Organic Chemistry Help (Sodium Borohydride Reduction)

[twt]

Homework Statement

Hi all,

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

 

[chemisttree]

The solvent system for most borohydride reductions is an alcohol. It must be a solvent that solvates the sodium borohydride. The oxygen from the ketone does not usually bond to the borane but rather the solvent does. The proton from the solvent protonates the alkoxide generated from the ketone to produce an alcohol. That is where the reduction ends for that particular borohydride anion... a new molecule of BH4- is required for each ketone reduced.

 

[chemisttree]

...That is where the reduction ends for that particular borohydride anion... a new molecule of BH4- is required for each ketone reduced.

Whoops! Wrong again! All 4 hydrogens can reduce the carbonyl and the stoichiometry is 4 carbonyls to 1 borohydride...