originally posted by tnphysics

What about heavier alcohols?

As for making Na and Li, I would try making them by using K to displace from the chlorides. They are

Anyone tried Al lately?

meant K + NaCl -> KCl + Na etc.

blogfast25 - 2-1-2011 at 06:52

Quote: Originally posted by tnphysics

meant K + NaCl -> KCl + Na etc.

Yep. I know. Still doesn't work...

[Edited on 2-1-2011 by blogfast25]

garage chemist - 4-1-2011 at 14:51

Today I finally have the chance to try this, as the supplier first sent me the wrong kind of Shellsol and then closed down over christmas and New Years. Now the Shellsol D70 arrived and I started a batch according to Pok's original procedure.

I used 3g of fine reagent grade magnesium powder (I don't have any filings), 6g of KOH flakes and 0,6g of t-Butanol.
40ml of the solvent, KOH and Mg were combined and heated over a free flame. Not even a minute after starting to heat (the solvent was barely warm), the mixture suddenly violently reacted, evolving a large amount of gas and white smoke!
There was visible "fire" under the solvent, as part of the KOH and magnesium apparently reacted in a thermite-like reaction.
Perhaps fine Mg powder wasn't the right thing to use... but I have no turnings, so I'll have to live with this for now.
I heated to a boil, and added the t-BuOH mixed with 10ml Shellsol over ca. 20 minutes.
It's currently refluxing. Tiny globules of K seem to form, but they're not coalescing as far as I can see. I'll report how it looks after 4 hours of reflux. There's still a large amount of unreacted Mg powder.

What's very annoying is that evaporated t-BuOH is solidifying on the cooling coil of the dimroth condenser. A large crystal crust is forming there.
I use very cold water in the condenser to keep the volatile t-BuOH from getting lost, but that's not how I meant it to be.
There isn't sufficient Shellsol evaporating to wash down the solid t-BuOH.
Did someone encounter this problem as well?

NurdRage - 4-1-2011 at 14:56

Yeah, this is why i eventually moved competely to t-amyl alcohol which has a lower melting point than t-butanol.

For the times i did t-butanol i just turned off the cooling water for a few minutes and let the t-butanol melt back down before restarting it again. It was crude but it worked for me.

I think some experimentors just blow air straight into the condesor instead of cold water.

garage chemist - 4-1-2011 at 15:19

I just turned off the water completely and let convective air cooling do the job. If too much alcohol gets lost, I'll add some more later.

The reaction products have formed a hard porous mass that encloses the K globules and prevents coalescence, so I interrupted the heating and broke it up with a spatula. I'll further look after it in an hour.
Manual agitation seems to be required at some point.

garage chemist - 4-1-2011 at 16:41

I added 0,2g extra t-BuOH and resumed the water cooling.
No more alcohol is solidifying on the condenser- a sign that most of it has reacted with the K.
The solvent has become very turbid and opaque. There is still no coalescence, and the reaction mass is hard and crunchy, just in smaller pieces.

ThatchemistKid - 4-1-2011 at 17:39

If someone is really interested in obtaining tetralin birch reduction of naphthalene yields tetralin and 1,2-dihydronaphthalene

But, this might be beside the point if access to Lithium is limited.

I do not have access to this article but the abstract on the first page seems to be relevant.

Chem. Eng. News, 1966, 44 (51), pp 70–72
DOI: 10.1021/cen-v044n051.p070
Publication Date: December 1966
Copyright © 1966 AMERICAN CHEMICAL SOCIETY

Sedit - 4-1-2011 at 19:11

Electroreduction of aromatics using magnesium electrodes in aprotic solvents containing alcoholic proton donors

DOI: 10.1016/S0013-4686(03)00259-7

Quote:

Regioselective electroreduction of the aromatics, including methoxybenzenes, by using LiClO4 as a supporting electrolyte in an aprotic solvent (THF) containing alcoholic proton donors such as t-BuOH was successfully achieved to afford the corresponding 1,4-cyclohexadienes regioselectively in high yield. The electrolysis can be performed under constant current conditions at ambient temperature. The effect of electrode materials was remarkable, that is, the use of Mg electrodes gave the best result. Moreover in the presence of t-BuOD instead of t-BuOH, the deutrated 1,4-cyclohexadienes were obtained in high deuterium incorporation at 1- and 4-positions. A direct electron transfer to the aromatics is unlikely, and it is reasonable that the solvated Li(0), which is generated by the cathodic reduction of LiClO4, is intermediately involved in the electron transfer with being assisted by the anodically generated Mg2+ as an electron transfer catalyst.

Author Keywords: Cathodic reduction; Aromatic compounds

If someone could kindly aquire this paper is would eliminate the issue with Lithium avalibility. Abstract link embedded in the title.

trilobite - 4-1-2011 at 19:40

Cool stuff!

Here is an idea regarding how to reduce the amount of magnesium needed (and the amount of Mg(II) byproducts formed, maybe even helping purification if the magnesium alcoholate is soluble enough in a purification solvent) by preforming the potassium tert-alcoholate in stoichiometric amount.

Water-immiscible alcohols form an azeotrope with water, so distilling the alcohol with KOH would displace the equilibrium of alcohol + KOH <==> K-alcoholate + H2O to the right. This is an established method of preparing sodium 1-butoxide. If the solubility of the alcohol in water is low enough, it might even be possible to perform the reaction with a Dean-Stark apparatus to return the alcohol back to the distilling flask while tapping off water. Tert-amyl alcohol is soluble in 8 parts of water, so it is not as favourable as it could be but adding toluene might have a positive effect, also in formation of a tertiary azeotrope. Alternatively one could use a higher-boiling tertiary alcohol.

