 Solution

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 $\mathrm{HOOC}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$, 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 $\mathrm{HOOC}-\mathrm{C}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$, 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 their ordering is tentative from the supplied shift list alone; a correlation experiment would resolve it.

For D, written $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHC}(=\mathrm{O})\mathrm{OCH}=\mathrm{C}{\mathrm{H}}_2$, 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 $\mathrm{C}{\mathrm{H}}_2$ is at $97.4$ ppm. The two sides of the ester oxygen therefore produce quite different vinylic environments.