walruslover69
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Potassium metal synthesis Alkoxide solubility and NaOH/Mg reaction mechanism
I have recently turned my attention to adapting the Potassium metal synthesis using Magnesium, KOH and alkoxide catalyst into a optimized procedure
for Sodium metal.
I have read on the forum and I believe Nurdrage even mentioned it in one of his videos that the hypothesized reason the reaction doesn't work with
NaOH/Na is because of the lower solubility of the sodium alkoxide in the solvent. This has spurned a search for more soluble longer chained tertiary
alkoxides.
Does anyone have a source for this? Solubility data is pretty scare on alkoxides but I found this source that seems to contradict the current
consensus. http://www.callery.com/userAssets/Alkoxides-Summary-2-28-201...
It claims that the sodium t-butoxide is almost an order of magnitude more soluble in hexane and toluene than the potassium alkoxide. Does anyone have
any further information on this?
If not solubility what else inhibits the sodium reaction?
Finally I have also seen people claim that primary and secondary alkoxides are not stable under these conditions. What decomposition reactions do they
go through?
I recently ordered 500g of lab grade magnesium and I have a wide range of alcohols I plan to experiment on. I will post the results as I work through
them.
[Edited on 9-5-2018 by walruslover69]
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Foeskes
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This is off topic but, where did you get the magnesium and is it in a powdered form?
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walruslover69
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I ordered turnings off labsupplyoutlaw. Turnings are what I had the best luck with doing grignards in undergrad. I have heard that the powdered form
can be much harder to activate once it gets oxidized.
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walruslover69
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looking at Yb(NO3)3's post using Aluminum to reduce KOH. Does anyone have any experience or insight into if aluminum can be substituted for magnesium
in this low temp alkoxide catalyed Potassium synthesis? It seems to be thermodynamically possible on paper.
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VSEPR_VOID
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Are you using high quality, clean, magnesium? I know some reactions, such as grignards, require clean magnesium.
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walruslover69
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What I plan on using is fairly high quality. I also have some low quality fire start shavings left over from something else that I might try, just to
see how selective the reaction is. My intuition says that it wont be as picky with regard to magnesium quality because I can increase the temperature
pretty high in order to initialize the reaction compared to only heating a grignard to the boiling point of ether.
I was thinking that aluminum might be interesting to experiment with because some of the alkoxides are soluble. My understanding is that the alkoxide
acts as a phase transfer catalyst, solubilizing the potassium ions. If that is true then also solubilizing the redundant might have promising results.
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AJKOER
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Actually, as per my comments on the recent thread making Sodium metal (link: http://www.sciencemadness.org/talk/viewthread.php?tid=82406 ):
Quote: Originally posted by AJKOER  | I did some research, and per a source on the action of H2 on NaOH (which I think may parallel KOH), we have:
H2 + NaOH = NaH + H2O (see Eq 1 on page 2192)
and at 360 C :
NaH + Heat --> Na + 1/2 H2
Source: See 'Revisiting the Hydrogen Storage Behavior of the Na-O-H System' by Jianfeng Mao , et al., in Materials 2015, 8, 2191-2203;
doi:10.3390/ma8052191 . Link to download file: https://www.mdpi.com/1996-1944/8/5/2191/pdf . To quote:
"The DTA profile for NaH shows one endothermic peak at ca. 360 °C, which can be assigned to the decomposition of NaH to Na metal and hydrogen."
