AS Chemistry: Carboxylic Acids and Esters — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Carboxylic Acids and Esters revision notes and the full study guide, which covers all three preparation routes and the LiAlH₄ reduction in more depth.

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Section A

1. Explain why carboxylic acids are described as weak acids. [2]

2. Name the products, and state one observation, when ethanoic acid reacts with (a) sodium carbonate, (b) magnesium. [5]

Section B

3. Ethanoic acid reacts with ethanol.

(a) Name the product and the catalyst required. [2] (b) Explain why the yield is limited and how it can be increased. [3] (c) Explain how the ester can be separated from the mixture. [2]

4. Compare the acid and alkaline hydrolysis of ethyl ethanoate, giving products and stating which goes to completion. [5]

A Level extension

Acyl chlorides are not part of the AS 9701 syllabus (Topic 18.2 covers only esterification and hydrolysis) — they are introduced at A Level in Topic 33. The question below is included as extension material only; see the Carboxylic Acids and Acyl Chlorides study guide for the full A Level treatment.

5. Ethanoyl chloride reacts with ethanol.

(a) Give the products and state one observation. [3] (b) Explain why acyl chlorides are far more reactive than carboxylic acids towards nucleophiles. [3]

6. Explain why ethanoic acid has a much higher boiling point than propanone, despite a similar relative molecular mass. [3]

7. State three different methods by which a carboxylic acid can be prepared, giving the reagents or conditions for each. [3]

8. Ethanoic acid is reduced to ethanol using LiAlH₄.

(a) Write an equation for the reaction, using [H] to represent the reducing agent. [1] (b) Explain why NaBH₄ cannot be used for this reduction instead. [2]

9. Write a balanced equation for the reaction between ethanoic acid and sodium hydroxide, and name the type of reaction taking place. [2]

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Answers

1. They only partially dissociate in aqueous solution [1], so an equilibrium is set up and only a small proportion of the molecules release H⁺ ions [1], unlike a strong acid such as HCl, which dissociates fully.

2. (a) Sodium ethanoate, carbon dioxide and water [1] [1]; effervescence is seen [1] — award 2 for products. (b) Magnesium ethanoate and hydrogen [1] [1].

3. (a) Ethyl ethanoate [1]; a few drops of concentrated sulfuric acid as catalyst [1]. (b) The reaction is reversible and reaches equilibrium, so it does not go to completion [1]. Use an excess of one reactant [1], or remove the ester by distilling it off as it forms, which shifts the equilibrium to the right by Le Chatelier’s principle [1]. (c) Distil off the ester, which has the lowest boiling point of the mixture [1], then shake with sodium carbonate solution to remove acid impurities, separate, and dry before redistilling to obtain the pure product [1].

4. Acid hydrolysis — heat under reflux with dilute sulfuric acid, giving ethanoic acid and ethanol [1] [1]; it is reversible and does not go to completion, reaching an equilibrium mixture instead [1]. Alkaline hydrolysis — heat under reflux with sodium hydroxide, giving sodium ethanoate and ethanol [1]; it goes to completion, because the carboxylate ion formed is negatively charged and not attacked by the alcohol, so nothing pulls the reaction backwards [1].

5. (a) Ethyl ethanoate [1] and hydrogen chloride [1]; steamy/misty white fumes are seen, and the reaction is vigorous even at room temperature, unlike the carboxylic acid equivalent [1]. (b) The chlorine is highly electronegative and there are two electronegative atoms on the carbonyl carbon [1], making the carbon more strongly δ+ and more open to nucleophilic attack [1]; in addition Cl⁻ is a very good leaving group, whereas in a carboxylic acid the OH group is a poor one and is not readily displaced [1].

6. Ethanoic acid molecules form hydrogen bonds with one another through the O–H group [1], and in the pure liquid they form dimers held by two hydrogen bonds [1]. Propanone can only form weaker permanent dipole–dipole attractions, as it has no O–H, so much less energy is needed to separate its molecules [1].

7. Any three of: oxidation of a primary alcohol or aldehyde with acidified K₂Cr₂O₇ or KMnO₄ under reflux [1]; hydrolysis of a nitrile with dilute acid or dilute alkali, followed by acidification [1]; hydrolysis of an ester with dilute acid or dilute alkali and heat, followed by acidification [1].

8. (a) CH₃COOH + 4[H] → CH₃CH₂OH + H₂O [1]. (b) A carboxylic acid is harder to reduce than an aldehyde or ketone, so a stronger reducing agent is needed [1]; LiAlH₄ is strong enough, but NaBH₄ is not [1].

9. CH₃COOH + NaOH → CH₃COONa + H₂O [1]; this is a neutralisation reaction [1].

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Where marks are usually lost

• Saying carboxylic acids are weak because they are dilute.
• Forgetting that esterification is reversible.
• Naming the ester the wrong way round — the alcohol supplies the alkyl part.
• Giving only electronegativity, not the leaving group, for acyl chloride reactivity.
• Trying to reduce a carboxylic acid with NaBH₄ — it isn’t strong enough; only LiAlH₄ can do it.
• Naming acid hydrolysis and alkaline hydrolysis as if they give the same products — acid hydrolysis gives the free carboxylic acid, alkaline hydrolysis gives the carboxylate salt.
• Forgetting that reaction with a reactive metal is a redox reaction (producing hydrogen), while reaction with an alkali is neutralisation — the same acid, two different reaction types depending on what it reacts with.