AQA A-Level Chemistry Paper 2, June 2025: Question 1

12 marks · Medium difficulty · Long Answer

Identify mechanisms, reagents, conditions, and write equations for the synthesis of 1,2-dichlorobutane via free-radical substitution and elimination-addition pathways.

Practise this question

Question

A multi-part organic chemistry exam question about different ways to make 1,2-dichlorobutane. A reaction scheme shows 2-chlorobutane converting to 1,2-dichlorobutane via Reaction 1 (direct), or via Reaction 2 to form but-1-ene, which then converts to 1,2-dichlorobutane via Reaction 3. Sub-questions 01.1 to 01.6 ask for mechanism names, conditions, equations for the free-radical substitution mechanism, reasons for low yield, reagents and conditions for elimination, and identifying alternative products.
Question text

01 This question is about different ways to make 1,2-dichlorobutane.

01.1 Give the name of the mechanism for reaction 1.

State one essential condition needed for reaction 1.

[2 marks]

Name of mechanism

Essential condition

01.2 Give equations to show three essential steps in the mechanism for reaction 1.

[3 marks]

01.3 Suggest two reasons why using reaction 1 is likely to give a low yield of

1,2-dichlorobutane.

[2 marks]

Reason 1

Reason 2

01.4 Give a suitable reagent and conditions needed for reaction 2.

[2 marks]

Reagent

Conditions

01.5 Identify the reagent and name the mechanism for reaction 3.

[2 marks]

Reagent

Name of mechanism

01.6 Use of the reagent for reaction 2 followed by the reagent for reaction 3 gives another

organic product.

Give the structure of the other organic product.

[1 mark]

Mark scheme

Show the mark scheme The mark scheme for question 01. 01.1 lists free radical substitution and UV light. 01.2 shows the initiation step forming chlorine radicals, and propagation steps forming the chlorinated radical and then 1,2-dichlorobutane. 01.3 lists reasons for low yield including polysubstitution and substitution at different positions. 01.4 lists KOH or NaOH in hot, ethanolic conditions. 01.5 lists chlorine and electrophilic addition. 01.6 shows the structure of 2,3-dichlorobutane as the other product.

Question Answers Additional comments/Guidelines Mark

M1 (Free) radical substitution Allow high temperature 2

01.1

M2 UV (light / radiation) Ignore heat/warm/pressure (2 x AO1)

– A-LEVEL CHEMISTRY – 7405/2 –

Apply list principle to M3

Ignore alternative termination steps

01.2

(3 x AO2)

M2 and M3 can be scored with non-skeletal structures

M2 CH CH CHClCH + Cl• → [CH CH CHClCH ]• + HCl

32 3 3 2 2

M3 [CH CH CHClCH ]• + Cl → CH CH CHClCH Cl + Cl•

32 2 2 3 2 2

OR

M3 [CH CH CHClCH ]• + Cl• → CH CH CHClCH Cl

32 2 3 2 2

– A-LEVEL CHEMISTRY – 7405/2 –

12 Any two from: Allow

Reaction 1 could give rise to further substitution (polysubstitution) chain reactions

01.3

Reaction 1 could give rise to substitution in different positions other isomers are possible (2 x AO1)

Combination of radicals to get longer chain product

M1 KOH or NaOH (Formulae or names allow other strong alkalis)

01.4 M2 Hot / reflux / ethanolic Ignore warm (1 x AO1,

1 x AO2)

Can score both marks on same line provided no

contradiction elsewhere

– A-LEVEL CHEMISTRY – 7405/2 –

M1 Chlorine / Cl2 Allow chlorine water

01.5

M2 Electrophilic addition Apply list principle (2 x AO3)

Any correct structure showing 2,3-dichlorobutane Allow CH3CH(OH)CHClCH3 or

CH3CH2CHClCH2OH or CH3CH2CH(OH)CH2Cl 13

e.g. CH3CHClCHClCH3 or

01.6

(1 x

AO3)

How to answer it

Synthesis Routes & Mechanisms of Halogenoalkanes

What this question tests

This question assesses your understanding of organic synthesis pathways, specifically focusing on:

  • Free-radical substitution mechanisms (initiation, propagation, termination) and conditions.
  • The limitations of radical substitution in organic synthesis (low yields due to isomeric mixtures and polysubstitution).
  • Elimination reactions of halogenoalkanes to form alkenes, including reagents and specific conditions.
  • Electrophilic addition of halogens to alkenes.
  • Identifying structural isomers formed via elimination pathways (positional isomerism).
Part 01.1

Mechanism and Conditions for Reaction 1

Converting 2-chlorobutane to 1,2-dichlorobutane

Correct Answers

  • Name of mechanism: Free-radical substitution (or Radical substitution)
  • Essential condition: UV light (or ultraviolet radiation / high temperature)
[2 marks] 1 mark for mechanism, 1 mark for condition.

