My book asks me to consider reacting 1-butene with NBS and light.

I know that NBS and light leads to radical formation, and this is an allylic/benzylic bromination problem.

However, my book only provides one product.

Aren't both products expected? Actually, I would expect one product in excess (the book actually listed my minor product as the sole product). There are at least two resonance contributors, and the more major one has the more substituted radical.

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2 Answers 2


Yes you are correct, at least two products are formed (ref. 1, ref. 2). Unfortunately, neither of these references provide the actual product distribution between the isomers.

I say that "at least two products are formed" because depending on your reaction conditions 3 products might be expected. NBS brominations can be carried out over a range of temperatures, from cold to somewhat above room temperature. The activation energy required to rotate about an allylic bond is only ~15-16 kcal/m (see here). Therefore, your 1-bromo isomer could exist as a cis-trans mixture, unless the reaction is performed at very cold conditions.

In the analogous case of NBS bromination of 1-octene, a mixture of 3-bromo-1-octene (~18%) and 1-bromo-2-octene (~82%) - both cis and trans are produced.

$\ce{RCH2CH=CH2 + NBS \rightarrow RCHBrCH=CH2 + RCH=CHCH2Br}$ (cis & trans)


Your textbook is wrong. Both products are possible. You are oversimplifying your analysis into which product is the major product.

While a secondary radical may usually be more stable than a primary radical due to inductive stabilization from alkyl groups, this is an allyl radical and the pi bond is important.

Remember Zaitsev's Rule. Pi bonds are also stabilized by alkyl groups. So now when we compare the two resonance contributors to see which one is "more important", we need to compare two competing factors:

                           Radical        Alkene
your "minor" radical       primary        internal
your "major" radical       secondary      terminal

However, as our computational chemists will tell us, this sort of analysis is goofy because the two radicals we are comparing are the same radical. Thus a more complete picture would require some computational calculations to tell us where the "radical density" is or how the distribution of the SOMO (singly-occupied molecular orbital) compares to the van der Waals surface (in other words, which position is more sterically accessible).

I bet you forgot sterics! I do not have the capability right now to plug the butenyl radical into a modeling program and generate the convincing pictures, but I can predict what we would learn:

  1. The molecular orbitals and electron density distribution would confirm that the pi bond has more density between carbon atoms 2 and 3 than between carbon atoms 1 and 2 (in other words, the pi bond's position is more important than the radical's position).
  2. The "radical density" and the SOMO will be more concentrated at carbon atom 1 than at carbon atom 3.
  3. The radical density and the SOMO will extend outside of the van der Waals surface at carbon atom 1 and they will remain inside the van der Waals surface at carbon atom 3 (in other words, carbon atom 1 is more sterically accessible).

I would love to see some MO pictures or electron/radical density maps compared to van der Waals surfaces to confirm this.


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