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Topic: Reduction by NaBH4 : mechanism  (Read 27810 times)

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Offline faust

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Reduction by NaBH4 : mechanism
« on: April 14, 2006, 09:42:02 AM »
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

Offline Albert

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Re: Reduction by NaBH4 : mechanism
« Reply #1 on: April 14, 2006, 12:42:06 PM »
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...

Offline faust

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Re: Reduction by NaBH4 : mechanism
« Reply #2 on: April 14, 2006, 12:54:34 PM »
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 HxB(OMe)4-x which react with the carbonyle. But also, the first proposition, which some BH4 directly react with carbonyle.


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Re: Reduction by NaBH4 : mechanism
« Reply #3 on: April 14, 2006, 06:21:01 PM »
I definitely favor the second mechanism because of the reaction with alcohol or water solvent.

Offline faust

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Re: Reduction by NaBH4 : mechanism
« Reply #4 on: April 15, 2006, 04:23:41 AM »
thank you for your answer.


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