What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD ?
I already know the mechanism the question is the water molecules that is attacked by the isocyanide is it already present in the solution or is the one formed within the reaction in situ?, secondly If we are given the reaction is carried out in Br2 and KOD then do i have to assume that it is actually KOD in D2O because my teacher says there would be no crossover product that if the reaction is carried out in aforesaid conditions the product would be RND2but i think if the water molecule in the reaction mechanism are the ones generated in situ then there is a fair chance of other products like RNHD or even RNH2.
So the main question are :-
1) Are the water molecules the ones generated in situ ?
2) Is KOD equivalent to KOD in D2O ?
3) Would RNHD and RNH2 also form in the given conditions ?