Choosing Between Ester Isomers from IR and NMR Evidence
Four groups propose structures for an unknown with molecular formula C4H8O2: butanoic acid, ethyl acetate, methyl propanoate, and 1,4-dioxane. The available data are: IR has a strong sharp absorption at 1740 cm−1 and no broad absorption from 2500 to 3300 cm−1; 1H NMR shows a triplet at δ 1.25 (3H), a singlet at δ 2.05 (3H), and a quartet at δ 4.10 (2H); the compound is neutral, dissolves in water to give a solution of pH about 6, and does not release CO2 when solid sodium bicarbonate is added. One group argues that the 1740 cm−1 band proves a carboxylic acid; another argues that the singlet at δ 2.05 is an aldehyde proton. The teacher wants students to use all the evidence, not just one peak, before choosing a structure.
1. Evaluate each proposed structure against the IR, NMR, and simple chemical evidence.
Read the worked answer
Butanoic acid is excluded: it would show a broad O-H stretch from about 2500 to 3300 cm−1 and would release CO2 with NaHCO3; the NMR would not show the observed 3H singlet at δ 2.05 or the 2H quartet at δ 4.10. 1,4-Dioxane is excluded because it has no carbonyl, so no 1740 cm−1 band, and its NMR is much simpler. Methyl propanoate has an ester carbonyl, but its OCH3 would be a singlet near δ 3.7 and its CH2 next to the carbonyl would be a quartet near δ 2.3 coupled to a triplet near δ 1.1; the observed δ 4.10 quartet and δ 2.05 singlet do not fit. Ethyl acetate fits the ester carbonyl, the absence of O-H, the neutral pH, the negative bicarbonate test, and all three NMR signals. The δ 2.05 singlet is the acetyl CH3, not an aldehyde proton, which would appear near δ 9–10.
2. Explain how the splitting pattern and integration distinguish ethyl acetate from methyl propanoate.
Read the worked answer
In ethyl acetate, CH3COOCH2CH3, the OCH2 protons are coupled to the terminal CH3, so they appear as a quartet; the acetyl CH3 has no neighboring hydrogens, so it is a singlet. The integration 3:3:2 matches the two methyl groups and the CH2. In methyl propanoate, CH3CH2COOCH3, the OCH3 is an isolated singlet, and the CH2 is coupled to a CH3 to give a quartet at a lower chemical shift (near δ 2.3). The observed quartet at δ 4.10 is characteristic of an OCH2, which points to the ethyl ester.
3. What additional simple chemical test would you use, and what result would support ethyl acetate?
Read the worked answer
Hydrolyze a small sample with NaOH and warm it, then neutralize and apply the iodoform test to the alcohol product. Ethyl acetate gives ethanol, which gives a yellow precipitate of CHI3 with I2/NaOH; methyl propanoate gives methanol, which does not. A positive iodoform test therefore supports ethyl acetate. The bicarbonate test is also useful as a negative control: it rules out butanoic acid but does not distinguish the two esters.