Azeotropes of t-amyl alcohol (bp 102,25C)
t-AmOH + 27,5 weight-% water bp. 87,35C
t-AmOH + 44 weight-% toluene bp. 100,5C
t-AmOH + water + toluene bp. 82C (no composition data)

After distilling the reaction water, one would add the high-boiling hydrocarbon and distill off all the lower boiling components. Some of the alcohol may stay retained in the potassium tert-alcoholate as alcohol of crystallization, but I suppose it would distill when heated enough. At least sodium t-butoxide may be freed of the alcohol by distillation with xylene. The residue would then be ready for reduction with Mg(0). Recovery of the alcohol would depend on separation of potassium metal from the Mg alcoholate, which could be decomposed with water to give Mg(OH)2 and alcohol.

Failure (Negative Result Report)

MagicJigPipe - 4-1-2011 at 20:49

1.58 g of Mg "shavings" from a firestarter and 3.06 g of reagent (though quite old) KOH hemispheres were added to a 125 mL flask containing approx. 50 mL of Kroger brand *mineral oil "laxative". The flask had a rubber stopper with a long glass tube in it. At the top of the tube was a balloon with a few small "needle-holes" poked in it. A wet paper towel was wrapped around the tube near the top. The flask sat in some sand in a 600 mL beaker. The stopper was placed on and the sand was heated (I can't find my damn glass thermometer so I couldn't measure the temp. directly) to a temperature of 200*C over a period of about 30 mins. Then, a hypodermic needle containing 0.6 g of mineral oil and 0.6 g of Reagent t-butanol (the alcohol did slowly dissolve with just a bit of stirring) was thrust through the stopper and injected into the solution over about 25 mins. I noticed "instant" boiling of the alcohol/oil soln. as soon as it hit the main mixute. Once the temp. got up to around 250*C there was sporadic boiling. S

tarting at about 200*C (I'm not quite sure on this exactly and will have to check my notes at home) there was quite a bit of gas evolution and it stayed steady for several hours. The t-butanol did seem to increase the amount of bubbles but it could've just been boiling. The reaction did not seem to proceed at all and after about 3 hours a few black specks were noticed floating around. Then the Mg started to turn black and eventually I started to notice a white "precipitate" on the bottom of the flask. It seems as if the reaction did not occur and the KOH just settled at the bottom and the Mg was oxidized (by H2O/air?) I desperately added quite a bit more t-butanol (unmeasured) and just got refluxing t-butanol in the rxn mixture. I finally stopped heating after 6 hours.

My guess is that there was not enough mixing due to the sporadic boiling. Also, the KOH could've been poorly stored and could thus be mostly the carbonate. It could also be the Mg but I doubt it. I simply shaved off Mg from the block with the tool that was included (and damn that took a long time and a lot of elbow grease; the battery to my drill was dead).

And before anyone says anything: Yes, I know there are different and better ways to set this up.

*The mineral oil said it contained vitamin E as a stabilizer.

The setup:

About an hour after adding t-butanol and 1.5 hours into the rxn:

The results:

Thanks for your time!

benzylchloride1 - 4-1-2011 at 22:33

Naphthalene will not work in this reaction since it reacts very easily with potassium to form the naphthalinide. I have prepared potassium naphthalinide several times for use in forming exotic organometallic anions. Upon mixing potassium and naphthalene in a glove box and shaking for several minutes a greenish color develops. After adding tetrahydrofuran, the mixture turns to a very dark green color and the potassium soon dissolves, I would assume that no potassium would be obtained if naphthalene was used as a solvent. I am very interested in this work since I would like to have some potassium of my own for syntheses when sodium metal cannot be used. I plan on using my glove box for the isolation procedure though for added safety.

[Edited on 5-1-2011 by benzylchloride1]

smuv - 5-1-2011 at 00:54

MJP, Looking at your setup, you very well may have lost your t-butanol to the atmosphere.

About the t-butanol crystallization problem, how about adding a touch of hexanes?

[Edited on 1-5-2011 by smuv]

woelen - 5-1-2011 at 03:55

I did not have the problem of t-butanol crystallizing. I blew air through the Liebig cooler at a speed of 2 liters per minute, using a small aquarium pump. This provides sufficient cooling not to loose any t-BuOH, while at the same time this does not solidify. My column was warm when touched, but not hot.

Sedit - 5-1-2011 at 05:41

MJP while I don't feel that stirring is to good of an idea in this reaction I do believe that during the addition of the alcohol it would be best for one to stir to react the alcohol as rapidly as possible with the KOH. The of the reaction goes thru a different mechanism that im not sure if stirring would be best.

garage chemist - 5-1-2011 at 07:05

Here's the rest of the report on the first batch:

Last night, I was too tired to look after the reaction after the 4 hours of reflux were over, so I let it run overnight, for a total of 12 hours reflux after addition of the t-BuOH.
The next day, after cooling down, it was still turbid and there was a large amount of brownish powder and crumbly substance under the solvent, with no visible potassium any more.
I emptied the flask into a beaker and sorted through the crud with a spatula and tweezers.
There was a large irregular dirty lump of what first appeared to be byproducts, but upon cutting turned out to be massive potassium metal! The whole lump was softer than sodium and could easily be deformed between two fingers.
It weighs 2,1g and is shown on the first picture, under clean fresh Shellsol. Anyone familiar with potassium will recognize the bluish appearance of the recently cut surface.
I could have cleaned its surface with 1-butanol, but since it rapidly regrows the oxide layer upon storage, this was omitted.
There was another smaller lump of potassium, weighing 0,4g, and several more 3mm pieces that were dropped onto ice outside, making a nice show of flames and sparks.

The K pieces were recovered from the byproducts by manual sorting based on appearance (the K pieces were dark colored on the outside, contrary to the white/grey Mg oxide byproduct). On the second picture, the reaction mixture minus the first lump of K is shown.

