= Solution
\b[B and C are the two geometrical <ketoxime> <isomers>], formed by replacing the reactive ring C=O with C=N–OH. The separate <spiroacetal> remains intact. Define $K$ as the original <ketone> <carbon>, $R_a$ as the adjacent ring path towards the <aromatic ring>, and $R_s$ as the adjacent path towards the <spiroacetal>-bearing junction. These are the two <carbon> <substituents> directly bonded to $K$, not detachable independent fragments.
In \b[B, N–OH is anti to the $K{-}R_a$ bond] across C=N; in \b[C, N–OH is anti to $K{-}R_s$]. Those geometries predict exactly the respective <nitrogen> insertions in D and E. Both <product (chemistry)>s have formula $\mathrm{C_{18}H_{21}NO_5}$: condensation adds NH2OH to $\mathrm{C_{18}H_{20}O_5}$ and removes H2O. An alternative ring regioisomer or reaction at the protected <ketone> would not represent the pair of <oxime> <geometrical isomers>.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2023/paper-1-beckmann-geometries.png]
{title=Original local structures of the two ketoximes and their anti-migrating ring bonds; the remaining scaffold joins the two labelled ring paths}
Back to article page
