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Inquiry Question 5: How are esters formed, what are their properties, and how are they used?

Investigate the structural formulae, properties, applications, formation by esterification, and hydrolysis (including saponification) of esters

A focused answer to the HSC Chemistry Module 7 dot point on esters. Naming as alkyl alkanoates, the equilibrium esterification with concentrated H2SO4 catalyst, acid and base hydrolysis (saponification), applications as flavours, fragrances and biodiesel, and worked HSC past exam questions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-22

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What this dot point is asking
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In plain English

Think of an ester as a handshake between an acid and an alcohol, where the two molecules join and shake off a drop of water as they connect. To make that handshake happen in the lab, you warm the acid and alcohol together with a splash of concentrated sulfuric acid, which does two jobs at once: it nudges the molecules together (catalyst) and it soaks up the water they produce, like a sponge sitting in the flask, so the handshake is less likely to come undone (Le Chatelier). Because the reaction can run in reverse, only about two-thirds of the starting acid and alcohol actually end up as ester, no matter how long you wait, unless you keep mopping up the water or crowd in extra alcohol. Once made, the ester can be pulled apart again two ways: gently, with dilute acid and water, which just reverses the handshake and settles back to an equilibrium mixture, or permanently, with sodium hydroxide, which grabs the acid half so tightly (as a stable salt) that it can never shake hands with the alcohol again. That second, one-way split is exactly what happens when animal fat or vegetable oil is boiled with sodium hydroxide to make soap: the "acid" halves of the fat molecule come away as sodium salts, which are soap, and the "alcohol" half comes away as glycerol.

What this dot point is asking

NESA wants you to name esters as alkyl alkanoate, write the esterification equation between an alcohol and a carboxylic acid with concentrated $H_2SO_4$ catalyst, describe the reflux procedure, list common applications of esters, and write both acid and base hydrolysis equations including saponification of long-chain esters to make soap.

The answer

Structure and naming

An ester contains the linkage $-COO-$ . The general formula is $R-COO-R'$ , where $R$ comes from the carboxylic acid and $R'$ comes from the alcohol.

Esters are named in two words: alkyl (from the alcohol) alkanoate (from the acid, including the carbonyl carbon).

Ester	Acid + alcohol	Smell/use
methyl ethanoate $CH_3COOCH_3$	ethanoic acid + methanol	solvent
ethyl ethanoate $CH_3COOC_2H_5$	ethanoic acid + ethanol	nail polish remover, pear
methyl butanoate $CH_3CH_2CH_2COOCH_3$	butanoic acid + methanol	apple
pentyl ethanoate $CH_3COOC_5H_{11}$	ethanoic acid + pentan-1-ol	banana
octyl ethanoate $CH_3COOC_8H_{17}$	ethanoic acid + octan-1-ol	orange

To work backwards from an ester to its parent acid and alcohol: split at the single-bonded $C-O$ , add $H$ to the acid oxygen and add $OH$ to the alcohol carbon.

Esterification (Fischer esterification)

An alcohol reacts with a carboxylic acid in the presence of a concentrated $H_2SO_4$ catalyst, heated under reflux, to give an ester and water:

R-COOH + R'-OH \underset{}{\overset{H_2SO_4, \text{reflux}}{\rightleftharpoons}} R-COO-R' + H_2O

The reaction is reversible and reaches equilibrium, typically with 60 to 70% conversion. Concentrated $H_2SO_4$ has two roles:

Catalyst: protonates the carbonyl oxygen of the acid, making the carbonyl carbon more electrophilic and easier for the alcohol to attack.
Dehydrating agent: absorbs the water byproduct, shifting equilibrium right by Le Chatelier's principle.

Reflux is essential because the reactants are volatile (boiling points 65 to 120 degrees C). Reflux returns evaporated reactants to the flask, so the mixture stays at temperature for long enough to reach equilibrium without losing material.

Purification. Pour the cooled mixture into saturated sodium hydrogencarbonate to neutralise unreacted acid (effervescence stops when complete). Separate the organic layer, dry over anhydrous $MgSO_4$ , then distil at the boiling point of the ester.

Physical properties of esters

Esters have a polar $C=O$ but no $O-H$ , so they cannot hydrogen-bond to each other. Boiling points are lower than the parent acid and alcohol of similar molar mass, comparable to ketones. Small esters are volatile liquids with characteristic fruity smells.

