Free Radical Reactions
By James Ashenhurst
Bonus Topic: Allylic Rearrangements
Last updated: March 22nd, 2019
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:
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!
What’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.
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:
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?
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]