Edexcel International A Level Chemistry: Carboxylic acids and their derivatives (YCH11) – 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.

This set covers parts 15C (Carboxylic acids) and 15D (Carboxylic acid derivatives) of Topic 15 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 15.9 to 15.16. The content is International A2 (IA2); Pearson assesses it in Unit 4: Rates, Equilibria and Further Organic Chemistry. When stuck, return to the carboxylic acids study guide or the shorter carboxylic acids revision notes. Log progress on the YCH11 checklist and the YCH11 course hub. A free diagnostic finds wider gaps.

Use these values throughout. Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5, Na = 23.0. Molar gas volume at RTP: 24.0 dm³ mol⁻¹.

Questions

1. Name each compound.

(a) (CH₃)₃CCOOH [1] (b) CH₃CH₂COCl [1] (c) CH₃COOCH₂CH₂CH₃ [1]

2. Use these values. Butanoic acid boils at 164 °C, pentan-1-ol at 138 °C and methyl propanoate at 80 °C. All three have Mr = 88.0.

(a) Explain, by comparing the forces between molecules, why the three compounds boil in this order. [3] (b) Butanoic acid is miscible with water, but decanoic acid, C₉H₁₉COOH, is almost insoluble. Explain this difference. [2]

3. 2-methylpropanoic acid, (CH₃)₂CHCOOH, can be prepared in two ways.

(a) It can be made from 2-methylpropan-1-ol. Give the reagents and conditions, write an equation using [O], and state the colour change seen. [4] (b) It can also be made by hydrolysing a nitrile. Name the nitrile and give the reagent and condition for the hydrolysis. [2]

4. 2.22 g of propanoic acid is added to excess sodium carbonate solution.

(a) Give a balanced equation for what happens. [1] (b) Calculate the volume, in cm³, of carbon dioxide produced at room temperature and pressure. [3]

5. This question is about 2-methylbutanoic acid, CH₃CH₂CH(CH₃)COOH.

(a) Name the organic product formed with lithium tetrahydridoaluminate(III), and state why water must be kept out of the reaction mixture. [2] (b) Write an equation for the reaction in (a), using [H] for the reducing agent. [1] (c) Write an equation for the reaction with phosphorus(V) chloride and state what you would see. [2]

6. A student warms 3.08 g of 3-methylbutan-1-ol, (CH₃)₂CHCH₂CH₂OH, with excess ethanoic acid and a few drops of concentrated sulfuric acid. 3.64 g of ester is collected.

(a) Name the ester and give its structural formula. [2] (b) Calculate the percentage yield. [3] (c) Suggest why the yield is below 100% even when no product is lost. [1]

7. For each reaction of ethanoyl chloride, write an equation and name the organic product.

(a) with water [2] (b) with 2-methylpropan-1-ol, (CH₃)₂CHCH₂OH [2] (c) with concentrated ammonia [2] (d) with methylamine, CH₃NH₂ [2]

8. Ethanoyl chloride is much more reactive towards nucleophiles than ethanoic acid.

(a) Explain this difference in reactivity. [2] (b) State one advantage of making an ester from an acyl chloride rather than from a carboxylic acid. [1]

9. Methyl 2-methylpropanoate, (CH₃)₂CHCOOCH₃, is hydrolysed.

(a) Name the products when it is heated under reflux with dilute sulfuric acid. [2] (b) Write an equation for its hydrolysis by sodium hydroxide solution. [1] (c) Hydrolysis with sodium hydroxide goes to completion. Explain why. [1] (d) Calculate the volume, in cm³, of 1.50 mol dm⁻³ sodium hydroxide that reacts with 3.06 g of the ester. [2]

10. Terylene is a polyester used for clothing fibres.

(a) Name the type of polymerisation that forms Terylene and the small molecule lost when the monomers are a diol and a dicarboxylic acid. [2] (b) Name the two monomers. [2] (c) The repeat unit is –OC–C₆H₄–COO–CH₂CH₂–O–. Calculate its Mr, and the number of repeat units in a chain of Mr 57 600. [2] (d) Suggest why Terylene is broken down by hot sodium hydroxide solution, but poly(ethene) is not. [1]

11. A chemist has a pineapple-scented ester, X, with molecular formula C₆H₁₂O₂. She refluxes X with excess sodium hydroxide solution and distils off an alcohol, B. Acidifying the residue releases a monoprotic carboxylic acid, A. In a titration, 0.352 g of A is neutralised by 16.00 cm³ of 0.250 mol dm⁻³ sodium hydroxide.

