In a sodium borohydride reaction, one of the hydride in BH4- will attack the ketone, and the O- in the ketone will bond with the BH3. This would continue until all the hydride ions are used up.
My question is..what would be the driving force that makes the B - H break and attack the ketone.
Also, after you have one or more ketone bonded with the Boron, can't the O - B bond break and attack the ketone instead of the B - H bond?
The Attempt at a Solution
My group have no idea for the first question...the most possible answer we came up with is: "it just happen".
The second question...we just take a guess that the O - B is much harder to break compared with B - H, and maybe it has to do with the electronegativity.
Thank you very much.
I will add some more questions later when we finish figuring out the exact mechanism for NaBH4