Reaction mechanism of imine reduction to amine using an aluminium amalgam

Question

I can not find any good resources on the Internet except for some shady drug syntheses. I was hoping that someone could provide me with some more educational information.

I'm interested in imine-amine reduction methods, but it was hardly discussed in my class. It would be very nice if someone could explain the exact reaction mechanism of the imine-amine reduction in the presence of an aluminium amalgam.

I also found that the very dangerous compound Mercury(II) chloride was used a lot in this method and I was wondering why for example Mercury sulfide is not used, as I thought it would be a lot safer since it is hardly soluble in water and is therefore less likely to do any physical harm.

Answer 1

The mechanism is electron transfer. The imine will take one electron from metal to form a radical anion ($\ce{.CR2-NR'^{-}}$) intermediate; the $\ce{-NR'^{-}}$ will grab one proton from solvent to give you the neutral radical ($\ce{.CR2-NR'H}$); the carbon radical grab one hydrogen atom ($\ce{.H}$) from solvent to form your amine product and a solvent radical ($\ce{.OH}$ if water used as solvent); and the resulting solvent radical will take another electron from aluminum to finish the whole process. The reason to add mercury is to destroy the relatively less reactive aluminum oxide layer. Mercury also serve as the medium for electron transfer.

$\ce{HgCl2}$ is not that dangerous, since it is not volatile at all like metal mercury or alkylmercury. After reaction, simply add some sulfide or sulfur to transfer all of them to $\ce{HgS}$ which is not soluble and not reactive in non-oxidative condition. This is also the reason that you can not use $\ce{HgS}$ instead, because it is neither soluble nor reactive.