Addition of NaBH4 to aldehydes to give primary alcohols

by James

Description:  Addition of sodium borohydride (NaBH4) to aldehydes gives primary alcohols (after adding acid)

Examples:

Notes: Lots of different acids can be used in the last step. It’s not important which specific acid is used, just that something is present that can form the alcohol.

Mechanism:

NaBH4 is a source of hydride (H-) and the reaction begins with the addition of hydride to the carbonyl to the aldehyde (Step 1, arrows A and B). Upon addition of acid, the oxygen is protonated (Step 2, arrows C and D) to give the neutral primary alcohol.

Notes: The choice of H2O / H2SO4 as acid isn’t crucial – this is just an example. Any source of proton (including water) will do.

{ 6 comments… read them below or add one }

Thomas July 12, 2013 at 10:43 am

Thank you, very helpful and complete, answered every question!

Reply

TJ July 26, 2013 at 4:44 pm

There is a problem in my book I am having a small issue with. It states “Draw the products formed when CH3COCH2CH2CH=CH2 is treated with NaBH4 (excess) in CH3OH”. I understand that NaBH4 selectively reduces C=O while leaving C=C inert, but what protonates the oxygen in the final step to give the alcohol? Does NaBH4 do that because it is in excess, or does methanol? I don’t 100% understand methanol’s role in the reaction, just that it is typically used with NaBH4.

Reply

prabhu September 16, 2013 at 12:52 am

in prescence of meoh, the MeOH acts as a proton source and also there will be formation of methoxy borohydride (mono,di,tri) which has higher susceptibility of donating the hydride. because of these two reasons MeOH or EtOH are often used as co-solvents in NaBH4 reduction

Reply

James Ashenhurst September 17, 2013 at 10:32 am

True, but actually methoxyborohydrides are not as nucleophilic as sodium borohydride itself.

Reply

Jason Mathias February 26, 2014 at 1:58 pm

Hi James,
It seems like the BH4 in step 1 is nucleophilic, and the H-B bond is moving (Electrons and H) to the C=O, which is why the C-O bond ‘jumps’ up. Is this not a substitution reaction? Or perhaps my definition of an Addition reaction is a bit narrow?

Reply

randy March 8, 2014 at 10:06 am

Could you explain why in some reaction condition there were also iodine added?

I’m wonder whether its goes through IBH2 reactant or this author(http://totallymechanistic.wordpress.com/2006/12/11/sodium-borohydrideiodine-reduction/) states that the reductant is the borane (BH3) which is generated in situ. Yet some people think it’s the BH3-THF complex being reductant, and then I’m not sure what’s the Iodine for.

Thank you!!

Reply

Leave a Comment