OCR A Level Chemistry A: Carboxylic acids and esters (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These original questions cover section 6.1.3, Carboxylic acids and esters, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions test every learning outcome, 6.1.3(a) to (f), from Module 6, Organic chemistry and analysis. H432 is the full A level. Questions 10 and 11 combine several outcomes with mole calculations.

Revise first with the Carboxylic acids and esters study guide and the Carboxylic acids and esters revision notes. Other topics are on the OCR chemistry hub; record your progress on the printable checklist. For a wider check, try the free diagnostics.

Where a calculation needs data, take relative atomic masses as O = 16.0, C = 12.0, H = 1.0, Na = 23.0 and Cl = 35.5, unless the question gives others.

Questions

1. Butanoic acid, CH₃CH₂CH₂COOH, mixes with water in all proportions.

(a) Describe one hydrogen bond between a butanoic acid molecule and a water molecule, stating where the lone pair and partial charges are. [2] (b) Explain why octanoic acid, CH₃(CH₂)₆COOH, is much less soluble in water than butanoic acid. [2]

2. Methanoic acid, HCOOH, reacts in aqueous conditions with each substance below.

(a) Calcium. Write the equation and name the salt formed. [2] (b) Lithium carbonate. Write the equation. [1] (c) Zinc oxide. Write the equation. [1] (d) Potassium hydroxide. Write the ionic equation. [1]

3. Two unlabelled solutions contain propanoic acid and 3-methylphenol.

(a) State a reagent that identifies the propanoic acid, and what you would see with each solution. [2] (b) 3-methylphenol is acidic. Explain why it does not give the same result as propanoic acid in (a). [1]

4. A student dissolves 2.20 g of a carboxylic acid, RCOOH, in water to make 250.0 cm³ of solution. Each 25.0 cm³ sample needs a mean of 20.0 cm³ of 0.125 mol dm⁻³ sodium hydroxide.

(a) Write the equation for the reaction in the titration, using RCOOH. [1] (b) Calculate the relative molecular mass of the acid. [3] (c) Draw or write structural formulae for two carboxylic acids that fit your answer to (b). [2]

5. Propyl methanoate is made by heating methanoic acid and propan-1-ol with concentrated sulfuric acid.

(a) Write the equation for the reaction, using structural formulae. [2] (b) State the role of the sulfuric acid and explain why the yield of ester is below 100% even with no losses during purification. [2]

6. Methyl butanoate can also be made from butanoic anhydride, (CH₃CH₂CH₂CO)₂O.

(a) Write the equation for butanoic anhydride reacting with methanol. [2] (b) Give two advantages of this route over heating butanoic acid with methanol and an acid catalyst. [2]

7. Pentyl propanoate, C₂H₅COOC₅H₁₁, is hydrolysed in two separate experiments.

(a) Write the equation for hydrolysis by heating under reflux with dilute hydrochloric acid. [2] (b) Write the equation for hydrolysis by heating under reflux with aqueous sodium hydroxide, and name the organic products. [2] (c) Explain why the reaction in (b) goes to completion but the reaction in (a) does not. [1]

8. Butanoyl chloride reacts separately with reagents A, B, C and D. The organic products are:

• A: butanoic acid
• B: methyl butanoate
• C: butanamide
• D: N-methylbutanamide

(a) Identify reagents A, B, C and D. [4] (b) Write the equation for the reaction with reagent C. [1] (c) State whether N-methylbutanamide is a primary or secondary amide, and give a reason. [1]

9. 4-methylphenyl ethanoate, CH₃COOC₆H₄CH₃, is made from 4-methylphenol, CH₃C₆H₄OH.

(a) Write the equation for its formation from 4-methylphenol and ethanoyl chloride. [2] (b) Explain why ethanoyl chloride is used rather than ethanoic acid with an acid catalyst. [1] (c) State what you would see in the air above the reaction mixture. [1]

10. 2.04 g of an ester X is heated under reflux with 50.0 cm³ of 1.00 mol dm⁻³ NaOH, an excess. The leftover NaOH is then neutralised by 30.0 cm³ of 1.00 mol dm⁻³ HCl. Each mole of X reacts with one mole of NaOH.

