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)


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.


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.

{ 12 comments… read them below or add one }

Thomas July 12, 2013 at 10:43 am

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


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.


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


James Ashenhurst September 17, 2013 at 10:32 am

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


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?


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( 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!!


James July 22, 2015 at 10:17 pm

Iodine is used to form BH3 in situ from NaBH4 . It’s used, for instance, in the reduction of carboxylic acids to alcohols.


Vanessa March 5, 2015 at 1:56 pm

Hey, what happens when there is an excess of NaBH4?


James July 22, 2015 at 10:16 pm

Nothing bad should happen. Once the aldehyde is reduced to the alcohol, there’s nothing else that can happen (assuming you just have one aldehyde).


aditya krishna July 6, 2015 at 3:46 am

what if we use NaBD⁴ in H²0² with OH·-· to reduce propene ? The answer is ⛪ CH³CHDCH²CH²OH.. please give the mechanism


Haydan White December 7, 2015 at 7:23 am

Got a question regarding reduction of P-nitrobenzealdehyde to p-nitrobenzyl alcohol. We mixed sodium hydroxide and potassium borohydride and can’t figure out how the mechanism takes place. Does sodium hydroxide have anything to do with it, or is it only the aldehyd + potassium?



James December 7, 2015 at 11:11 am

I’m not sure why you added NaOH with KBH4. By the way why did you use KBH4 – you usually have to make it, and NaBH4 is much cheaper?

The mechanism for reduction with a borohydride is simply that the B-H bond breaks, donating two electrons to form a new bond with the carbonyl carbon, and the C-O pi bond breaks. You end up with an alkoxide, which is then protonated to give the alcohol.


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