= Solution
The secondary hydroxylamine condenses with formaldehyde. <nitrogen> attacks the <carbonyl> <carbon>, giving a carbinolamine; proton transfer and loss of water generate the tethered <nitrone> \b[$\mathrm{R-N^+(O^-)=CH_2}$], where $R$ is the entire bicyclic substituent attached to <nitrogen> in D. Molecular sieves remove water, favouring this condensation.
The <nitrone> undergoes an intramolecular <1,3-dipolar cycloaddition> with the retained cyclohexene double bond. To define E precisely, label the former oxime <carbon> $C_A$, the adjacent upper ring junction $C_B$, the other junction $C_C$, and the <alkene> <carbons> $C_D$ and $C_E$ such that the chain from $C_A$ to the more remote <alkene> terminus is $C_A-C_B-\mathrm{CH_2}-C_D=C_E-\mathrm{CH_2}-C_C$. Let $C_F$ be the new formaldehyde-derived <nitrone> methylene. The <cycloaddition> forms \b[$C_F-C_D$ and O–$C_E$], retains N–O and N–$C_A$, and changes N=$C_F$ and $C_D=C_E$ into single bonds. The new five-membered heterocycle is \b[N–$C_F$–$C_D$–$C_E$–O]. These connections, together with the original bicycle, specify \b[the bridged <isoxazolidine> E].
The inwardly tethered <cycloaddition> is <suprafacial> at the <alkene> and preserves the existing <carbon> <stereocentres>; the C–C and C–O additions occur on the same accessible <alkene> face. The pre-existing N-substituent orientation restricts which fold can bring both reacting ends together without crossing the bridge. This is a genuine two-new-bond <cycloaddition>, not a Mannich reaction followed by unrelated oxygenation.
\Image[/past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3-isoxazolidine.png]
{title=Original atom-labelled connectivity model of the nitrone cycloaddition product E}
Back to article page
