Preparation of Alkenes

E2 elimination reactions are the main method for preparing alkenes, although under some conditions E1 reactions are useful.
Elimination reactions involve the removal of a hydrogen atom and a leaving group to form an alkene. Most elimination reactions proceed through an E2 mechanism, although some proceed through an E1 mechanism. E2 reaction mechanisms occur in a single step, although there are several components to the actual reaction: the beta hydrogen, the leaving group, and the carbon substrate. When a base deprotonates the beta hydrogen on a carbon, a double bond will form as a halide or leaving group leaves with a pair of electrons. Rearrangements do not occur in E2 reactions.


A base (B-) will remove a hydrogen from a carbon, forming a π\pi bond and causing the leaving group (LG) to leave with a pair of electrons.
E1 reaction mechanisms are multistep. In the first step, the molecule loses a halide or leaving group, forming a carbocation. In the second step, the carbocation quickly loses a proton in the presence of a base, and a double bond forms, creating the alkene.

E1 Alcohol Mechanism

Elimination of alcohols occurs via a protonation of the alcohol, forming water (H2O), which is a good leaving group. This is followed by an E1 mechanism, resulting in a double bond and the loss of a hydroxyl group.
E2 reactions yield either a Zaitsev or Hofmann product, depending on the base used. Older texts may refer to Zaitsev as Saytzeff. Zaitsev's rule is the rule that states that the alkene formed is more highly substituted because it is more stable. This rule is used to predict the favored alkene product of an elimination reaction. In most E2 reactions, the more stable the alkene, the faster it forms. The more-substituted product is the more-stable product.

Beta Hydrogen versus Alpha Hydrogen

E2 elimination reactions occur with a base and result in Zaitsev orientation, where a more-substituted product is formed.
Hoffman eliminations of quaternary amines, which are amines bonded to four substituents, do not follow Zaitsev's rule, because the least-substituted product is favored. Hofmann's rule states that the steric effects have the most considerable influence on outcomes, specifically the loss of a beta hydrogen from the least substituted position. This means the predominant alkene product will be the least substituted.

Hofmann Elimination

When quaternary amines undergo a Hoffman E2 elimination, the least-substituted product, which is the product with the fewest number of substituents surrounding an alkene, will form as the major product.
Elimination reactions can yield either E or Z alkenes. If the starting material has two beta hydrogen atoms on the same carbon that can be lost, then both E and Z alkenes form, depending on which hydrogen is removed. When there is only one beta hydrogen on a carbon, only one stereoisomer (E or Z) forms instead of a mixture of E and Z alkenes. The stereoisomer that forms depends on the orientation of the hydrogen, the leaving group, and the substituents around the carbon atoms that form the alkene. The leaving group and the hydrogen atom that is being removed must be anti-periplanar, sometimes called anti-coplanar, which means they are in the same plane but opposite each other. When the hydrogen and leaving group are anti-periplanar, the orientation of the substituents will determine the stereochemistry of the product. When (1R,2R)-1,2-dibromo-1,2-diphenylethane undergoes E2 elimination in the presence of potassium tert-butoxide (tBuOK), the hydrogen and bromine must be anti-periplanar in order to eliminate and form the alkene. The elimination forms (Z)-1-bromo-1,2-diphenylethene.
The stereochemistry of the product of an E2 elimination is determined by the orientation of the substituents in the anti-periplanar conformation. When (1R,2R)-1,2-dibromo-1,2-diphenylethane is rotated into an anti-periplanar conformation, the hydrogen (H) and bromine (Br) that are removed are in the same plane but face opposite directions. In one case, the rotation places Br and one phenyl are on the same side of the alkene, with H and the other phenyl on the other side. Alternatively, the rotation places Br and H on the same side of the alkene.