Anti periplanar geometry elimination reactions

A final piece of evidence involves the stereochemistry of elimination. (Jnlike the E2 reaction, where anti periplanar geometry is required, there is no geometric requirement on the El reaction because the halide and the hydrogen are lost in separate steps. We might therefore expect to obtain the more stable (Zaitsev s rule) product from El reaction, which is just what w e find. To return to a familiar example, menthyl chloride loses HC1 under El conditions in a polar solvent to give a mixture of alkenes in w hich the Zaitsev product, 3-menthene, predominates (Figure 11.22). 

Although anti periplanar geometry is preferred for E2 reactions, it isn t absolutely necessary. The deuterated bromo compound shown here reacts with strong base to yield an undeuterated alkene. Clearly, a svn elimination has occurred. Make a molecular model of the reactant, and explain the result. 

First identify the H and the leaving group (L) that are eliminated in the reaction. Remember that E2 elimination requires a conformation that has the H and the L in an anti-periplanar geometry. If they are not in such a conformation as drawn, redraw the molecule so that they are. When the elimination occurs, groups that aie on the same side of the plane defined by H-C-C-L in the reactant become cis in the product alkene. 

As can be seen in Figure 9.5, neomenthyl chloride has two hydrogens in an anti-periplanar geometry with the chlorine. Either of these hydrogens can be lost in the elimination reaction, resulting in the formation of two alkenes. (All of the examples presented up to this point were chosen so that only one alkene could be formed.) Let s now address this issue of the direction of the elimination and learn how to predict which will be the major product when more than one alkene can be formed. 

Problem 8.17 Draw the anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base. Then draw the product that results after elimination of HBr. 

We just saw that to achieve the anti-periplanar geometry favored in an E2 reaction, the two groups being eliminated must be parallel (Section 10.6). For two groups on a cyclohexane ring to be parallel, they both must be in axial positions. 

With the strong base OCH2CH3, the mechanism is E2, whereas with dilute base, the mechanism is El. E2 elimination proceeds with anti periplanar arrangement of H and X. In the El mechanism there is no requirement for elimination to proceed with anti periplanar geometry. In this case the major product is always the most stable, more substituted alkene. Thus, C is the major product under El conditions. (In Chapter 9, we will learn that additional elimination products may form in the El reaction due to carbocation rearrangement.)... 

Anti periplanar fSection 8.8A) In an elimination reaction, a geometry where the P hydrogen and the leaving group are on opposite sides of the molecule. 

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