= Solution
\b[The carbon spectra are A → 5, B → 8, C → 6, and D → 7.] Number each acyclic acid or ester chain from its carbonyl carbon. The assignments below account for every <carbon-13 NMR> environment.
For A, both <ketone> carbonyl carbons have $\delta=208.3$ ppm and all four methylene carbons have $\delta=36.5$ ppm. Molecular symmetry and rapid conformational averaging explain why six carbons give only two resonances.
For B, written $\mathrm{HOOC-CH=CH-CH=CH-CH_3}$, assign $\delta(\mathrm C_1)=172.8$, $\delta(\mathrm C_2)=118.1$, $\delta(\mathrm C_3)=147.4$, $\delta(\mathrm C_4)=129.7$, $\delta(\mathrm C_5)=140.8$, and $\delta(\mathrm C_6)=18.1$ ppm. The alternation among vinylic <chemical shifts> reflects polarization of the conjugated chain, especially the deshielded carbon beta to the carboxyl group.
For C, written $\mathrm{HOOC-C\equiv C-CH_2-CH_2-CH_3}$, assign the carbonyl at $158.5$, the adjacent sp carbon at $72.9$, the other sp carbon at $92.6$, and the terminal methyl at $13.4$ ppm. The two methylene resonances are $20.7$ and $21.0$ ppm. A plausible assignment is $\mathrm C_4=20.7$ and $\mathrm C_5=21.0$ ppm, but \b[their ordering is tentative from the supplied shift list alone]; a correlation experiment would resolve it.
For D, written $\mathrm{CH_3CH=CHC(=O)OCH=CH_2}$, the carbonyl is at $163.2$ ppm. The crotonyl beta-CH, alpha-CH and methyl carbons are at $148.0$, $121.7$ and $18.2$ ppm, respectively. The O-bound vinyl CH is at $141.3$ ppm and its terminal $\mathrm{CH_2}$ is at $97.4$ ppm. The two sides of the <ester> oxygen therefore produce quite different vinylic environments.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1-carbon-assignments.png]
{title=Original structural maps of all supplied <carbon-13 NMR> assignments}
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