E2 eliminations have anti-periplanar transition states

E2 eliminations have anti-periplanar transition states 

In an E2 elimination, the new 7t bond is formed by overlap of the C-H a bond with the C-X a antibonding orbital. The two orbitals have to lie in the same plane for best overlap, and now there are two conformations that allow this. One has H and X syn-periplanar, the other anti-periplanar. The anti-periplanar conformation is more stable because it is staggered (the syn-periplanar conformation is eclipsed) but, more importantly, only in the anti-periplanar conformation are the bonds (and therefore the orbitals) truly parallel. 

E2 eliminations therefore take place from the anti-periplanar conformation. We shall see shortly how we know this to be the case, but first we consider an E2 elimination that gives mainly one of two possible stereoisomers. 2-Bromobutane has two conformations with H and Br anti-periplanar, but the one that is less hindered leads to more of the product, and the -alkene predominates. 

H and Br must be anti-periplanar for E2 elimination two possible conformations 

There is a choice of protons to be eliminated—the stereochemistry of the product results from which proton is anti-periplanar to the leaving group when the reaction takes place, and the reaction is stereoselective as a result. 

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E2 ELIMINATIONS HAVE ANTI-PERIPLANAR TRANSITION STATES... 

In the next example, there is only one proton that can take part in the elimination. Now there is no choice of anti-periplanar transition states. Whether the product is E or Z> the E2 reaction has only one course to follow. And the outcome depends on which diastereoisomer of the starting material is used. When the first diastereoisomer is drawn with the proton and bromine anti-periplanar, as required, and in the plane of the page, the two phenyl groups have to lie one in front and one behind the plane of the paper. As the hydroxide attacks the C-H bond and eliminates Br , this arrangement is preserved and the two phenyl groups end up trans (the alkene is E). This is perhaps easier to see in the Newman projection of the same conformation. 

A single-step E2 elimination would have to go via an anti-periplanar transition state and would be stereospecific. You will be able compare this stereoselective Julia olefination with the stereospecific Peterson elimination shortly. 

Anti periplanar geometry for E2 eliminations has specific stereochemical consequences that provide strong evidence for the proposed mechanism. To take just one example, meso-l,2-dibromo-l,2-diphenylethane undergoes E2 elimination on treatment with base to give only the pure E alkene. None of the isomeric Z alkene is formed because the transition state leading to the Z alkene would have to have syn periplanar geometry. 

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