= Solution
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1-q3-cyclization.png]
{title=Bromonium formation and intramolecular opening give the six-membered cyclic bromoether}
The <alkene> <pi bond> attacks Br$_2$ as the Br-Br bond breaks to bromide, forming a bridged, unsymmetrical \b[<bromonium ion>]. The <carbon> bearing the two <methyl groups> has greater carbocation-like character.
The tethered <alcohol> <oxygen> attacks that <carbon> from the side opposite the bromine bridge. Its <lone pair> forms the new O-C bond while the attacked C-Br bridge bond opens, giving a six-membered cyclic oxonium ion. Bromide or <solvent> removes the O-H <proton> to give the neutral bromoether.
This explains why O becomes attached at the more substituted <alkene> <carbon> while Br remains at the other <carbon>. Polar <solvent> assists charge separation, and intramolecular capture competes effectively with external bromide attack. Opening is anti; attack on the two faces of the achiral starting material gives a racemic product at the newly formed CHBr stereocentre.
The ring-forming step is <bromoetherification>.
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