= Solution
\b[D is racemic trans-2-methylcyclohexan-1-ol. E is racemic 1-methyl-2-phenylcyclohexan-1-ol with OH and Ph trans, and Me and Ph cis.] Number the starting alkene carbon carrying methyl as $\mathrm C_1$ and its partner as $\mathrm C_2$ in the reaction explanation below; the product name of D renumbers the OH-bearing carbon as 1.
For D, <hydroboration> proceeds through a concerted four-centre transition state: the alkene pi pair donates to boron while B–H supplies the new C–H bond. Boron attaches mainly to less substituted $\mathrm C_2$ and hydrogen to $\mathrm C_1$, on the same face. Under peroxide/base conditions, hydroperoxide attacks B; migration of the C–B bond to the adjacent oxygen, with O–O cleavage, replaces C–B by C–O with retention at carbon. Hydrolysis/proton transfer releases the alcohol. Thus the newly added H and OH are syn, making the pre-existing methyl and OH trans. Attack on either alkene face gives the enantiomeric pair.
For E, the peracid transfers its terminal oxygen across the alkene in concerted <epoxidation>, through cyclic electron flow that also breaks O–O and transfers the acidic proton to the peracid carbonyl oxygen. The resulting <epoxide> retains the relative configuration of the alkene substituents. The phenyl equivalent from <phenylmagnesium bromide> then attacks less substituted $\mathrm C_2$ from the face opposite the epoxide oxygen. Its C–O bond breaks onto oxygen, an SN2-like ring opening; the oxygen stays attached to $\mathrm C_1$. Water workup protonates that alkoxide. Consequently OH at $\mathrm C_1$ and Ph at $\mathrm C_2$ are anti, while the methyl at $\mathrm C_1$ lies on the same relative face as Ph.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1-stereochemical-products.png]
{title=Original product structures showing the relative stereochemistry of hydroboration–oxidation and epoxide opening}
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