AQA A-Level Chemistry Paper 2, 2023: Question 4

15 marks · Medium difficulty · Long Answer

Synthesis of Butane-1,4-Diamine and Nylon 4,6 This question involves naming isomer W, drawing isomers from Reaction 1, identifying reagents and conditions for Reactions 2 and 3, deducing values for x and y in nylon 4,6 formation, and drawing sections of nylon 4,6 polymer to show hydrogen bonding.

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AQA A-Level Chemistry Paper 2, 2023: Question 4
Question text

04 Acrylonitrile, H2C=CHCN, can be used as a starting material for the synthesis of

butane-1,4-diamine, as shown in this reaction scheme.

04.1 Use IUPAC rules to name isomer W.

[1 mark]

04.2 Reaction 1 produces a mixture of W and two other isomers.

Draw the structures of the two other isomers.

Explain, by considering the mechanism of this reaction, why all three isomers are

formed. 10

[6 marks]

The reaction scheme is repeated here.

04.3 Identify the reagent that is warmed with isomer W in reaction 2.

State the other reaction condition needed.

[2 marks]

Reagent

Condition

04.4 State the reagent and reaction conditions needed for reaction 3.

Give an equation for reaction 3.

[2 marks]

Reagent and conditions

Equation

04.5 An incomplete equation for the formation of nylon 4,6 from five molecules of

butane-1,4-diamine and five molecules of hexanedioic acid is shown.

Deduce the values of x and y in this equation.

[2 marks]

x y

04.6 Figure 6 shows a section of the nylon 4,6 polymer molecule.

Figure 6

Draw, on Figure 6, another section of nylon 4,6 polymer showing two hydrogen bonds

between the two sections.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2023: Question 4

Question Answers Additional Comments/Guidelines Mark

3-bromopropanenitrile Allow 3-bromopropane-1-nitrile 1

04.1

(AO1)

This question is marked using Levels of Response. Refer to the Mark Indicative Chemistry content

Scheme Instructions for Examiners for guidance.

Stage 1 Types of Isomers formed

Level 3 All stages are covered and each stage is generally 1a CH CHBrCN

5–6 marks correct and virtually complete.

Answer is communicated coherently and shows a logical 1b Exists as two Optical isomers / enantiomers

progression from Stage 1 to Stages 2 and 3.

Stage 2 Mechanism

Level 2 All stages are covered but stage(s) may be incomplete or 2a 2 curly arrows

3–4 marks may contain inaccuracies

OR two stages are covered and are generally correct

and virtually complete.

Answer is communicated mainly coherently and shows a 6

04.2 logical progression from Stage 1 to Stages 2 and 3. (3 x AO1,

2b Intermediate structure primary carbocation OR 3 x AO3)

Level 1 Two stages are covered but stage(s) may be incomplete

1–2 marks or may contain inaccuracies OR only one stage is H H

covered but is generally correct and virtually complete.

H C C CN

Answer includes isolated statements but these are not

presented in a logical order. H

2c Alternative Intermediate structure secondary

0 mark Insufficient correct chemistry to gain a mark. carbocation OR

H H

H C C CN

16 H

M1 KCN or NaCN Penalise acid in M1

04.3 17

(2 x AO1)

M2 Aqueous AND ethanol (alcohol)

M1 H2 and Ni/Pt/Pd Allow LiAlH4 and (Dry) ether BUT not NaBH4 2

04.4 (ignore heat and pressure) (1 x AO1,

M2 NCCH2CH2CN + 4H2 → H2N(CH2)4NH2 Allow with 8[H] 1 x AO2)

M1 x = 5

04.5

(2 x AO1)

M2 y = 9

Structure shown on the left of the given structure.

The correct answer is the same irrespective of

C O whether it’s drawn on the left or right of the

polymer section.

(H2C)4

Deduct a mark(s) for error(s)/omission(s)

C O

Must have the following:

C O H N 19

• Minimum correct structure

C O

(H2C)4 (CH2)4

C O H N (H2C)4

H N C O 2

04.6

(2 x AO2)

(CH ) Or

24 H N

H N

(CH2)4

H N

• Lp on O or N

• 2 Linear dashed lines from O or N to H

Allow alternative connection below

C O C O

(H2C)4 (H2C)4

C O C O

H N H N

(CH2)4 (CH2)4

H N H N

How to answer it

Chemistry Exam Commentary: Understanding the Mark Scheme

Question (a): Naming Isomer W

To achieve full marks here, you must apply IUPAC naming rules correctly.

The mark scheme allows for two names:

  • 3-bromopropanenitrile (preferred)
  • 3-bromopropane-1-nitrile (alternative)
Tip: Always identify the longest carbon chain, prioritise functional groups like nitriles (-CN), and number carbons to give substituents (e.g., bromine) the lowest possible numbers.

Question (b): Drawing Isomers and Mechanism

Breaking Down the Marking Levels

The mark scheme uses levels of response. To score the highest marks (5-6), you need:

  • A clear and coherent answer covering all stages.
  • Logical progression between Stage 1 (isomers), Stage 2 (mechanism), and Stage 3 (isomerism).

Stage 1: Types of Isomers

Isomer W: 3-bromopropanenitrile
Isomer 2: 2-bromopropanenitrile
Isomer 3: 1-bromopropanenitrile
Tip: Understand how the HBr addition can produce different products depending on carbocation stability.

Stage 2: Mechanism

Focus on showing:

  • Curly arrows to demonstrate electrophilic addition.
  • Primary and secondary carbocation intermediates.

The mark scheme accepts either carbocation structure:

Primary: H-C(+)-C-CN
Secondary: H-C-C(+)-CN

Stage 3: Optical Isomerism

To achieve maximum marks, explain that:

  • The 2-bromo isomer is chiral because the central carbon has four different groups.
  • This results in two enantiomers due to non-superimposable mirror images.

Question (c): Reagent and Condition for Reaction 2

To score full marks, you must state:

  • Reagent: KCN or NaCN
  • Condition: Aqueous ethanol (heat under reflux)
Warning: Don’t mention “acid” here, as the mark scheme penalises this.

Question (d): Reduction of Nitriles

The mark scheme requires both the reagent and equation:

  • Reagent: H2 with Ni/Pt/Pd catalyst
  • Equation: NC-(CH2)2-CN + 4H2 → H2N-(CH2)4-NH2
Tip: Learn that reduction of nitriles produces primary amines. Avoid mentioning LiAlH4 unless specified.

Question (e): Deduce x and y for Polymer Formation

The mark scheme gives clear values:

  • x = 5
  • y = 9
Tip: Count the molecules of each reactant carefully in the polymerisation equation.

Question (f): Hydrogen Bonds in Nylon 4,6

The mark scheme expects:

  • Two linear dashed lines showing hydrogen bonds.
  • The bond must be between O (from C=O) and H (from N-H) .

Alternative orientations are acceptable as long as the connections are correct.

Tip: Double-check hydrogen bonding diagrams to ensure accuracy.

Topics

Organic Chemistry · 3.3.4 Alkenes · 3.3.14 Organic Synthesis · 3.3.3 Halogenoalkanes · 3.3.12 Polymers · 3.3.11 Amines

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.