Anti periplanar geometry

Anti periplanar geometry (staggered, lower energy)... 

Anti periplanar geometry for E2 reactions is particularly important in cyclohexane rings, where chair geometry forces a rigid relationship between the substituents on neighboring carbon atoms (Section 4.8). As pointed out by Derek Barton in a landmark 1950 paper, much of the chemical reactivity of substituted cyclohexanes is controlled by their conformation. Let s look at the E2 dehydro-halogenation of chlorocyclohexanes to see an example. 

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. 

Molar absorptivity. 502 Molecular ion (M+), 410 Molecular mechanics. 130 Molecular model, dopamine, 930 acetaminophen, 29 acetylene, 18 adenine, 67 adrenaline, 323 alanine, 28, 1016 alanylserine, 1028 rr helix, 1039 p-aminobenzoic acid, 25 anti periplanar geometry, 387 a recoline, 79 aspartame, 29 aspirin. 17 ball-and-stick, 61 /3-pleated sheet, 1039 p-bromoacetophenone, 449 bromocyclohexane, 121 butane, 80... 

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. 

Convert this drawing into a Newman projection, and draw the conformation having anti periplanar geometry for -H and -Br. 

This alkyl halide gives the less substituted cycloalkene (non-Zaitsev product). Elimination to form the Zaitsev product does not occur because the -Cl and -H involved cannot assume the anti periplanar geometry preferred for E2 elimination. 

Both Newman projections place -H and -Cl in the correct anti periplanar geometry for E2 elimination. 

Zaitsev s rule E2 reaction syn periplanar geometry anti periplanar geometry deuterium isotope effect El reaction ElcB reaction... 

The more stable dehydration product is 1-methylcyclopentene, which can be formed only via syn elimination. The product of anti elimination is 3-methylcyclopentene. Since this product predominates, the requirement of anti periplanar geometry must be more important than formation of the more stable product. 

C—X bond. This is the simplest perception available to the organic chemist to account for why E2 reactions take place with anti-periplanar geometry. 

All evidence suggests that E2 elimination occurs most often in the anti periplanar geometry. 

Anti periplanar geometry is the preferred arrangement for any alkyl halide undergoing E2 elimination, regardless of whether it is cyclic or acyclic. This stereochemical requirement has important consequences for compounds containing six-membered rings. 

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. 

Taking into account anti periplanar geometry, predict the major E2 product formed from each starting material. CI, o D... 

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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