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356 Organic Chemistry Solutions Manual CH₂ groups. The bromine must be on one of them and the ester oxygen on the other. The extra DBE is a ring. 1735 cm⁻¹ 4.18 H 0 H 0 CO₂H 0 0 Br Br H H H H 1645 cm 5.00 6.18 Since both methyl groups are functionalized, the unstable A must have one Br on each methyl group. The peroxide produces benzoate radicals that abstract a proton from allylic positions to give stabilized radicals and these attack bromine molecules providing bromine atoms to continue the chain reaction. In base, the carboxylate ion cyclizes on to the cis CH₂Br group. initiation: PhCO₂ H CO₂H CO₂H Br CO₂H Br Br thereafter: Br H Br Br₂ Br CO₂H Br CO₂H Br CO₂H A 0 Br Br 0 base Br Br CO₂H Br 0 B A Problem 14 The product formed in Problem 9 of Chapter 20 was actually used to make this cyclic ether. What is the mechanism? 0 AcO AcO CI CI OH Purpose of the problem A slightly unusual method to make radicals and a remote functionalization reaction. Suggested solution Lead tetraacetate provides radicals on heating (maybe or lead radicals) and these abstract hydrogen atom from the OH group. The oxygen radical is in an ideal position to abstract hydrogen atom from the methyl group (conformational diagram essential). H 0 0 AcO AcO AcO CI CI CI 0 AcO AcO AcO CI CI CI