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Chem210-1 F05 Ex3 p05a

Chem210-1 F05 Ex3 p05a - "lactone&quot...

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5. When compound 'H' (shown below) was treated with a solution of molecular iodine, in the presence of a weak base like bicarbonate ion, product 'Z' was formed. a) Provide a step-by-step mechanism for this process; be sure to include any and all intermediates that are likely to occur to any significant extent. (8 points) b) A cyclic ester group (such as that found as part of the structure of compound 'Z') is called a
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Unformatted text preview: "lactone" group. Provide the structure of a lactone isomer of 'Z', which could at least be hypothetically considered as a possible product in the transformation above. (2 points) OH O I 2 / HCO 3 O O I (= ' Z ') (compound ' H ') O O I I O O HCO 3 I 2 O O I O O I O O I = O O I O O I (question dropped due to ambiguity) (result of attack at less-crowded carbon of "iodinium ion" intermediate)...
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