The E2 elimination, bimolecular reaction

This is a concerted (one-step) mechanism. The C=C bond begins to form at the same time as the C-H and C-X bonds begin to break. The reaction is second order (or bimolecular), as the rate depends on the concentration of the base and the alkyl halide. 

The elimination requires the alkyl halide to adopt an antiperiplanar shape (or conformation), in which the H and X groups are on the opposite sides of the molecule. (Synperiplanar conformation is when the H and X groups are on the same side of the molecule.) The antiperiplanar arrangement is lower in energy than the synperiplanar arrangement, as this has a staggered, rather than an eclipsed, conformation. 

The elimination is stereospecific, as different stereoisomers of the alkyl halide give different stereoisomers of the alkene. This is because the new Jt-bond is formed by the overlap of the C-H o-bond with the C-X o -bond. These orbitals must be in the same plane for the best overlap, and they become p-orbitals in the Jt-bond of the alkene. 

A different alkene stereoisomer is obtained from each diastereoisomer of the alkyl halide. 

For elimination in cyclohexanes, both the C-H and C X bonds must be axial. 

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