= Solution
\b[C gives spectrum 3.] For <allyl acetate>, $\mathrm{CH_3C(=O)OCH_2CH=CH_2}$, assign
$$
\delta_C(\mathrm{C=O})=170.6,\quad
\delta_C(\mathrm{COCH_3})=20.9,\quad
\delta_C(\mathrm{OCH_2})=65.1,\quad
\delta_C(\mathrm{CH=})=132.3,\quad
\delta_C(\mathrm{=CH_2})=118.1\ \mathrm{ppm}.
$$
The 65.1-ppm <resonance> is the oxygen-bound methylene. Its <attached proton test> phase matches the <terminal alkene> methylene and the nonprotonated <carbonyl>, while the <alkene> CH and acetyl methyl have the opposite phase. The <carbonyl> is not conjugated with the <alkene> because <oxygen> and a saturated methylene interrupt the carbon–carbon conjugated path. This explains its higher <carbonyl> shift relative to B without requiring the two <esters> to have identical electronic environments.
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