[jammydodger]

Dear all,

I have been led to believe that as the borohydride group has four hydrogens, it can therefore reduce four ketone carbonyl groups (to form the secondary alcohol), i.e. giving up four hydride ions, as it were. What happens to the boron left at the end!? Is the three plus charge stabilised perhaps by three alkoxide groups (from the alcohol solvent)?

Thanks,
JD

[stewie griffin]

Both NaBH4 and LiBH4 reductions are run with MeOH. After the first hydride attacks the carbonyl, the boron has a neutral charge and three things on it (two H's left and now the O of the carbonyl). So then methanol attacks the boron to give again a boron minus. Now with four things on the boron, the next hydride can take off and attack a carbonyl. This repeats until you have at then end replaced all of the boron-hydrogen bonds with boron-oxygen bonds. The reaction workup liberates your reduced carbonyl from the boron.
A similar situation is found with hydroborations where the boron-hydrogen bonds of BH3 are replaced by boron-carbon bonds to make BR3. Then in the reaction work-up the carbon groups are liberated.

[jammydodger]

Cool, thanks. That clears it up.