--------------------------------------------------
Actually, in my opinion, the actual chemistry is likely a function of the highly anodic metal (like Mg, Zn, Al, Fe, Sn,.., see http://www.zygology.com/cms/upload_area/pdf/Zyg-Anodic-Index... where Zn and Al are the frequent cited sources of so called 'nascent hydrogen'
with HCl) reacting with water as follows:
H2O = H+ + OH-
Al --> Al3+ + 3 e-
e- + H+ = .H = 1/2 H2 (as .H + .H = H2)
Hence, the cited Equation (1) above could actually be the net of the following reactions:
.H + OH- = e- + H2O = e-(aq) (see Eq 37 and Table 2 at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... )
e- + .H = H- (see http://pubs.rsc.org/-/content/articlelanding/1966/tf/tf96662... )
Net of the two above:
2 .H + OH- = H- + H2O
or, seemingly (but not so at very high temperatures as H2 + Heat = .H + .H):
H2 + OH- = H- + H2O
Further:
H- + Heat --> .H + e- = 1/2 H2 + e- (as .H + .H = H2)
and:
e- + Na+ = Na
My point is if one is going to give credit to the hydrogen, and ignore the role of the highly anodic metal, it is really the mono-atomic hydrogen
radical (.H formed courtesy of the anodic metal or high temperatures), or as historically sometimes referred to, in my estimation, as 'nascent'
hydrogen. Here is a recent account of the surface chemistry of adsorbed .H on aluminum and applications, see https://books.google.com/books?id=1etfSdk55SYC&pg=PA817&... . This last reference per Equation (5) appears to support my reaction presentation
in terms of 2 .H (which is adsorbed on an aluminum surface) and not H2, to quote:
" PbS + 2 .H = Pb + H2S (5) "
[Edited on 3-5-2018 by AJKOER]
[Edited on 3-5-2018 by AJKOER] |
Continuing with the anodic metal and also the surface chemistry theme, one of my prior comments and references (where one can largely replace Al with
Mg, I suspect):
| Quote: Originally posted by AJKOER | In my defense, what we have here, in my opinion, is not a classic metal/O2 battery, but a related lesser known rendition with (well, at least prior to
revision, see Thesis link below) the same anodic half reaction cited in the Al/O2 battery.
For those interested, one of my favorite thesis is from 2008, "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial
phenomena", by Saikat Adhikari, link: https://www.google.com/url?sa=t&source=web&rct=j&...
I believe a similar electrochemical reaction scheme is occuring with Magnesium. Some extracts of interest for the brave, to quote:
"In addition to being a primary corrosion process, dissolution behavior of aluminum and its alloys in alkaline solutions is of
considerable interest because it is the anode reaction in aluminum-air batteries.[4] ......The anodic half-reaction at the Al electrode
is
Al + 4 OH − → Al (OH)4− + 3 e− (1.1)
which exhibits an electrode potential of -2.35 V in alkaline solutions(vs. NHE).
"2 Al + 6 H2O → 2 Al (OH)3 + 3 H2 (1.2)"
....."dissolution of aluminum in alkaline solutions at open-circuit also leads extremely high rates of H-absorption into the metal,
[9-14] ".....
"Another study of the dissolution of aluminum in aqueous solutions by Perrault revealed that the open circuit potential of aluminum
in strongly alkaline solutions corresponds closely to the Nernst potential for oxidation of aluminum hydride to aluminate ions [25]
AlH3 + 7 OH− (aq) → Al (OH)4− + 3 H2O ( aq ) + 6 e− (1.3)
This suggests a role of surface aluminum hydride as a reaction intermediate in the dissolution process. Additional evidence for the
presence of aluminum hydride was provided by Despic and co-workers.[26, 27] They found that aluminum hydride formation was one of
the major processes apart from aluminum dissolution and hydrogen evolution, during the cathodic polarization of aluminum. Titanium
corrosion in alkaline solutions is also thought to proceed through a hydride mediated mechanism.[28-30] "
"He found that the open-circuit potential in strongly alkaline media was determined by the equilibrium of the reaction
AlH3 + 7 OH− (aq) → Al (OH )4- + 3 H2O ( aq ) + 6 e− (3.7)
He obtained a standard chemical potential of 25 kcal/mol for AlH3 from his data, which was in reasonable agreement with prior
thermochemical calculations done by Sinke et al who obtained a value of 11.1 kcal/mol for the chemical potential.[80] ...."
"The anodic reaction 3.7 is accompanied by the cathodic reduction of water to form hydrogen
H2O + e- → OH- + H (3.8)
and the reaction of hydrogen with aluminum to from hydride
Al + 3 H → AlH3 (3.9)"
So, a bit technical but related half cell reactions I would guess for the likes of Magnesium.