(a) Calculate the Mr of A and identify A. [3] (b) Deduce the identity of B and name X. [2] (c) X can be made from A in two steps through an acyl chloride. Give the reagent for each step and write an equation for the second step. [3] (d) Give a simple chemical test that gives a positive result with A but not with X. [1]

Answers

1. (a) 2,2-dimethylpropanoic acid [1] (b) Propanoyl chloride [1] (c) Propyl ethanoate [1] Examiner insight: in (c) the alkyl word comes from the alcohol and is written first; “ethyl propanoate” is a different compound, so check which side of the ester oxygen each chain is on.

2. (a) Methyl propanoate has no O–H, so it has only London forces and permanent dipole–dipole forces between molecules [1]. Pentan-1-ol forms hydrogen bonds between molecules through its O–H group [1]. Butanoic acid forms hydrogen-bonded dimers using O–H and the C=O oxygen, so the most energy is needed to separate its molecules [1]. (b) The –COOH group forms hydrogen bonds with water molecules [1]. In decanoic acid the long hydrocarbon chain cannot hydrogen bond with water, so it does not mix [1]. Examiner insight: with equal Mr values, an answer that only says “bigger London forces” misses the point; name the force each substance has and say which needs most energy to overcome.

3. (a) Excess potassium dichromate(VI) acidified with dilute sulfuric acid [1]; heat under reflux [1]. (CH₃)₂CHCH₂OH + 2[O] → (CH₃)₂CHCOOH + H₂O [1]. Orange to green [1]. (b) 2-methylpropanenitrile, (CH₃)₂CHCN [1]; dilute hydrochloric acid, heated under reflux [1]. Examiner insight: “distil” in place of “reflux” describes how to collect the aldehyde, so it is unlikely to earn credit for the condition when the acid is the target.

4. (a) 2CH₃CH₂COOH + Na₂CO₃ → 2CH₃CH₂COONa + H₂O + CO₂ [1] (b) Mr(propanoic acid) = 74.0; n = 2.22 ÷ 74.0 = 0.0300 mol [1]. n(CO₂) = 0.0300 ÷ 2 = 0.0150 mol [1]. Volume = 0.0150 × 24 000 = 360 cm³ [1]. Examiner insight: the 2 : 1 ratio from the balanced equation is easy to miss; check your equation before you use it, and give the unit asked for.

5. (a) 2-methylbutan-1-ol [1]. LiAlH₄ reacts violently with water, so water must be excluded [1]. (b) CH₃CH₂CH(CH₃)COOH + 4[H] → CH₃CH₂CH(CH₃)CH₂OH + H₂O [1] (c) CH₃CH₂CH(CH₃)COOH + PCl₅ → CH₃CH₂CH(CH₃)COCl + POCl₃ + HCl [1]. Steamy (misty) fumes [1]. Examiner insight: “white smoke” describes solid NH₄Cl forming with ammonia, a different observation; say steamy or misty fumes for HCl.

6. (a) 3-methylbutyl ethanoate [1]; CH₃COOCH₂CH₂CH(CH₃)₂ [1] (b) Mr(C₅H₁₂O) = 88.0, n = 3.08 ÷ 88.0 = 0.0350 mol [1]. Mr(C₇H₁₄O₂) = 130.0, maximum mass = 0.0350 × 130.0 = 4.55 g [1]. Yield = 3.64 ÷ 4.55 × 100 = 80.0% [1]. (c) The reaction is reversible, so an equilibrium mixture forms and not all the alcohol is converted [1]. Examiner insight: base the yield on the reagent not in excess (the alcohol), and show the maximum mass as a separate step.

