 Solution

B should have a lower carbonyl stretching frequency because its ester carbonyl is conjugated with a C=C bond. A typical diagnostic comparison is a conjugated ester carbonyl around $1715$–$1730\mathrm{c}{\mathrm{m}}^{-1}$ for B versus a nonconjugated ester carbonyl around $1740$–$1750\mathrm{c}{\mathrm{m}}^{-1}$ for C; the precise positions depend on the measurement conditions. conjugation distributes the pi bonding over the acyl fragment and lowers the effective carbonyl bond force constant.

C also has a terminal alkene, giving characteristic out-of-plane C–H bending features near 990 and $910\mathrm{c}{\mathrm{m}}^{-1}$. B has an internal, trans-disubstituted alkene, for which a band near $965\mathrm{c}{\mathrm{m}}^{-1}$ is useful. Both compounds contain ester C–O bonds and alkene C–H bonds, so merely observing those general features would not distinguish them. The carbonyl comparison plus the alkene substitution pattern is the stronger identification.