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How do carboxylic acids behave as weak acids, and how are esters formed and broken down?

Describe the properties and reactions of carboxylic acids, and explain esterification and hydrolysis of esters

A focused answer to the WACE Year 12 Chemistry dot point on carboxylic acids and esters, covering carboxylic acid acidity and reactions, the esterification equilibrium, ester hydrolysis, and the uses of esters, with a worked example and common exam mistakes.

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

A carboxylic acid contains the carboxyl group, -COOH. An ester contains the -COO- linkage and is formed from a carboxylic acid and an alcohol. The two families are closely related through esterification and hydrolysis.

Properties and reactions of carboxylic acids

Carboxylic acids are weak acids: they only partially ionise in water.

CH3COOHβ‡ŒCH3COOβˆ’+H+\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+

They show typical acid reactions:

  • With bases: neutralisation to give a salt and water.
  • With reactive metals: producing a salt and hydrogen gas.
  • With carbonates and hydrogen carbonates: producing a salt, water and carbon dioxide (effervescence), a useful test for the -COOH group.

The -COOH group hydrogen bonds strongly, so carboxylic acids have high boiling points and the smaller ones are very soluble in water.

Esterification

An ester forms when a carboxylic acid reacts with an alcohol, with concentrated sulfuric acid as catalyst:

CH3COOH+C2H5OHβ‡ŒCH3COOC2H5+H2O\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}

Because esterification is an equilibrium, the yield is incomplete. By Le Chatelier's principle, using an excess of one reactant or removing the water (or ester) as it forms shifts the equilibrium towards the ester and improves the yield. This is a direct link back to the equilibrium chemistry of Unit 3.

Hydrolysis of esters

Hydrolysis is the reverse of esterification, splitting the ester with water.

  • Acid hydrolysis: with water and an acid catalyst, the ester reverts to the carboxylic acid and the alcohol (an equilibrium).
  • Base hydrolysis (saponification): with hot aqueous sodium hydroxide, the ester is split into the sodium salt of the carboxylic acid (a carboxylate) and the alcohol. This goes to completion because the carboxylate ion does not re-esterify, and it is the basis of soap making.

Why carboxylic acids are weak yet detectable

The acidity of carboxylic acids comes from the stability of the carboxylate ion left after the proton leaves: the negative charge is spread (delocalised) over both oxygen atoms, which stabilises the ion and favours ionisation more than for an alcohol, whose O-H\text{O-H} does not ionise appreciably. Even so, carboxylic acids are only weak acids, so they partially ionise and have pKa\text{p}K_a values around 4 to 5. A reliable distinguishing test exploits their reaction with carbonates: only a compound acidic enough, such as a carboxylic acid, fizzes with sodium hydrogen carbonate to release carbon dioxide, CH3COOH+NaHCO3β†’CH3COONa+H2O+CO2\text{CH}_3\text{COOH} + \text{NaHCO}_3 \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2. Phenols and alcohols do not, so the effervescence test confirms the -COOH\text{-COOH} group.

Uses and why this matters

Esters are responsible for many fruit flavours and floral scents and are used as solvents and in fragrances; larger esters form fats, oils and biodiesel. The esterification equilibrium ties organic synthesis to the equilibrium principles of Unit 3, and ester hydrolysis underlies soap manufacture and the digestion of fats.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20216 marksEthyl ethanoate is prepared by refluxing ethanoic acid with ethanol and a few drops of concentrated sulfuric acid. (a) Write the equation and name the ester. (b) Explain the role of the sulfuric acid. (c) Using Le Chatelier's principle, describe two ways the yield of ester could be increased.
Show worked answer β†’

A 6 mark question rewards the equation, the catalyst role, and two justified yield strategies.

(a) CH3COOH+C2H5OHβ‡ŒCH3COOC2H5+H2O\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}. The ester is ethyl ethanoate.

(b) Concentrated sulfuric acid acts as a catalyst, speeding up the attainment of equilibrium without being consumed. It is also a dehydrating agent that helps remove water, which (see below) shifts the position towards the ester.

(c) Because the reaction is an equilibrium, by Le Chatelier's principle the yield rises if you: (i) use an excess of one reactant (for example excess alcohol), shifting the equilibrium to the right; and (ii) remove a product, such as removing water with the sulfuric acid or distilling off the ester, again shifting the system towards more ester.

Markers reward the reversible equation and name, the catalyst (and dehydrating) role, and two distinct Le Chatelier strategies with directions.

WACE 20235 marksAn ester X\text{X} is hydrolysed. With dilute acid it gives propanoic acid and ethanol; with hot sodium hydroxide it gives a different organic product. (a) Identify and name ester X\text{X}. (b) Write the equation for the base hydrolysis (saponification) of X\text{X}. (c) Explain why base hydrolysis goes to completion whereas acid hydrolysis does not.
Show worked answer β†’

A 5 mark question rewards the ester identity, the saponification equation, and the completion reasoning.

(a) The acid part is propanoic acid (propanoate) and the alcohol part is ethanol (ethyl), so X\text{X} is ethyl propanoate, C2H5COOC2H5\text{C}_2\text{H}_5\text{COOC}_2\text{H}_5.

(b) C2H5COOC2H5+NaOH→C2H5COONa+C2H5OH\text{C}_2\text{H}_5\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{COONa} + \text{C}_2\text{H}_5\text{OH} (sodium propanoate and ethanol).

(c) Acid hydrolysis is the reverse of esterification and reaches the same equilibrium, so it does not go to completion. Base hydrolysis produces the carboxylate ion (C2H5COOβˆ’\text{C}_2\text{H}_5\text{COO}^-), which is stable and does not re-esterify with the alcohol, so the reaction is effectively irreversible and proceeds to completion.

Markers reward ethyl propanoate, the saponification equation giving the carboxylate salt, and the irreversible-carboxylate explanation.

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