# What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD?

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 ?

• @orthocresol if the reaction is carried out in some non polar solvent as DMS then would the replacement be by the H2O or HOD formed in situ – Nitro phenol Apr 21 '17 at 3:56

This isn't the most well-thought-out scenario, because if you dissolve $\ce{KOD}$ in $\ce{H2O}$ all the deuteriums will be lost immediately anyway.

$$\ce{KOD + H2O <=>> KOH + HOD}$$

Assuming that the amount of $\ce{KOD}$ is much less than the amount of $\ce{H2O}$ (after all $\ce{H2O}$ is the solvent) then the position of this equilibrium will lie very, very far to the right. So, at the end of the day, if you are using $\ce{H2O}$ as the solvent, the use of $\ce{KOD}$ won't have any effect. If you want to see any difference in deuteration, you have to at least use $\ce{KOD/D2O}$. However, there is a second problem with this:

If you dissolve an amine in $\ce{D2O}$, then all the amine hydrogens will be very rapidly replaced with deuteriums. It does not even matter mechanistically what the pathway is. Regardless of whether the initial product is $\ce{RNH2}$, $\ce{RNHD}$, or $\ce{RND2}$, as long as there is $\ce{D2O}$ in the mixture, all three will become $\ce{RND2}$. The same can be said of dissolving them in $\ce{H2O}$.

(1) No. In general, protonation can and will occur by any solvent molecule that happens to be close by. If you are running the reaction in $\ce{D2O}$ solvent, then there is much, much more $\ce{D2O}$ than there will ever be $\ce{H2O}$ produced. So, protonation by $\ce{H2O}$ is statistically close to impossible.

(2) None of us can answer that, as we cannot presume what your teacher was thinking of when they set the question.

(3) Depends on what your solvent is, see comments above on dissolving amines in $\ce{D2O}$ or $\ce{H2O}$.