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len1
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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.
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garage chemist
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@ 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.
  
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blogfast25
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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]
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garage chemist
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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.
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NurdRage
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I have had success using straight grignard turnings or powders. I have never mixed both.
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blogfast25
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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"?
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MagicJigPipe
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| 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]
"There must be no barriers to freedom of inquiry ... There is no place for dogma in science. The scientist is free, and must be free to ask any
question, to doubt any assertion, to seek for any evidence, to correct any errors. ... We know that the only way to avoid error is to detect it and
that the only way to detect it is to be free to inquire. And we know that as long as men are free to ask what they must, free to say what they think,
free to think what they will, freedom can never be lost, and science can never regress." -J. Robert Oppenheimer
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len2
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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.
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blogfast25
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| 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]
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mr.crow
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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?
Double, double toil and trouble; Fire burn, and caldron bubble
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garage chemist
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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.
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mr.crow
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| 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]
Double, double toil and trouble; Fire burn, and caldron bubble
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rrkss
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| 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]
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Cuauhtemoc
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Well, that's an advantage of living in Brazil
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blogfast25
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One pot experiment
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.
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Nicodem
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| 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.
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blogfast25
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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#...
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len1
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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]
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blogfast25
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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]
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watson.fawkes
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| 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.
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blogfast25
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| 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]
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Eclectic
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!!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] |
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DJF90
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Don't mix halogenated solvents with alkali metals, no matter what the temperature is. Its asking for trouble.
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watson.fawkes
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| 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.
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blogfast25
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| 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]
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