Chad's Organic Chemistry Videos
|Less Substituted||More Substituted|
|Two Groups Added||X (Cl orBr)||X (Cl orBr)|
|Reagents Added|| X2 (Cl2 or Br2)|
CH2Cl2 or CCl4 (inert solvents)
|Other Characteristics||Intermediate is a 3-membered ring (halonium ion).|
Halogenation of Alkenes
Halogenation results in the addition of two halogen atoms (Cl or Br) across an alkene forming a vicinal dihalide. The reaction proceeds with Anti stereospecificity, and as the intermediate is NOT a carbocation rearrangements are NOT possible. The reagents are either Cl2 or Br2 in an inert solvent (usually CH2Cl2 or CCl4).
Halogenation of Alkenes Mechanism
Step 1: The pi electrons of the alkene attack one of the bromine atoms which attacks one the of the alkene carbons back forming a three-membered ring with Br referred to as a bromonium ion (generically referred to as a halonium ion and specifically a chloronium ion in the reaction with Cl2). The bond between the bromine atoms also breaks in this step forming a bromide ion.
While the bromonium ion is more stable than a carbocation it is still a rather high energy intermediate, and this first step is the rate determining step (slow step) of the reaction. Also, the bromonium intermediate does not undergo rearrangements such as are possible with carbocations.
Step 2: Ring-Opening of the Bromonium Ion
The bromide ion carries out back-side attack (as in SN2) on the more substituted carbon of the bromonium ion (the two carbon atoms are equally substituted in this example) resulting in the bond to the Br in the bronomium ion breaking (analogous to loss of a leaving group in SN2 even though it is still attached to the adjacent carbon atom) and the opening of the 3-membered ring.
The back-side attack results in inversion of configuration at the carbon attacked explaining why the two bromine atoms add to opposite sides of the alkene (i.e. anti addition).