Bonus Topic: Allylic Rearrangements

by James

in Alkenes, Organic Chemistry 1

In this series on free radical reactions we’ve mostly covered the basics. In this post (and the next one) we’re going to go into a little bit more detail on certain topics that until now I haven’t had time to dive into. 

Today’s topic flows right from the subject of the last post, on allylic bromination

The examples I used in the allylic bromination post were actually quite simple. For example, if you take cyclopentene and treat with NBS and light (hv) in carbon tetrachloride solvent (CCl4), you get this product: 

1-cyclopentene Let’s extend the complexity of the substrate just a bit. Just a bit – one methyl group! and do the same reaction. Here, we get not one product, but two. And note how our major product is different!

1-methylcyclopenteneWhat’s going on here?

Let’s think about the mechanism of this reaction. What’s the first thing to happen (after initiation, of course)?

 Removal of the weakest C-H bond by the bromine radical! This leaves us with an allylic radical, which can then react with Br2 to give us product A. 
2-product a

 Simple enough. But how do we explain the formation of product B? 

 Look again at the free radical that is produced. Notice anything special about it? 

We can draw a resonance form! 


This means that there are two carbons on this molecule which can potentially participate in free radical reactions. Therefore we can also draw a reaction mechanism which shows Br2 reacting at this bottom (tertiary) carbon: 
3-product b
See how the tertiary radical forms a new π bond while the other π bond breaks? This leaves one of the electrons of the “top” alkene to form a new bond with bromine, giving us a new C-Br bond. If you analyze the bonds that form and break in this reaction (always a useful exercise), note that this reaction has an extra pair of events – one C-C π bond breaks, and one C-C  π bond forms. The net effect is that it looks like the  π bond has moved. This phenomenon is called “allylic rearrangement”. 

Note: as commenter Keith helpfully points out, remember that resonance forms are “hybrids”. When drawing the mechanism, it’s best to show it all happening in one step (as in the middle drawing, above) rather than to draw the resonance form and then draw bromination. 

Last question. Can you think of a reason why product B might be more favoured, especially under conditions of high temperature?

5- a vs b

Think back to Zaitsev’s rule (if you’ve covered this) : the more substituted an alkene is, the more stable it is (why? the reason is complex and usually not covered in introductory textbooks – it has to do with a phenomenon called “hyper conjugation”) . The alkene in product A is what we’d call “disubstituted” – it is directly attached to two carbon atoms and two hydrogen atoms. The alkene in product B is “trisubstituted” – it is directly attached to three carbon atoms and one hydrogen atom. Therefore there is good reason to expect that product B will be a significant product in this case. [I’m hedging on the exact ratio because I don’t have a literature reference. You shouldn’t completely believe me without firm data from a literature reference; I’ll try to dig one up]. 

Next Post: Free Radical Addition Of HBr To Alkenes

[Note, Dec 5, 2013 – significantly revised from previous version. Thanks to commenter Keith for constructive criticism] 

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{ 9 comments… read them below or add one }


Excellent post, but I really dislike the term “more stable resonance form”. Resonance structures are not discrete entities, and all resonance forms must have exactly the same energy since there is only one true structure. The better way to express this concept is, as you said, that one form makes a greater contribution to the resonance hybrid.

I am also surprised that compound B is the major product since (1) it contains a less highly substituted double bond (disfavored thermodynamically via the Hammond postule) and (2) halogenation is occurring at the more hindered site (presumably disfavored kinetically). Moreover, I would expect compound B to rearrange under the reaction conditions via the SN1 mechanism to form compound A. Is there a specific literature reference for this reaction or can you provide the actual product ratio? Thanks, and keep up the good work with the blog!



You have written an excellent post about radical substitution. It would be perfect if you give some explanation about why the compound does not follow E1 instead of SN1. Thank you for your hard work :)



Nice explanation
But I just have one doubt regarding the above mechanism.
Why didn’t we take into consideration the stability of the free radical?
In my opinion it should be a significant factor as the whole thing follows the free radical mechanism.
Looking forward to your explanation.


Andrea Jurado

It depends on the temperature!

I think the main point why we covered the thermodynamically product is because it’s conceptually the point of the rearrangement. The most stable alkyl radical would be the major product at low temperatures, because it is kinetically favored due to the faster rate determining step. However at HT, thermodynamics trump kinetics to give you the more substituted alkene/ (mark pdct).

Good point!



should we check the stability of the free radical intermediate or the final alkene?
in B the free radical is unstable as compared to A…
so shouldnt A be the major product?



Love the site! Just want put in encouragement to finish this conversation. Both a and b seem like they could be the major product for mutually exclusive reasons. Looking for a way to reconcile all the points of view and get the definitive answer.



Product B is major because the double bond is more substituted, which is ultimately more stable (similar to why eliminations tend to occur in a way to give the most substituted alkene – Zaitsev’s rule).



Hey James. Is it possible for the bromine to also bond with the CH3 hanging off the double bond in product B?



Hi – if one had an excess of NBS (>1 equivalents) then a second allylic bromination could occur, and that CH3 hanging off the double bond in B is one position where it could happen.

The rate of that reaction will be proportional to the concentration of NBS and the concentration of B, so as the reaction proceeds (and the concentration of B goes up) we should expect to see a little bit of it happening.


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