E2 eliminations occur when the leaving group and the proton are in the anti-periplanar conformation not syn-periplanar. This is beyond a merely a preference as leads to E2 reactions being stereoselective. However, when it comes to vinyl bromides for example (involving a double bond) the reaction can occur in the Z- and E- isomers (i.e both syn-periplanar and anti-periplanar). How can this be the case?
Elimination from vinyl halides occurs faster when the halide and proton have a trans relationship. When the halide and proton are located on the same side of the double bond (cis), the elimination to form an alkyne is much slower. The faster elimination from the trans arrangement is consistent with a preferred anti-periplanar arrangement in the E2 reaction.
The fact that elimination still occurs, albeit at a slower rate, when the groups being eliminated are arranged cis to one another suggests that a different mechanism is involved in the "cis" case. Two possible mechanistic alternatives are
- cis elimination from a higher energy (higher energy, therefore less preferred, slower) syn-periplanar conformation, or
- an E1CB mechanism where the proton is removed in a first step and a carbanion is generated, and this is followed by halide ejection in a second step.
Experiments in analogous molecules suggest that the E1CB pathway is most often followed in "cis" elimination from vinyl halides.