phendrol
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Oximes from nitroalkenes with FeCl2 ?
Has the possibility of reducing nitroalkenes to oximes with FeCl2 alone, crossed someones mind?
Dissolving metal reductions can be used to form a ketone from a nitroalkene. Iron or tin with HCl for an example. The nitrocompound is reduced to the
oxime, which is hydrolyzed in the acidic environment. SnCl2 can be used instead of the tin/acid pair to yield mostly the oxime in high yield.
Could FeCl2 give similar results? Maybe FeCl3?
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blogfast25
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Certainly Fe(III) cannot be a reducing agent because (in these circumstances) it cannot be further oxidised.
That leaves FeCl2, a fairly weak reducing agent via:
Fe<sup>2+</sup> === > Fe<sup>+3</sup> + e<sup>-</sup>, Eox = - 0.771 V
Compare to SnCl2: Sn<sup>2+</sup> === > Sn<sup>4+</sup> + 2 e<sup>-</sup>, Eox = - 0.15 V
Ferrous ions are the stronger reducing agent of the two. For what that's worth...
[Edited on 8-11-2014 by blogfast25]
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phendrol
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What would have to be the molar ratio of FeCl2 to the nitroalkene ?
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blogfast25
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The reduction, 'on paper', of a secondary nitroalkane to a ketoxime would require 2 protons (H<sup>+</sup> and 2 electrons.
Since as the oxidation of ferrous to ferric ions releases only one electron per mol, stoichiometrically 2 mol of ferrous ions and 2
mol of HCl (or one of H2SO4) per mol of sec. nitroalkane would be needed. That would be the stoichiometric minimum (in reality you'll need more).
But I'm in NO WAY implying this will work. I simply don't know that.
If you are going to test it do it on a really small scale, nitros being what they are.
[Edited on 9-11-2014 by blogfast25]
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phendrol
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Thank you for your responses.
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