My wish is to evaluate in this thread the stoichiometry of
1) the Mg / HCOONH
4 reduction of oximes
https://www.thevespiary.org/rhodium/Rhodium/chemistry/oxime2amine.mg-af.html
2) the Zn / HCOONH
4 reduction of nitroalkanes
https://www.thevespiary.org/rhodium/Rhodium/chemistry/nitro2amine.zn-formate.html
EDIT: Ehh, well: a ammonium formate Pd /C - CTH reduction will be mentioned too in the threadTheory:The Zn / Mg ammonium formate reduction of nitroalkanes or oximes is a redox reaction, where the oxidation of Zn / Mg and ammonium formate generates the electrons that are consumed in the reduction.
The
oxidation half reactions are:
Zn --> Zn
2+ + 2 e
-or Mg --> Mg
2+ + 2 e
-and HCOONH
4 --> NH
3 + CO
2 + 2 H
+ + 2 e
-The
reduction half reaction is
1) in the case of oximes
>C=NOH + 4 H
+ + 4 e
- --> >CH-NH
2 + H
2O
2) in the case of nitroalkanes
R-NO
2 + 6 H
+ + 6 e
- --> R-NH
2 + 2 H
2O
Evaluation:I have a theoretical problem with the nitro-reduction in
https://www.thevespiary.org/rhodium/Rhodium/chemistry/nitro2amine.zn-formate.html
According to the lab procedure, they react 5 mmol nitro compound with 6 mmol zinc and 0,5 g ammonium formate. 0,5 grams of ammonium formate corresponds to 7,92 mmole. Let 's be generous and consider it 8 mmole. So we have
oxidations8 mmoles * ( HCOONH
4 --> NH
3 + CO
2 + 2 H
+ + 2 e
- )
6 mmoles * ( Zn --> Zn
2+ + 2 e
-)
reduction5 mmoles * ( R-NO
2 + 6 H
+ + 6 e
- --> R-NH
2 + 2 H
2O )
Now, what 's wrong with this picture:
1) even if the oxidations go 100 % to completion, they deliver 28 mmoles electrons and the reduction needs (at least) 30 mmoles electrons.
2) the proton balance is even more screwed up (14 mmoles H
+ missing).
For the reduction of the nitrpropanol, the article where of more synthetic value if Barium 's procedure was published. I 'm sure he wouln't mind me checking out the Barium-stoichiometry (cfr.:
Post 370871
(Barium: "2,4-dimethoxyphenylpropanolamine", Novel Discourse)). Under the above redox assumptions 80 mmol Zn and 100 mmol produce 360 mmol electrons and 200 mmol protons. The reduction of the nitropropanol needs 246 mmol electrons and an equal amount of protons. The reaction is done in a 90 % EtOH solution, so 46 mmol water dissociates to generate the further 46 mmol protons needed in the reduction of the nitropropanol. I guess 50 mmol Zn will complex with the 100 mmol NH
3 generated in the oxidation of the formate, 23 mmol Zn(OH)
2 will be formed and 7 mmol Zn
2+ ions remain in solution.
Bandil and Barium have done ground breaking research in testing out the reaction on usefull compounds. If these are the correct redox reactions, mabey the yields could go up a bit more ammonium formate, like in
Post 352269
(Barium: "Wow", Chemistry Discourse) where 35,1 mmol ammonium formate would theoretically be needed to generate the 6 x 11,7 mmol protons in the nitropropane reduction.
The reaction principle is the same for the Mg / HCOONH
4 reduction of oximes. In this article however, the chemists did their homework before doing the experimental. They react 10 mmol of oxime with 30 mmol of ammonium formate and 41 mmol of Mg. This corresponds to:
oxidations30 mmoles * ( HCOONH
4 --> NH
3 + CO
2 + 2 H
+ + 2 e
- )
41 mmoles * ( Mg --> Mg
2+ + 2 e
-)
reduction10 mmoles * ( >C=NOH + 4 H
+ + 4 e
- --> >CH-NH
2 + H
2O )
The oxidations generate 142 mmoles of electrons and 60 mmoles protons, the reduction needs 40 mmoles of electrons and 40 mmoles of protons. The main difference between the 2 above articles is that in the Mg / ammonium formate procedure the redox reaction is equilibrated, and that in the Zn / ammonium formate reduction, it isn't
.