Reagent Friday: NBS (N-Bromo Succinimide)

by James

in Alkenes, Alkyl Halides, Organic Chemistry 1, Organic Reagents

In a blatant plug for the Reagent Guide, each Friday  I profile a different reagent that is commonly encountered in Org 1/ Org 2. 
If you’ve ever had the “pleasure” of working with bromine (Br2), you’ll know that this dense orange liquid is a pain in the butt for two reasons.  First of all, it fumes like a bastard. Once you open the bottle, orange fumes start spewing everywhere, and if you haven’t put the bottle deep into the fume hood, you will soon be savoring the unforgettable aroma of Br2 (mixed with HBr) in your nostrils. Secondly, it’s extremely dense (d=3.19) and therefore drops of it tend to fall from whatever you’re using to dispense it with, Jackson Pollock style,  leaving little violently fuming orange puddles behind.

In contrast, NBS (N-bromo succinimide) is a gleaming white crystalline solid and easy as pie to work with.  But don’t be deceived. It packs a punch. It will do many of the same reactions as bromine – attached to the electron-withdrawing nitrogen of succinimide, the bromine has a partial positive charge and is therefore electrophilic.

There are two major reactions NBS is used for in Org 1/ Org 2: allylic bromination (the most common) and also as a replacement for Br2 in the formation of bromohydrins.

Allylic bromination is the replacement of a hydrogen on a carbon adjacent to a double bond (or aromatic ring, in which case it’s called benzylic bromination). NBS is used as a substitute for Br2 in these cases since Br2 tends to react with double bonds to form dibromides. The advantage of NBS is that it provides a low-level concentration of Br2, and bromination of the double bond doesn’t compete as much.

Once Br2 is formed, the reaction proceeds much like other free-radical halogenation reactions: homolytic cleavage of the Br2 with light or head (initiation), followed by abstraction of the allylic H (propagation step #1) and subsequent reaction of this radical with another equivalent of Br2 to give the desired product. The remaining Br radical then reacts with another equivalent of the hydrocarbon in this chain reaction until the limiting reagent is consumed.

NBS can also serve as a replacement for Br2 in formation of halohydrins.

Recall that alkenes react with Br2 to form “bromonium ions”, which are 3-atom rings with a positive charge on the bromine. Well, NBS will also form bromonium ions with alkenes. When water (or an alcohol) is used as a solvent, it will attack the bromonium ion, resulting in formation of the halohydrin. Note that the stereochemistry is always “trans”.

There are tons of other uses for NBS beyond what you see in Org 1/Org 2, of course, but those are the basics.

P.S. You can read about the chemistry of NBS 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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{ 11 comments… read them below or add one }

mamid June 11, 2011 at 6:01 am

“The advantage of Br2 is that it provides a low-level concentration of Br2″… It should read “The advantage of NBS is that it provides a low-level concentration of Br2″…


James June 11, 2011 at 6:38 am

Fixed. Thanks for the correction!


Alex February 25, 2014 at 11:25 pm

Great! Exactly what I was looking for and explained beautifully.


Felipe February 23, 2015 at 8:51 am

Is it possible to carry out electrophilic aromatic substitutions using NBS as the brominating agent like we do with Br2?


James February 28, 2015 at 2:56 pm

Belated to answer your question but yes, Lewis acids are compatible with NBS and make it more reactive. It can be used to brominate aromatics. Usually works best for electron rich aromatics like phenol and pyrrole. For electron poor species I’d stick with Br2 and FeBr3 or similar.

Sorry for the late response, hope it’s helpful James


Felipe February 28, 2015 at 7:30 pm

Very helpful! Thank you!


Sarah February 28, 2015 at 8:10 pm

When 4-methyl-hex-2-ene reacts with NBS, you get two products. How do you know which of the products is the major product? My teacher has 2-bromo-4-methyl-hex-3-ene as the major product.


James March 8, 2015 at 3:02 pm

Hi – that’s not an easy question to answer. Often the product depends delicately on the reaction conditions, especially the temperature. I would certainly expect that the hydrogen on C-3 is abstracted first (as opposed to C-1) giving a tertiary allylic radical (vs. a primary allylic radical) but where the c-Br bond forms is hard to predict. One thing to note is that forming the C-3 C4 double bond is more substituted (more stable) than the C2-C3 double bond (less substituted, less stable) and higher temperatures would tend to favor that product.
I don’t have a simple answer for you unfortunately. Thanks for writing, James


MVS Saketh May 2, 2015 at 11:51 pm

Does the radical intermediate undergo rearrangement to form a more stable free radical (if possible) , I have read that radicals have little tendency to rearrange unlike carbocations,
But since the radical here is in conjugation with the double bond, shouldnt there be rearrangement (because of the contribution of second canonical form)


James May 3, 2015 at 10:29 am

Hi, they don’t undergo H or alkyl shifts like carbocations do, but they do undergo allylic rearrangements.


james L November 5, 2015 at 9:26 am

How about mono-bromination of cyclopentylphenylmethanone to (1-bromo-cyclopentyl)-phenyl-methanone ?

Using n-bromo succinimide?

I wanted to insert images into this post but it does not seem possible to do so.

Thanks for your attention!


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