Part 2: Ester formation, hydrolysis and product mapping

Identify where each part of an ester came from and predict different products in acidic and alkaline hydrolysis.

Split the ester at the correct bond

An ester has R–C(=O)–O–R′. The acid supplies RCO– and the alcohol supplies the O–R′ portion. In the name, the alcohol-derived alkyl group comes first and the acid-derived carboxylate name comes second.

CH₃CH₂COOCH₃ is methyl propanoate: the methoxy-side carbon comes from methanol, while the three-carbon propanoate portion comes from propanoic acid. CH₃COOCH₂CH₃ is ethyl ethanoate. Both have four carbons but are different structural isomers.

For hydrolysis predictions, draw a division between the carbonyl carbon and its single-bonded oxygen. Restore OH on the acyl side and H on the O–R′ side for the neutral acid/alcohol products. This is a product-mapping tool, not a claim that every hydrolysis mechanism breaks bonds in one identical elementary step.

Equilibrium explains the conditions and limitations

Heating a carboxylic acid with an alcohol and an acid catalyst, commonly concentrated sulfuric acid, forms an ester and water. The reaction is reversible. Reflux allows heating without continually losing volatile reactants; it does not itself remove water or guarantee complete conversion.

An excess of one reactant or removal of a product can shift the equilibrium towards ester. A catalyst speeds both forward and reverse reactions and does not alter the equilibrium constant at a fixed temperature. Connect this preparation to Module 5 equilibrium rather than memorising isolated conditions.

An acid anhydride also reacts with an alcohol to give an ester, with a carboxylic acid as the other product. For ethanoic anhydride plus methanol, products are methyl ethanoate and ethanoic acid. Do not write water as the by-product simply because the organic product is an ester.

The reaction medium controls the acid-side product

Acid hydrolysis heats the ester with water and an acid catalyst, producing a carboxylic acid and alcohol in a reversible reaction. Use excess water to favour hydrolysis. The acid catalyst is not consumed overall.

Alkaline hydrolysis heats the ester with aqueous alkali and forms a carboxylate salt plus an alcohol. The carboxylic acid that would otherwise form is deprotonated by alkali, so the overall reaction is effectively driven towards the products under these conditions. OH⁻ is consumed; calling it only a catalyst loses the stoichiometry.

If a question asks for products before acidification, show COO⁻ with its counterion where appropriate. If a subsequent acidic work-up is specified, show COOH. State both stages when required rather than silently assuming that a neutral acid emerges from a strongly alkaline mixture.

Worked example: an ester with two different fragments

Ethyl butanoate is CH₃CH₂CH₂COOCH₂CH₃. Acid hydrolysis produces butanoic acid, CH₃CH₂CH₂COOH, and ethanol, CH₃CH₂OH. Alkaline hydrolysis with NaOH gives sodium butanoate and ethanol instead. The four-carbon acid fragment and two-carbon alcohol fragment are conserved.

For an original quantitative example, 5.80 g of ethyl butanoate, Mr 116.0, is 0.0500 mol. Complete alkaline hydrolysis needs at least 0.0500 mol NaOH by the 1:1 equation and gives a theoretical 0.0500 mol ethanol, or 2.30 g using Mr 46.0. A measured isolated mass may be lower because conversion and recovery are separate issues.

In a molecule containing multiple ester links, mark and hydrolyse each link. Retain unrelated C–C bonds and identify whether a diol, a diacid or additional functional groups remain. This prepares the same reasoning needed for polyesters.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources