OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 24

9 marks · Medium difficulty · Structured Questions

Draw repeat units, assess disposal, write a combustion equation, outline the electrophilic addition mechanism with HCl, and explain carbocation stability for alkenes.

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Question

Question text

24 This question is about alkenes.

(a) 2-Chloropropene can form an addition polymer.

(i) Draw two repeat units of the polymer formed by 2-chloropropene.

[1]

(ii) 2-Chloropropene can be disposed of by combustion. This process releases the greenhouse gas

carbon dioxide.

State one other problem with this method of disposal.

… [1]

(b) Pentene, C5H10, can undergo complete combustion to produce the same products as the

combustion of an alkane.

Write the equation for the complete combustion of C5H10.

… [1]

Compound A is the major product.

(i) Outline the mechanism for the formation of the major product, compound A.

The structure of (CH3)2C=CH2 has been provided.

Include curly arrows and relevant dipoles, the structure of compound A and the name of the

mechanism.

H3C H

C C

H3C H

Compound A

Name of mechanism …

[5]

(ii) Explain why compound A is the major product.

… [1]

Mark scheme

Show the mark scheme

Question Answer Mark Guidance

24 (a) (i) 1 ALLOW structural OR displayed OR skeletal formula

OR mixture of the above (as long as unambiguous)

ALLOW:

Two correct repeat units for 2-chloropropene

(n and brackets not required) ✓ … but CH3 and Cl must be on same C atom

CH3 and Cl can be shown either side of the structure

NOTE: ‘side bonds’ ARE required on either side of

repeat unit from C atoms

IGNORE connectivity of methyl group

IGNORE brackets

IGNORE n or subscript numbers

24 (a) (ii) Formation of HCl/hydrochloric acid/ OR chlorine 1 ALLOW Cl or Cl2 for chlorine

ALLOW produces CO (from incomplete combustion)

IGNORE toxic waste products

Response must reflect chlorine or carbon monoxide in

some way

24 (b) C5H10 + 7½ O2 → 5CO2 + 5H2O ✓ 1 ALLOW 2C5H10 + 15O2 → 10CO2 + 10H2O

IGNORE state symbols (even if incorrect)

24 (c) (i) Curly arrows can be straight, snake-like, etc. 19 5 1st curly arrow must

but NOT double headed or half headed arrows • go to the H atom of H–Cl

AND

1. Curly arrow from C=C to HCl and H–Cl 2 marks • start from, OR be traced back to

H3C H any point across width of C=C

C C DO NOT ALLOW

partial charge on C=C

H3C H

H δ+

Cl δ–

Curly arrow from C=C bond to H of H–Cl ✓ 2nd curly arrow must

• start from, OR be traced back to

any part of +H–Cl – bond

Correct dipole shown on H–Cl

AND curly arrow that breaks H–Cl bond ✓ AND

• go to Cl –

Use of Cl-Cl or H-Br loses first marking point but

gains ECF for second marking point

2. Curly arrow from Cl– to carbocation 1 mark ALLOW formation of minor carbocation, i.e.

DO NOT ALLOW

+ on C of

carbocation

Curly arrow from

Cl– to C+ of carbocation (carbocation must follow (formation of major product assessed below)

structure from step 1 i.e. 2 methyl groups) ✓

ALLOW ECF from use of Cl2 OR HBr

3rd curly arrow must

• go to the C+ of carbocation

AND

• start from, OR be traced back to any point

across width of lone pair on :Cl–

• OR start from – charge of Cl– ion

(Lone pair NOT needed if curly arrow shown from –

3. Product 1 mark charge of Cl– ion)

ALLOW ECF for correct tertiary product following use

of HBr ONLY (i.e. Cl in structure shown replaced

with Br)

✓

4. Name of mechanism 1 mark

Electrophilic addition ✓

24 (c) (ii) More stable (tertiary) carbocation is formed as 1 NOTE must refer to stability of carbocation

intermediate

OR IGNORE references to Markownikoff’s rule

The minor product forms a less stable (primary) IGNORE secondary

carbocation ✓

How to answer it

Alkenes: Polymers, Combustion & Electrophilic Addition

WHAT THIS QUESTION TESTS

This question assesses core organic chemistry fundamentals for AS-Level alkenes:

Drawing polymer repeat units accurately from substituted alkene monomers.
Environmental hazards associated with polymer disposal by incineration.
Balancing hydrocarbon complete combustion equations.
Drawing the step-by-step mechanism for electrophilic addition of hydrogen halides to asymmetric alkenes (curly arrows, charges, and dipoles).
Explaining regioselectivity and major product formation via carbocation stability.

PART (a)(i) — 1 MARK

Drawing Addition Polymers

Draw two repeat units of the polymer formed by 2-chloropropene

✅ Correct Answer

Draw a backbone of 4 carbon atoms linked by single bonds, with open "continuation" bonds on both ends:

Note: The methyl group (-CH₃) and chlorine (-Cl) must be bonded to the same carbon atom in each unit. Brackets and 'n' are not required.

❌ Common Errors

Drawing only one repeat unit: The question specifically asked for two repeat units (4 carbons in the backbone chain).
Missing open extension bonds: Leaving off the side bonds pointing out from the end carbons forfeits the mark.
Retaining double bonds: Addition polymerisation breaks the C=C double bond into a single bond backbone.

PART (a)(ii) — 1 MARK

Environmental Issues with Polymer Disposal

State one other problem with the disposal of 2-chloropropene by combustion

✅ Correct Answer

Any one of the following:

Formation of HCl (hydrogen chloride / hydrochloric acid).
Formation of toxic chlorine gas (Cl₂).
Produces CO / carbon monoxide (toxic gas formed via incomplete combustion).

🧠 Exam Technique: Be Chemically Specific

Examiners routinely reject generic phrases such as "causes pollution" or "produces toxic waste". You must identify a specific hazardous molecule resulting directly from the chemical composition of the monomer (e.g., chlorine content leading to acidic/toxic HCl gas).

PART (b) — 1 MARK

Combustion of Alkenes

Write the equation for the complete combustion of pentene, C₅H₁₀

✅ Balanced Chemical Equation

C₅H₁₀ + 7.5O₂ → 5CO₂ + 5H₂O

OR with whole-number balancing:

2C₅H₁₀ + 15O₂ → 10CO₂ + 10H₂O

📐 Balancing Method

Balance Carbons: 5 carbons in C₅H₁₀ → 5CO₂
Balance Hydrogens: 10 hydrogens in C₅H₁₀ → 5H₂O (since 5 × 2 = 10)
Count Oxygens on Right: (5 × 2) + (5 × 1) = 15 O atoms
Balance Oxygen Molecule: 15 ÷ 2 = 7.5O₂ (or multiply entire equation by 2).

Mark Scheme Note: State symbols are not required and are ignored even if incorrect. Fractional coefficients like 7½ O₂ are fully credited.

PART (c)(i) — 5 MARKS

Mechanism: Electrophilic Addition

Outline the mechanism for the formation of major product A from methylpropene and HCl

💡 Step-by-Step Diagram Requirements

Step 1: Electrophilic Attack

Draw the partial charges on hydrogen chloride: H(δ+)—Cl(δ-) .
Arrow 1: Starts strictly from the electron-rich C=C double bond and points directly to the H(δ+) atom.
Arrow 2: Starts from the covalent H—Cl bond and points to the electronegative Cl(δ-) atom.

Step 2: Intermediate & Nucleophilic Attack

Draw the tertiary carbocation intermediate: (CH₃)₂C⁺—CH₃ (ensure the formal positive charge is positioned clearly on the central carbon atom).
Draw the chloride ion with a lone pair and full negative charge: :Cl⁻ .
Arrow 3: Starts from either the lone pair or the negative charge on :Cl⁻ and points directly to the positive carbon atom C⁺ .

✅ Identity of Compound A & Mechanism Name

Structure of Compound A:

CH₃
|
H₃C--C--CH₃ (2-chloro-2-methylpropane)
|
Cl

Name of Mechanism:

Electrophilic addition

❌ Strict Examiner Rules for Mechanism Marks

Arrow starting points matter: The first curly arrow must start directly on or inside the C=C bond. If it starts from a carbon atom or space, 0 marks.
No dipoles on alkene: Do NOT put δ+ or δ- on the C=C double bond carbons.
Intermediate Charge: Never write δ+ on the carbocation intermediate; it carries a full + formal charge.
Arrow destination: The arrow from :Cl⁻ must terminate clearly at the positive carbon atom, not vaguely near the molecule.

Total 5 Marks Breakdown: [1] Arrow from C=C to H · [1] Correct dipoles on H—Cl AND arrow breaking H—Cl bond to Cl · [1] Arrow from :Cl⁻ to C⁺ of correct tertiary carbocation · [1] Correct structure of Compound A · [1] Mechanism name: Electrophilic addition.

PART (c)(ii) — 1 MARK

Explaining Carbocation Stability

Explain why compound A is the major product

✅ Model Answer

It is formed via the more stable tertiary carbocation intermediate.

(Alternatively: "The minor product forms via a less stable primary carbocation intermediate.")

❌ Common Error: Quoting Markownikoff's Rule

Stating "Because Markownikoff's rule says hydrogen adds to the carbon with more hydrogens" receives 0 marks. Markownikoff's rule is purely descriptive; OCR requires the chemical reason: the comparative stability of carbocations due to electron-donating alkyl groups.

🧠 The Golden Rule for Major Product Questions

Always structure your answer using the template:

[Type of product] is formed via the more stable [tertiary / secondary] carbocation intermediate.

You must explicitly use the word carbocation (or carbonium ion). Simply saying "it forms a more stable intermediate" or "a tertiary molecule" is insufficient.

Topics

Module 4: Core organic chemistry · Module 2: Foundations in chemistry · 4.1 Basic concepts and hydrocarbons · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.