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blogfast25
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Yes, like all fatty acid esters (IPM = the isopropyl ester of myristic acid - tetradecanoic acid), the process is described on the thread:
http://www.sciencemadness.org/talk/viewthread.php?tid=15171&...
| Quote: | | And isopropyl myristate as the organic phase of ‘Dentyl’ mouth wash can be double Grignarded (that becomes expensive) with MeI to
2-methyl-2-alkanol or with EtBr to 3-ethyl-3-alkanol. Very long chain alcohols… |
But I’ve no real experience with Grignards. Also, my gut feeling is that the resulting alcohol may already be too long to be very effective. C6, C7,
maybe even C8 would be ideal, I think.
So, it's a bit low on the list of priorities...
[Edited on 10-5-2011 by blogfast25]
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Random
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Beeswax should contain ester between CH3(CH2)28CH2-OH and CH3(CH2)14COOH. This is very long chain alcohol, so could that be substitute for those
tertiary alcohols. Maybe there could be some other use too as it's easy to prepare with hydrolysis.
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blogfast25
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Quote: Originally posted by Random  | | Beeswax should contain ester between CH3(CH2)28CH2-OH and CH3(CH2)14COOH. This is very long chain alcohol, so could that be substitute for those
tertiary alcohols. Maybe there could be some other use too as it's easy to prepare with hydrolysis. |
These super long alcohols are almost certainly too long for my purpose (solubility becomes an issue). Also, primary alkoxides don’t appear to be
stable enough at around 200 C.
To separate them from the beeswax mix is also very difficult.
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LanthanumK
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Why do these threads inevitably drift?
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Megamarko94
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Did anyone have any progress on this preparation?
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maxidastier
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Did anyone try it with Dioxane meanwhile?
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Bitburger
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| Quote: Originally posted by blogfast25 | Thermodynamically it makes sense though: NaOH HoF 298 K = - 426 kJ/mol, KOH HoF 298 K = - 425 kJ/mol, MgO HoF 298 K = - 606 kJ/mol:
NaOH (KOH) + Mg - Na (K) + MgO + ½ H2 HoR 298 = -602 + 426 = -176 kJ/mol. Quite a bit of heat... This would be typical of quite a slow
reaction which is what we're seeing...
It can be shown that the Rielke reactions take place too with the same logic.
Other example: reduction of AlCl3 with K (or Na) but also reduction of NaOH with Al powder or Mg...
If the t-butyl alcohol somehow solubilises the KOH then that's half the work done...
[Edited on 2-12-2010 by blogfast25] |
Where did you find these values, references?
How did you done this calculation?
This reaction is not done under standard conditions, so delta S and delta G are totally different from delta S° and delta H° and so deltaG°.
Have you calculated the phase transitions?
Good to be wrong
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blogfast25
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Values of HoF in standard conditions can be found in NIST Webbook.
The calculation is elemental, basic undergraduate stuff but you need to understand Hess' Law.
The correction for higher temperature can be shown to quite neglible in this case (and many like it). Total Delta H at 200 C is almost identical to
Delta H at RT, perhaps 10 - 20 % difference...
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Omniata
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Just out of curiosity more than anything, don't flame me for asking but I've noted a few things reading this thread in its entirety and would like a
few answers to things I've been thinking about, or a discussion at the very least...
I used to do quite a lot of Grignard chemistry some time ago and a few ideas came to light...
1) Has anyone tried using Iodine as an activator for the Magnesium? Potassium reacts with Magnesium Iodide to form Reike Magnesium (far more reactive)
3) The product of reacting t-BuOH with Potassium is, obviously, Potassium tertiary butoxide. As this can be readily purchased and is a salt/solid,
would it not be more efficient considering the reaction conditions to add the Potassium salt as opposed to the alcohol thus reducing the amount of
water or hydrogen produced?
Therefore, the reaction could be begun by heating the KOH and Mg to about 100°C to begin the dehydration then as the temperature starts to rise add a
crystal/bead of Iodine to activate the Mg further, then add the t-BuOK salt to the reaction...
Just an idea, I know the process works as it stands but one of the limiting factors seems to be creating a "dry" reaction... So far this appears to
have only been achieved by adding extra Mg or using a high boiling point solvent...
[Edited on 21-8-2011 by Omniata]
[Edited on 21-8-2011 by Omniata]
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blogfast25
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Omniata:
The idea of adding the catalyst as a K-salt probably works but potassium t-butoxide is far more expensive (and harder to get) than t-butanol...
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Omniata
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| Quote: Originally posted by blogfast25 | Omniata:
The idea of adding the catalyst as a K-salt probably works but potassium t-butoxide is far more expensive (and harder to get) than t-butanol...
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I'm inclined to disagree...
It would cost me £25 for 500g of t-BuOK and £80 for 500ml of t-BuOH...
It's also far easier for me to get the K salt, hence my query...
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blogfast25
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Well, try it by all means, it should work. It would further confirm the reaction path theory we've developed if it did work...
Have you compared price on mol/mol basis? And where would you buy the K t-butoxide?
[Edited on 22-8-2011 by blogfast25]
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Omniata
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Apollo scientific stock it but can be "haggled" with...
Another supply was through eurolabsupplies ...
I'd like to know about using Iodine as a catalyst...
I'm setting up to do a run in a few weeks, just wanted to see if others had tried these options first before I go ahead blind...
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blogfast25
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Thanks!
Iodine? I understand the idea but from my experience with this method most Mg powders work well, no need for iodine activation. Considering there is a
lot of alkali present it may not work as it (allegedly) does with Grignards. I use a decent reagent grade Mg, not even very fine.
The patent also mentions reducing butoxides, rather than straight alkali metal hydroxides (look it up near the top of this thread, if you haven't
already)...
There is also this very related thread, in case you've missed it:
http://www.sciencemadness.org/talk/viewthread.php?tid=15171
[Edited on 22-8-2011 by blogfast25]
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Omniata
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The other thing that caught my attention with the halogens was Reike Magnesium...
It's produced by reacting an Mg halogen, eg: MgI2, with K under specific conditions, wiki link
This could promote the reaction considerably...
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blogfast25
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| Quote: Originally posted by Omniata | The other thing that caught my attention with the halogens was Reike Magnesium...
It's produced by reacting an Mg halogen, eg: MgI2, with K under specific conditions, wiki link
This could promote the reaction considerably... |
Rieke has been touched upon briefly higher up. That reacion is almost the opposite of what we're doing here. Not sure how that 'could promote the
reaction'...
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Omniata
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I was thinking in terms of the reactivity... MgI would be produced then consumed by the pure K thus producing highly active Mg... For a small amount
of I it should drive the reaction as the active Mg would drive the parent reaction...
Although not a healthy thing to do my other consideration was to using a VERY small amount of Mercury to activate the Mg but then it would form an
amalgam with the K
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blogfast25
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Hmmm... my personal feeling is that more is to be gained by looking at the catalyst, in particular for producing sodium. As the patent indicates,
that's a very long process with t-butanol. Several of us here believe it may be due to poor solubility of the sodium alkoxide and that longer chain
2-methyl-alkan-2-ols might resolve that problem. t-alcohols are of course made (among other ways) by means of Grignard alkylation (see relevant
thread).
Try perhaps simply replication one of our runs? We could do with some fresh activity here!
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Omniata
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I'm not too sure that it will just be down to the solubility of the Sodium alkoxide...
Sodium has a significantly higher melting point to the tune of almost 40°C greater than potassium and a higher density by 0.1gm/cm³, that's
approximately 10% greater than Potassium...
Also the properties of Sodium, shear stresses, etc, are significantly greater compared to potassium...
This would to me indicate that to produce Sodium from the alkoxide you'll need a higher boiling point solvent and more aggressive conditions in the
favour of the pure metal to drive the reaction, ie: as I'm sure you already know, conditions that the alkoxide doesn't like and the pure metal does...
In layman's terms from my point of view the production of pure Sodium is far more energetic than Potassium, I'm not talking reactivity here, just that
it would seem to take a lot more energy to "construct" solid sodium than Potassium...
Additionally Sodium can form a negative ion due to the reduced nuclear shielding compared to Potassium and this may interfere with the reaction...
Edit: Also as Mg is higher in the periodic table than K and so selectively more reactive due to nuclear shielding, it may require Beryllium by theory
to produce the Sodium metal by this method? Just speculation of course...
[Edited on 23-8-2011 by Omniata]
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blogfast25
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| Quote: Originally posted by Omniata |
In layman's terms from my point of view the production of pure Sodium is far more energetic than Potassium, I'm not talking reactivity here, just that
it would seem to take a lot more energy to "construct" solid sodium than Potassium...
Additionally Sodium can form a negative ion due to the reduced nuclear shielding compared to Potassium and this may interfere with the reaction...
Edit: Also as Mg is higher in the periodic table than K and so selectively more reactive due to nuclear shielding, it may require Beryllium by theory
to produce the Sodium metal by this method? Just speculation of course...
