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9.31 A cyclobutane ring is relatively rigid and is nearly eclipsed about its C-C bonds. It is easier for a syn elimination from an eclipsed conformation to occur that it is for the molecule to attain a staggered conformation so that an anti elimination can occur. As a result, the compound shown prefers to lose the D, which is cis to the leaving group. 9.32 For syn elimination occur the hydrogen and the leaving group must be in a syn coplanar arrangement. This requires the molecule to be in a higher- energy eclipsed conformation. Although the cyclopentane ring is nonplanar to relieve its torsional strain, a fully staggered geometry is not easily attained. Therefore the molecule can achieve the eclipsed geometry needed for syn elimination about as readily as the geometry needed for anti elimination. 9.33 a) The reactant has the leaving group on a tertiary carbon, there is no strong base present, and the solvent is polar, so the reaction follows the SN1 and E1 mechanisms. b) OH OEt and c) and major d) lodide ion is a better leaving group than bromide ion, so the rate of reaction will be faster with 2-iodo-2-methylbutane. However, the amount of substitution and elimination products will not change because the leaving group is not involved in the second step of the mechanism, where the partitioning between substitution and elimination occurs. 9.34 Because of the relatively acidic hydrogen on the carbon adjacent to the carbonyl group, this CH₂CH₃ compound reacts by the E1cb mechanism to produce the alkene with the double bond CH₂CH₃ conjugated with the carbonyl group. 144