Past exam of the chemistry B course of the University of Cambridge / 2025 / Paper 1 / B3 / c

At 400 MHz, a separation $\Delta \delta$ in ppm corresponds to $J=400\Delta \delta$ Hz. The two multiplets near 5.10 and 4.47 ppm each have a quoted separation of 0.02 ppm, giving $J\simeq 8$ Hz. The multiplets near 4.625 and 3.865 ppm each have separation 0.01 ppm, giving $J\simeq 4$ Hz. These are approximate values from the rounded peak labels.

The fused unsaturated/aromatic framework restricts the five-membered ring, as in the comparison in (a). Assign the larger coupling to cis and the smaller to trans. Matching equal couplings and integrals pairs the 5.10/4.47 signals with one another, and the 4.625/3.865 signals with one another. The integral ratio isWith an achiral substrate and achiral catalytic environment, each diastereomer is expected as a racemic enantiomeric pair: cis $(2R,3S)/(2S,3R)$ and trans $(2R,3R)/(2S,3S)$. The reported spectrum requires both diastereomers; the reagent label alone cannot override that evidence. Their ratio does not establish whether the selectivity arose in the initial surface addition or after product epimerization.

Only the low-coupling pair remains, so the detected product is the trans diastereomer, with approximate coupling 4 Hz. The reaction is highly diastereoselective within the spectrum's detection limit. Its possible absolute configurations are $(2R,3R)$ and $(2S,3S)$.

A conventional proton NMR spectrum in an achiral environment cannot distinguish those enantiomers: they have the same shifts and couplings. Thus a chiral catalyst plus one set of resonances does not establish an enantiomeric excess or an absolute configuration. Chiral chromatography, an appropriate chiral NMR method, or a calibrated stereochemical comparison would be needed.