7. (a) CH₃COCl + H₂O → CH₃COOH + HCl [1]; ethanoic acid [1] (b) CH₃COCl + (CH₃)₂CHCH₂OH → CH₃COOCH₂CH(CH₃)₂ + HCl [1]; 2-methylpropyl ethanoate [1] (c) CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl [1]; ethanamide [1] (d) CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl⁻ [1]; N-methylethanamide [1] Examiner insight: in (c) and (d), use two molecules of ammonia or amine; an equation that releases HCl does not show the ammonium or alkylammonium salt that actually forms.

8. (a) In ethanoyl chloride, both the oxygen and the chlorine withdraw electrons from the carbonyl carbon, so it carries a larger δ+ charge and attracts nucleophiles more strongly [1]. Chloride is a good leaving group, so the C–Cl bond breaks readily [1]. (b) Any one: the reaction is not reversible, so the yield is higher; it is fast at room temperature; no acid catalyst is needed [1]. Examiner insight: a vague “acyl chlorides are more reactive” only repeats the question; give the reasons, the size of the δ+ charge on the carbon and how easily Cl⁻ leaves.

9. (a) 2-methylpropanoic acid [1] and methanol [1] (b) (CH₃)₂CHCOOCH₃ + NaOH → (CH₃)₂CHCOONa + CH₃OH [1] (c) Excess hydroxide converts the acid into its carboxylate ion, and that ion cannot react with methanol to re-form the ester [1]. (d) Mr(C₅H₁₀O₂) = 102.0; n = 3.06 ÷ 102.0 = 0.0300 mol, ratio 1 : 1 [1]. Volume = 0.0300 ÷ 1.50 = 0.0200 dm³ = 20.0 cm³ [1]. Examiner insight: giving 2-methylpropanoic acid as a product of the alkaline reaction in (b) is wrong; with excess hydroxide the product is the sodium salt.

10. (a) Condensation polymerisation [1]; water [1] (b) Ethane-1,2-diol [1] and benzene-1,4-dicarboxylic acid [1] (c) C₁₀H₈O₄: Mr = 120.0 + 8.0 + 64.0 = 192.0 [1]. Number of repeat units = 57 600 ÷ 192.0 = 300 [1]. (d) The ester links in Terylene are hydrolysed by hydroxide ions, but the poly(ethene) chain contains only C–C and C–H bonds, which hydroxide ions do not attack [1]. Examiner insight: when a question says “name”, give names rather than formulae, and include the positions 1,2 and 1,4 in full.

11. (a) n(NaOH) = 16.00 ÷ 1000 × 0.250 = 0.00400 mol = n(A) [1]. Mr = 0.352 ÷ 0.00400 = 88.0 [1]. CₙH₂ₙ₊₁COOH: 14n + 46 = 88, n = 3, so A is butanoic acid, CH₃CH₂CH₂COOH [1]. (b) X has six carbons and A has four, so B has two: B is ethanol [1]. X is ethyl butanoate, CH₃CH₂CH₂COOCH₂CH₃ [1]. (c) Step 1: phosphorus(V) chloride, PCl₅ [1]. Step 2: ethanol at room temperature [1]. CH₃CH₂CH₂COCl + CH₃CH₂OH → CH₃CH₂CH₂COOCH₂CH₃ + HCl [1]. (d) Add sodium hydrogencarbonate (or carbonate) solution: A gives effervescence of CO₂, X gives none [1]. Examiner insight: in (b), link the carbon count to the evidence rather than just naming the alcohol.

Where marks are usually lost

• Swapping the two halves of an ester name, as in “ethyl propanoate” for CH₃COOCH₂CH₂CH₃.
• Explaining boiling temperatures with London forces alone when the Mr values are the same.
• Writing “distil” when the product needed is the carboxylic acid.
• Writing 2[H] instead of 4[H] for reducing a carboxylic acid with LiAlH₄.
• Describing HCl as “white smoke” rather than steamy fumes.
• Showing HCl as a product with ammonia or an amine instead of the ammonium or alkylammonium salt.
• Giving the free acid when sodium hydroxide is the reagent; with excess alkali you get the sodium salt.

Next steps

• Carboxylic acids revision notes
• Carboxylic acids study guide
• YCH11 course hub
• YCH11 printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 4, Topic 15, parts 15C Carboxylic acids and 15D Carboxylic acid derivatives, outcomes 15.9 to 15.16.