Answer: D (methanol and butanoic acid). To find the alcohol and acid that make an ester, split the ester at the –CO₂– linkage. In CH₃CH₂CH₂CO₂CH₃, the part on the C=O side (the acyl group) comes from the acid, and the part on the single-bonded oxygen side comes from the alcohol.
The acid part is CH₃CH₂CH₂CO–, which comes from CH₃CH₂CH₂COOH, that is butanoic acid, written C₃H₇CO₂H. The alcohol part is the –CH₃ attached to the oxygen, which comes from CH₃OH, methanol.
Esterification then gives \(\text{CH}_3\text{OH} + \text{C}_3\text{H}_7\text{CO}_2\text{H} \to \text{CH}_3\text{CH}_2\text{CH}_2\text{CO}_2\text{CH}_3 + \text{H}_2\text{O}\). So the pair is methanol and butanoic acid, option D. The common error is to split the ester the wrong way round and pick ethanol with a different acid; remember the carbon chain on the C=O side belongs to the acid.
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