Free Radical Reactions of Alkanes and Alkenes, Initiation Propagation, and Master Organic...

The basics have been covered in the series on free radical reactions.We are going to go into a little bit more detail on certain topics that I haven't had time to think about.

The subject of the last post was allylic bromination.It is on allylic bromination reactions.

You get this product if you take cyclopentene and treat it with light and light emitting particles.

The complexity of the substrate should be extended a little bit.Do the same thing if you replace one of the C–H bonds with a CH3 bond.

Think about the mechanism of the reaction.What is the first thing to happen after initiation?

The weakest C-H bond was removed by the bromine radical.This leaves us with an allylic radical, which can give us product A.

There are two carbons on this molecule which can be involved in free radical reactions.We can draw a reaction mechanism which shows the reaction at the bottom.

The tertiary radical forms a new bond while the other bond breaks.A new bond with bromine is formed by one of the electrons from the top alkene.If you analyze the bonds that form and break in this reaction, you should notice that there are at least one C-C bond forms.It looks like the bond has moved.The phenomenon is called anallylic rearrangement.

Remember that resonance forms are also called hybrid forms.Rather than drawing the resonance form and then drawing bromination, it is better to show the mechanism in one step.

The last question.Is there a reason why product B is more preferred under high temperature conditions?

The more substituted an alkene is, the more stable it is.

Why?The reason is complex and not covered in introductory textbooks.

Di-substituted alkene is what we call it and it is attached to two carbon atoms and two hydrogen atoms.The alkene in product B is made up of three carbon atoms and one hydrogen atom.Product B will be a significant product in this case.I don't have a literature reference so I'm hedging on the exact ratio.I will try to find firm data from a literature reference.

The note was significantly revised from the previous version.Thanks to the person who gave constructive criticism.

Excellent post, but I don't like the term "more stable resonance form".All resonance forms must have the same energy since there is only one true structure.One form making a greater contribution to the resonance hybrid is the better way to express this concept.

I am surprised that compound B is the major product since it contains a less highly substituted double bond and halogenation is occurring at the more hindered site.I would expect compound B to be rearranged under the reaction conditions.Is there a specific literature reference for this reaction?Keep up the good work, thanks!

You wrote an excellent post about radical substitution.It would be great if you could explain why the compound doesn't follow E1 instead of SN1.Thank you for your hard work.

I have one doubt about the above mechanism.The stability of the free radical was not taken into account.The whole thing follows the free radical mechanism, so it should be a significant factor.Looking forward to your explanation.

The main point is that the product is conceptually the point of the rearrangement.The major product at low temperatures is the alkyl radical and it is favored due to the faster rate determining step.The more substituted alkene you get, the better.

The hunger of the sp2 carbon atom is satisfied by the hyperconjugative effects of more substituted alkenes.Is Bromine more hungry than sp2 carbon?Wouldn't product A be more stable if bromine was on a tertiary carbon atom?

The free radical intermediate or the final alkene should be checked.The free radical is unstable as compared to A.

I love the site!Just want to finish the conversation.It seems like a and b could be the major product.Looking for a way to get the definitive answer.

Product B is major because the double bond is more stable, similar to why eliminations tend to occur in a way to give the most substituted alkene.

Hey, James.Is it possible for the CH3 to hang off the double bond in product B?

CH3 hanging off the double bond in B is one position where a second allylic bromination could occur if one had an excess ofNBS.

As the concentration of B goes up, we should expect to see a little bit of the reaction happening.

Hi!If one started with the unbrominated version of Product B, would the National Bureau of Standards favor bromination of the group?Thank you so much!

If we look at the transition state leading to B and A, we find that A has a tertiary free radical which is more stable than B.