Solubility in water decreases sharply with chain length: methyl ethanoate is somewhat soluble, ethyl ethanoate has limited solubility (about 8% w/w), and esters above C6 are essentially insoluble. Esters are good solvents for non-polar organic compounds (paints, varnishes, glues).

Applications

Flavours and fragrances: short-chain esters give fruits their characteristic smells. Synthetic esters are added to lollies, drinks, perfumes and air fresheners.
Solvents: ethyl ethanoate is the active solvent in nail polish remover; butyl ethanoate is used in lacquers.
Biodiesel: methyl esters of long-chain fatty acids ( $C_{16}$ to $C_{18}$ ) are biodiesel, made by transesterification of vegetable oil with methanol and a NaOH catalyst.
Plasticisers: dialkyl phthalate esters soften PVC.
Fats and oils: triesters of glycerol with fatty acids (triglycerides) are biological lipids.

Hydrolysis: the reverse reactions

Acid hydrolysis. Reflux the ester with dilute sulfuric acid; the reverse of esterification:

R-COOR' + H_2O \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} R-COOH + R'-OH

Reversible, reaches equilibrium. Useful to identify an unknown ester by isolating its acid and alcohol fragments.

Base hydrolysis (saponification). Reflux the ester with aqueous NaOH:

R-COOR' + NaOH \rightarrow R-COONa + R'-OH

Irreversible because the carboxylate anion $R-COO^-$ is unreactive and cannot recombine with the alcohol. The reaction goes to completion, giving the sodium salt of the carboxylic acid and the alcohol.

When the ester is a triglyceride (the natural form of fats and oils), saponification with NaOH gives glycerol plus three sodium fatty-acid salts, which are soap:

\text{Triglyceride} + 3 NaOH \rightarrow 3 R-COONa + \text{glycerol}

This is the chemistry of soap-making: hot fat or oil is stirred with sodium hydroxide solution, the soap is salted out with brine, washed, and pressed into bars. KOH gives softer potassium soaps (liquid soaps).

Mechanism in brief

Fischer esterification is acid-catalysed nucleophilic acyl substitution. The acid is protonated on the carbonyl O, the alcohol O attacks the carbonyl C, proton transfers occur, water leaves, and a proton is lost from the protonated ester. You do not need the full curly-arrow mechanism for HSC, but you do need to know what each reactant contributes.

An owned illustrative curve shows how ester concentration climbs to a sub-100% plateau as the reversible esterification approaches equilibrium:

Esterification and hydrolysis worked examples

Worked example 1: preparing methyl butanoate (apple-smelling ester) and calculating yield.

Reflux 3.70 g of methanol ( $M = 32.04\ \text{g mol}^{-1}$ ) with excess butanoic acid and concentrated $H_2SO_4$ .
$n(\text{methanol}) = 3.70 / 32.04 = 0.1155\ \text{mol}$ , so theoretical $n(\text{ester}) = 0.1155\ \text{mol}$ (1:1 ratio).
Theoretical mass of methyl butanoate ( $M = 102.13\ \text{g mol}^{-1}$ ): $0.1155 \times 102.13 = 11.80\ \text{g}$ .
If the actual mass recovered is 7.67 g, percentage yield $= (7.67 / 11.80) \times 100 = 65.0\%$ , consistent with a typical uncontrolled Fischer esterification.

Marker's note: always identify the LIMITING reagent first (here, methanol, since butanoic acid is stated as excess) before computing theoretical yield; a common error is basing the calculation on whichever reagent's mass happens to be given first.

Worked example 2: distinguishing acid vs base hydrolysis of methyl propanoate.

Acid hydrolysis (dilute $H_2SO_4$ , reflux): $CH_3CH_2COOCH_3 + H_2O \rightleftharpoons CH_3CH_2COOH + CH_3OH$ . Reversible; a mixture of ester, acid and alcohol remains at equilibrium.
Base hydrolysis (aqueous NaOH, reflux): $CH_3CH_2COOCH_3 + NaOH \rightarrow CH_3CH_2COONa + CH_3OH$ . Irreversible; all the ester is converted because the carboxylate anion cannot re-esterify.
If asked to fully hydrolyse a sample of ester for quantitative analysis, base hydrolysis is preferred precisely because it goes to completion, giving a reliable stoichiometric result.

Marker's note: full marks require stating WHY base hydrolysis is irreversible (carboxylate anion is unreactive/stable), not just asserting that it is.

