Vibrating_Lights (Hive Addict)
08-23-02 10:46
No 348049
      H2O Al/Hg stoich check.  Bookmark   

How much H2O is needed to complete the Al2O3 formation in an al/Hg with 1 mole of Imine to amine And three moles Of nitromethane to MeAm. So

(3*2)+1= 7 total O to reduce.  How much water is needed here.
3MeNO2 + 6Al(6H+)+ _?H2O[1]  --> 3MeNH2 +  3Al2O3
RO-MeNH2)R +1Al(1H+)_?H2O[2] --> RMeNH2R +  AlO3

Would the H2O for [1] = 3
and [2] = 1
so a totalk of 4 moles of H20 to complete the reduction. two excess moles of MeNH2 to be collected after the Rxn.
does that sound right.  The excess of MeAm i snot a problem.

So much game I could sell a hooker some pussy
(Old P2P Cook)
08-23-02 20:39
No 348175
      Aluminum reductant stoichiometry.  Bookmark   

I would make the assumption that the aluminum goes to the hydroxide not the oxide.

Then I would model the reduction of nitromethane as being the sum of these two reactions:

A) Al + 3H2O __> Al(OH)3 + 3H

B) CH3NO2 + 6H __> CH3NH2 + 2H2O

Then, combining 2A + B, we obtain the overall reaction:

C) 2Al + 4H2O + CH3NO2  __> CH3NH2 + 2Al(OH)3

which requires 4 moles of water for every mole of nitromethane.

If, instead of reaction A, we assume that the aluminum goes to the oxide rather than the hydroxide (a bad assumption in my guestimation) then we have:

D) 2Al(OH)3 __> Al2O3 + 3H2O

Then, combining D + C, we have an overall reaction:

E) 2Al + H2O + CH3NH2 __> CH3NH2 + Al2O3

which requires only 1 mole of water per mole of nitromethane but, as I said before, I would expect reaction C to be more likely thus requiring the 4 moles of water per mole of nitromethane.
(Old P2P Cook)
08-23-02 20:59
No 348185
      Imine/Aluminum reduction stoichiometry.  Bookmark   

For the reaction of the methylamine with ketone to produce imine we have:

A) R2CO + CH3NH2 __> R2CNCH3 + H2O

For the reduction of the imine we have:

B) R2CNCH3 + 2H __> R2CHNHCH3

And, as before, the reaction of aluminum with water:

C) Al + 3H2O __> Al(OH)3 + 3H

Then, combining 3A + 3B + 2C, we have the overall reaction:

D) 3R2CO + 3CH3NH2 + 2Al + 3H2O __> 3R 2CHNHCH3 + 2Al(OH)3

which requires 1 mole of water per mole of ketone.
(Old P2P Cook)
08-23-02 22:12
No 348205
      Methanol as hydrogen donor.  Bookmark   

One other consideration that might reduce the amount of water needed in reductions performed with aluminum in methanol is that the methanolic OH is also quite reactive and could possible take the place of water thusly:

Al + 3CH3OH __> Al(OCH3)3 + 3H