= Solution
The <oxygen> substituent changes position through a <halohydrin>–<epoxide> sequence. First, <hydroxide> attacks the acetate <carbonyl> of $\mathrm{BrCH_2C(Me)_2OAc}$, giving a <tetrahedral intermediate>. Collapse expels the substrate <alkoxide>, and <proton transfer> gives $\mathrm{BrCH_2C(Me)_2OH}$ plus acetate. This is <saponification>, breaking the acyl–<oxygen> bond rather than replacing <bromide> directly with <hydroxide>.
A small equilibrium population of $\mathrm{BrCH_2C(Me)_2O^-}$ undergoes intramolecular <SN2 reaction> at the adjacent primary <carbon>: the O <lone pair> attacks $\mathrm{CH_2Br}$ and its C–Br pair leaves with <bromide>. The result is \b[<2,2-Dimethyloxirane>]. Addition of HBr protonates the ring <oxygen>. <bromide> then attacks preferentially at the more substituted <carbon>, where the protonated ring has greater <carbocation> character, breaking that C–O bond and leaving <oxygen> attached to the former primary <carbon>:
$$
\mathrm{BrCH_2C(Me)_2OH\ \longrightarrow\ (CH_2)O(CMe_2)\ \xrightarrow{HBr}\ HOCH_2C(Br)(Me)_2}.
$$
The middle formula denotes a three-membered <epoxide>, not an open-chain ether. Acidic opening of this highly unsymmetrical <epoxide> has the opposite regioselectivity to a simple basic nucleophilic opening at its less substituted <carbon>.
Finally, the primary <alcohol> <oxygen> attacks <acetyl chloride>. The <tetrahedral intermediate> collapses with <chloride> departure, and <triethylamine> removes the O–H <proton> and captures HCl. The product is \b[$\mathrm{AcOCH_2C(Br)(Me)_2}$]. <oxygen> is carried through the <epoxide> to the other <carbon>; it is not necessary to invent a methyl migration or a carbon-skeleton rearrangement.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-epoxide-regioselectivity.png]
{title=Original electron-flow drawing for bromohydrin cyclization, acid-directed epoxide opening and reacylation}
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