Examples in context

Example 1. Biodiesel manufacture from tallow at Australian Renewable Fuels Picton. The Picton plant southwest of Sydney converts beef tallow into biodiesel by base-catalysed transesterification: triglyceride plus three methanol gives glycerol plus three fatty acid methyl esters (FAMEs). Each FAME molecule contains the ester linkage $R-COO-CH_3$ . The reaction uses NaOH at 60 degrees C and is fast, irreversible and high-yielding. Quality control engineers check ester content by GC-MS against the AS 3572 fuel-grade standard. The HSC esterification mechanism extended to triglyceride substrates explains the entire industrial flowsheet, and students applying mass balance can predict that 1.0 tonne of tallow yields about 1.05 tonnes of biodiesel.

Example 2. NSW HSC depth study synthesis of ethyl ethanoate. A common Stage 6 depth study has students reflux 5 mL of ethanol with 5 mL of glacial ethanoic acid and 1 mL of concentrated sulfuric acid for 20 minutes. After cooling and pouring into sodium carbonate solution, students smell the characteristic pear-drop aroma of ethyl ethanoate: $CH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O$ . Yield rarely exceeds 70 percent because the equilibrium is unfavourable; Le Chatelier reasoning explains why excess alcohol or removal of water by distillation would push yield higher. The synthesis appears almost every year as a 5 to 7 mark Section II question.

Try this

Q1. Name the ester formed when propan-1-ol reacts with ethanoic acid, and write the balanced equation including the catalyst. [3 marks]

Cue. Propyl ethanoate; $CH_3COOH + CH_3CH_2CH_2OH \rightleftharpoons CH_3COOCH_2CH_2CH_3 + H_2O$ with conc $H_2SO_4$ catalyst.

Q2. A 9.20 g sample of ethanol is reacted with excess butanoic acid to form ethyl butanoate. If the percent yield is 60 percent, calculate the mass of ester produced. [3 marks]

Cue. $n(\text{ethanol}) = 9.20 / 46.07 = 0.200$ mol; theoretical $n(\text{ester}) = 0.200$ mol; actual $= 0.120$ mol; mass $= 0.120 \times 116.16 = 13.9$ g.

Q3. Sodium hydroxide is used to hydrolyse ethyl ethanoate. (a) Write the balanced equation. (b) State why this reaction is described as saponification when applied to a triglyceride. (c) Contrast this with acid hydrolysis. [2+2+1 marks]

Cue. (a) $CH_3COOCH_2CH_3 + NaOH \rightarrow CH_3COONa + C_2H_5OH$ . (b) Triglyceride saponification gives soap (sodium salts of fatty acids). (c) Acid hydrolysis is reversible; base hydrolysis is irreversible because the carboxylate ion is stable.

Exam-style practice questions

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

2022 HSC5 marksDescribe an experimental procedure to prepare ethyl ethanoate in the laboratory. Include a balanced equation, the catalyst, the reflux setup, and how the product is isolated and identified.

Show worked answer →

A 5 mark answer needs the equation, the role of the catalyst, the apparatus, and the purification and identification steps.

Equation:

CH_3COOH_{(l)} + C_2H_5OH_{(l)} \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_{5(l)} + H_2O_{(l)}

Procedure.

In a round-bottomed flask, mix about 10 mL ethanoic acid and 10 mL ethanol. Add 5 mL concentrated $H_2SO_4$ slowly while swirling and cooling. Add anti-bumping granules.
Attach a reflux condenser and heat with a water bath or heating mantle at about 70 degrees C for 20 minutes. Reflux returns volatile reactants to the flask while the reaction reaches equilibrium.
After cooling, transfer to a separating funnel and add saturated $NaHCO_3$ solution to neutralise excess acid (until no more $CO_2$ effervescence).
Discard the aqueous layer. Dry the organic layer over anhydrous $MgSO_4$ .
Distil at 77 degrees C (the boiling point of ethyl ethanoate) to isolate the pure product.

Identification. Sweet, fruity, pear-drop smell. Confirm by boiling point and refractive index.

Role of catalyst. Concentrated $H_2SO_4$ catalyses the reaction by protonating the carbonyl and also absorbs the water byproduct, shifting equilibrium right (Le Chatelier).

Markers reward (1) the balanced equation with reversible arrows, (2) reflux setup with diagram or description, (3) the catalyst role (both proton donor and dehydrator), (4) purification with bicarbonate wash, (5) identification by smell and boiling point.

2019 HSC3 marksOutline the difference between acid hydrolysis and base hydrolysis (saponification) of an ester, using methyl ethanoate as the example. Write equations for both.

Show worked answer →

Acid hydrolysis is the reverse of esterification. Dilute $H_2SO_4$ catalyst, reflux:

CH_3COOCH_3 + H_2O \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOH + CH_3OH

The reaction is reversible and reaches equilibrium. Products are the carboxylic acid and the alcohol.

Base hydrolysis (saponification) uses aqueous NaOH, reflux:

CH_3COOCH_3 + NaOH \rightarrow CH_3COONa + CH_3OH

The reaction is irreversible because the carboxylate salt $CH_3COO^-$ cannot react back with the alcohol. Products are the sodium salt of the acid (a soap if the acid is a long fatty acid) and the alcohol.

Key differences. Acid hydrolysis is an equilibrium that gives the free acid; base hydrolysis goes to completion and gives the carboxylate salt. The name saponification reflects soap-making from fats and oils (long-chain esters) with NaOH.

Markers reward (1) both balanced equations with correct reversibility arrows, (2) acid hydrolysis as equilibrium, (3) saponification as irreversible because of the salt.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksName the ester formed from butan-1-ol and ethanoic acid, draw its skeletal structure, and write the balanced esterification equation including the catalyst.

Show worked solution →

Naming: Alkyl (from the alcohol, butan-1-ol) + alkanoate (from the acid, ethanoic acid) gives butyl ethanoate.
Skeletal structure: $CH_3-C(=O)-O-CH_2CH_2CH_2CH_3$ , i.e. a methyl group attached to a carbonyl carbon, single-bonded to O, then a four-carbon chain.
Equation

CH_3COOH + CH_3CH_2CH_2CH_2OH \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOCH_2CH_2CH_2CH_3 + H_2O

Marking criteria: 1 mark for the correct name (order: alkyl then alkanoate), 1 mark for a correct skeletal or condensed structure showing the $-COO-$ linkage, 1 mark for the balanced equation with reversible arrows and the $H_2SO_4$ catalyst shown.

foundation3 marksState the reagent and condition for each conversion: (a) ethanoic acid + ethanol to ethyl ethanoate; (b) ethyl ethanoate to ethanoic acid + ethanol (equilibrium); (c) ethyl ethanoate to sodium ethanoate + ethanol (irreversible).

Show worked solution →

(a) Esterification: Concentrated $H_2SO_4$ catalyst, reflux.
(b) Acid hydrolysis: Dilute $H_2SO_4$ , reflux (reverse of esterification, reaches equilibrium).
(c) Base hydrolysis (saponification): Aqueous NaOH, reflux (goes to completion, irreversible).

Marking criteria: 1 mark per step for BOTH the correct reagent and its matching condition; giving "sulfuric acid" without specifying concentrated/dilute loses the mark for (a) or (b).

core5 marksA student reacts 6.90 g of ethanol with excess butanoic acid under reflux with concentrated

H_2SO_4

. The percentage yield of ethyl butanoate is 65.0%. Calculate the actual mass of ester produced, to 3 significant figures. (

M(\text{ethanol}) = 46.07\ \text{g mol}^{-1}

M(\text{ethyl butanoate}) = 116.16\ \text{g mol}^{-1}

Show worked solution →

Step 1: write the equation (1:1 stoichiometry, ethanol is limiting as the acid is in excess).

CH_3CH_2CH_2COOH + C_2H_5OH \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3CH_2CH_2COOC_2H_5 + H_2O

Step 2: moles of ethanol (the limiting reagent).

n(\text{ethanol}) = \frac{m}{M} = \frac{6.90\ \text{g}}{46.07\ \text{g mol}^{-1}} = 0.14976\ \text{mol}

Step 3: theoretical moles of ester (1:1 ratio).

n(\text{ester, theoretical}) = 0.14976\ \text{mol}

Step 4: theoretical mass of ester.

m(\text{theoretical}) = n \times M = 0.14976\ \text{mol} \times 116.16\ \text{g mol}^{-1} = 17.396\ \text{g}

Step 5: apply the percentage yield.

m(\text{actual}) = 17.396\ \text{g} \times \frac{65.0}{100} = 11.307\ \text{g}

Step 6: round to 3 significant figures (matching 6.90 g and 65.0%, both 3 s.f.).

m(\text{ethyl butanoate}) = 11.3\ \text{g}

Marking criteria: 1 mark for the correct balanced 1:1 equation, 1 mark for correct moles of ethanol, 1 mark for the theoretical mass calculation, 1 mark for correctly applying the percentage yield, 1 mark for the final answer to 3 significant figures with units.

core5 marksThe graph below is an owned illustrative curve of ester concentration versus time for the esterification of ethanoic acid with excess ethanol under reflux with concentrated

H_2SO_4

, reaching a plateau at

t \approx 45

minutes. (a) Describe the shape of the curve. (b) Explain, using Le Chatelier's principle, why the curve plateaus rather than reaching 100% conversion. (c) Suggest ONE practical change that would raise the plateau height, and justify it.

Show worked solution →

(a) Description: Ester concentration rises quickly at first, then the rate of increase slows, and the curve flattens into a horizontal plateau from about $t = 45$ minutes onward.
(b) Why it plateaus: Esterification is a reversible reaction. As ester and water accumulate, the reverse (hydrolysis) reaction speeds up while the forward reaction slows as reactants are used up. The plateau is reached when the forward and reverse rates become equal, that is, dynamic equilibrium, NOT when the reaction stops. Because $K$ for this reaction is modest (typically around 4), a substantial amount of unreacted acid and alcohol remains at equilibrium, so the plateau sits below 100% conversion.
(c) Practical change: Use a large excess of ethanol (already used here) or continuously remove water as it forms (e.g. with excess concentrated $H_2SO_4$ acting as a dehydrating agent, or a Dean-Stark trap). Removing a product or adding excess reactant shifts the equilibrium position further to the right by Le Chatelier's principle, raising the height of the plateau (the final ester concentration), even though $K$ itself is unchanged.

Marking criteria: (a) 1 mark for describing the fast-rise-then-plateau shape. (b) 1 mark for identifying the reaction as reversible/an equilibrium, 1 mark for explaining the plateau as equal forward/reverse rates (not reaction stopping), 1 mark for linking the sub-100% plateau height to a modest $K$ value. (c) 1 mark for a valid Le Chatelier-justified strategy (excess reactant or water removal) with correct reasoning.

exam6 marksDistinguish between acid hydrolysis and base hydrolysis (saponification) of ethyl ethanoate, and evaluate why saponification, rather than acid hydrolysis, is used industrially to manufacture soap from triglycerides.

Show worked solution →

This is a 6-mark DISTINGUISH-and-EVALUATE: markers reward a clear contrast AND a reasoned industrial judgement, not just two equations.

Band 6 plan.

State both equations for ethyl ethanoate with correct reversibility arrows.
Contrast: acid hydrolysis is an equilibrium (partial conversion, gives the free acid); saponification is irreversible (complete conversion, gives the carboxylate salt).
Extend to triglycerides and explain why irreversibility matters industrially (yield, product identity, no free fatty acid corrosivity).
End with an explicit judgement.

Model answer.

Acid hydrolysis reflux with dilute $H_2SO_4$ is the reverse of esterification and is an equilibrium:

CH_3COOCH_2CH_3 + H_2O \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOH + CH_3CH_2OH

Base hydrolysis (saponification) with aqueous NaOH goes to completion:

CH_3COOCH_2CH_3 + NaOH \rightarrow CH_3COONa + CH_3CH_2OH

The key distinction is reversibility: acid hydrolysis reaches a partial-conversion equilibrium because the carboxylic acid and alcohol can recombine, whereas base hydrolysis is irreversible because the carboxylate anion $CH_3COO^-$ is a poor electrophile and stable, so it cannot react back with the alcohol.

Applied to a triglyceride, saponification with NaOH gives glycerol plus three sodium fatty-acid salts (soap):

\text{Triglyceride} + 3NaOH \rightarrow 3R-COONa + \text{glycerol}

Industrially, saponification is preferred over acid hydrolysis for soap manufacture for three reasons: (1) irreversibility drives the reaction to completion, maximising yield of soap from a fixed batch of fat; (2) the product is directly the sodium salt (soap), whereas acid hydrolysis would give the free fatty acid, which would then need a separate neutralisation step with a base to form soap anyway; (3) acid hydrolysis requires handling and neutralising strong acid on an industrial scale, which is more corrosive and costly than working with aqueous NaOH.

Marker's note: top-band responses (1) give both correctly balanced equations with the right reversibility arrows, (2) explicitly name the carboxylate anion's stability as the reason base hydrolysis is irreversible, (3) extend the reasoning to the triglyceride case with the 3:1 stoichiometry, and (4) close with a clear, justified industrial judgement rather than simply restating that saponification is "better".

