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AS · Practice questions · Esters

Join it, then split it.

Six original Cambridge-style questions on esters.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Acid plus alcohol, reversible; alkali goes all the way.

01
Recall
[3 marks]

Ethanoic acid is warmed with propan-1-ol and a few drops of concentrated sulfuric acid. Write the equation, name the ester and state the role of the sulfuric acid.

  • CH3COOH + CH3CH2CH2OH ⇌ CH3COOCH2CH2CH3 + H2O. ✓
  • The ester is propyl ethanoate. ✓
  • Concentrated H2SO4 is a catalyst. ✓
02
Analysis
[2 marks]

Give the names of the carboxylic acid and the alcohol needed to make (a) methyl butanoate, (b) ethyl methanoate, HCOOC2H5.

  • (a) Butanoic acid and methanol. ✓
  • (b) Methanoic acid and ethanol. ✓
03
Explain
[3 marks]

Compare the hydrolysis of methyl ethanoate by dilute sulfuric acid and by aqueous sodium hydroxide. Give the products of each and explain why only one goes to completion.

  • Dilute acid, reflux: ethanoic acid and methanol, in an equilibrium mixture (reversible). ✓
  • NaOH(aq), reflux: sodium ethanoate and methanol. ✓
  • The ethanoate ion cannot react with methanol, so there is no reverse reaction and alkaline hydrolysis goes to completion. ✓
04
Calculation
[3 marks]

6.00 g of ethanoic acid (Mr 60.0) is heated with excess ethanol and concentrated sulfuric acid. 5.72 g of ethyl ethanoate (Mr 88.0) is obtained. Calculate the percentage yield and suggest why it is below 100%.

  • n(acid) = 6.00 ÷ 60.0 = 0.100 mol; maximum ester = 0.100 × 88.0 = 8.80 g. ✓
  • Yield = 5.72 ÷ 8.80 × 100 = 65.0%. ✓
  • The reaction is reversible and reaches equilibrium, and some ester is lost during separation and purification. ✓
05
Calculation
[3 marks]

17.8 g of glyceryl tristearate (Mr 890) is boiled with excess NaOH(aq). Write a word equation and calculate the mass of NaOH used and the mass of sodium stearate (Mr 306) formed.

  • Glyceryl tristearate + sodium hydroxide → glycerol + sodium stearate (1 : 3 : 1 : 3). ✓
  • n(fat) = 17.8 ÷ 890 = 0.0200 mol, so n(NaOH) = 0.0600 mol = 0.0600 × 40.0 = 2.40 g. ✓
  • n(soap) = 0.0600 mol; mass = 0.0600 × 306 = 18.4 g. ✓
06
Recall
[2 marks]

State two uses of esters and explain why unsaturated oils are liquids at room temperature while saturated fats are solids.

  • Uses: flavourings and perfumes (fruity smells); solvents, e.g. ethyl ethanoate in glues and nail varnish remover. ✓
  • C=C double bonds put kinks in the chains, so oil molecules cannot pack closely; intermolecular forces are weaker, so melting points are lower. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.