So I conclude that Pok's method has worked for me and made 2,5g of useable potassium from 6g KOH and 3g Mg powder in 50ml of Shellsol D70.
Deviations from the procedure were that I added 0,2g more t-BuOH after losing some to evaporation, and boiling for 12h instead of 4h.
The most important observed difference was that I did not get any shiny potassium balls, but large lumps with oxidised surface.

blogfast25 - 5-1-2011 at 07:47

garage chemist:

2.7 g potassium is a yield of 71 % (based on 10 % water in the KOH). And one has to account also for the K that’s bound up as K alkoxide (as the Mg has reacted completely away). Not bad at all!

I think you and magic may also have been using a little too much head space over the reacting mix: that with intensive cooling would promote t-butanol to crystallise out in the upper part of the reactor.

What someone really importantly should do is recover all post reaction solvent: as it should contain all the catalyst and is bone dry, it should theoretically be directly reusable, no fresh catalyst needed!

And another important test would be to, say, double the amount of t-butanol/t-amyl to check effect on reaction time...

[Edited on 5-1-2011 by blogfast25]

blogfast25 - 5-1-2011 at 07:58

Quote: Originally posted by Sedit

MJP while I don't feel that stirring is to good of an idea in this reaction I do believe that during the addition of the alcohol it would be best for one to stir to react the alcohol as rapidly as possible with the KOH. The of the reaction goes thru a different mechanism that im not sure if stirring would be best.

Sedit, if you've ever seen with your own eyes what happens when you add the aclcohol to the hot solvent, you wouldn't be writing that. The moment the alocohol/solvent pre-mix starts hitting the hot solvent, very intensive refluxing starts and this continues mostly during addition of the alcohol. Vigorous boiling of the solvent/alcohol mixture means that stirring is a complete waste of time.

I'm contemplating using much longer chain alcohols like 2-methyl-2-octanol, they have much higher boiling points and there some gentle swirling may be advisable. But with Shellsol D70 and gentle boil is achieved eventually, making stirring unnecessary.

[Edited on 5-1-2011 by blogfast25]

garage chemist - 5-1-2011 at 08:20

With the next batch, I will try to use selfmade Mg turnings from scrap Mg using a drill press.
I will also make myself some 2-methyl-2-hexanol via n-butylmagnesium bromide and acetone. Like Nicodem, I think that this alcohol may be one of the best to use- relatively easy to make, suitably high boiling point (141°C) and not too long-chained.
The longer the carbon chain of the alcohol, the higher is its molar mass, and the more you will need of it to achieve a given molar concentration in the solvent. With the "margarine tert-alcohols" proposed in the tert-alcohol thread, it may become unattractive since you will probably need quite a large amount of your precious grignard-made alcohols.

Sedit - 5-1-2011 at 09:10

Sedit, if you've ever seen with your own eyes what happens when you add the aclcohol to the hot solvent, you wouldn't be writing that. The moment the alocohol/solvent pre-mix starts hitting the hot solvent, very intensive refluxing starts and this continues mostly during addition of the alcohol. Vigorous boiling of the slovent/alcohol mixture means that stirring is a complete waste of time.

I'm contemplating using much longer chain alcohols like 2-methyl-2-octanol, they have much higher boiling points and there some gentle swirling may be advisable. But with Shellsol D70 and gentle boil is achieved eventually, making stirring unnecessary.

[Edited on 5-1-2011 by blogfast25]

Really, if I ever seen it with my own eyes I wouldn't be writting this? Have you hacked my brain and seen what my eyes have seen?

Nope not a chance else you would know my reason for writting it was because the reflux is mostly superficial and the surface rapidly boils with no complete mixing of the reactants. You have a bunch of settled solids with a heavy boiling at the top. Its akin to what causes bumping only in this case the lower boiling substance is on the top and not the bottom.

Would it be so damn hard to just stir the reaction on the addition of alcohol since it would ensure more alcohol reacts with the KOH in a decent time frame?

blogfast25 - 5-1-2011 at 09:35

Quote: Originally posted by Sedit

Ok, calm down already.

Firstly it’s understood that len1’s experiment of years back failed, probably due to too fast stirring. pok’s experiment involved occasional swirling of the flask and that is what most here have been doing successfully, including myself.

Secondly, refluxing isn’t ‘superficial’. Assuming you have total reflux and not a huge amount of head space, most of the alcohol is in the liquid phase, in homogeneous solution. If that wasn’t the case you wouldn’t get reaction at all. The ENTIRE solution boils, not just the surface, as can very clearly be seen during the reaction and as logic dictates. On top of that you’ve got hydrogen generation and heat generation (quite a bit too: 2 KOH + 2 Mg === > 2 K + 2 MgO + H2 is highly exothermic mainly due to the formation of the high lattice energy MgO). And the alcohol is added as a premix of alcohol and solvent. Combine all this and you have a heterogeneous solid-liquid system but with a highly homogeneous liquid phase.

[Edited on 5-1-2011 by blogfast25]

blogfast25 - 5-1-2011 at 09:46

With the next batch, I will try to use selfmade Mg turnings from scrap Mg using a drill press.
I will also make myself some 2-methyl-2-hexanol via n-butylmagnesium bromide and acetone. Like Nicodem, I think that this alcohol may be one of the best to use- relatively easy to make, suitably high boiling point (141°C) and not too long-chained.
The longer the carbon chain of the alcohol, the higher is its molar mass, and the more you will need of it to achieve a given molar concentration in the solvent. With the "margarine tert-alcohols" proposed in the tert-alcohol thread, it may become unattractive since you will probably need quite a large amount of your precious grignard-made alcohols.

