BTW:
I have found an article with described procedure of reduction of some oxime to amine. But conditions are strange (to me):
oxime+ dry THF + Al(Hg)
According to:
....C-N=OH + 3H ->....C-NH2 + H2O
and
Al + 3H2O -> Al(OH)3 + 3H (simplified, of course)
, for each molecule of oxime two molecules of H2O are needed.
Am I right or I overlooked something and guys from article are right ? Is THF a H-donor ??
In attachment page from cited article with this procedure. |