E2 eliminations from cyclohexanes

The stereospecificity of the reactions you have just met is very good evidence that E2 reactions proceed through an anti-periplanar transition state. We know with which diastereoisomer we started, and we know which alkene we get, so there is no question over the course of the reaction. 

More evidence comes from the reactions of substituted cyclohexanes. You saw in Chapter 18 that substituents on cyclohexanes can be parallel with one another only if they are both axial. An equatorial C-X bond is anti-periplanar only to C-C bonds and cannot take part in an elimination. For unsubstituted cyclohexyl halides treated with base, this is not a problem because, although the axial conformer is less stable, there is still a significant amount present (see the table on p. 462), and elimination can take place from this conformer. 

These two diastereoisomeric cyclohexyl chlorides derived from menthol react very differently under the same conditions with sodium ethoxide as base. Both eliminate HC1 but diastereoisomer A reacts rapidly to give a mixture of products, while diastereoisomer B (which differs only in the configuration of the carbon atom bearing chlorine) gives a single alkene product but very much more slowly. We can safely exclude El as a mechanism because the same cation would be formed from both diastereoisomers, and this would mean the ratio of products (though not necssarily the rate) would be the same for both. 

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E2 Elimination Where The Only Eligible Axial Cyclohexane P-Hydrogen is From a Less Stable Conformer... 

Almost all cyclohexane systems are most stable in the chair conformations. In a chair, adjacent axial positions are in an anfi-coplanar arrangement, ideal for E2 eliminations. Adjacent axial positions are said to be in a tran -diaxial arrangement. E2 reactions only proceed in chair conformations from... 

The stereochemical requirement of E2 elimination is anti-coplanar in cyclohexanes, this translates to rra/i.y-diaxial. Both dibromides are trans, but because the r-butyl group must be in an equatorial position, only the left molecule can have the bromines diaxial. The one on the right has both bromines locked into equatorial positions, from which they cannot undergo E2 elimination. 

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