R)-(+)-Glyceraldehyde can also be transformed into (-)-malic acid. This synthesis begins with the conversion of (R)-(+)-glyceraldehyde into (-)-tartaric acid, as shown in Problem 17.41, parts (e) and (g). Then (-)-tartaric acid is allowed to react with phosphorus tribromide in order to replace one alcoholic iOH group with iBr. This step takes place with inversion of configuration at the carbon that undergoes attack. Treating the product of this reaction with dimethyl sulfide produces (-)-malic acid.
(a) Outline all steps in this synthesis by writing stereochemical structures for each intermediate.
(b) The step in which (-)-tartaric acid is treated with phosphorus tribromide produces only one stereoisomer, even though there are two replaceable iOH groups. How is this possible?
(c) Suppose that the step in which (-)-tartaric acid is treated with phosphorus tribromide had taken place with "mixed" stereochemistry, that is, with both inversion and retention at the carbon under attack. How many stereoisomers would have been produced?
(d) What difference would this have made to the overall outcome of the synthesis?