-------------------------
One should accept that if some electric current is generated in situ, it is possible for electrons to be solvated in an appropriate organic medium.
This could lead to the decomposition of associated organics, resulting in the release of Potassium. Note, even in aqueous settings, long ago there are
reports of unexplained reactions of various salts with Mg metal, including converting nitrate into nitrite and finally ammonia (see, for example, p.
314 at "Chemical News and Journal of Physical Science, Volumes 87-88, 1903, p. 312-316, link: https://books.google.com/books?id=jvjmAAAAMAAJ&pg=RA1-PA... ).
-----------------------------------------------------------------------
[Edit] Here is an interesting article citing a temperature and Magnesium alloy presence associated with hydride formation: "absorption of hydrogen by
magnesium based alloys", in METAL 2014, link: https://www.google.com/url?sa=t&source=web&rct=j&...
This short discussion on preparing the Mg for interaction with H2 may be of value: https://books.google.com/books?id=NR3OxpSiA60C&pg=PA491&...
[Edited on 24-5-2017 by AJKOER]
[Edited on 24-5-2017 by AJKOER] |
----------------------------------------------------
There is also, in my opinion, some good experience to be taken from the Make Potassium thread (see http://www.sciencemadness.org/talk/viewthread.php?tid=14970 ).
[Edited on 10-5-2018 by AJKOER]
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SWIM
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In the potassium thread Nicoderm says beta hydride elimination reactions are the problem with alcohols with an alpha hydrogen.
Makes acetone, which would itself be pretty reactive in those conditions.
Quite a surprise about the solubility of the alkoxides.
Are they ionized in a nonpolar solution like that, or just floating around?
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walruslover69
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| Quote: Originally posted by SWIM | In the potassium thread Nicoderm says beta hydride elimination reactions are the problem with alcohols with an alpha hydrogen.
|
That makes sense now that you say it, my organic is fairly rusty.
As for ionization, I can not comprehend that there is any dissociation. I found that Potassium tert-butoxide forms tetrameric cubane-type clusters.
One of the reasons for sodium's inhibition might be the structure of the alkoxide complexes.
[Edited on 10-5-2018 by walruslover69]
[Edited on 10-5-2018 by walruslover69]
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AJKOER
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Some further remarks on my comments above. First, per Wikipedia on Potassium tert-butoxide (https://en.wikipedia.org/wiki/Potassium_tert-butoxide):
"Potassium tert-butoxide is the chemical compound with the formula K+(CH3)3CO−. This colourless solid is a strong base (pKa of conjugate acid around
17), which is useful in organic synthesis. It exists as a tetrameric cubane-type cluster. It is often seen written in chemical literature as potassium
t-butoxide. The compound is often depicted as a salt, and it often behaves as such, but it is not ionized in solution."
Repeating my cited reaction above:
" PbS + 2 .H = Pb + H2S (5) " (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&... )
which is more meaningful, in my opinion, upon expressing .H as the reversible radical reaction equation: .H = H+ + e- , then more clearly:
PbS + 2 ( H+ + e- ) = Pb + H2S
Further, in the case of Potassium tert-butoxide, a possible corresponding reaction:
K+(CH3)3CO− + ( H+ + e- ) = K + .(CH3)3COH
So, in a manner described per my reference on Eq 5 above, one could create, in a separate step, hydrogen adsorbed onto Mg or Al as a temporary source
of .H, which could then be added to previously created Potassium tert-butoxide (or, as created in an equilibrium reaction between KOH and the alcohol)
to possibly liberate metal potassium.
At least, that is the theory, which I am unfortunately not equipped to test in the near term.
[Edited on 17-5-2018 by AJKOER]
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walruslover69
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I might able to do some experiments to test if that is the mechanism, or atleast rule it out if it isn't. Running with the reaction with K2O instead
of KOH and using already prepared potassium tert-butoxide instead of tertbutanol. Doing it this way there would be no evolution of hydrogen.
Does that seem valid to you?
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Σldritch
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Has anyone tried the reaction with phenol?
