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How are esters formed and broken down?

Describe the formation of esters by esterification and their breakdown by hydrolysis, writing equations and naming products.

Forming esters from a carboxylic acid and an alcohol, naming esters, the equilibrium nature of esterification, acid and base hydrolysis (saponification), and worked SACE-style naming and yield calculations.

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  1. What this dot point is asking
  2. Lead worked calculation
  3. Esterification
  4. Naming esters
  5. Driving the equilibrium
  6. Hydrolysis: the reverse
  7. Why it matters

What this dot point is asking

SACE expects you to write equations, name esters and their products, explain the equilibrium and how to drive the yield, and distinguish acid from base hydrolysis.

Lead worked calculation

Esterification

The concentrated sulfuric acid plays two roles: it catalyses the reaction and it acts as a dehydrating agent, removing water to shift the equilibrium toward the ester.

Naming esters

An ester name has two words. The first is the alkyl group from the alcohol (methyl, ethyl, propyl, ...). The second is the acid stem with -oate (methanoate, ethanoate, propanoate, ...). So ethanol plus ethanoic acid gives ethyl ethanoate. Reading an ester structure R-COO-R’\text{R-COO-R'}: the R’\text{R'} on the single-bonded oxygen came from the alcohol (alkyl, first word), and the R-COO\text{R-COO} part came from the acid (-oate, second word).

Driving the equilibrium

Because esterification is an equilibrium, the yield can be raised using Le Chatelier's principle:

Hydrolysis: the reverse

Hydrolysis splits an ester apart using water. There are two routes:

  • Acid hydrolysis: water with a dilute acid catalyst reverses esterification, giving the carboxylic acid and the alcohol. It is an equilibrium, so it does not go to completion. R-COO-R’+H2Oβ‡ŒR-COOH+R’-OH\text{R-COO-R'} + \text{H}_2\text{O} \rightleftharpoons \text{R-COOH} + \text{R'-OH}.
  • Base hydrolysis (saponification): warming with aqueous NaOH\text{NaOH} gives the carboxylate salt and the alcohol, and goes essentially to completion (irreversible) because the carboxylate ion does not react back. R-COO-R’+NaOHβ†’R-COOβˆ’Na++R’-OH\text{R-COO-R'} + \text{NaOH} \rightarrow \text{R-COO}^-\text{Na}^+ + \text{R'-OH}.

Saponification is the industrial process for making soap from fats (esters of long-chain acids).

Why it matters

Esters give fruits and flowers their scents and are used as flavourings, solvents and in biodiesel. The forward reaction (condensation) and reverse (hydrolysis) are the same chemistry that builds and breaks down the lipids and proteins studied next, and saponification underpins soap manufacture.

Exam-style practice questions

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

SACE 20215 marksEthanoic acid reacts with propan-1-ol in the presence of concentrated sulfuric acid. (a) Write the equation and name the ester product. (b) State the roles of the concentrated sulfuric acid. (c) The reaction is an equilibrium; state one way to increase the yield of ester and justify it with Le Chatelier's principle.
Show worked answer β†’

(a) CH3COOH+CH3CH2CH2OHβ‡ŒCH3COOCH2CH2CH3+H2O\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}; the ester is propyl ethanoate. (2 marks)

(b) Concentrated H2SO4\text{H}_2\text{SO}_4 acts as a catalyst (speeds the reaction) and as a dehydrating agent that removes water, shifting the equilibrium toward the ester. (2 marks)

(c) Removing water (or using excess of one reactant) shifts the equilibrium to the right by Le Chatelier's principle, increasing the ester yield. (1 mark)

SACE 20194 marksAn ester, ethyl ethanoate (CH3COOCH2CH3\text{CH}_3\text{COOCH}_2\text{CH}_3), is hydrolysed. (a) Write the equation for its **acid** hydrolysis and name the products. (b) Write the equation for its **base** hydrolysis (with NaOH\text{NaOH}) and name the organic products. (c) State one difference between the two hydrolysis routes.
Show worked answer β†’

(a) CH3COOCH2CH3+H2Oβ‡ŒCH3COOH+CH3CH2OH\text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH}; products are ethanoic acid and ethanol. (1 mark)

(b) CH3COOCH2CH3+NaOH→CH3COONa+CH3CH2OH\text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{CH}_3\text{CH}_2\text{OH}; products are sodium ethanoate and ethanol. (2 marks)

(c) Acid hydrolysis is an equilibrium (reversible) and gives the carboxylic acid; base hydrolysis goes essentially to completion (irreversible) and gives the carboxylate salt rather than the free acid. (1 mark)

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