Although E1 reactions show some stereo and regioselectivity, the level of selectivity in E2 reactions can be much higher because of the more stringent demands on the transition state for E2 elimination. In an E2 elimination, the new π bond is formed by overlap of the C–H σ bond with the C–X σ* 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 preferentially 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 E alkene predominates.

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.

مواضيع ذات صلة

Distinguishing acid derivatives by carbon NMR is difficult

E1cB eliminations in context

Anion-stabilizing groups allow another mechanism—E1cB

The regioselectivity of E2 eliminations

E2 elimination from vinyl halides: how to make alkynes

E2 eliminations from cyclohexanes

E2 eliminations can be stereospecific

E1 reactions can be stereoselective

The role of the leaving group

Substrate structure may allow E1

E1 and E2 mechanisms

Size