Free Radical Reactions
Bonus Topic: Allylic Rearrangements
Last updated: December 7th, 2022 |
Allylic Rearrangements – Allylic Bromination With Rearrangement
- In allylic bromination reactions, the C-H bond of an allylic carbon breaks, and a new C-Br bond is formed.
- This reaction sometimes occurs with allylic rearrangement. That is, the new C-Br bond forms on the end carbon of the pi-bond, and the pi bond moves over such that the original sp3-hybridized C-H carbon becomes an alkenyl carbon.
- This can be favored in cases where a more substituted double bond results (i.e. Zaitsev’s rule)
Table of Contents
- An Allylic Bromination That Gives Two Products
- “Product B” Is An Example Of Allylic Bromination Accompanied By Allylic Rearrangement
- Why Might “Product B” Be Favored Here? Remember Zaitsev’s Rule!
- (Advanced) References and Further Reading
1. An Allylic Bromination That Gives Two Products
Today’s topic flows right from the subject of the last post, on allylic bromination. It’s on allylic bromination reactions that happen with a rearrangements.
In the last post on allylic bromination, the examples used 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:
There’s no other possibility.
Now let’s extend the complexity of the substrate a little bit. Just replace one of the C–H bonds with a C–CH3 bond, and do the exact same reaction.
Here, we get not one product, but two. And note how our major product is different!
What’s going on here?
2. “Product B” Is An Example Of Allylic Bromination Accompanied By Allylic Rearrangement
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.
3. Why Might “Product B” Be Favored Here? Remember Zaitsev’s Rule!
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 “hyperconjugation”) .
The alkene in product A is what we’d call “di-substituted” – 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
(Advanced) References and Further Reading
- Brominations with N-Bromosuccinimide and Related Compounds. The Wohl-Ziegler Reaction.
Chemical Reviews 1948, 43 (2), 271-317
An old review by noted chemist Carl Djerassi, whose major contribution to global health was the development of norethindrone – the first female contraceptive. This review contains several examples of allylic systems that rearrange under free-radical conditions (bromination with NBS).
- Substitution And Rearrangement Reactions Of Allylic Compounds
R. H. DeWolfe and W. G. Young
Chemical Reviews 1956, 56 (4), 753-901
This early review mentions free-radical allylic rearrangements briefly – see pg. 761. Prof. W. G. Young helped build up the UCLA department of chemistry to what it is today, and there is a building named after him on the UCLA campus.
- Cyclic Polyolefins. II. Synthesis of Cycloöctatetraene from Chloroprene
Arthur C. Cope and William J. Bailey
Journal of the American Chemical Society 1948, 70 (7), 2305-2309
An early paper by Prof. Arthur C. Cope (of Cope rearrangement fame, and now the eponymous ACS Cope Scholar Award) on the synthesis of cyclooctatetraene. One of the steps involves bromination of 1,5-cyclooctadiene with NBS, which gives a monobromo compound. Cope mentions that they did not ascertain the exact location of the Br in the product and that 2 possible compounds can be formed (IIIa and IIIb), due to rearrangement.
- THE PEROXIDE EFFECT IN THE ADDITION OF REAGENTS TO UNSATURATED COMPOUNDS. XIII. THE ADDITION OF HYDROGEN BROMIDE TO BUTADIENE
M. S. KHARASCH, ELLY T. MARGOLIS, and FRANK R. MAYO
The Journal of Organic Chemistry 1936, 01 (4), 393-404
Another early paper by M. S. Kharasch in which he studies the addition of HBr to butadiene, in which he attempts to rigorously separate the two modes of addition – electrophilic vs. radical. There is an example of an ‘allylic rearrangement’ here – Prof. Kharasch shows that reaction of 3-bromo-1-butene with HBr under free-radical conditions (peroxides) can give 1,3-dibromobutane. This would be obtained via the intermediate allyl radical, which can add a bromine radical at either the 1- or 3- position.
- Cyclobutane Derivatives. IV. Ziegler Bromination of Methylenecyclobutane
Edwin R. Buchman and David R. Howton
Journal of the American Chemical Society 1948, 70 (7), 2517-2520
Ziegler reaction of methylenecyclobutane gives a mixture of “normal” and rearranged allylic bromination products (normal dominates).
- CYCLOHEXENE-3,3,6,6-d4 A USEFUL COMPOUND FOR THE STUDY OF MECHANISM AND STRUCTURE
Saul Wolfe and P. G. C. Campbell
Can. J. Chem. 1965, 43 (5), 1184-1198
Allylic bromides can easily undergo rearrangement – while the initial “normal” product might be formed in the reaction mixture, workup can often lead to scrambling. This study found that allylic bromination occurred “normally” but upon workup, rearrangement occurred to give the isomeric allyl bromide.
14 thoughts on “Bonus Topic: Allylic Rearrangements”
Hello james! I thoroughly read your post
but I have a question..
compared to radical structures..
radical structure of A is TERTIARY allyl radical, radical structure of B is SECONDARY allyl radical.
the former is more stable radical structure..
But Major product is B.
Just Zaitsev’s rule(hyperconjugation) has more afffection than Radical Stability?
I Think I got your explanation of more substituted alkene B being more stable than A , but then 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 secondary free radical(for B) , so why shouldn’t A be the major product by this line of thought , even though I agree That the final product B Is more stable than A but shouldn’t we also compare the transition states leading to the two products … I am confused as to why we are not considering this free-Radical stability … Can you please elaborate on this part ?
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.
Hi! I was coming here to ask a similar question — specifically, if one started with 1-methylcyclohexene (the unbrominated version of Product B), would NBS favor bromination of the methyl group? Thanks so much!
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).
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?
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.
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).
More substituted alkenes are more stable because the inductive effect and hyperconjugative effects satisfies the hunger of the (relatively) electron hungry sp2 carbon atom.
But here, isn’t Bromine more electrong ‘hungry’ than sp2 carbon?
Wouldn’t product A where bromine is on a tertiary carbon atom be more stabilized?
Thanks James and Andrea Jurado.
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 :)
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!