A Fourth Alkene Addition Pattern – Free Radical Addition

by James

in AlkeneCC, Alkenes, Organic Chemistry 1, Stereochemistry

I’ve written that there are three major alkene reactivity patterns [carbocation, three membered ring, and concerted], but there are two minor pathways as well. This post discusses one of them.

As discussed previously, alkenes normally react with HBr to give products of “Markovnikov” addition; the bromine ends up on the most substituted carbon of the alkene, and the hydrogen ends up on the least substituted carbon. However, something interesting happens when the same reaction is performed in the presence of peroxides and  heat / light: the pattern of addition changes! Instead of Br ending up on the most substituted carbon of the alkene, it ends up on the least. [The stereochemistry of the reaction, however, is unchanged: it still gives a mixture of “syn” and “anti” products.]

1-alkene hbr

This so-called “anti-Markovnikov” addition is intriguing. What difference could the presence of peroxides, and furthermore heat (or light) make to this reaction?

To make a long story short [not much mention of free-radical addition reactions have been made yet on this blog, so you’ll have to pardon the lack of lead-in], this reaction occurs through a free-radical process.  Here are the essential details:

  • Peroxides contain a weak oxygen-oxygen bond [approximately 35 kcal/mol;  compare to C-H at approx 100 kcal/mol]
  • Heating leads to homolytic fragmentation of this bond – that is, the bond breaks such as to leave one unpaired electron on each atom. Strong sources of light [e.g. a floodlight or other source of light radiation which reaches into the near UV] can also serve to sever this bond.
  • The resulting highly reactive alkoxy radical can then abstract a hydrogen from H-Br, giving a bromine radical. This is the species that adds to the alkene.
  • Addition to the alkene will preferably occur in such a way that the most stable free radical is formed [in our case, the tertiary radical]. That’s why bromine ends up on the least substituted carbon of the alkene. 
  • This tertiary radical then removes hydrogen from H-Br, liberating a bromine radical, and the cycle continues.

Note that only a trace [catalytic] amount of peroxide is required to get the reaction started, although of course at least one molar equivalent of HBr is required to result in full addition of HBr to the alkene.

Here it is, Chemdrawed:

2-alkene hbr

This initiation step results in homolytic cleavage of O-O. The singly barbed arrows depict the movement of single electrons; two alkoxy radicals are formed. Common “peroxides” for this purpose are t-butyl peroxide or benzoyl peroxide. * [Note 1]. Alternatively other free-radical “initiators” such as AIBN can also be used.

Only a catalytic amount of peroxides are used to initate this reaction (typically 10-20 mole %, although more can be used, especially when added batchwise) , because the next step is for the peroxy radical to remove a hydrogen from H-Br:

3-hbr

Once formed, the bromine radical can then add to the alkene, from either face. Addition occurs in such a way as to give the most substituted radical (tertiary in this case, not secondary).

4-hbr
Finally, the tertiary radical then removes a hydrogen from another equivalent of H–Br, giving the final addition product. A bromine radical is generated by this process, which can then add to another equivalent of alkene.

Note that hydrogen here can attack either face of the free radical [note 2]. Therefore we obtain a mixture of syn and anti products.

5-hbr

This reaction pathway is most commonly observed (in Org 1 and Org 2, anyway) for addition of HBr, although a rich chemistry of radical addition reactions to alkenes exists (particularly for organostannanes).

NEXT POST: Ozonolysis of Alkenes

* Note. Benzoyl peroxide enjoys a common household use as an acne cleanser, and even makes an appearance in this classic ad.

**Note 2. The geometry of free radical carbons is that of a  shallow pyramid with a low barrier for inversion, allowing for reactivity on either face. The exception is in weird cases where inversion would be highly disfavored, such as on a bridgehead.

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{ 12 comments… read them below or add one }

mamid

Penultimate paragraph: “This reaction pathway is most commonly observed (in Org 1 and Org 2, anyway) for” – unfinished sentence.

Great job, as usual.

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james

Fixed. Thank you for reminding me to put in a brief shout out to organostannane chemistry.

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Sajeda

is there a video tutorial on this? I learn better by hearing and doing rather than reading :)

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james

Not at present, sorry!

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dr.klbajaj

Why hydrogen free radical does not add first to alkene instead of bromine free radical

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Oosman Beekawoo

Because it is not formed! See the comment below for a clarification of (or at least, an attempt to clarify) it.

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Jacob

Why does the tertiary radical have a preference to remove the hydrogen from the HBr instead of the bromine?

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James

Not sure what you mean – you mean react with bromine radical?

The concentration of bromine radical at any given moment is much less than the concentration of HBr.

If it reacted with bromine radical that would be a termination step!

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OO

In step 2, why does the bromine radical not abstract an allylic hydrogen (like in NBS halogenation reactions)? I presume that allylic radical would be almost as stable as a tertiary radical.

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James

I think it’s because the C-C pi bond strength is about 60-65 kcal/mol whereas C-H allylic bond strength is about 80-85 kcal/mol. Higher activation energy in other words

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Bruce

Could it be that allylic hydrogen abstraction does occur, but the resulting allyl radical has nothing to react with except H-Br? The allyl radical could reabstract a hydrogen from H-Br, to reform the alkene, which then eventually adds a bromine radical. Potentially the alkene could isomerize if this occurred, and I don’t know if this is observed. Just a thought…

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James

Thanks Bruce, I think that’s exactly what’s going on.

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