Personally I’m hesitating between 2-methyl-2-hexanol and 2-methyl-2-octanol, the former being cheaper to synth., the latter being even higher boiling (>180C) and the alkoxide possibly more soluble in C12-C15 based solvents (for both K, Na). The higher boiling point of 141C will still lead to considerable refluxing unless you add it at low temperature. But refluxing isn't a problem anyway, it's part of the solution (no pun intended). 2-methyl-2-hexanol is only 2 C longer than t-amyl alcohol, would that make a decisive difference? n-hexyl bromide, BTW, is surprisingly reasonably priced by Sigma-Aldrich...

As regards ‘The longer the carbon chain of the alcohol, the higher is its molar mass, and the more you will need of it to achieve a given molar concentration in the solvent’, that’s been dealt with in the organics thread. I don’t think it’s an attractive proposition to make longer chain 2-methyl-2-alkanols if it turns out they’re not considerably more active (than t-butanol or 2-methyl-2-butanol), thus requiring lower molarities for same or better reactivity. In particular higher solubility of the K alkoxide should speed up the first two steps of the (proposed) reaction mechanism. Personally I'm convinced it's possible to bring the whole procedure down to under an hour, as Nurdrage with his Tetralin/t-amyl alcohol qualitative experiment already showed.

But the main purpose of longer chain t-alkanols must remain IMHO the higher solubility of the corresponding sodium alkoxides, opening up the way to Na production.

[Edited on 5-1-2011 by blogfast25]

garage chemist - 5-1-2011 at 13:30

I cleaned up my potassium pieces by melting them in dioxane.
This works marvellously, much better than my old method with IPA in kerosene.
In hot dioxane, the molten potassium floats and the impurities either sink to the bottom (MgO and Mg) or dissolve (KOH), leaving absolutely clean shiny potassium spheres.
However, on weighing the cleaned potassium again, I found that the large 2,1g piece now only weighs 1,5g. There is a lot of brown crud in the dioxane.
So my yield is not as high as initially assumed, and the potassium directly from the Shellsol D70 method contains a lot of adhering or suspended impurities and should be cleaned up with hot dioxane.

blogfast25 - 5-1-2011 at 13:55

0.2 g t-butanol accounts for about 0.1 g of K (bound as alkoxide), so that doesn’t explain the lower yield.

But I think if you repeat the experiment, this time avoiding some of the butanol freezing out near the refluxer, you’ll find your results to be close to all of ours.

And since as you have dioxane you could try and help coalescence by replacing the Shellsol (after all hydrogen has stopped evolving and cooling) with dioxane and remelting the fine K…

rrkss - 5-1-2011 at 15:19

I cleaned up my potassium pieces by melting them in dioxane.
This works marvellously, much better than my old method with IPA in kerosene.
In hot dioxane, the molten potassium floats and the impurities either sink to the bottom (MgO and Mg) or dissolve (KOH), leaving absolutely clean shiny potassium spheres.
However, on weighing the cleaned potassium again, I found that the large 2,1g piece now only weighs 1,5g. There is a lot of brown crud in the dioxane.
So my yield is not as high as initially assumed, and the potassium directly from the Shellsol D70 method contains a lot of adhering or suspended impurities and should be cleaned up with hot dioxane.

If you have some pictures I would like to see. Did you prepare the dioxane yourself from condensation of ethylene glycol or is it purchased?

garage chemist - 5-1-2011 at 16:14

The dioxane was purchased. This is a quite expensive and difficult to obtain chemical, but it really pays off to get some if you're trying to make clean potassium.
The K looked even better than the large ball of liquid potassium in tetralin that NurdRage obtained. It was like floating mercury.
I didn't make any pictures, but I will make some the next time I purify a batch of potassium.

Dioxane can be prepared at home from ethylene glycol and sulfuric acid. A friend did this a few years ago and still remembers vividly how much work it was to get a pure product from the crude distillate. All kinds of impurities, aldehydes, unsaturated products, and so on are formed as byproducts in the synthesis.
Dioxane is also very difficult to obtain anhydrous. If you don't have any sodium or potassium to reflux it with, you'll never make an anhydrous product.
With the commercial product, specified at max. 0,1% water, there was clearly visible hydrogen evolution at the molten K- but this did not impair the process, since KOH seems to be soluble in hot dioxane.

blogfast25 - 6-1-2011 at 02:41

Because some of us here will now be looking at other catalysts, I’ve rewritten the proposed reaction mechanism presented on woelen’s page in a format that represents what we believe happens in a clearer manner (I think). Just four reactions are involved, arranged here into two SETS, where ‘(sol)’ stands for ‘dissolved in inert, aprotic, apolar solvent’ and ROH is either t-butanol or 2-methyl 2-butanol.

SET 1:

1.1. Alkoxide formation:

2 KOH(s) + 2 ROH(sol) < == > 2 KOR(sol) + 2 H2O(sol)

1.2. (= 2.2.) Redox reaction:

2 KOR(sol) + Mg(s) < == > 2 K(l) + Mg(OR)2(sol)

1.3. (= 2.3.) Hydrolysis:

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

Sub Overall 1:

2 KOH(s) + Mg(s) < == > 2 K(l) + MgO(s) + H2O(sol)

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

SET 2:

2.1. Hydrogen generation:

K(l) + 2 ROH(sol) < == > 2 KOR(sol) + H2(g)

2.2. (= 1.2.) Redox reaction:

2 KOR(sol) + Mg(s) < == > 2 K(l) + Mg(OR)2(sol)

2.3. (= 1.3.) Hydrolysis:

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

Sub Overall 2:

Mg(s) + H2O(sol) < == > MgO(s) + H2(g)

Overall = Sub Overall 1 + Sub Overall 2:

2 KOH(s) + 2 Mg(s) < == > 2 K(l) + 2 MgO(s) + H2(g)

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

This scheme relies on the same reactions woelen’s does but eliminates one step I’ve always found highly contentious (I quote from woelen's scheme):

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

I just don’t see this happening. My scheme doesn’t have to call on it while still respecting catalyst conservation and overall stoichiometry.

