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In an E2 eliminations, the new π bond is formed by overlap of the C-H σ bond with the C-X σ* anti-bonding orbital. The two orbitals have to lie in the same plane for best overlap. E2 elimination takes place from the anti-periplanar conformation, as this is the most stable conformation due to its staggered nature.

As you can see from the reaction below, when 2-bromobutane undergoes an E2 reaction, two possible stereoisomers are formed. This is because 2-bromobutane has two conformations with H and Br anti-periplanar, but the one that is less hindered forms the major product, so the E-alkene predominates.