(a) Write a general equation for the reaction of an ester, RCOOR’, with sodium hydroxide. [1] (b) Calculate the relative molecular mass of X. [4] (c) The hydrolysis products of X are sodium methanoate and an alcohol that acidified dichromate oxidises to a ketone. Deduce the structure and name of X. [3]

11. A student makes phenyl hexanoate in two steps.

Step 1: hexanoic acid + SOCl₂ → hexanoyl chloride Step 2: hexanoyl chloride + phenol → phenyl hexanoate

The student starts with 4.64 g of hexanoic acid. Step 1 has a yield of 80.0% and step 2 has a yield of 75.0%.

(a) Write the equation for step 1. [1] (b) Write the equation for step 2. [1] (c) Calculate the mass of phenyl hexanoate, C₅H₁₁COOC₆H₅, obtained. [4] (d) Calculate the overall percentage yield. [1] (e) Explain why the student did not make phenyl hexanoate in one step by heating hexanoic acid with phenol and concentrated sulfuric acid. [1]

Answers

1. (a) The acid’s O–H hydrogen bonds to a lone pair on a water oxygen (or water’s H to a lone pair on the C=O oxygen) [1]; δ+ on the H, δ− on the O with the lone pair [1] (b) Octanoic acid has a much longer non-polar chain [1]; the chain cannot hydrogen bond with water, so only the –COOH end is attracted to water [1] Examiner insight: “Butanoic acid is polar” is unlikely to earn credit; the answer needs the words hydrogen bond, the δ+ hydrogen and a lone pair on oxygen.

2. (a) 2HCOOH + Ca → (HCOO)₂Ca + H₂ [1]; calcium methanoate [1] (b) 2HCOOH + Li₂CO₃ → 2HCOOLi + H₂O + CO₂ [1] (c) 2HCOOH + ZnO → (HCOO)₂Zn + H₂O [1] (d) HCOOH + OH⁻ → HCOO⁻ + H₂O [1] Examiner insight: Check that each salt formula balances the metal’s charge, (HCOO)₂Ca not HCOOCa, and keep the weak acid as HCOOH in (d) rather than H⁺.

3. (a) Sodium carbonate solution [1]; propanoic acid fizzes (CO₂, turns limewater milky); 3-methylphenol shows no change [1] (b) Phenols are too weakly acidic to react with carbonate ions, so no CO₂ forms [1] Examiner insight: Give an observation for both solutions; “fizzes” for one alone does not distinguish them.

4. (a) RCOOH + NaOH → RCOONa + H₂O [1] (b) n(NaOH) = 0.0200 × 0.125 = 0.00250 mol, so n(acid) in 25.0 cm³ = 0.00250 mol [1]; n(acid) in 250.0 cm³ = 0.0250 mol [1]; Mr = 2.20 ÷ 0.0250 = 88.0 [1] (c) CH₃CH₂CH₂COOH (butanoic acid) [1]; (CH₃)₂CHCOOH (2-methylpropanoic acid) [1] Examiner insight: Forgetting to scale up from 25.0 cm³ to 250.0 cm³ gives Mr = 880, which should prompt a check.

5. (a) HCOOH + CH₃CH₂CH₂OH ⇌ HCOOCH₂CH₂CH₃ + H₂O: correct structures [1]; water as the second product and the equation balanced [1] (b) Sulfuric acid is a catalyst [1]; the reaction is reversible and reaches equilibrium, so some acid and alcohol remain unreacted [1] Examiner insight: “Loss on transfer” does not answer (b), since the question rules out losses; the point needed is equilibrium.

6. (a) (CH₃CH₂CH₂CO)₂O + CH₃OH → CH₃CH₂CH₂COOCH₃ + CH₃CH₂CH₂COOH: ester correct [1]; butanoic acid as the second product [1] (b) Not reversible, so a higher yield [1]; no acid catalyst needed (or faster) [1] Examiner insight: Water in place of butanoic acid in (a) is a common error; balance the atoms and you will see the second acyl group must leave as the acid.

