(Hive Bee)
05-06-04 15:12
No 505289
(Rated as: good read)
My wish is to evaluate in this thread the stoichiometry of
1) the Mg / HCOONH4 reduction of oximes
../rhodium /oxime2a
2) the Zn / HCOONH4 reduction of nitroalkanes
../rhodium /nitro2a
EDIT: Ehh, well: a ammonium formate Pd /C - CTH reduction will be mentioned too in the thread
Theory:
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 --> Zn2+ + 2 e-
or Mg --> Mg2+ + 2 e-
and HCOONH4 --> NH3 + CO2 + 2 H+ + 2 e-
The reduction half reaction is
1) in the case of oximes
>C=NOH + 4 H+ + 4 e- --> >CH-NH2 + H2O
2) in the case of nitroalkanes
R-NO2 + 6 H+ + 6 e- --> R-NH2 + 2 H2O
Evaluation:
I have a theoretical problem with the nitro-reduction in ../rhodium /nitro2a
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
oxidations
8 mmoles * ( HCOONH4 --> NH3 + CO2 + 2 H+ + 2 e- )
6 mmoles * ( Zn --> Zn2+ + 2 e-)
reduction
5 mmoles * ( R-NO2 + 6 H+ + 6 e- --> R-NH2 + 2 H2O )
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 NH3 generated in the oxidation of the formate, 23 mmol Zn(OH)2 will be formed and 7 mmol Zn2+ 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 / HCOONH4 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:
oxidations
30 mmoles * ( HCOONH4 --> NH3 + CO2 + 2 H+ + 2 e- )
41 mmoles * ( Mg --> Mg2+ + 2 e-)
reduction
10 mmoles * ( >C=NOH + 4 H+ + 4 e- --> >CH-NH2 + H2O )
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
.
(The Archetypical "Good Guy")
05-06-04 15:35
No 505295
Indeed you are right!
It has always pondered my mind how the heck they could use less formate than a Pd/C reduction. Perhaps this is the reason...
I never got any measured yields from the reduction method using Zn/formate, but they seemed kind of low usually. Some other bees has also reported very low yields (20%) using this method. Perhaps this is the simple reason?
It would be nice to try it out with a 5X molar eq.(as is done in the Pd/C reduction) of potassium formate and see what that did for the yields? Im out of Zn (or so i think) so i can't try it just yet. Any volanteers?
Regards
Bandil
Nuke the whales!
(Hive Bee)
05-06-04 16:13
No 505302
It would be nice to try it out with a 5X molar eq.(as is done in the Pd/C reduction) of potassium formate and see what that did for the yields? Im out of Zn (or so i think) so i can't try it just yet. Any volanteers?
I UTFSE, and think nobee has tryed out the Zn / HCOOK reduction yet. It would indeed be a good idea to check thing out with an excess of potassium formate I guess some water would be nice in the reaction medium, for the proton balance.
It seems more convenient to prepare potassium formate than ammonium formate. If the redox reaction works nicely with potassium formate, that would be great.
(Heavyweight Chempion(eer))
05-06-04 17:05
No 505315
I have tried that system. I know I have but I just can't remember the details.
Severe Aztecoholic and President of Sooty's fanclub - Sooty for President!!
(Hive Bee)
05-06-04 19:01
No 505336
In the Encylopedia of Reagents for Organic Synthesis (section Zinc-Acetic acid) they say that unsaturated nitro compounds can be reduced with Zn/AcOH to the oxime (ref.: Baer, H.H., Rank, W.; CJC (1972) vol 50 p 1292). More vigorous reaction leads to the corresponding ketone (ref.: Anagnostopoulos C.E. Fieser, L.F.; JACS (1954), vol 76, p 532).
In the CJC article, a Zn / AcOH reduction is done in ether (1 h reflux), generating 87 % of oxime.
With further reduction (for example Zn / ammonium formate), it should be possible to convert a nitropropene to an amine with just zinc, magnesium and a formate.
I have tried that system. I know I have but I just can't remember the details.
This is rather regretable. Bummer
(Hive Bee)
05-06-04 20:48
No 505362
These are some results of a nitropropanol CTH Pd/C reduction with 5 eq. of ammonium formate:
Patent US509967
EXAMPLE 1
Reduction of Threo 1-Phenyl-2-Nitro-Propan-1-ol
To a solution of threo 1-phenyl-2-nitro-propan-1-ol (0.61 g, 0.0037 moles) in methanol (10 ml), was added 10% palladium on carbon (0.3 g) followed by ammonium formate (1.06 g, 5 equivalents). The resulting mixture was stirred for 1 hour then poured into Et.sub.2 O (200 mL) and filtered. The ether solution was washed with 0.1 M sodium hydroxide (50 mL), dried and evaporated in vacuo. The residue was taken up in Et.sub.2 O and treated with 2 mL of saturated HCl in Et.sub.2 O giving the crystalline hydrochloride salt (0.27 g, 44%).
It is nice to see that the HCOOK Pd/C CTH developped @ the Hive gives better results.
(Heavyweight Chempion(eer))
05-06-04 22:53
No 505383
Me and Ritter discuss that patent/paper in "the most interesting CTH" thread in the novel forum. I got crap when I tried it, if my memory serves me. The catalyst just dies a bit into the reduction. I can't understand how they got the yields they state. They must have some damn magic catalyst.
Severe Aztecoholic and President of Sooty's fanclub - Sooty for President!!