Organic Reagents

By James Ashenhurst

Reagent Friday: Sodium Borohydride (NaBH4)

Last updated: January 29th, 2020 |

Sodium Borohydride (NaBH4) For Reduction Of Aldehydes And Ketones (And Demercuration)

In a blatant plug for the Reagent Guide, each Friday  I profile a different reagent that is commonly encountered in Org 1/ Org 2. Version 1.2 just got released, with a host of corrections and a new page index. 

Having just talked about the oxidation ladder, it makes sense to start going into reagents for oxidation and reduction reactions.

Sodium borohydride (NaBH4)


What it’s used for: Sodium borohydride is a good reducing agent. Although not as powerful as lithium aluminum hydride (LiAlH4), it is very effective for the reduction of aldehydes and ketones to alcohols. By itself, it will generally not reduce esters, carboxylic acids, or amides (although it will reduce acyl chlorides to alcohols).  It is also used in the second step of the oxymercuration reaction to replace mercury (Hg) with H.

Similar to: lithium aluminum hydride (LiAlH4) although less reactive.

Reduction Of Aldehydes And Ketones With NaBH4

For our purposes, sodium borohydride is really useful for one thing: it will reduce aldehydes and ketones. In this sense it traverses one rung on the oxidation ladder. Here are some examples of it in action.


Notice the pattern: we are breaking a C-O bond and replacing it with a C-H bond. This is what helps us classify the reaction as a reduction.

Note that we also form an O-H bond. This is where textbooks and other sources are sometimes not as clear as they should be: in order to make the alcohol, the oxygen needs to pick up a proton (H+) from either water or acid that is added after the reaction is complete (note: this is often referred to as the workup).

Sodium Borohydride For “Demercuration” (The Second Step Of Oxymercuration-Demercuration)

NaBH4 also makes an appearance in the oxymercuration reaction. Specifially, NaBH4 is used in the second step of the reaction, to break the C-Hg bond and turn it into a C-H bond.


Mechanism For The Reduction Of Aldehydes And Ketones With NaBH4

How it works: The mechanism of the reaction of sodium borohydride with aldehydes and ketones proceeds in two steps. In the first step, H(–) detaches from the BH4(–) and adds to the carbonyl carbon (an example of [1,2]-addition). This forms the C-H bond, and breaks the C-O bond,  resulting in a new lone pair on the oxygen, which makes the oxygen negatively charged (FYI: we call these negatively charged oxygens alkoxides, as they are deprotonated alcohols).  In the second step, a proton from water (or an acid such as NH4Cl) is added to the alkoxide to make the alcohol. This is performed at the end of the reaction, a step referred to as the workup.



I suppose I should also mention that NaBH4 will reduce acyl halides to alcohols, but things are a little lengthy here already.

Note: this mechanism assumes a polar protic solvent. What if you use a slightly different solvent? You have a slightly different mechanism, see note below. 

I also won’t go into the detailed arrow pushing for NaBH4 in the oxymercuration reaction. But the key point is that the carbon-mercury bond is broken, and a new carbon hydrogen bond is formed, and it is NaBH4 which performs this reaction. It works out like this:


P.S. You can read about the chemistry of NaBH4 and more than 80 other reagents in undergraduate organic chemistry in the “Organic Chemistry Reagent Guide”, available here as a downloadable PDF.


A note about the mechanism.

The mechanism drawn above works in a polar protic solvent like methanol, which can protonate the alkoxide. What happens if you use a non-protic solvent like DMF? Well, since you don’t have a proton source, you form a salt! It would look something like this:


This article has more detail:

Reduction of ketones by sodium borohydride in the absence of protic solvents. Inter versus intramolecular mechanism.
Kayser, M., Eliev, S., & Eisenstein, O.
Tetrahedron Letters1983 24(10), 1015–1018.
DOI: 10.1016/S0040-4039(00)81590-7


(Advanced) References and Further Reading:

  1. First example
    Brown, H.C.; Wheeler, O.H.; Ichikawa, K.
    Tetrahedron 1:214 (1957)
    DOI: 10.1016/0040-4020(57)88041-7
    Early paper by Nobel Laureate H. C. Brown describing the reactivities of simple aldehydes and ketones to reduction by NaBH4, in which it is shown that aldehydes are more reactive than ketones to nucleophilic reactions.
  2. Mechanistic studies
    Brown, H. C.; Ichikawa, K.
    Tetrahedron 1:221 (1957)
    DOI: 10.1016/0040-4020(57)88042-9
    This paper and the above are both mechanistic studies on the reduction of carbonyls – this paper investigates the effect of ring size on the reduction of cyclic ketones (e.g. reduction of cyclobutanone vs. cyclopentanone, cyclohexanone, etc.).
  3. Reference To An Experimental Procedure
    Antonio Bermejo Gómez, Nanna Ahlsten, Ana E. Platero-Prats and Belén Martín-Matute
    Org. Synth. 2014, 91, 185
    DOI: 10.15227/orgsyn.091.0185
    The first step in this procedure uses NaBH4 to reduce a cinnamyl ketone to the alcohol.


Comment section

56 thoughts on “Reagent Friday: Sodium Borohydride (NaBH4)

  1. Hi James: in the ACS study guide (for the ACS organic exam) they state that NaBH4 “reduces ketones to secondary alcohols but does not react with alkenes.” My textbook (McMurry) says that “unsaturated ketones often undergo overreduction with NaBH4 to give a mixture of both unsaturated alcohol and saturated alcohol.”

    I will go with the ACS guide’s interpretation of reality for the exam, I guess, but which source is correct?

    Thank you!

    1. In Org 2 you learn about “conjugate addition” (like the Michael reaction) where nucleophiles add to alkenes adjacent to a carbonyl. NaBH4 can do “conjugate addition” as well. However conjugate addition only occurs when there is a good electron withdrawing group adjacent to the alkene. It doesn’t work for, let’s say, butene.

    1. Pretty contrived question, but reduction of the ketone gives a secondary alcohol. After that, how can you possibly differentiate between the alcohol you want to reduce and the alcohol on the 5-membered ring? You need a procedure that reduces them both at once. I hate the question, but given the options, the Wolff Kishner is the way to go.

    1. No, it can only reduce aldehydes and ketones. To reduce benzoic acid to benzaldehyde, you’d need to either: 1) treat with LiAlH4 to get the alcohol, then treat with PCC, Swern, or DMP to get the aldehyde. Or 2) use SOCl2 to make the acid halide, then reduce to the aldehyde with LiAlH(OtBu)3

  2. I have C=C double bond in conjugation with ester in Biginelli compounds. If this reagent NaBH4 will work for reduction of C=C bond?

    1. It’s going to be tougher because the ester enolate intermediate is a stronger base than the ketone enolate. Conjugate reduction works for alpha-beta unsaturated nitriles, so it’s reasonable to think that it should work for alpha-beta unsaturated esters too.

    1. What’s to stop it going back is that you’d have to break a C-H bond, which is not easy to do. These reactions are not reversible. Once you get the alcohol, however, you could always re-oxidize to the aldehyde or ketone with something like PCC.

  3. Hello James
    I’m studying the reduction of aldehyde with NaBH4. If we use MeOH as the solvent, the protonation step will generate a MeO-
    It’s a good nucleophile and I wondering whether it will attack the carbonyl group and make side product

  4. What should be the molar ratio while using NaBH4 on commercial production of a molecule to be converted to alcohol?
    e.g acetophenone to corresponding alcohol

  5. My professor said NaBH4 does not reduce esters but can reduce lactones which are cyclic esters. What is the reason for that?

  6. Thank you for your practical tips, I’m a first year graduate student and this website is an amazing resource.

  7. Hi, I want to reduce ketone in the presence of primary alcohol is there any effect of free of alcohol with NaCNBH4 and ZnI2

    1. An ordinary alcohol should not interfere with the reduction of the ketone, but then you will have two alcohols and will have to differentiate them somehow. How do you plan to do that?

      1. One of them is from a ketone, so it’s bound to be secondary. You have another primary alcohol, so I suggest Lucas test(Dil.HCl + Zinc Chloride)

        Hope I helped!

