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Organic Reagents

By James Ashenhurst

Reagent Friday: NBS (N-Bromo Succinimide)

Last updated: May 24th, 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. 
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.

1-nbs

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.

2-nbs 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.

3-nbs

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

4-nbsRecall 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”.

5-nbsThere 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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33 thoughts on “Reagent Friday: NBS (N-Bromo Succinimide)

  1. “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″…

    1. When cyclohexanone reacts with NBS/CCl4 +aq KOH we get cyclopentanoic acid as final product .I found this on a test. Can you please tell me what happened here

    1. 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

  2. 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.

    1. 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

  3. 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)

  4. 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!

  5. regarding your statement in “NBS (N-Bromo Succinimide)”

    “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.” The reason for this is not density, but bromine’s high vapor pressure. Drawing it into a pipet immediately expels it because the pressure builds in the pipet. Even the warmth of your hand on a pipet filled with bromine will pressurize the pipet and expel the bromine.

  6. James,

    NBS generates a bromine radical which then abstracts the allylic Hydrogen on propene because the allylic c-h bond is the weakest c-h bond due to resonance of the resulting allylic radical.

    However, I was also taught that bromine radicals generated from hbr reacting with RO radicals (generated by homolytic cleavage of a peroxide bond) will react with the pi bond generating the 1-broom-2-propyl radical as a propagation step of antimarkovnikov addition of hbr in the presence of peroxides.

    So where do br radicals react, at the pi bond or at the allylic c—h bond?

    1. The radicals will add to the double bond, but the radical created by that (carbon-based radical) needs something to react with to continue the chain reaction. When HBr is present, it reacts with H-Br to regenerate Br• and the cycle continues. However with allylic bromination, there isn’t a source of HBr present that can continue the chain. Therefore if radical addition happens, it just reverts back to the double bond and bromine radical. In other words, addition to the double bond is a dead end.

  7. into what is nbs converted as it reacts in allylic bromination reactions?
    – i thought it would be bromine but that is incorrect. Could it be bromine radical ?

  8. Hydrogen peroxide and bromine also gives radical bromine why does only n bromo succinamide give allylic product

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