Aldehydes and Ketones
Sodium Borohydride (NaBH4) Reduction of Aldehydes and Ketones
Last updated: February 16th, 2023 |
Sodium borohydride (NaBH4) For the Reduction of Aldehydes and Ketones
- Sodium borohydride (NaBH4) is a convenient source of hydride ion (H-) for the reduction of aldehydes and ketones.
- Aldehydes are reduced to primary alcohols and ketones are reduced to secondary alcohols.
- Esters (including lactones) and amides are not reduced.
- As a source of hydride ion, NaBH4 will also act as a strong base, deprotonating water, alcohols, and carboxylic acids.
- NaBH4 also sees use in the reduction of organomercury bonds after oxymercuration reactions.
Table of Contents
- Sodium Borohydride, NaBH4
- NaBH4 For The Reduction of Aldehydes and Ketones
- Mechanism For the Reduction of Aldehydes and Ketones by Sodium Borohydride
- NaBH4 Will Not Reduce Esters or Amides
- Reduction of Hemiacetals
- Reduction of Organomercury Compounds with NaBH4
- Quiz Yourself!
- (Advanced) References and Further Reading
1. Sodium Borohydride (NaBH4 )
Sodium borohydride (NaBH4) can be made through the addition of sodium hydride (NaH) to our old friend borane (BH3 – See post: Hydroboration-Oxidation of Alkenes) in an appropriately chosen solvent [Note 1]. We generally don’t think of the hydride ion (NaH) as being a very good nucleophile, but the empty p-orbital of BH3 makes this addition much easier.
In contrast to BH3, which is a highly air-sensitive liquid requiring special inert-atmosphere (Schlenk line) techniques, sodium borohydride NaBH4 is a white crystalline solid generally dispensed in the form of pellets, very easily handled and weighed on the benchtop.
It’s worth a reminder about the properties of the B-H bond because this can be a common source of confusion.
NaBH4 has a tetrahedral arrangement of hydrogen atoms about the central boron atom, and a formal charge of -1 on the boron.
That negative charge on boron does not represent a lone pair on boron, however!
Because hydrogen is more electronegative (2.20) than boron (2.04) the electrons in the B–H bond are polarized towards the hydrogen.
So where are the electrons, if they’re not on the boron?
They’re on the hydrogens!
True to its name, sodium borohydride acts as a source of hydride ion, H(-).
You may recall that hydride is the conjugate base of hydrogen (H2) (pKa about 36), making it a very strong base. NaBH4 reacts with water and other weak acids (such as methanol) to generate hydrogen gas (H2).
See if you can draw an arrow-pushing mechanism for the formation of H2 :
(It’s actually quite common to use methanol (CH3OH) as the solvent for sodium borohydride reductions. As long as the temperature is kept low (a dry ice / acetone cold bath at –78°C is common) the bubbling can be kept under control. It’s nowhere near as reactive towards water as lithium aluminum hydride (LiAlH4), which requires rigorously dry solvents to be used).
2. NaBH4 For The Reduction of Aldehydes and Ketones
The most important reaction of NaBH4 is its use in the reduction of aldehydes and ketones to give alcohols.
(You may recall that in organic chemistry, reduction generally refers to a process whereby a C-H bond is formed at the expense of a C-O bond. This results in a decrease in the oxidation state of the carbon – see article: Oxidation and Reduction in Organic Chemistry)
The reduction of aldehydes with sodium borohydride gives primary alcohols. Note the bonds that form and break here – a new C-H bond is formed, and a C-O (pi) is broken. An additional O-H bond forms during during a workup step with mild acid.
The reduction of ketones follows a similar pattern, and results in the formation of secondary alcohols. A C-H bond is formed and a C–O (pi) bond is broken.
Note that if the two R groups flanking the C=O bond are different, a new chiral center will be created. In the case of a simple ketone such as acetophenone (phenyl methyl ketone) this will result in a racemic mixture. (See article: What’s A Racemic Mixture?)
If the molecule already contains one or more stereogenic centers, a mixture of diastereomers will form. One notable example is the reduction of the bicyclic ketone [2.2.1]bicycloheptanone (above). Addition of the hydride ion occurs preferentially from the least hindered face (where there is only one bridging carbon) to give an 86:14 ratio of diastereomers.
Chiral reducing agents similar in reactivity to NaBH4 have been developed that are capable of performing enantioselective reductions of ketones. One prominent example is the CBS (Corey-Bakshi-Shibata) family of reagents.