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

Some might still wonder how come the potassium isn’t attacked by water and I believe it can be explained as follows. Firstly we must realise that we’re in an aprotic, apolar medium. For the potassium – water reaction we normally have:

2 H2O(l) < == > H3O+(aq) + OH-(aq) and K(s) + H3O+(aq) == > K+(aq) + ½ H2(g) + H2O(l), the latter which proceeds very quickly due to K’s small first ionisation energy, K+ solvation enthalpy and escape of the hydrogen. This of course then drives the dissociation of water to the right. But in an apolar, aprotic solvent, water isn’t dissociated.

Secondly we have to assume that Mg(OR)2(sol) + H2O(sol) < == > MgO(s) + 2 ROH(sol) does proceed very quickly with the very high lattice energy of the MgO being the ΔG driver here. Kinetically the reaction somewhat resembles hydrolysis of a Grignard Reagent: although the water isn’t dissociated, water molecules still have a partial positive charge δ+ on their hydrogen ends and a partial negative charge δ- on the oxygen end. Similarly the Mg(OR)2 moiety may not be dissociated but contains a partial positive charge δ+ on the Mg part and partial negative charges δ- on the oxygen ends of the alkoxide part. This makes rearranging H2O + Mg(OR)2 into MgO + 2 ROH very likely and very fast, much like the hydrolysis of Grignard Reagents.

In any case we don’t have much choice but to assume that reaction water from 1.1. regenerates the catalyst, by reacting with Mg(OR)2 rather than reacting with K, as otherwise no K could be formed AT ALL.

I believe Mg(OR)2(sol) + H2O(sol) < == > MgO(s) + 2 ROH(sol) are the ‘four cylinders’ which drive much of the whole edifice, at least thermodynamically speaking.

I also see no good reason why the preliminary drying step has to be carried out without the catalyst present. As long as a small excess of Mg is accounted for as H2O + Mg === > MgO + H2 the catalyst doesn’t seem to interfere or be interfered with.

[Edited on 6-1-2011 by blogfast25]

Magpie - 6-1-2011 at 10:02

Very nice, blogfast. I was really starting to wonder what was driving this remarkable synthesis. When you first look at it, it doesn't seem possible that a less active metal (Mg) will act to reduce a more active metal (K+). Then I assumed that the generation of H2 and the solid MgO were drivers and that the kinetics were favorable enough even at the relatively low temperature.

I was going to calculate the delta G for the overall reaction when I saw your post. Have you done this yet?

blogfast25 - 6-1-2011 at 10:46

Quote: Originally posted by Magpie

Very nice, blogfast. I was really starting to wonder what was driving this remarkable synthesis. When you first look at it, it doesn't seem possible that a less active metal (Mg) will act to reduce a more active metal (K+). Then I assumed that the generation of H2 and the solid MgO were drivers and that the kinetics were favorable enough even at the relatively low temperature.

I was going to calculate the delta G for the overall reaction when I saw your post. Have you done this yet?

Thanks Magpie… Kinetic barriers are lowered by the catalyst, that's its definition, basically...

ΔH @ 298 K = + 428 kJ/mol – 601 kJ/mol = - 173 kJ/mol K, quite respectable…

[Edited on 6-1-2011 by blogfast25]

NurdRage - 6-1-2011 at 12:28

I also see no good reason why the preliminary drying step has to be carried out without the catalyst present. As long as a small excess of Mg is accounted for as H2O + Mg === > MgO + H2 the catalyst doesn’t seem to interfere or be interfered with.

Because the temperature at which drying takes place is well above the boiling point of the alcohols.

The alkoxides don't form until the system is dry so by the time it reaches that temperature most of the alcohol will be in the vapor phase. When drying starts the fast hydrogen production pushes the alcohol vapor out of the condensor and is gone.

The condensor won't recapture all of it since the vapor pressure is obviously less than 100% (since it's being carried by the hydrogen).

Only when the alkoxide is produced will it become non-volatile.

If we can start working with alcohols that have boiling points higher than the solvent then we might be able to do everything in one pot.

Alternatively i suppose you could simply load up with LOTS of alcohol and hope enough is retained to form the non-volatile alkoxide

But that's just my theory, experimentally i find performance better when i add the alcohol later. but i haven't performed a rigorous test of the theory yet.

Just like i tested aromatics and it worked, if you want to go ahead and test your alcohol theory i'd love to hear the results. You might find out that my "vapor loss" theory is a negligible contributor.

Experiment > Theory

blogfast25 - 6-1-2011 at 13:14

No, that doesn’t convince me. Add the volatile alcohol at the start. Now temperature is raised and the alcohol/solvent mixture starts boiling. The vapour phase is of course much richer in alcohol than the liquid phase but in terms of moles there’s far more liquid phase than vapour phase. Now the chemical drying process starts and independently alcohol in the liquid phase starts reacting with the KOH, drawing more alcohol into the solvent as alcoholate.

In the case of separate, post drying addition of the alcohol, introduced at higher temperature, some/much of it stays in the vapour phase, also to be drawn into the solvent by KOH converting it to alcoholate.

I can’t see it make one iota of difference, provided reflux is total (apart from the permanently volatile hydrogen, of course) and the drying reaction and alcoholate formation do not interfere with each other… The equilibria, reactions and liquid/vapour, are what they are.

