Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 1 / c

Use the proton labels in the original drawing below. Let $a$ denote the three methyl hydrogens, $b$ the crotonyl beta-CH, $c$ the crotonyl alpha-CH, $d$ the O-bound vinyl CH, and $e,f$ the terminal vinyl hydrogens trans and cis to $d$, respectively.The integrals sum to eight hydrogens. Multiplet centres give the chemical shifts; a peak label at one edge of a multiplet is not its centre. The terminal vinyl hydrogens are chemically nonequivalent because one is cis and the other trans to the O-bound vinyl hydrogen. Each crotonyl CH couples to the other and to the methyl group; the resolved long-range interaction makes the methyl a doublet of doublets rather than a simple doublet.

At $300\mathrm{MHz}$, $J/\mathrm{Hz}=300\Delta \delta /\mathrm{ppm}$. The six resolved scalar spin-spin couplings areFor example, the two quartets of $c$ are $\{5.842,5.848,5.854,5.860\}$ and $\{5.893,5.899,5.905,5.911\}$ ppm. Their origins differ by $0.051$ ppm, while the spacing within either quartet is $0.006$ ppm. This separates the trans alkene interaction from the smaller allylic methyl interaction.

The four $d$ lines independently give $(7.377-7.330)300=14.1$ and $(7.377-7.356)300=6.3\mathrm{Hz}$. The large interaction identifies $e$ as trans to $d$; $f$ has the smaller cis interaction. The double-headed arrows in the structural map identify the actual interacting nuclei. Coupling to $a$ means the same coupling to each of its three equivalent hydrogens.