madscientist - 7-9-2002 at 15:31
I know for a fact that esters of carboxylic acids will react with ammonia to form an amide and an alcohol. I'm wondering if there would be a similar
reaction with hydrazine - my first guess would be that there isn't. But still, the idea interests me. Can anyone enlighten me?
Perhaps something like this would happen?
(COOCH3)2 + N2H4 ----> (CONH)2 + 2CH3OH
Not Likely In That Way
PrimoPyro - 8-9-2002 at 04:44
I bet hydrazine would react with esters, but I think it highly unlikely for it to easily form a cyclic compound like you have pictured there.
I think that one hydrazine moietywill react with one moiety of ester, and if enough hydrazine is present for further reaction, another moiety of
hydrazine might then react with the other moiety of ester in that compound, giving the diamide-like structure.
I seriously doubt that cyclization would so readily occur because for it to do so, the hydrazine-amide of one ester would have to ionically react with
the other ester, and the hydrazine amide is unlikely to bear a negative charge needed to do so through common mechanisms. Even if it did work, other
free hydrazine molecules would have far better odds of reacting with the remaining ester function first, and so the yields of truly cyclic diamides
would be minimal.
There may of course be some reaction conditions that do promote that reaction. Anything is possible, I am just saying that the reaction of putting the
two together is not likely to produce that compound without some sort of "external influence of will" of some sort, on the chemist's behalf. 
PrimoPyro
PHILOU Zrealone - 10-4-2003 at 15:18
1)CH3-O2C-CO2-CH3 + H2N-NH2 --> NH2-NH-CO-CO2-CH3 + CH3OH
2)NH2-NH-CO-CO2-CH3 + H2N-NH2 --> (NH2-NH-CO)2 + CH3OH
3)NH2-NH-CO-CO2-CH3 + CH3-O2C-CO2-CH3 --> CH3-O2C-CO-NH-NH-CO-CO2-CH3 + CH3OH
1) equimolar
2) exces NH2-NH2
3) excess ester
CH3-O2C-CH2-CO2-CH3 + H2N-NH2 -heat-> CH3-O2C-CH2-CO-NH-NH2 + CH3OH -heat-> cyclo(-NH-CO-CH2-CO-NH-) + CH3OH
