AQA A-Level Chemistry Paper 2, 2024: Question 8

Mark scheme

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Question Answers Additional Comments/Guidelines Mark

M1 Electrophilic Addition

08.1

M2 HBr (2 x AO1)

M1 Butan-2-ol or correct structure eg CH3CH(OH)CH2CH3 2

08.2 (1 x AO1,

M2 NaOH AND (warm) aqueous (Allow cold if stated) 1 x AO2)

– A-LEVEL CHEMISTRY – – JUNE 2024

M1 HCN or KCN/H2SO4

M2 Arrow from lone pair on C to C of C=O

M3 Arrow from C=O to O

08.3 M4 Structure of intermediate including negative on O (1 x AO1,

4 x AO2)29

M5 Arrow from lone pair on O to H+

OR

M2 M4

M3 M5

– A-LEVEL CHEMISTRY – –

M1 Allow planar carbonyl group

08.4

(3 x AO2)

M2 Equal chance (50/50) attack from above/below OWTTE Either side

M3 Giving equal amounts of both optical isomers/enantiomers

How to answer it

Four-step synthesis: from alkene to hydroxy nitrile

What this question tests

Recognising mechanisms across an organic synthesis sequence: electrophilic addition, nucleophilic substitution, oxidation and nucleophilic addition. The question also probes understanding of optical isomerism and why certain steps yield racemic mixtures.

Step 1 • Alkene → Haloalkane

Electrophilic Addition

✅ Mechanism name

Electrophilic addition

Reagents and conditions

Hydrogen bromide (HBr) gas or conc. solution
Room temperature, no catalyst needed

Product: CH₃CH₂CHBrCH₃ (2-bromobutane)

💡 Key point

The double bond acts as a π electron source for the electrophile (H⁺ from HBr).

2 marks

Step 2 • Haloalkane → Nitrile

Nucleophilic Substitution (SN2)

✅ Compound J

CH₃CH₂CH(CN)CH₃ (2-methylbutanenitrile / butanenitrile isomer)

Reagents and conditions

Reagent: KCN (or NaCN) in ethanol
Condition: heat under reflux

💡 Mechanism

CN⁻ attacks the δ⁺ carbon bonded to Br, displacing Br⁻ via S_N2.

2 marks

Step 3 • Nitrile → Ketone

Hydrolysis

✅ Product

CH₃CH₂COCH₃ (butan-2-one)

Reagents and conditions

Reagent: acidic hydrolysis (H⁺/H₂O)
Heat under reflux to form the ketone

1 mark (identified as intermediate in question)

Step 4 • Ketone → Hydroxy nitrile

Nucleophilic Addition to a Carbonyl

✅ Reagent(s)

HCN (hydrogen cyanide)
or NaCN + dilute H₂SO₄ (to generate HCN in situ)

💡 Mechanism

Step 1: CN⁻ attacks the δ⁺ carbon in the C=O bond.
Step 2: The π bond breaks, forming a negatively charged O⁻ intermediate.
Step 3: Protonation by H⁺ → forms the hydroxynitrile CH₃CH₂C(OH)(CN)CH₃.

🧠 Examiner tip

Curly arrows should show:
• From CN⁻ to the carbon of the C=O bond
• From the C=O double bond to the oxygen
• Then from O⁻ to a proton (H⁺)

5 marks

Step 4 outcome • Racemic mixture

Why a racemic mixture forms

✅ Explanation

The carbonyl carbon (C=O) in butan-2-one is planar.
The CN⁻ nucleophile can attack from either side of this plane with equal probability.
This forms two enantiomers — mirror images of each other — in equal amounts.

→ The product is a racemic mixture (optically inactive overall).

3 marks

❌ Common errors

Stating that attack happens from “above and below” but omitting the idea of equal probability.
Forgetting to mention planar carbonyl or enantiomers.

Final takeaways

💡 Key knowledge

Electrophilic addition adds HX to alkenes.
Nucleophilic substitution replaces halogens with CN⁻ (chain length +1).
Nucleophilic addition to a carbonyl forms a hydroxynitrile, often chiral.

🧠 Exam technique

Always label curly arrows and intermediates.
Use correct functional group names in each step.
Highlight chiral centre formation and symmetry when discussing racemates.

Topics

Organic Chemistry · 3.3.4 Alkenes · 3.3.3 Halogenoalkanes · 3.3.14 Organic Synthesis