= Solution
Base deprotonates the <sulfonamide>. Its <nitrogen> attacks one <epoxide> <carbon> from the back, breaking the C–O bond at that <carbon> and producing an adjacent <alkoxide>. The <nitrogen> and <oxygen> attachments on the <cyclopentane> are therefore \b[trans]. The <alkoxide> then attacks the aromatic <carbon> bearing the <fluorine> ortho to the sulfonyl group. Intramolecular <nucleophilic aromatic substitution> expels fluoride and closes a seven-membered <heterocycle> with the path $\mathrm{Ar{-}S{-}N{-}C_1{-}C_2{-}O{-}Ar}$. The para <fluorine> remains.
This tandem epoxide-opening/ring-closing pathway is required by the formula of F: simple <epoxide> addition would retain both <fluorines>, whereas F contains only one. In the second stage, <pyrrolidine> replaces the remaining para <fluorine> by another <nucleophilic aromatic substitution>, producing G. The mass balance is F plus <pyrrolidine> minus HF, as required by the given formula of G.
The meso <epoxide> has <enantiotopic> ends. With achiral reagents, opening at either end gives an enantiomeric product, so \b[F is a racemic pair with the same trans relative configuration; G retains that configuration].
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2025/paper-1-epoxide-cascade.png]
{title=Trans epoxide-opening cascade, seven-membered ring closure, and replacement of the remaining aryl fluoride}
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