= Solution
Both free primary allylic alcohols undergo <Sharpless asymmetric epoxidation>. With slightly more than two equivalents of <hydroperoxide>, the expected fully oxidized product is \b[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 \b[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.<The first practical method for asymmetric epoxidation by Tsutomu Katsuki and K. Barry Sharpless>[JACS 102, pp. 5974–5976: tartrate-controlled allylic-alcohol epoxidation]{ref}
\Image[/past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3-stereochemistry.png]
{title=Original facial-selectivity and axial-projection solution models}
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