Exam Technique

Always write the full term "Free-radical substitution" rather than just "substitution" to secure the mark. For the condition, "UV light" is the standard expected answer. Do not write "light" or "sunlight" on its own as they are too vague.

Part 01.2

Mechanism Steps for Reaction 1

Equations for the three essential steps

Correct Equations

Step 1: Initiation

Cl₂ → 2Cl•

Step 2: Propagation 1 (Radical attacks the hydrocarbon chain at C1)

CH₃CH₂CHClCH₃ + Cl• → CH₃CH₂CHClCH₂• + HCl

Step 3: Propagation 2 (Carbon radical reacts with a chlorine molecule)

CH₃CH₂CHClCH₂• + Cl₂ → CH₃CH₂CHClCH₂Cl + Cl•

Alternative Step 3: Termination (Combination of radicals)

CH₃CH₂CHClCH₂• + Cl• → CH₃CH₂CHClCH₂Cl
[3 marks] 1 mark for each correct step. Skeletal structures are fully acceptable.

Key Knowledge

To form 1,2-dichlorobutane, the chlorine radical must abstract a hydrogen atom from Carbon-1 (the methyl group next to the CHCl group). This forms a primary radical intermediate on C1, which then reacts with Cl₂ to place the second chlorine atom on C1.

Common Errors

  • Missing radical dots (•): Ensure the dot is clearly visible on the chlorine atoms ( Cl• ) and on the specific carbon atom carrying the unpaired electron ( CH₂• ).
  • Writing HCl•: Hydrogen chloride is a stable molecule, NOT a radical. Do not put a dot on HCl!
Part 01.3

Why Reaction 1 Gives a Low Yield

Limitations of Free-Radical Substitution

Correct Answers (Any two of the following)

  • Further substitution can occur (polysubstitution / formation of tri- or tetra-chlorobutanes).
  • Substitution can occur at different positions on the carbon chain, yielding positional isomers (e.g., 2,2-dichlorobutane or 2,3-dichlorobutane).
  • Termination steps can combine hydrocarbon radicals, yielding longer-chain hydrocarbon impurities.
[2 marks] 1 mark for each distinct reason.

Examiner Commentary

Top-performing students use precise terminology like "polysubstitution" and "positional isomers". Simply saying "other reactions happen" is too vague to score.

Part 01.4

Reagent and Conditions for Reaction 2

Elimination of 2-chlorobutane to form but-1-ene

Correct Answers

  • Reagent: Potassium hydroxide ( KOH ) or Sodium hydroxide ( NaOH )
  • Conditions: Hot, reflux, and ethanolic (dissolved in ethanol)
[2 marks] 1 mark for reagent, 1 mark for conditions.

The "Aqueous vs Ethanolic" Trap

Crucial Distinction:

  • KOH(aq) (aqueous) promotes Nucleophilic Substitution, converting the halogenoalkane into an alcohol.
  • KOH(alc) (ethanolic) acts as a base to promote Elimination, forming an alkene.

If you write "aqueous" or fail to specify "ethanolic/in ethanol", you will lose the conditions mark!

Part 01.5

Reagent and Mechanism for Reaction 3

Converting but-1-ene to 1,2-dichlorobutane

Correct Answers

  • Reagent: Chlorine ( Cl₂ ) or chlorine water
  • Name of mechanism: Electrophilic addition
[2 marks] 1 mark for reagent, 1 mark for mechanism.

Key Knowledge

Alkenes are electron-rich due to the high electron density of the C=C double bond. They undergo addition reactions where the double bond opens up. The attacking species is an electrophile (induced dipole on Cl-Cl).

Part 01.6

Identifying the Alternative Organic Product

Isomerism in Elimination Reactions

Correct Structure

The other organic product is 2,3-dichlorobutane:

CH₃CHClCHClCH₃

Skeletal Structure Representation:

Draw a 4-carbon butane chain. Place one chlorine atom on Carbon-2 and another chlorine atom on Carbon-3.

[1 mark] For any correct structural, displayed, or skeletal formula of 2,3-dichlorobutane.

How did this product form?

In Reaction 2, elimination of 2-chlorobutane can remove a hydrogen from C1 (forming but-1-ene) OR from C3 (forming but-2-ene). When but-2-ene reacts with Cl₂ in Reaction 3, electrophilic addition occurs across the C2=C3 double bond, yielding 2,3-dichlorobutane.

Examiner Tip

Always look at the symmetry of your starting material. Halogenoalkanes like 2-chlorobutane are unsymmetrical, meaning elimination can yield positional isomers of the alkene. Always trace both pathways to find alternative products!

Topics

Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.2 Alkanes · 3.3.3 Halogenoalkanes · 3.3.4 Alkenes

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.