Personally I think the separate addition of alcohol is a bit of a ‘chemist’s superstition’. All to play for, I guess…

The other theory I have is that assuming reflux is total and the alcohol doesn’t chemically degrade (or separate out as solid alcoholate), then separating off the solvent from the K and MgO should make it directly reusable with a fresh batch of KOH/Mg, no extra catalyst needed. Conservation of mass! That some or all of the alcohol is present as alkoxide shouldn’t make a blinding bit of difference. You could even see it as the ultimate empirical test of the proposed mechanism… This would start to really play when we're using 'super catalysts' of higher cost.

[Edited on 6-1-2011 by blogfast25]

NurdRage - 6-1-2011 at 13:59

I'm pretty sure i included the statement where i acknowledged i didn't rigorously test where alcohol additions were best. I also believed I stated that this should be tested experimentally.

blogfast25 - 6-1-2011 at 14:03

I'm pretty sure i included the statement where i acknowledged i didn't rigorously test where alcohol additions were best. I also believed I stated that this should be tested experimentally.

I wasn't criticising Nurd

...

What do you think of my second theory?

NurdRage - 6-1-2011 at 14:15

As Homer Simpson would say: "Less yapping more zapping!"

A thorough experiment to test if alcohols can be used in a "one-pot" reaction vs. "post drying" reaction would be a good idea. I personally would like to be wrong on my vapor loss theory and things would be so much easier if we could do this one-pot.

As for the alkoxide, an issue is that alkoxide might solidify into the MgO crust/sand when the solvent cools to room temperature. Since working with 200 degree solvent is rather dangerous i think the alcohol experiment should be done first.

A more controlled experiment is to add pure alkoxide to a fresh reaction directly.

I think we've had enough theory, lets go back to lab and gets some data now that we know what to test for.

blogfast25 - 6-1-2011 at 14:25

As Homer Simpson would say: "Less yapping more zapping!"

Fair enough. And I say: 'It doesn't hurt to theorise!' :cool:

MagicJigPipe - 6-1-2011 at 16:31

I forgot to mention that I did swirl occasionally throughout the "reaction". About every 5 minutes or so. Also, there was no reaction upon addition of water to the flask after rxn. (and I didn't expect there to be because the Mg didn't even seem to react at all). However, the white 'precipitate' at the bottom was extremely insoluble in water. Mg(OH)2?

Since I don't think anyone has had this result before, could anyone offer any insight? I definitely saw alcohol refluxing throughout the solvent after I added a bit more later on in the reaction so something tells me it's probably not that...

Ah screw it. I'll just suck it up and buy some Shellsol. But I would really like to find a solvent and Mg source that people could very easily obtain and would still work (like what I tried). I suppose I could try a few more times and eliminate some unknowns like the age of the KOH (and whether or not it is mostly carbonate). I guess a titration could be in order.

I don't have internet at the moment so it might take me a few days to respond. Good luck to everyone.

garage chemist - 6-1-2011 at 16:52

I made some turnings today from a magnesium printing/stamping plate that I got from ebay a long time ago, using a drill press with a 4mm drill.
The turnings that resulted were 1-3mm in particle size, mostly 3mm. They burn fine when lit.
I realize that printing plates aren't pure magnesium. One alloy that I found via google is AZ31. Ullmann lists this as being 3% Al, 1% Zn and 0,3% Mn. Other Mg based alloys contain 6 to 9% Al.
Just a few grams of turnings were made, enough for at least one experiment, starting tomorrow.
In order to stay true to the principle of changing one variable at a time, I will stay with the t-BuOH for my experiments with different sources and forms of magnesium.

More Negative results

NurdRage - 6-1-2011 at 18:16

Quick update:

Since aromatics are back on the table (a limited number of them anyway) for solvents. I thought i'd also check out some simple aromatic alcohols:

Phenol - failure

Butylated hydroxytoluene, also known as "BHT" - Failure

So i guess phenols (at least these ones) aren't very promising.

I should also update that while cleaning up i smashed my only ground glass flask. So it looks like i'll be outta the Potassium Race until further notice.

Take care.

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

Since i'm outta the race until i get another flask (this could be awhile), i thought I'd summarize all my findings on this topic:

Notable contributions:

"Floating coalescence" - when high density solvent is used (>0.86g/mL) the resulting molten potassium floats as it forms. The gentle agitation of the hydrogen bubbling knocks the potassium together and causes coalescence into very large balls of potassium. This negates the need for a second coalescence procedure.

Aromatic solvents can work - Successful results with 1,2,3,4-tetrahydronaphthalene ("tetralin") disprove the conjecture that only aliphatic solvents may be used.

Magnesium quality is an issue - Probably obvious but i thought gathering some direct data is better than just going with anecdotal. Magnesium treated with water and heating to produce a thick layer of passivated magnesium oxide did not work. This same type of magnesium works perfectly if untreated. So surface quality is a big factor in this reaction.

Solvents I've personally tested:

Candle Wax - Successful - but produces K-sand that solidifies into the wax, the resulting wax is extremely dangerous in my opinion since it isn't deactivated by alcohol and can spontaneously catch fire during cleaning or handling. Only extremely careful heating with alcohol will properly destroy it. I do not recommend using candle wax if alternatives are available.

Paraffin Oil - IR Grade - Successful - high density (0.86g/mL) IR grade paraffin produces high quality potassium with an approximate run time of 4hr. "Floating coalescence" enabled.

1,2,3,4-tetrahydronaphthalene - "Tetralin" - VERY successful - FAST run time of ~1hr when used with t-amyl alcohol. "Floating coalescence" enabled.

Toluene - Failure - I was hoping to try a low-temperature approach.