    1. Yes, it can. The way around that is to add anhydrous CeCl3, which will lead to a regioselective reduction of the carbonyl and it will leave the alkene alone. This is called the Luche reduction.

    1. Much milder reagent. Only reduces aldehydes and ketones, so you don’t have to worry about your esters getting reduced. Also, the workup is very simple. Usually just add saturated NH4Cl or similar, extract and dry. With LiAlH4, the aluminum salts are a pain, so you have to stir with Rochelle’s salt or do the Fieser workup protocol.

  8. Hi, I’ve got a question. I’m supposed to identify a carbonyl compoud with the formula c3h6o and one of the clues is that “reaction with NaBH4 in the presence of water produced a colourless liquid”. What does this mean, and what compound could it be? (I know it’s not an aldehyde and I already know that one of the possible structures is a ketone)

    1. The product is likely some kind of alcohol. The water was likely there to provide a proton source so that you get the neutral alcohol product (not the alkoxide, which would be a salt). Working backwards, you have a carbonyl compound of some kind. C3H6O only gives you two places to put it – on C-1 (which would be propanal) or C-2 (which would be acetone).

    1. Hi MIchael – in general NaBH4 will reduce the alkene in alpha,beta unsaturated ketones (“conjugate reduction”) unless a strong Lewis acid such as CeCl3 is used (“Luche reduction”).

      NaBH4 will also do conjugate reductions of alpha beta unsaturated nitriles.

      With conjugated imines I am not 100% sure. The imine you’ve drawn is a simple one with an N-H substituent. These types of imines are not very stable and easily lose water.

      My first guess is that NaBH4 in the absence of a Lewis acid would also perform conjugate reduction but I am not sure.

      Is this for laboratory work or is this exam question related?


  9. Hi, we did a reduction of ketone with NaBH4. We had ketone in n-hexanne, we added methanol and NaBH4, we stirred, left for couple minutes, than added 0.1M EDTA. What is the purpose of adding EDTA? Thank you for an answer.

    1. Haven’t heard of that workup before. The byproduct of the reaction is BH3 and the sodium salt of the alcohol (alkoxide). EDTA is a weak acid and also a metal complexing agent. When you add the aqueous solution of EDTA, you’ll protonate the alcohol and any Na+ ions will end up coordinated to it, as well as any residual BH3/borate salts

  10. Hi, in the reduction with water, what becomes of the OH- ion. Does the solution just become more alkaline?

    PS thanks for your great website.

    1. Yes, NaBH4 plus H2O gives H2 , BH3, and NaOH. The solution does become basic. For that reason the reaction is often quenched with a very mild acid (saturated ammonium chloride) which protonates the conjugate base of the alcohol and sequesters any BH3.

  11. Hi, I have a question as to why NaBH4 is good for ketones/aldehydes but why it won’t work for esters?

  12. Hey,I have a doubt. Why can’t NaBH4 reduce nitriles when it can reduce imines? Shouldn’t triple bond be easier to reduce compared to double bond?

    1. Generally requires a stronger reducing group like LiAlH4. The problem is that addition of hydride to nitrile forms C=N(-) and the negatively charged nitrogen is quite basic. Can’t find a pKa number at the moment unfortunately.

  13. Would you have some suggestions on how to clean up after nitrile reduction to amines? I thought to use acid to break down the NaBH4 first, then add base, and extract with solvent, then dry with NaSO4 (anh). Is this about right? Thanks!

    1. That sounds about right, but instead of trying to re-invent the wheel, I would look for a literature procedure and use that. Supporting information in papers is usually free to access.

  14. Two more questions:
    I saw H2 evolving after I added NaBH4 to MeOH?. Is this because of residual moisture in the solvent?
    Can I run the NaBH4 reaction in Ammonium-MeOH?

    Thank you!

    1. Yes, H2 is evolved. That’s because NaBH4 is basic (the hydride ion H-) , and will react with weak acids (like MeOH) to give H-H (hydrogen gas). The reaction of NaBH4 with MeOH is slower than the reaction of NaBH4 with aldehydes/ketones, especially at low temperatures. It’s common to run NaBH4 reductions at very low temperatures (e.g. -78) for this reason.

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