Marking criteria: 2 marks for the two correct equations, 1 mark for correctly identifying which is reversible and which is irreversible, 1 mark for the mechanistic reason (carboxylate anion stability), 1 mark for the triglyceride extension, 1 mark for a justified industrial evaluation.

foundation2 marksState two physical properties of small esters (such as ethyl ethanoate) and explain briefly why esters have lower boiling points than the parent carboxylic acid of similar molar mass.

Show worked solution →

Two physical properties. Small esters are volatile liquids at room temperature with characteristic fruity smells; solubility in water decreases sharply as chain length increases, with esters above about $C_6$ essentially insoluble.

Explanation of lower boiling point. Esters have a polar $C=O$ bond but no $O-H$ bond, so they cannot hydrogen-bond to each other (only weaker dipole-dipole and dispersion forces act between molecules). The parent carboxylic acid, by contrast, has an $O-H$ bond and can hydrogen-bond (even forming hydrogen-bonded dimers), giving it a substantially higher boiling point at similar molar mass.

Marking criteria: 1 mark for two correct physical properties, 1 mark for correctly explaining the lower boiling point via the absence of hydrogen bonding (no O-H bond) in the ester compared with the acid.

exam7 marksA biodiesel plant produces fatty acid methyl esters (FAMEs) by base-catalysed transesterification of vegetable oil with methanol and a NaOH catalyst at 60 degrees C. Explain how this industrial process differs chemically from the Fischer esterification studied for small esters, and evaluate why base-catalysed transesterification, rather than acid-catalysed Fischer esterification, is used at industrial scale for biodiesel production.

Show worked solution →

This is a 7-mark EXPLAIN-and-EVALUATE: markers reward a clear chemical distinction plus a reasoned industrial judgement.

Band 6 plan.

Explain transesterification: the alcohol group already on an EXISTING ester (the triglyceride) is swapped for methanol, rather than an ester being built from a free acid and a free alcohol.
Contrast with Fischer esterification (acid + alcohol, conc $H_2SO_4$ , reversible equilibrium, 60 to 70% conversion).
State the transesterification equation and note it is base-catalysed and effectively irreversible (glycerol, a poor leaving group nucleophile once formed, does not readily re-esterify under these conditions and is removed as a separate, denser layer).
Evaluate industrial reasons: rate, safety, and glycerol separation/by-product value.

Model answer.

In Fischer esterification, a free carboxylic acid and a free alcohol combine under concentrated $H_2SO_4$ catalysis to form a new ester and water, and the reaction is a reversible equilibrium that typically converts only 60 to 70% of the limiting reagent. Transesterification is chemically different: it starts from an EXISTING ester, the triglyceride (glycerol tri-ester of long-chain fatty acids), and swaps the alcohol portion for methanol under base catalysis (NaOH):

\text{Triglyceride} + 3CH_3OH \xrightarrow{NaOH} 3R-COOCH_3 + \text{glycerol}

No free carboxylic acid is ever formed or consumed, so the mechanism (base-catalysed nucleophilic attack of methoxide on the ester carbonyl) is distinct from the acid-catalysed nucleophilic acyl substitution of Fischer esterification.

Base-catalysed transesterification is preferred industrially for three reasons: (1) it proceeds rapidly at a mild 60 degrees C without needing the strongly corrosive concentrated $H_2SO_4$ and extended reflux times typical of Fischer esterification, reducing equipment corrosion and energy cost at scale; (2) glycerol, being far denser and more polar than the FAME product, separates cleanly into a lower layer, so the reaction effectively goes to a high, near-complete conversion once glycerol is continuously removed, avoiding the equilibrium ceiling seen in Fischer esterification; (3) the glycerol by-product itself has commercial value (used in pharmaceuticals and cosmetics), improving the overall process economics.

Marker's note: top-band responses (1) correctly identify transesterification as swapping an alcohol on an EXISTING ester rather than building an ester from a free acid, (2) give a correctly balanced transesterification equation, (3) explain at least two genuine industrial advantages (rate/mildness, separation/near-complete conversion, or by-product value) rather than simply asserting transesterification is "better", and (4) do not confuse the base-catalysed mechanism with acid-catalysed Fischer esterification.

Marking criteria: 2 marks for correctly distinguishing transesterification from esterification (existing ester vs free acid), 1 mark for the correct balanced transesterification equation, 1 mark for identifying the base catalyst and mild conditions, 1 mark for explaining near-complete conversion via glycerol removal, 1 mark for a further genuine industrial advantage, 1 mark for a clear, justified evaluative judgement.