Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 1 / b / ii / Solution

Use the NMR coupling constant conversion $J/\mathrm{Hz}=(\Delta \delta /\mathrm{ppm})({\nu }_0/\mathrm{MHz})$. At $300\mathrm{MHz}$, separating the overlapping quartets givesThe ${\mathrm{H}}_b$ doublet separation independently gives $(6.513-6.461)300=15.6\mathrm{Hz}$. For ${\mathrm{H}}_a$, the two quartets are $\{6.728,6.750,6.772,6.794\}$ and $\{6.780,6.802,6.824,6.846\}$ ppm; simply taking adjacent lines in their combined ordering would mix different scalar spin-spin couplings.

The methylene quartet gives an average adjacent spacing $(4.317-4.244)/3=0.02433$ ppm, or $7.3\mathrm{Hz}$. Thus its printed rounded positions agree with the methyl-derived value to about $0.1\mathrm{Hz}$; quote the ethyl NMR coupling constant as approximately $7.2$–$7.3\mathrm{Hz}$. The large vinylic value supports the trans arrangement. The arrows below join the interacting nuclei; each arrow to ${\mathrm{F}}_3$ represents the same interaction with each equivalent fluorine nucleus.