AQA A-Level Chemistry Paper 3, 2017: Question 21

1 mark · Medium difficulty · Multiple Choice

Identify which curly arrow does not appear in the mechanism for the reaction of propanone with hydroxide ions to form an enolate ion.

Practise this question

Question

The question presents a chemical equilibrium reaction: HO⁻ + H–CH₂–C(=O)–CH₃ ⇌ H₂O + H–CH=C(O⁻)–CH₃. It asks: 'Which curly arrow does not appear in the mechanism of this reaction?' Four options are given: A shows an arrow from the C–H bond to the C–C bond; B shows an arrow from the C–C bond to the C=O bond; C shows an arrow from the lone pair on HO⁻ to the H atom; D shows an arrow from the C=O double bond to the O atom.
Question text

21 In concentrated alkali, propanone reacts with hydroxide ions to form an equilibrium

mixture as shown.

Which curly arrow does not appear in the mechanism of this reaction?

[1 mark]

A

B

C

D

Mark scheme

Show the mark scheme Mark scheme table displaying question number 21 with the correct answer as B.

21 B

How to answer it

Enolate Formation: Curly Arrow Mechanisms

📋 What this question tests

This question assesses your ability to deduce a plausible, electron-accurate mechanism for an unfamiliar reaction involving carbonyl compounds and base-catalysed equilibrium:

  • Understanding what curly arrows represent: the movement of a pair of electrons.
  • Tracking bond-breaking and bond-forming steps from reactants to products.
  • Recognising acid-base proton transfer (deprotonation of an α-hydrogen) by a hydroxide ion (HO⁻).
  • Delocalisation and formation of an enolate ion containing a C=C double bond and an O⁻ ion.
Question 21 • 1 Mark

Identifying the Incorrect Curly Arrow

Analysis of the step-by-step mechanism

✅ Correct Answer: B

Option B does not appear in the mechanism.

The curly arrow in B starts from the C–C single bond and points towards the C=O bond. This would mean breaking the carbon–carbon backbone, which does not happen in this reaction.

💡 The Real Mechanism (Step-by-Step)

Comparing the reactant (propanone) and products (water + enolate ion):

  • Arrow C: A lone pair on the hydroxide ion (:OH⁻) attacks the acidic α-hydrogen to form H₂O (proton removal).
  • Arrow A: The pair of electrons in the breaking C–H bond moves into the C–C bond to form the new C=C double bond.
  • Arrow D: A pair of electrons from the C=O π-bond shifts onto the electronegative oxygen atom, forming the O⁻ group.

🧠 Exam Technique: "Spot the Bond Changes"

  • Compare Reactant vs Product:
    • C–H bond is broken.
    • O–H bond (in H₂O) is formed.
    • C–C single bond becomes a C=C double bond.
    • C=O double bond becomes a C–O⁻ single bond.
  • Trace Electron Flow: In base-promoted reactions, curly arrows flow sequentially: from the base, through the breaking C–H bond into the C–C system, and onto the electronegative oxygen.
  • Rule out impossible movements: An arrow starting on a C–C single bond (Option B) represents cleaving the alkyl chain, which would contradict the molecular formula of the enolate product.

❌ Common Errors & Traps

  • Missing the word "NOT": A common multiple-choice pitfall is identifying the first arrow that does happen (e.g. Option C) and marking it down without reading carefully.
  • Confusing Arrow Tails and Heads: Remember, the tail shows the origin of the electron pair (a lone pair or a covalent bond), and the head shows where the electrons are moving.
  • Assuming Nucleophilic Addition: Students familiar with aldehydes/ketones often expect HO⁻ to attack the carbonyl carbon (nucleophilic addition). However, the equation clearly shows deprotonation at the α-carbon, forming an enolate.
Mark Scheme: Award 1 mark for selecting option B. No intermediate working or explanations needed for Section A/MCQ papers.

Topics

Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.8 Aldehydes and Ketones

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.