[Edited on 23-8-2011 by Omniata] |
At 200 C reaction temperature, the MP of solid Na is almost certainly not at play here: the metal is well above its MP...
As regards the energies and reactivities:
for 2 MOH(s) + 2 Mg(s) == > 2 M(l) + 2 MgO(s) + H2(g)
... the reaction enthalpy for M=K and M=Na is almost identical, because the Heats of Formation of NaOH and KOH are almost identical. And not much
different for Rb and Cs. These small differences are the unlikely to be the cause of sodium being slower to produce. Lower concentration of the Na
alkoxide would lower the reaction speed (see proposed reaction mechanism higher up in the thread), at least that would fit what we know.
Of K and Na, K is the more reactive one here, with reactivity increasing further to Rb and Cs (see for instance their reactivities towards water). But
that difference doesn't seem to play here because the reaction is driven by lattice energy of the precipitating MgO and by the escaping hydrogen.
Be is unaccessible to almost anyone, at least in significant quantities. Also it may not form stable enough alkoxides
[Edited on 23-8-2011 by blogfast25]
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Omniata
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I'm aware of the reactivity of the alkali metals, I'm fully aware of that from studying high school chemistry, albeit many moons ago, and they aren't
at dispute here...
My query was with relevance to the positioning of Mg relative to K in the periodic table and the structural properties of Na vs. K.
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blogfast25
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| Quote: Originally posted by Omniata | My query was with relevance to the positioning of Mg relative to K in the periodic table and the structural properties of Na vs. K.
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Neither of these things are really relevant here. What matters for any chemical reaction is the change in Gibbs Free Energy ΔG = ΔH -
TΔS when going from left to right (state 2 - state 1). ΔG < 0 means the reaction is thermodynamically speaking possible, ΔG > 0
not.
Secondly, even if ΔG < 0, that still doesn’t mean the reaction will proceed spontaneously in the given conditions because there may be
thermal hindrances to be overcome. That is actually the case here: simply heating a mixture of KOH and Mg in a non-polar, aprotic solvent to 200 C
does not cause reaction. That’s where the catalyst comes into it: it lowers the thermal hindrance so that that stuff can go on at about 200 C.
That’s why looking at potentially better catalysts is of real interest here.
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blogfast25
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| Quote: Originally posted by Omniata | Be careful, if a nuclear property is measurable it is applicable...
That is default by basics laws of physics...
[...]
Sometimes it's good practice to take a step back and look at the whole picture, not just what you want to see  , or what you think you are seeing...
[...]
No offence intended... |
Hmmm… the only nuclear property that is at play here is the atomic number because it determines the quantum mechanics of the electron cloud of the
elements involved and thus their chemistry. This kind of chemistry is determined by the properties of the electron cloud, not by nuclear properties.
The whole picture? I don’t see you looking at the whole picture at all. Your dismissal of the thermodynamics of the reaction is telling.
As regards:
<i>”I started the "fresh look" based on looking through the smokescreen and what was actually taking place, NOT what was being SEEN by
blinkered eyes!!!”</i>
Wow! And what have you actually contributed with this ‘open mind’ of yours? ‘Blinkered eyes’? Sure, it’s those blinkered eyes of many here
that have carried this project forward!
No offence was taken or intended here but you need to get down and dirty and run some tests before you can make much of any kind of claim. You’re
looking through ‘the smoke screen’? O-kay: what have you seen so far?
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watson.fawkes
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| Quote: Originally posted by Omniata | Be careful, if a nuclear property is measurable it is applicable...
That is default by basics laws of physics... |
No. Such a difference might be applicable, and it might not be
applicable. Not every difference leads to a distinction. The essence of good scientific theory is the elimination of extraneous causes, or at the very
least ranking causes in order of significance of their effects. | Quote: | | We're not discussing Gibbs free energy or thermodynamics, we're looking at basic principles and analysing the results... |
You'll be doing bad science if you simply toss away what others have already been thinking about this system. What you are saying,
with this comment, is that you are ranking whatever is in your head above everything discussed before. This is pure arrogance, and I don't see yet
even a shred of justification for it. | Quote: | | Sometimes it's good practice to take a step back and look at the whole picture, not just what you want to see , or what you think you are seeing... |
In general, that's true. Yet all I've
seen is that you've observed differences only, without positing any mechanism about how these differences would be significant.
All the preceding is the polite and precise way of stating what I first thought upon reading your post: Put up or shut up.
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blogfast25
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Definitely better put than I did, dear Watson.
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