3. Mechanism for the Reduction of Aldehydes and Ketones With NaBH4
The mechanism for these reductions follows the very common two-step addition-protonation pattern often found in reactions of aldehydes and ketones (See article: The Common Two-Step Pattern for Addition to Aldehydes and Ketones)
The first step in this reaction is nucleophilic addition to the carbonyl carbon. (See article: Nucleophilic Addition to Carbonyls), forming a C-H bond and breaking a C–O (pi) bond.
The addition is followed by protonation of the oxygen with a mild acid (leading to the formation of O–H).
In practice, this reaction is usually performed in an alcoholic solvent like CH3OH and the reaction is quenched with a mild acid such as a saturated solution of ammonium chloride (NH4Cl)
4. NaBH4 Will Not Reduce Esters or Amides
NaBH4 will not generally reduce esters or amides. (Note 2).
(This reaction can be done by LiAlH4 , however. See article – Lithium Aluminum Hydride LiAlH4)
Why are esters and amides so unreactive? After all, shouldn’t these functional groups be more reactive than aldehydes and ketones since the carbonyl is attached to the electronegative oxygen and nitrogen atoms?
It’s actually the opposite! The lone pairs from oxygen and nitrogen are capable of donating electron density to the carbonyl carbon through forming a pi bond. This makes the carbonyl carbon less electrophilic and less reactive with nucleophiles.
(You might recall that this is the exact same reason why OH and NH2 are activating groups in electrophilic aromatic substitution reactions – See Article: Understanding Ortho, Para and Meta Directors)
What about anhydrides and acid halides?
NaBH4 will reduce anhydrides and acid halides, but in practice, these functional groups will react with the solvent (CH3OH) before they have a chance to react with NaBH4.
5. Reduction of Hemiacetals
Aldehydes and ketones, check. Esters and amides, no go. So what else can be reduced by sodium borohydride.
The molecule below might look familiar. It’s glucose!
When treated with NaBH4, glucose is reduced to the alcohol sorbitol.
Hold on for a second. What happened here? There’s no aldehyde or ketone. Or is there?
There actually is an aldehyde present here, but it is in equilibrium with a cyclic hemiacetal. (In cyclic molecules such as sugars, this equilibrium process is known as ring-chain tautomerism – See Ring Chain Tautomerism in Sugars)
Although the open-chain aldehyde form only comprises 0.02% of an aqueous mixture of glucose at equilibrium, NaBH4 will quickly reduce any aldehyde that is present to give sorbitol. Via Le Chatelier’s principle, equilibrium between the cyclic hemiacetal and the aldehyde will eventually result in all of the cyclic hemiacetal being reduced to the alcohol.
See if you can draw the mechanism:
(Sodium borohydride can also be used in the reductive amination of iminium ions to give amines. For more on that, see this article on Reductive Amination. )
6. Reduction of Organomercury Compounds
There’s one more use of NaBH4 worth noting.
You may recall that alkenes can undergo oxymercuration when treated with water (or alcohols) in the presence of mercuric acetate Hg(OAc)2 or similar (See article – Oxymercuration of Alkenes) which results in net Markovnikov addition of water to an alkene.
The resulting organomercury compound can then be treated with NaBH4 to give an alcohol.
The precise details of this “demercuration” step are often skipped over, but for completeness we’ll briefly go through it here.
The first step is addition of hydride to mercury, giving NaOAc and a new Hg-H bond. Carbon-mercury bonds are extremely weak (this is part of the reason why organomercury compounds are extremely toxic) and upon homolytic cleavage of Hg-C, the resulting carbon radical is then reduced with Hg-H to give C-H and metallic mercury (Hg0). On large enough scale, this results in a little puddle of mercury forming at the bottom of the reaction flask.
- Sodium borohydride will reduce aldehydes to primary alcohols and ketones to secondary alcohols.
- This proceeds via a two-step mechanism consisting of 1) nucleophilic addition, followed by 2) protonation.
- Esters and amides are not reduced by NaBH4 under normal conditions. (They can be reduced by lithium aluminum hydride (LiAlH4) however).
- NaBH4 is also used in the demercuration step of oxymercuration-demercuration.
Note 1. In practice NaBH4 is made on industrial scale by the treatment of trimethyl borate [B(OCH3)3] with sodium hydride at high temperatures (250°C).
