What is the needed amount of Al to reduce one mole of nitromethane? I've searched for some time using the search engine but the only information I
could find was regarding the reductions of imines, nitropropenes or oximes. Is this equation correct?
2 Al + 4 H2O + CH3NO2 > CH3NH2 + 2 Al(OH)3
Also what would be the expected yield?
I want to procure some EtOH/MeNH2 solution by doing the reduction in a closed system and then decanting as much of the ethanol as possible. Knowing
the yield I could roughly evaluate what is the concentration of it. |