 Solution

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 ${\mathrm{C}}_j$.