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 Incoming links: Spiroacetal

Past exam of the chemistry B course of the University of Cambridge / 2023 / Paper 1 / B4 / a / i / Solution 2026-10-20

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, N–OH is anti to the $K - R_{a}$ bond across C=N; in C, N–OH is anti to $K - R_{s}$ . Those geometries predict exactly the respective nitrogen insertions in D and E. Both product s have formula

C_{18} H_{21} N O_{5}

: condensation adds NH2OH to

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.

Figure 1.
Original local structures of the two ketoximes and their anti-migrating ring bonds; the remaining scaffold joins the two labelled ring paths
. **Details**.

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Past exam of the chemistry B course of the University of Cambridge / 2025 / Paper 1 / B4 / b / i / Solution 2026-10-20

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Protonation of a spiroacetal oxygen makes the adjacent C–O bond easier to cleave. Reversible opening gives an oxocarbenium ion, in which the spiro carbon is locally planar. The tethered alcohol can attack either face; deprotonation closes the acetal and regenerates acid. Repetition interconverts the spiro epimers while retaining the other stereogenic centres. Methanol can participate in reversible proton transfer or trapping, but there is no need to invert the remote carbon centres.

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