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Organic Reagents

By James Ashenhurst

Reagent Friday: Sodium Borohydride (NaBH4)

Last updated: March 27th, 2019

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.  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).

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.


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.

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.

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Comment section

37 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?


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