The reaction with LiH and BH3 in ether works well to make lithium borohydride, LiBH4. However the same reaction between NaH and BH3 requires using either THF or diglyme (ethylene glycol dimethyl ether) as the solvent, not diethyl ether.
Prior to the development of NaBH4, aldehydes and ketones were reduced either with sodium amalgam , sodium in alcohol solvent [Ref] , or through reductions such as the Meerwein-Pondorff-Verley reduction, all of which have various drawbacks. [Ref] Having a convenient crystalline solid as a bench-stable reducing agent has made reductions of aldehydes and ketones much more efficient.
Note 2. The reaction of esters with NaBH4 is extremely slow. However, lithium borohydride (LiBH4) will successfully reduce esters, owing to the greater Lewis acidity of the lithium ion that helps to activate the carbonyl oxygen towards attack. (See article – Acid Catalysis In Addition-Elimination Reactions)
Note 3. If protic solvents are rigorously excluded and the reaction is run in a polar aprotic solvent like DMF, the resulting product of sodium borohydride reduction will be a boronic ester.
This article has more detail.
Note 4. The rate of reduction of cyclobutanone is about 3300 times faster than the reduction of cyclooctanone. This is due to the relief of ring strain upon going from sp2-hybridization (bond angle 120 °) to sp3 hybridization (109 .5 °) . Reduction of cyclo-octanone is considerably slower due to the presence of transannular strain in eight-membered rings.
(Advanced) References and Further Reading
A useful (graduate-level) PDF handout on reducing agents can be found in these course notes from Prof Andrew G. Myers’ Chemistry 115 class at Harvard
Sodium borohydride was discovered in 1942 as part of a wartime research program toward finding volatile compounds of uranium that would enable isotopic enrichment through centrifugation. [Uranium (IV) borohydride, a volatile green solid, was synthesized in pound-scale quantities for this purpose].
- Forty Years of Hydride Reductions
Herbert C. Brown and S. Krishnamurthy
Tetrahedron, 1979, 35, 567-607
A very accessible review on the history and development of hydride reducing agents, including NaBH4 and its many relatives.
- The Preparation of Sodium Borohydride by the High Temperature Reaction of Sodium Hydride with Borate Esters
H. I. Schlesinger, Herbert C. brown, and A. E. FinholtJournal of the American Chemical Society 1953 75 (1), 205-209
NaBH4 was discovered in 1942 but this work was not declassified until 1953. The prepration of NaBH4 through the method described in this paper is still in use today.
- Reaction of Sodium Borohydride With Carbonyl Groups
Brown, H.C.; Wheeler, O.H.; Ichikawa, K.
Tetrahedron 1:214 (1957)
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.
- Mechanistic studies
Brown, H. C.; Ichikawa, K.
Tetrahedron 1:221 (1957)
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.).
- Reference To An Experimental Procedure
Antonio Bermejo Gómez, Nanna Ahlsten, Ana E. Platero-Prats and Belén Martín-Matute
Org. Synth. 2014, 91, 185
The first step in this procedure uses NaBH4 to reduce a cinnamyl ketone to the alcohol.
- Reduction of ketones by sodium borohydride in the absence of protic solvents. Inter versus intramolecular mechanism.
Kayser, M., Eliev, S., & Eisenstein, O.
Tetrahedron Letters, 1983 24(10), 1015–1018.
- Lanthanides in organic chemistry. 1. Selective 1,2 reductions of conjugated ketones
Jean Louis Luche
Journal of the American Chemical Society 1978 100 (7), 2226-2227
Using sodium borohydride for the reduction of unsaturated ketones sometimes results in the side reaction of conjugate reduction, i.e. the reduction of the double bond instead of the carbonyl. This paper shows that selectivity for 1,2-addition (to the carbonyl) can be greatly increased by treating the alpha,beta unsaturated ketone with 1.1 equivalent of the Lewis acid cerium chloride. This makes the carbonyl carbon more electrophilic and allows for a “chemoselective” reduction of the carbonyl. This procedure has become known as the Luche reduction.
68 thoughts on “Sodium Borohydride (NaBH4) Reduction of Aldehydes and Ketones”
What does the reaction between sodium borohydride and copper II oxide look like?
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?
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.
Thank you, James!
I’m sorry, This link would be more appropriate…
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.
Can anyone explain reduction of esters using NaBH4?
Esters don’t generally reduce. Esters can sometimes be reduced with LiBH4, the Li+ is more Lewis-acidic and can coordinate to the carbonyl oxygen making it more electrophilic.
