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Esterification and hydrolysis (SACE Stage 2 Chemistry)

Syllabus dot point

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

8 min read

Quick answer

Esterification joins a carboxylic acid and an alcohol, with concentrated sulfuric acid as catalyst, to form an ester plus water: R-COOH+R’-OH⇌R-COO-R’+H2O\text{R-COOH} + \text{R'-OH} \rightleftharpoons \text{R-COO-R'} + \text{H}_2\text{O}. It is a reversible (equilibrium) reaction. Hydrolysis is the reverse, splitting an ester back into its components: acid hydrolysis regenerates the acid and alcohol (equilibrium), while base hydrolysis (saponification) goes to completion, giving the carboxylate salt and the alcohol.

Jump to a section
  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
  8. Exam-style questions

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

Naming and writing an esterification

Write the equation and name the ester from methanol and butanoic acid, with concentrated sulfuric acid.

Identify the two parts of the ester name

The ester name is "(alcohol-derived alkyl) (acid-derived -anoate)". From methanol the alkyl part is methyl; from butanoic acid the acyl part is butanoate.

Write the equation

CH3CH2CH2COOH+CH3OH⇌CH3CH2CH2COOCH3+H2O\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{CH}_3\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_3 + \text{H}_2\text{O}

Name the product

The ester is methyl butanoate. The alcohol always supplies the alkyl group written first; the acid supplies the -anoate part written second. Water is the by-product, and the equilibrium arrow signals reversibility.

Esterification

Esterification

Esterification is the reaction of a carboxylic acid with an alcohol, catalysed by concentrated sulfuric acid, to form an ester and water. It is a condensation reaction (a small molecule, water, is eliminated) and it is reversible, reaching equilibrium rather than going to completion.

R-COOH+R’-OH⇌R-COO-R’+H2O\text{R-COOH} + \text{R'-OH} \rightleftharpoons \text{R-COO-R'} + \text{H}_2\text{O}

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:

Increasing ester yield

Shift the equilibrium to the right by removing water (the concentrated sulfuric acid does this) or by using an excess of one reactant (usually the cheaper alcohol or acid). Removing the ester as it forms, or distilling out water, also pulls the equilibrium toward products. The catalyst speeds the approach to equilibrium but does not change the position.

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’+H2O⇌R-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’+NaOH→R-COO−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).

Naming the ester the wrong way round

The most common SACE error is swapping the two parts of the ester name. The alcohol gives the first word (the alkyl group), and the acid gives the second word (the -oate). Ethanol plus methanoic acid is ethyl methanoate, not methyl ethanoate. A second error is writing esterification with a single (irreversible) arrow; it is an equilibrium. A third is forgetting that base hydrolysis gives the carboxylate salt, not the free acid, and goes to completion.

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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2021-style5 marks
Ethanoic 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 →

(a) CH3COOH+CH3CH2CH2OH⇌CH3COOCH2CH2CH3+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 2019-style4 marks
An 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 →

(a) CH3COOCH2CH3+H2O⇌CH3COOH+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)

Practise this

Sources & how we know this

ExamExplained