2018 H2 Chem P1 Q28 — Carboxylic Acid Solution

The answer is C.
The \(^{18}\text{O}\) from the water ends up in the acid, \(\text{CH}_3\text{C}(\text{=O})^{18}\text{OH}\), while the \(^{16}\text{O}\) that was bonded to the ethyl group ends up in the alcohol, \(\text{C}_2\text{H}_5{}^{16}\text{OH}\).
So the water oxygen attacks the \(\delta+\) carbonyl carbon and forms a new C-\(^{18}\text{O}\) bond there, and the bond that breaks is the acyl C-O single bond between the carbonyl carbon and the \(^{16}\text{O}\) of the \(\text{OC}_2\text{H}_5\) group, in the \(\text{-C}(\text{=O})\text{-O-}\) group:
\(\text{CH}_3\text{C}(\text{=O})\text{-}{}^{16}\text{OC}_2\text{H}_5 + \text{H}_2{}^{18}\text{O} \rightarrow \text{CH}_3\text{C}(\text{=O})^{18}\text{OH} + \text{C}_2\text{H}_5{}^{16}\text{OH}\)
• A is wrong: isotopes have the same electronic configuration and form bonds of essentially the same strength; in any case the observation shows which bond breaks, not how strong it is.
• B is wrong: if the O-\(\text{C}_2\text{H}_5\) bond broke, the ethyl group would pick up the water oxygen and the alcohol would be \(\text{C}_2\text{H}_5{}^{18}\text{OH}\), which is not observed.
• D is wrong: the water is attracted to the \(\delta+\) carbonyl carbon, not a carbon of the \(\text{C}_2\text{H}_5\) group; attack at the ethyl carbon would again give \(\text{C}_2\text{H}_5{}^{18}\text{OH}\).