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Texium
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If anyone with experience running the potassium reaction would like some longer chain tertiary alcohols to experiment with, I can make
2-methyl-2-octanol and 2-methyl-2-decanol fairly easily if there is any interest. I would rather not experiment with the reduction myself, simply
because I already have a decent amount of sodium metal and I'd rather spend my time doing more organic chemistry, but I'm curious to see if it works,
and it seems like this project could be a good inter-member collaboration.
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walruslover69
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Phenol might have some potential since it won't undergo beta hydride elimination. Synthesizing phenol has been near the top of my to do list for a
while. I might have to finally get around to it and try it out.
My goal for the project is to be able to optimize the procedure and try to find some easier to obtain catalysts than long chain tertiary alcohols or
even tertiary alcohols at all.
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AJKOER
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| Quote: Originally posted by walruslover69 | I might able to do some experiments to test if that is the mechanism, or atleast rule it out if it isn't. Running with the reaction with K2O instead
of KOH and using already prepared potassium tert-butoxide instead of tertbutanol. Doing it this way there would be no evolution of hydrogen.
Does that seem valid to you? |
Per Eq 1, scaling by 2:
2 H2 + 2 NaOH = 2 NaH + 2 H2O
2 NaOH = Na2O + H2O
So:
2 H2 + Na2O + H2O = 2 NaH + 2 H2O
Or:
2 H2 + Na2O = 2 NaH + H2O
I would rewrite as:
2 .H + 1/2 Na2O = NaH + 1/2 H2O
In the presence of the highly anodic metal (Mg or Al) forming e-, replacing the KOH with K2O may reduce available hydrogen atoms, and, as such, I
would suspect a reduction (but not elimination) in the level of diffusible hydrogen (see, for example, page 41 at http://www.jwri.osaka-u.ac.jp/publication/trans-jwri/pdf/421... ) and a lower yield of the target metal.
[Edited on 19-5-2018 by AJKOER]
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walruslover69
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| Quote: Originally posted by AJKOER | | Quote: Originally posted by walruslover69 | I might able to do some experiments to test if that is the mechanism, or atleast rule it out if it isn't. Running with the reaction with K2O instead
of KOH and using already prepared potassium tert-butoxide instead of tertbutanol. Doing it this way there would be no evolution of hydrogen.
Does that seem valid to you? |
Per Eq 1, scaling by 2:
2 H2 + 2 NaOH = 2 NaH + 2 H2O
2 NaOH = Na2O + H2O
So:
2 H2 + Na2O + H2O = 2 NaH + 2 H2O
Or:
2 H2 + Na2O = 2 NaH + H2O
I would rewrite as:
2 .H + 1/2 Na2O = NaH + 1/2 H2O
In the presence of the highly anodic metal (Mg or Al) forming e-, replacing the KOH with K2O may reduce available hydrogen atoms, and, as such, I
would suspect a reduction (but not elimination) in the level of diffusible hydrogen (see, for example, page 41 at http://www.jwri.osaka-u.ac.jp/publication/trans-jwri/pdf/421... ) and a lower yield of the target metal.
[Edited on 19-5-2018 by AJKOER] |
Where would you be getting any hydrogen at all from? My idea was to try it that way and eliminate all hydrogen completely in order to see if that is
actually the mechanism.
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Ozone
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NurdRage appears to have gotten this to work (reported 30%, but he thinks it can be optimized, and it doesn't involve fire). Check it out soon in case
YouTube kills his channel.
https://www.youtube.com/watch?v=kJ2C2w8Ntt4
Cheers,
O3
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
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SWIM
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I've been wanting to try this with xylenols or cresols.
Cheap, high boiling, looks good from a steric standpoint.
But phenol is an interesting one too.
The intermediate acidity and the stability at high temps.
Would the additional methyls on xylenols stabilize the ogygen atom more and make for weaker complexation?
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VSEPR_VOID
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Make sure to add what ever you make to the SM list
https://docs.google.com/document/d/1AoI2VA5L4bmFw2HwXS2OVYTV...
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