Ethyl ethanoate preparation: limiting reagent, equilibrium and purity
A student prepares ethyl ethanoate by refluxing 10.0 mL of ethanol (density 0.789 g/mL) with 10.0 mL of glacial ethanoic acid (density 1.05 g/mL) and 2.0 mL of concentrated sulfuric acid as catalyst. After reflux, the mixture is distilled, washed with water and saturated sodium hydrogencarbonate solution, dried over anhydrous sodium sulfate, and redistilled. The final product mass is 8.50 g. The student records an IR spectrum of the final product with a strong sharp band at 1740 cm-1 but also a weak broad absorption between 2500 and 3300 cm-1. A 1H NMR spectrum shows a quartet at 4.1 ppm, a singlet at 2.0 ppm, and a triplet at 1.2 ppm, plus a small quartet at 3.7 ppm. The student must decide whether the product is pure enough to report as ethyl ethanoate and how the yield should be explained.
1. Calculate the limiting reagent, the theoretical yield of ethyl ethanoate, and the percentage yield.
Read the worked answer
Mass of ethanol = 10.0 mL x 0.789 g/mL = 7.89 g. Moles of ethanol = 7.89/46.07 = 0.1713 mol.
Mass of ethanoic acid = 10.0 mL x 1.05 g/mL = 10.5 g. Moles of ethanoic acid = 10.5/60.05 = 0.1749 mol.
The reaction is 1:1, so ethanol is the limiting reagent because 0.1713 mol < 0.1749 mol. The theoretical mass of ethyl ethanoate is
0.1713 mol x 88.11 g/mol = 15.09 g.
Percentage yield = (8.50/15.09) x 100 = 56.3%.
The yield is below 100% partly because esterification is reversible and the equilibrium does not lie completely to the right, and partly because product is lost during washing, drying and distillation.
2. Interpret the IR and NMR evidence. Is the final product pure ethyl ethanoate? What impurities are indicated?
Read the worked answer
The strong sharp IR band at 1740 cm-1 is characteristic of the ester carbonyl group, so ethyl ethanoate is present. However, the weak broad absorption between 2500 and 3300 cm-1 indicates an O-H stretch, most likely from residual ethanoic acid or possibly water that was not fully removed. Pure ethyl ethanoate should not show a broad O-H absorption.
The NMR data support the main ester: the quartet at 4.1 ppm is the OCH2CH3 methylene group coupled to the CH3 triplet at 1.2 ppm, and the singlet at 2.0 ppm is the CH3CO group. The extra small quartet at 3.7 ppm is characteristic of the CH2 of ethanol, indicating residual ethanol. Therefore the product is not pure; it contains ethanol and probably ethanoic acid. The sodium hydrogencarbonate wash should remove acid as sodium ethanoate, but the small O-H band and the ethanol quartet show that the purification was incomplete.
3. Explain the purpose of each purification step and why simple reflux without removal of water gives a limited yield.
Read the worked answer
Reflux provides energy and keeps volatile components from escaping while the reaction approaches equilibrium. Distillation removes the ester from the higher-boiling acid and water, but ethanol can co-distil, so further washing is needed. Washing with water removes water-soluble ethanol and some acid. Saturated sodium hydrogencarbonate neutralises residual ethanoic acid, converting it to sodium ethanoate, which is water-soluble; the wash must be vented because CO2 is produced. A final water wash removes sodium salts, and anhydrous sodium sulfate removes traces of water. Redistillation separates the ester from non-volatile salts and high-boiling impurities.
The esterification reaction is reversible: ethanoic acid + ethanol <=> ethyl ethanoate + water. As water accumulates, the reverse reaction becomes significant and the equilibrium yield is limited. Removing water or distilling the ester as it forms shifts the equilibrium to the right. The observed 56% yield is therefore consistent with an equilibrium process plus mechanical losses, not with a simple failure to react completely.