Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 1 / c

The six carbon environments are assigned as follows, with chemical shifts at the centres of the fluorine-split multiplets:The supplied carbon-13 spectrum is interpreted as proton-decoupled NMR, as indicated by its unsplit ethoxy signals. Proton decoupling removes C–H splitting but leaves C–F scalar spin-spin coupling. Three equivalent spin-$\frac{1}{2}$ fluorine nuclei give each resolved quartet its $1:3:3:1$ intensity pattern. The carbonyl's much smaller four-bond C–F interaction is unresolved in the plotted spectrum; “singlet” describes what is observed, rather than proving an exactly zero interaction.

The wide quartet is centred at $(127.48+116.74)/2=122.11$ ppm. The other two centres are $(132.04+130.63)/2=131.335$ and $(129.09+128.85)/2=128.97$ ppm. The signal near $77$ ppm is the deuterated chloroform solvent, not another carbon of the analyte. The structure labels skip ${\mathrm{C}}_i$: the ethoxy methylene is ${\mathrm{C}}_j$.

At a carbon-13 resonance frequency of $75\mathrm{MHz}$, the three resolved NMR coupling constants areEach is the spacing of adjacent lines within its own quartet. The lower panel of the original coupling diagram in 1(b)(ii) marks the corresponding carbon–fluorine interactions. The bond superscripts count bonds along the connecting path; they are not additional numerical factors multiplying $J$.