7. (a) C₂H₅COOC₅H₁₁ + H₂O ⇌ C₂H₅COOH + C₅H₁₁OH: correct products [1]; reversible arrow and balanced [1] (b) C₂H₅COOC₅H₁₁ + NaOH → C₂H₅COONa + C₅H₁₁OH [1]; sodium propanoate and pentan-1-ol [1] (c) In alkali the acid is converted to its carboxylate ion, which the alcohol cannot turn back into ester, so (b) is one-way; (a) reaches equilibrium [1] Examiner insight: In (b) the organic product is the salt, not propanoic acid; naming the acid there does not match the conditions.

8. (a) A: water [1]; B: methanol [1]; C: ammonia [1]; D: methylamine, CH₃NH₂ [1] (b) CH₃CH₂CH₂COCl + 2NH₃ → CH₃CH₂CH₂CONH₂ + NH₄Cl [1] (c) Secondary: the nitrogen carries one alkyl (methyl) group as well as the acyl group [1] Examiner insight: In (b), one NH₃ with HCl as a product is not chemically sensible alongside excess ammonia; show two NH₃ and NH₄Cl.

9. (a) CH₃COCl + CH₃C₆H₄OH → CH₃COOC₆H₄CH₃ + HCl: ester correct [1]; HCl as the other product [1] (b) Phenols are not readily esterified by carboxylic acids, but react readily with the more reactive acyl chloride [1] (c) Misty (steamy) fumes of hydrogen chloride [1] Examiner insight: With an acyl chloride the second product is HCl, never water; water here is a slip carried over from acid-catalysed esterification.

10. (a) RCOOR’ + NaOH → RCOONa + R’OH [1] (b) n(NaOH) added = 0.0500 × 1.00 = 0.0500 mol [1]; n(NaOH) left = n(HCl) = 0.0300 × 1.00 = 0.0300 mol [1]; n(NaOH) used = n(X) = 0.0200 mol [1]; Mr = 2.04 ÷ 0.0200 = 102 [1] (c) Sodium methanoate means the acid part is HCOO– [1]; the alcohol gives a ketone, so it is secondary; HCOO– is 45, leaving 57 for C₄H₉, so the alcohol is butan-2-ol [1]; X is HCOOCH(CH₃)CH₂CH₃, butan-2-yl methanoate [1] Examiner insight: In (b) the moles of HCl give the NaOH left over, not the NaOH that reacted; subtract before dividing into the mass.

11. (a) C₅H₁₁COOH + SOCl₂ → C₅H₁₁COCl + SO₂ + HCl [1] (b) C₅H₁₁COCl + C₆H₅OH → C₅H₁₁COOC₆H₅ + HCl [1] (c) Mr(hexanoic acid) = 116.0, so n = 4.64 ÷ 116.0 = 0.0400 mol [1]; n(hexanoyl chloride) = 0.0400 × 0.800 = 0.0320 mol [1]; n(phenyl hexanoate) = 0.0320 × 0.750 = 0.0240 mol [1]; Mr = 192.0, mass = 0.0240 × 192.0 = 4.61 g [1] (d) 0.800 × 0.750 × 100 = 60.0% [1] (e) Phenol is not readily esterified by carboxylic acids, so very little ester would form [1] Examiner insight: Overall yield is the product of the step yields; 77.5% or 155% in (d) shows the yields were averaged or added.

Where marks are usually lost

• Writing water as a by-product when an anhydride or acyl chloride forms an ester (questions 6 and 9).
• Giving propanoic acid rather than sodium propanoate as the product of alkaline hydrolysis.
• Using one NH₃ or one amine per acyl chloride instead of two.
• Describing solubility as “polar” without naming hydrogen bonds, the δ+ hydrogen and the oxygen lone pair.
• Treating the HCl in a back titration as the NaOH used, rather than the NaOH left over.
• Averaging step yields instead of multiplying them.

Next steps

• Recap the tables and steps: Carboxylic acids and esters revision notes
• Full explanations and worked examples: Carboxylic acids and esters study guide
• Check the whole course: all free 10-minute diagnostics
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.1.3 Carboxylic acids and esters, learning outcomes (a) to (f). Published by OCR.