[faust]

Hello to you all!

Just a little question...

What is the real mechanism for the reduction of a cetone or aldehyde by NaBH4?

one says that NaBH4 reaction with RCHO to forme RCH2-OBH3 than It could react with another carbonyle compound etc...

But some say that NaBH4 react first with the solvent (Methanol or Ethanol) in order to do these species :

H3B(OMe), H2B(OMe)2 and HB(OMe)3. Which will reduce the carbonyle. The hydride becomes more nucleophilic because of the methoxy groups which are electron donnating groups.

So follows you, which is the correct mechanism?

I have tried to put NaBH4 in methanol... And I have noticed bubble of H2... so may be the second mechanism is the good one no?

Thank you again for your help

[Albert]

one says that NaBH4 reaction with RCHO to forme RCH2-OBH3 than It could react with another carbonyle compound etc...

I'd opt for this one. This is also how LiAlH4 reacts with ketones (remember: ketones with LiAlH4 and aldehydes with NaBH4).

However, I'm not sure...

[faust]

thank you for your advice...

But with LiAlH4 we commely use a non protic solvent.

However for NaBH4 we could use MeOH. I have tried to put a large quantity of NaBH4 in MeOH, without any other reagent... And it react.. Slowly at first, then I clearly saw hydrogen bubles! So I wonder if , actually, there is not a generation of H_xB(OMe)_4-x which react with the carbonyle. But also, the first proposition, which some BH4 directly react with carbonyle.

[movies]

I definitely favor the second mechanism because of the reaction with alcohol or water solvent.

[faust]

thank you for your answer.