Question 10 of 13: Identifying Two Esters from Formula + 1 H NMR

Question 10: Identifying Two Esters from Formula + ¹H NMR

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

DoU check: $\mathrm{DoU}=(2(7)+2-14)/2=1$, consistent with a single ester C=O and no rings or other unsaturation in either compound — every remaining carbon is sp³.

1. (a) Isoamyl (3-methylbutyl) acetate — the classic "banana oil." The 2.04 ppm singlet (3H) with no neighbouring coupling is the tell-tale acetate CH₃–C(=O) methyl (an acyl methyl has no adjacent C–H to couple to). The rest of the spectrum is one continuous alkyl chain read outward from the ester oxygen: 4.10 (t, 2H, O–CH₂–, coupled only to the next CH₂) → 1.52 (m, 2H, –CH₂–) → 1.69 (m, 1H, the isopropyl methine) → 0.93 (d, 6H, the two equivalent methyls of a terminal isopropyl group). Assembling O–CH₂CH₂CH(CH₃)₂ on the alcohol side of an acetate ester gives CH₃CO₂CH₂CH₂CH(CH₃)₂, exactly C₇H₁₄O₂ — isoamyl acetate.
2. (b) Isobutyl propanoate — the classic "rum" ester. Here the acyl side shows its own ethyl group instead of an isolated singlet: 1.15 (t, 3H) coupled to 2.33 (q, 2H) is a classic CH₃CH₂–C(=O)– (propanoyl) fragment. The alcohol side is a simple isobutyl group: 3.86 (d, 2H, O–CH₂–, coupled only to the adjacent methine) → 1.91 (m, 1H, isopropyl methine) → 0.94 (d, 6H, the two equivalent methyls). Assembling CH₃CH₂CO₂CH₂CH(CH₃)₂ also gives C₇H₁₄O₂ — isobutyl propanoate.