smuv
National Hazard
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Why is methanol so inert to oxidation?
I had a spectacular run-away today while trying to make some ethyl nitrate, which made me consider the reactivity of alcohols...
Methanol up to moderate temperatures ~60c is not appreciably oxidized by nitric acid.
Solutions of TCCA in methanol are fairly stable, and oxidation is only noted upon refluxing the methanol.
Ethanol is immediately and violently oxidized by conc. nitric acid.
Solutions of TCCA in ethanol are unstable, slowly precipitating cyanuric acid.
I dont know the reactivity of nitric acid and isopropanol (i assume it is oxidized rather easily).
Solutions of TCCA in isopropanol deposit cyanuric acid much more quickly than those of ethanol.
Why is methanol so much more inert to oxidation than the higher alcohols? Is it due to greater electron deficiency of the alpha hydrogen in relation
to other alcohols that makes hydride transfer less favorable? Or maybe is it maybe a beta hydride transfer for higher alcohols and alpha hydride
transfer for methanol?
Any info would be much appreciated?
[Edited on 12-19-2009 by smuv]
"Titanium tetrachloride…You sly temptress." --Walter Bishop
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Sedit
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I wish I could help but I can't sorry.
I can however state the truth in what you say in that with KNO3/H2SO4 + EtOH the oxidation happens nicely when a temperature is reached and it
sustaines itself in a Soda like fizz for a few minutes or more.
Methanol on the other hand, which I assumed would be simpler to perform oxidation on, would not react to any significant amount enought to sustain the
oxidation reaction. The smell of aldahyde was there but not in the strength it should have been and the reaction would just not carry.
[Edited on 19-12-2009 by Sedit]
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Nicodem
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Both reactions, the oxidation of alcohols by TCCA and HNO3 are supposed to involve C-H bond homolysis. I do not know if the exact mechanism was
ever proposed, but the decomposition of both intermediates (R2CH-O-Cl and R2CH-O-NO2), probably first involves the O-Cl and O-N bonds homolysis
correspondingly, followed by the rearrangement of the R2CH-O* radical via hydrogen abstraction from the alpha position, and finally the oxidation of
R2C*-OH to the corresponding carbonyl compound R2CO. The process is probably concerted all the way to R2C*-OH which means the energy of hydrogen
abstraction from C-H is included in the activation energy. So you have a situation where isopropanol, ethanol and methanol have 1 vs. 2 vs. 3
alpha-hydrogens, so statistically methanol should be more prone to oxidation if the activation energy would be equal for all these alcohols. Yet the
energies for H abstraction differ significantly and follow the reverse order. This is just a simplification - for more exact explanation and H
abstraction energy values check the literature (the H-CH2OH homolysis energy is 94.1 kcal/mol; at the moment I can not check Beilstein for the values for EtOH and iPrOH).
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smuv
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Nicodem, I question whether this oxidation proceeds through an alkyl nitrate intermediate. At least in respect to the synthesis of ethyl nitrate, the
consensus of literature, is that nitrogen oxides in the nitric acid promote the oxidation of ethanol. How does this agree with the oxidation
proceeding through an alkyl nitrate intermediate?
"Titanium tetrachloride…You sly temptress." --Walter Bishop
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Nicodem
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You should consider that NO2 is formally a radical and that it is able to abstract labile hydrogens - such as the alpha-hydrogens of alcohols or the
benzylic ones.* As such, the presence of NO2 in HNO3 can catalyse the oxidation of alcohols by opening a new mechanistic route, based on NO + HNO3
<-> NO2 regeneration of the consumed NO2. Furthermore, NO2 does reacts with alcohols to give alkyl nitrites which are similarly prone to
homolysis just as alkyl nitrates are. So, yes, the oxidation of alcohols might start slower in the absence of NO2 where time is given for the ester to
form, but the reason for EtOH>MeOH oxidation tendency is the same due to both mechanisms involving C-H homolysis. In any case the product of MeONO2
decomposition are NO and NO2 thus initiating the autocatalytic process so common in the HNO3 based oxidations. These oxidations are mechanistically
messy and I do not know what route prevails or is more plausible, and if any was ever demonstrated, but the my reasoning for the oxidation stability
would be the same for all of them. Now, whether my reasoning is sound or not, I do not know, because this is not my field of research and I did no
specific literature search.
* I only consider NO2 because I think NO and N2O3, would get oxidized to NO2, if formed in a fairly concentrated HNO3. These could be more long lived
only in diluted and not too hot HNO3.
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