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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.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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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 H2SO4H_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 βˆ’COOβˆ’-COO-. The general formula is Rβˆ’COOβˆ’Rβ€²R-COO-R', where RR comes from the carboxylic acid and Rβ€²R' 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 CH3COOCH3CH_3COOCH_3 ethanoic acid + methanol solvent
ethyl ethanoate CH3COOC2H5CH_3COOC_2H_5 ethanoic acid + ethanol nail polish remover, pear
methyl butanoate CH3CH2CH2COOCH3CH_3CH_2CH_2COOCH_3 butanoic acid + methanol apple
pentyl ethanoate CH3COOC5H11CH_3COOC_5H_{11} ethanoic acid + pentan-1-ol banana
octyl ethanoate CH3COOC8H17CH_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 Cβˆ’OC-O, add HH to the acid oxygen and add OHOH to the alcohol carbon.

Esterification (Fischer esterification)

An alcohol reacts with a carboxylic acid in the presence of a concentrated H2SO4H_2SO_4 catalyst, heated under reflux, to give an ester and water:

Rβˆ’COOH+Rβ€²βˆ’OHβ‡ŒH2SO4,refluxRβˆ’COOβˆ’Rβ€²+H2OR-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 H2SO4H_2SO_4 has two roles:

  1. Catalyst: protonates the carbonyl oxygen of the acid, making the carbonyl carbon more electrophilic and easier for the alcohol to attack.
  2. 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 MgSO4MgSO_4, then distil at the boiling point of the ester.

Physical properties of esters

Esters have a polar C=OC=O but no Oβˆ’HO-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 (C16C_{16} to C18C_{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:

Rβˆ’COORβ€²+H2Oβ‡ŒH2SO4Rβˆ’COOH+Rβ€²βˆ’OHR-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:

Rβˆ’COORβ€²+NaOHβ†’Rβˆ’COONa+Rβ€²βˆ’OHR-COOR' + NaOH \rightarrow R-COONa + R'-OH

Irreversible because the carboxylate anion Rβˆ’COOβˆ’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:

Triglyceride+3NaOHβ†’3Rβˆ’COONa+glycerol\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.

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 Rβˆ’COOβˆ’CH3R-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: CH3CH2OH+CH3COOHβ‡ŒCH3COOCH2CH3+H2OCH_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; CH3COOH+CH3CH2CH2OHβ‡ŒCH3COOCH2CH2CH3+H2OCH_3COOH + CH_3CH_2CH_2OH \rightleftharpoons CH_3COOCH_2CH_2CH_3 + H_2O with conc H2SO4H_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(ethanol)=9.20/46.07=0.200n(\text{ethanol}) = 9.20 / 46.07 = 0.200 mol; theoretical n(ester)=0.200n(\text{ester}) = 0.200 mol; actual =0.120= 0.120 mol; mass =0.120Γ—116.16=13.9= 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) CH3COOCH2CH3+NaOHβ†’CH3COONa+C2H5OHCH_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 β†’

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

Equation:

CH3COOH(l)+C2H5OH(l)β‡ŒH2SO4CH3COOC2H5(l)+H2O(l)CH_3COOH_{(l)} + C_2H_5OH_{(l)} \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_{5(l)} + H_2O_{(l)}

Procedure.

  1. In a round-bottomed flask, mix about 10 mL ethanoic acid and 10 mL ethanol. Add 5 mL concentrated H2SO4H_2SO_4 slowly while swirling and cooling. Add anti-bumping granules.
  2. 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.
  3. After cooling, transfer to a separating funnel and add saturated NaHCO3NaHCO_3 solution to neutralise excess acid (until no more CO2CO_2 effervescence).
  4. Discard the aqueous layer. Dry the organic layer over anhydrous MgSO4MgSO_4.
  5. 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 H2SO4H_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 β†’

Acid hydrolysis is the reverse of esterification. Dilute H2SO4H_2SO_4 catalyst, reflux:

CH3COOCH3+H2Oβ‡ŒH2SO4CH3COOH+CH3OHCH_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:

CH3COOCH3+NaOH→CH3COONa+CH3OHCH_3COOCH_3 + NaOH \rightarrow CH_3COONa + CH_3OH

The reaction is irreversible because the carboxylate salt CH3COOβˆ’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.

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