PEG-3350 - Failure - catches fire! do not repeat.

catalysts i've tried:
(i'm calling these catalysts in case we find a non-alcohol that works)

t-butanol - confirmed successful - Has a tendency to freeze into my condenser since I use very cold water.

t-amyl alcohol - Successful - wide liquid range and slightly higher boiling point (102C) make this a better alcohol from a more practical perspective.

PEG-3350 - Failure

Phenol - Failure

butylated hydroxytoluene - failure

I think those are all my noteworthy findings.

Hopefully i'll get back in the race eventually.

But it seems i won't be making a video for my youtube channel on this until i replace my broken apparatus.

Who else is still in the race and what aspects are you pursuing?

[Edited on 7-1-2011 by NurdRage]

smuv - 6-1-2011 at 19:21

NurdRage I am glad you tested both of those, but you have really left me scratching my head. No way you can do a birch reduction on a phenolate! So the only conclusion is that the phenolates have poor solubility in the solvent system? Or something about the difference in solubility between the Mg and K salts is preventing the reaction from going to the production of K.

NurdRage - 6-1-2011 at 19:44

Quote: Originally posted by smuv

NurdRage I am glad you tested both of those, but you have really left me scratching my head. No way you can do a birch reduction on a phenolate! So the only conclusion is that the phenolates have poor solubility in the solvent system? Or something about the difference in solubility between the Mg and K salts is preventing the reaction from going to the production of K.

good questions.

The mixture rapidly becomes too cloudy to see whats happening so i don't really know what's up. Just that in both cases hydrogen evolution stops (as if drying is all it does and goes no further). I let the mixtures cool and then scoop out some of the solid crud and drop it into water to test for potassium flashes - there are none.

Whatever the reason, they don't work.

garage chemist - 7-1-2011 at 07:54

I started the batch with the selfmade scrap Mg turnings.
On initial heatup, there was some mild hydrogen evolution, the KOH seemed to melt and solidify again after it had given off its water.
After adding the t-BuOH, there was no further reaction. It's been boiling for one hour now and the solvent is still clear, with no sign of potassium production. This is a striking difference to the first successful run with pure Mg powder where the solvent soon became very turbid.
t-BuOH is again solidifying in the condenser, a Liebig this time.
The batch is looking exactly like MagicJigPipe's failed attempt on page 23 of this thread.

If this batch fails, then we have one more reason to assume that the purity of the magnesium is important, since my turnings were absolutely clean and oxide-free.

[Edited on 7-1-2011 by garage chemist]

blogfast25 - 7-1-2011 at 08:24

If this batch fails, then we have one more reason to assume that the purity of the magnesium is important, since my turnings were absolutely clean and oxide-free.

[Edited on 7-1-2011 by garage chemist]

Strange indeed!

Perhaps you could try a qualitative test on both grades of magnesium by subjecting equal amounts of them to quite dilute HCl and see which of the two reacts the most vigorously? Test tube test…

It’s well known that the reactivity of a metal can be altered significantly by small amounts of adjuvant metals to form corrosion resistant alloys. Perhaps your printing Mg alloy has been formulated to be corrosion resistant? Considering the application field that’s perhaps not such a far fetched guess…

blogfast25 - 7-1-2011 at 08:38

@Nurdrage:

Your contribution here has been invaluable but I do have question marks over the Tetralin result. No, not the result itself but rather what caused the extremely short reaction time: was it the solvent or did you measure out (‘eyeballed’ was the term you used) the alcohol in such a way that alcohol concentration was higher than usual? If the proposed reaction mechanism is correct then lower reaction times (but also lower yields) should result from increased catalyst concentration.

I’m on a knife’s edge to go and buy some Tetralin from Sigma but it would prove rather a waste of money if it turned out the alcohol concentration was mainly responsible for your fantastic result with Tetralin…

NurdRage - 7-1-2011 at 08:46

@Nurdrage:

Your contribution here has been invaluable but I do have question marks over the Tetralin result. No, not the result itself but rather what caused the extremely short reaction time: was it the solvent or did you measure out (‘eyeballed’ was the term you used) the alcohol in such a way that alcohol concentration was higher than usual? If the proposed reaction mechanism is correct then lower reaction times (but also lower yields) should result from increased catalyst concentration.

I’m on a knife’s edge to go and buy some Tetralin from Sigma but it would prove rather a waste of money if it turned out the alcohol concentration was mainly responsible for your fantastic result with Tetralin…

(can you explain to me, even in u2u, why you add three periods at the end of your posts? are you planning to edit the post later? is your keyboard broken? is it an unfinished thought that you wish to follow up on in a later edit?)

Since i can no longer perform the experiments i can't test how alcohol loading effects the rate. but you can, just start your reaction in whatever solvent you're currently using and add in more alcohol. If indeed the rate is wholly dependent on the alcohol then it shouldn't matter what solvent you use. No need to go out and buy expensive tetralin to test my particular conditions.

Jor - 7-1-2011 at 08:51

Maybe one of the reactants or intermediates are more soluble in tetralin?
Or it's density... It has a high density than molten potassium, meaning any formed K will go to the surface, meaning no K will coat the Mg preventing further reaction?

blogfast25 - 7-1-2011 at 08:57

Quote: Originally posted by MagicJigPipe

Since I don't think anyone has had this result before, could anyone offer any insight? I definitely saw alcohol refluxing throughout the solvent after I added a bit more later on in the reaction so something tells me it's probably not that...

Ah screw it. I'll just suck it up and buy some Shellsol. But I would really like to find a solvent and Mg source that people could very easily obtain and would still work (like what I tried). I suppose I could try a few more times and eliminate some unknowns like the age of the KOH (and whether or not it is mostly carbonate). I guess a titration could be in order.

