Reagent Friday: Sodium Borohydride (NaBH4)

by James

in Alcohols, Aldehydes, Ketones, Organic Chemistry 2

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)

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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{ 4 comments… read them below or add one }

Francis Onyejaka September 14, 2011 at 3:48 pm

What does the reaction between sodium borohydride and copper II oxide look like?

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john_k June 5, 2013 at 7:53 pm

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!

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james June 8, 2013 at 7:23 pm

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.

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john_k June 17, 2013 at 11:02 pm

Thank you, James!

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