Bromination of alkenes with Br2 to give dibromides

by James

Description: Treatment of alkenes with bromine (Br2) gives vicinal dibromides (1,2-dibromides). 

Notes: The bromines add to opposite faces of the double bond (“anti addition”). Sometimes the solvent is mentioned in this reaction – a common solvent is carbon tetrachloride (CCl4). CCl4 actually has no effect on the reaction, it’s just to distinguish this from the reaction where the solvent is H2O, in which case a bromohydrin is formed (see also).

Notes: Again note that in the first example CCl4 is merely the solvent and in these cases has no effect on the reaction (unlike when water or alcohols are the solvent – see bromohydrin formation page).


Attack of the alkene on bromine (Step 1, arrows A and B) gives the bromonium ion, which is attacked at the backside by bromide ion to give the trans-dibromo product. Note that the bromines are delivered to opposite sides of the alkene (“anti” addition).

Notes: Note that a 1:1 mixture of enantiomers will be formed in this reaction. The enantiomer formed will depend on which face of the alkene the bromine adds to.


{ 5 comments… read them below or add one }


Unlike Bromohydrin. Why isn’nt there any chlorohydrin?Is that not possible?



Great post here, one question- could you discuss what would happen when 1,2 dimethylcyclohexene reacts with Br2 (1 mole)/light? The answer key shows a double bond shift and only one Br attaching. Everywhere I’ve read, in an alkene reaction both Br’s attache to the ring in trans.. Could you show the mechanism involved?



I believe only 1 Br attaches because it is a radical reaction, hence the “light”. Being radical means only 1 attaches during propagation



Yeah I just looked this up because I was confused too. When you have light plus Br2 plus an alkene you get allylic bromination, which means a bromine radical removes a hydrogen (and one electron from its bond) from a carbon adjacent to the double bond (in allylic position), leaving only one electron on that carbon which reacts with another Br2 to take one of the bromines and one of its electrons to form a halogenated alkene and another bromine radical in the propagation step. If you have an alkene plus light plus HBr, the light forms a bromine radical which forms a bond to the least substituted carbon of the alkene and puts a radical on the most substituted carbon, which then grabs the hydrogen and one electron from another HBr molecule, forming another bromine radical and making it a propagation step.

When you don’t have light but have Br2 and an alkene and a non-nucleophilic solvent the bromines add to both carbons that formerly held the double bond- the first bromine is grabbed by the pi bond and leaves its electrons to form a ring with the alkane (halonium ion), which the other bromine (which has extra electrons and is thus nucleophilic) then attacks from the back, which breaks the ring and forms an alkane with two bromines attached to adjacent carbons. If you add HBr with a non-nucleophilic solvent and without light the same thing happens, but the first atom grabbed by the pi bond is the hydrogen without its electrons. (In this process the hydrogen doesn’t form a ring, just a carbocation, so rearrangement can occur to generate different di-halogenated alkanes.)

So basically radical addition of halogens to a pi bond (that gives the anti-Markovinov product) only happens if you have HBr and light (and peroxides to create the bromine radical), not Br2 and light. If you have Br2 and light, allylic bromination occurs, which adds Br not to the double bond but to the carbon adjacent to the doubly bonded carbons. Or the bromine can just add across the double bond to each of the formerly double bonded carbons, as our dear Ochem master says on his page:
Allylic bromination:
Free Radical Addition of HBr to Alkenes:



What would happen if bormination were to occur with a diene? There is no other intermediate besides the bromonium ion.
Specifically how does the rearrangement occur?


Leave a Comment