Can NaBH4 reduce benzoic acid to benzaldehyde?
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
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?
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.
Can these reactions go back? What’s stopping it from going back?
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.
Can NaBH4 reduce Beta-keto sulfones to B-hydroxy sulfones.
plz suggest mi
Yep, I would expect that it could. Although the ketone will be spending a considerable amount of time in the enol form.
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
No, even if it does it would make a hemiacetal and formation of that is reversible.
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
Slight excess, e.g. 1.1 equivalents.
My professor said NaBH4 does not reduce esters but can reduce lactones which are cyclic esters. What is the reason for that?
Thank you for your practical tips, I’m a first year graduate student and this website is an amazing resource.
Can NaBH4 and CuCl reduce alkene(olefine)?
Usually this combination makes copper hydride, which is a soft reductant for reducing alpha beta unsaturated ketones/esters etc. I don’t believe it works for normal alkenes.
Hi, I want to reduce ketone in the presence of primary alcohol is there any effect of free of alcohol with NaCNBH4 and ZnI2
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?
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!
If in a alkane chain there is a C=C and aldehyde group. Will NaBH4 reduce both?
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.
What are the advantages of using NaBH4 as reducing reagent in comparison to LiAlH4??
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.
What if NABD4 is used?
Then you will form a C-D bond instead. Can be a very useful method for incorporating deuterium.
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)
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).
I want reduce only aldehyde in the presence of esteric grp
Easy enough in most cases. NaBH4 should be fine.
Would NaBH4 reduce an alkene adjacent to an imine?
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?
Thank you for this valuable comment.
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.
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
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.
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.
Hi, I have a question as to why NaBH4 is good for ketones/aldehydes but why it won’t work for esters?
Any thoughts on why esters might be less reactive towards nucleophiles than ketones (or aldehydes) ?
Dose chlorobenzene reacts with NaBH4,and why?
No, absolutely not.
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?
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.
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!
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.
It is possible to reduce nitriles to amine with NaBH4 in the presence of Co (II).
Yes, cobalt hydrides are produced here and are great for reducing nitriles!
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?
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.
Can NaBH4 react with CH3CN while producing CH3CH2NH3?
No, NaBH4 will not react with nitriles. You’d need LiAlH4 for that.
Can NaBH4 react with CH3CN? From https://accespedia.com
No. NaBH4 will react with aldehydes and ketones, but not esters or nitriles.
Hi, thank you for the post, I did found it very interesting but I have one question:
I did myself one reduction with NaBH4 in the laboratory of organic chemistry, but the weird thing its that I was using MeOH as solvent for the ethylphenilketone to reduce it to 1-phenylpropan-1-ol, but the NaBH4 was prepared in water and was aded to the reaction mixture dropwise. How its posible that the NaBH4 don’t decomposes in protic solvents?
It does decompose in protic solvents (there are 4 equiv of hydride per mole of NaBH4) but the decomposition is relatively slow compared to, say, LiAlH4.
[edited your comment to say, “dropwise”]
The last comment I did its wrong, I wanted to say dropwise instead of drowse and I can’t edit it
Edited it to say dropwise
I have a fun project I am working on right now but I think I am running into some issues.
I am trying to make a di-ol derivative of caffeine. Initially, I would like to reduce both ketones to secondary alcohols. So far, I have tried NaBH4 (2 molar excess) + caffeine in MeOH (dry-ish / used some 3A dessicant beads) and stirred at 0C for 8 hours.
I do not seem to have made the product. Any suggestions?
Thanks in advance!!
Hi, there’s a fundamental problem with trying to do that. The functional groups are not actually ketones, they are flanked by nitrogens with lone pairs and this affects their reactivity significantly. Furthermore even if reduction were achieved the resulting product would not be particularly stable; it would likely dehydrate (lose water) to give the aromatic heterocycle known as “purine”. In fact if you look up the structure of purine you will note the similarity to caffeine (and DNA bases adenine and guanine).
In short it’s *possible* to make the reduced form(s) of caffeine but you would need some way of protecting the resulting OH groups, and furthermore you would need to prevent the molecule from dehydrating to purine.
Can Sodium Borohydride reduce Acid anhydrides? I have seen some sources saying that.can you explain it?
It actually can reduce cyclic anhydrides to esters. When reducing anhydrides however you have to be careful about the choice of solvent, as solvents under basic conditions (e.g. MeOH with NaBH4) will just undergo nucleophilic acyl substitution.