 Solution

Both free primary allylic alcohols undergo Sharpless asymmetric epoxidation. With slightly more than two equivalents of hydroperoxide, the expected fully oxidized product is the diepoxide, with both original C=C bonds converted to epoxides and both primary alcohols retained. The two central pre-existing stereocentres and their ether substituents are retained.

The stereochemical construction is unambiguous by applying the L-(+)-tartrate mnemonic separately at each end: orient that alkene horizontally with its CH${}_2$OH group at lower right, and deliver oxygen from the back face. In the source's original in-plane orientation this gives the left epoxide oxygen behind the plane and the right epoxide oxygen in front. The methyl and hydroxymethyl groups retain their original relative arrangement about each former double bond. One cannot simply draw both oxygens on the same face of the original extended chain: the two allylic-alcohol ends have opposite drawing orientations. titanium binding to each free hydroxyl directs oxygen transfer; the pre-existing stereocentres can create matched/mismatched rate differences, but the stated reagent control predicts the stereochemistry of both epoxidations. Molecular sieves maintain a dry active titanium–tartrate system.^[ref]