There’s quite a choice of solvents now, thanks to Nurdrage’s work. I’m convinced a clean grade of lamp oil (without adjuvants like ‘citronella’ or vitamins!) should work with a ‘good’ KOH and clean Mg (try a pyro grade, not extremely fine – perhaps a 100 mesh).

On the KOH, most of us have been using flakes, not moulded hemispheres (pellets). Although I’ve not tested both back-to-back, the flakes may have the advantage of being somewhat porous and hence of greater surface area (m2/g). High K2CO3 content would be detrimental because it gives the catalyst nothing to react with.

blogfast25 - 7-1-2011 at 09:10

Nurdrage, you seem a little hung up over my three periods! ;-) It’s a valid punctuation mark, you know, implying ‘more to follow’, a ‘pregnant pause’, ‘food for thought’ or even a bit of a cliff hanger. Don’t stare yourself blind on them… we need your observational skills! :cool:

I hope that explains it.

Your other point is completely accepted. I should be able to make another run on Sunday, this one a one-pot with 50 % increased catalyst (2-methyl-2-butanol) concentration.

I'm also ordering a sample of the tetrahydro myrcenol longer chain alcohol, discussed in the organics thread.

[Edited on 7-1-2011 by blogfast25]

woelen - 7-1-2011 at 11:33

On the KOH, most of us have been using flakes, not moulded hemispheres (pellets). Although I’ve not tested both back-to-back, the flakes may have the advantage of being somewhat porous and hence of greater surface area (m2/g). High K2CO3 content would be detrimental because it gives the catalyst nothing to react with.

I can say that my successful attempt was done with hemispheres of KOH, which are more than 20 years old and which are covered with some K2CO3. So, apparently the KOH is not that critical.

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

An off-topic request: Please try to combine your posts into a single one. I have some complaints of people about the double posting. This is a long thread, let's try to make things as compact as possible (of course without sacrificing any useful content, and yes, you post useful things)

[Edited on 7-1-11 by woelen]

not_important - 7-1-2011 at 12:12

At elevated temperature, >150 C I believe, you will see

K2CO3 + ROH <=> KHCO3 + ROK
KHCO3 <=> H2O (g) + KOH
and to a small degree
K2CO3 + ROH <=> KOH + RO(C=O)OK
RO(C=O)OK + ROH <=> KOH + RO(C=O)OR
RO(C=O)OK + ROH <=> RO(C=O)OH + ROK
RO(C=O)OH => ROH + CO2(g)

It would generally be expected to be slower to react at first, as the K2CO3 likely has low solubility in the solvent. It can be accelerated by starting off with a small amount of water present, and slowing raising the temperature. Works with Na2CO3 as well, again with high boiling alcohols so that the water can be driven off as the conversion progresses. SIf you can get the hydroxide, it's much quicker to convert to the alkoxide.

garage chemist - 7-1-2011 at 12:32

Results of the test batch with scrap Mg turnings:

After 4 hours of boiling, the surface of the turnings had become uniformly dark grey. Some ominous clumps of material had collected, but upon examination, they consisted of porous solid KOH. No potassium could be found in the mass, and upon adding it to water there was no reaction.
The turnings were only very superficially attacked and had seemingly become passivated, possibly by Al2O3.
The mixture still smelled strongly of tert-butanol, in contrast to the successful batch, which was practically odorless after the reaction.
So this, in conclusion, was a total failure, and further hints towards a necessity to use pure magnesium. Perhaps even firestarter magnesium is too impure.

The next step will be to make turnings from an ingot of pure magnesium, using the same drill as was used for the Mg scrap, to produce turnings of pure magnesium with the same particle size as the unreactive Mg scrap.
This will show whether the impure Mg was really the culprit, or if the turnings were simply too coarse.

blogfast25 - 7-1-2011 at 12:33

Woelen:

Agreed on the KOH: it’s probably not very critical. Interesting suggestions also by not_important.

Re. the ‘double posting’, there is none from my side. You do get people asking the same questions over and over again, typical of long threads. Cramming multiple responses into a single post provides only a negligible saving on the actual thread length (but sacrifices readability somewhat). This thread is long because the subject fascinates many. Having said that, I’ll try and ‘compact’ as much as possible.

Of your complainants at least one showed a mean-spiritedness that doesn't belong in a science forum.

blogfast25 - 7-1-2011 at 13:00

The next step will be to make turnings from an ingot of pure magnesium, using the same drill as was used for the Mg scrap, to produce turnings of pure magnesium with the same particle size as the unreactive Mg scrap.
This will show whether the impure Mg was really the culprit, or if the turnings were simply too coarse.

garage chemist:

Going back to your posts, so far you had one successful test with fine reagent Mg (here) and one unsuccessful one using turnings from a Mg rich alloy (here).

While I definitely have to commend your scientific diligence for wanting to now test pure Mg turnings to complete the picture, I wonder if it’s really necessary (considering the effort involved): most people can get hold of magnesium powder or filings of reasonable purity and there’s not much to be gained from size reducing pure Mg ingots at home, at least not for most of us.

Personally I think an experiment with home brewed 2-methyl-2-hexanol (2-methyl hexan-2-ol) as catalyst (as you suggested) and using the reagent grade Mg used in your first attempt would contribute more to our expanding knowledge base. Don't use too much head space and cool moderately in any event. Longer chain t-alcohols have lower MPs, t-butanol is a bit 'the odd one out'...

[Edited on 7-1-2011 by blogfast25]

woelen - 7-1-2011 at 13:10

I also can add that using impure Mg does not give any results. Some time ago I already posted results with using magnalium which is a mix of at least 50% Mg and at most 50% Al and at most 70% Mg and at least 30% Al. With this material I did have a violent reaction at the start of the experiment, but finally, I had no potassium at all.

Nicodem - 7-1-2011 at 13:18