Edexcel A-Level Chemistry Paper 2, June 2024: Question 7

17 marks · Hard difficulty · Synoptic Questions

A multi-part question covering polymers, IUPAC naming, ideal gas equation calculations, electrophilic addition mechanisms, intermolecular forces of alcohols, Avogadro constant calculations, and ester hydrolysis.

Practise this question

Question

Question text

7 This question is about polymers, an example of which is neoprene, a synthetic

rubber material.

It is formed by the polymerisation of chloroprene.

chloroprene

(a) What is the IUPAC name of chloroprene?

(1)

A 3‑chlorobuta‑1,3‑diene

B 2‑chlorobuta‑2,4‑diene

C 3‑chlorobuta‑2,4‑diene

D 2‑chlorobuta‑1,3‑diene

(b) Calculate the volume, in cm3, occupied by 10.0 g of chloroprene in the

gaseous phase, at 80.0°C and 205 kPa.

Give your answer to an appropriate number of significant figures.

[Gas constant (R) = 8.31 J mol–1 K–1]

(5)

removing hydrogen chloride from the product of this reaction.

Step 1 Cl Step 2 Cl

+ Cl2 + HCl

Compound A

(i) Explain how chlorine, Cl2, can act as an electrophile in Step 1 even though a

chlorine molecule is symmetrical.

You may find it helpful to include a diagram.

(2)

… 20

… *P76896A02032*

(ii) Draw the mechanism for Step 1.

Include curly arrows, and any relevant lone pairs and dipoles.

(3)

(iii) Chloroprene polymerises to form neoprene as shown.

H Cl H H Cl H

n C C C C C C C C

H H H H H H n

Give the name of the type of reaction that occurs when

chloroprene polymerises.

(1)

(d) Flexible and water‑resistant materials can be made by combining neoprene with

polyester fabric.

A polyester may be made by the reaction of benzene‑1,4‑dicarboxylic acid with*P76896A02132*

ethane‑1,2‑diol.

(i) Explain why ethane‑1,2‑diol is soluble in water.

A detailed description of the forces involved is not required.

(2)

(ii) The molar mass of a polyester is 8400 g mol–1.

Calculate the number of polymer molecules in 4.25mg of this polyester.

[Avogadro constant (L) = 6.02 × 1023 mol–1]

(2)

*P76896A02232*

(iii) A student spilled a small amount of 0.40 mol dm–3 sodium hydroxide solution

onto a polyester laboratory coat. A hole formed in the laboratory coat as the

result of a chemical reaction.

What type of reaction occurred?

(1)

A dehydration

B hydrolysis

C neutralisation

D redox

(Total for Question 7 = 17 marks)

Mark scheme

Show the mark scheme

Question

Answer Mark

Number

7(a) The only correct answer is D (2-chlorobuta-1,3-diene) (1)

A is not correct as the position of chlorine is incorrectly numbered

B is not correct as the position of double bonds are incorrectly numbered

C is not correct as the position of double bonds and chlorine are incorrectly numbered

Question

Answer Additional Guidance Mark

Number

7(b) Example of calculation (5)

• calculation of Mr of chloroprene (1) 88.5

• calculation of amount of chloroprene in mol (1) 10.0 ÷ 88.5 / = 0.11299 (mol)

(1) 353 (K) and 205 × 103 / 205000 (Pa)

• conversion of °C to K and kPa to Pa

M3 may be subsumed in M4

Allow pressure in kPa if volume is clearly shown as dm3 in

• re-arrangement of pV = nRT and calculation of V (1) V = (nRT) ÷ p

= (0.11299 × 8.31 × 353) ÷ 205000 = 1.6169 × 10–3 (m3)

M4 may be subsumed in M5

Ignore units in M4

3 (1) 1600 / 1620 (cm3)

• conversion to cm and rounding to 2 or 3 SF

1620 with no working scores 5

Comments

Allow use of correctly rounded values carried forward

Take care with awarding 5 for 1600, as it may be the result

of an incorrect value from M4, correctly rounded

Question

Answer Additional Guidance Mark

Number

7(c)(i) An explanation that makes reference to the following points: (2)

• chlorine (molecule) moves close to (electron rich)

double bond / pi bond (1)

(1)

M1 via diagram e.g

Allow C=C bond has a high electron density

• which induces partial positive charge on chlorine Accept repulsion between the electrons of chlorine

(atom) (nearest to double bond / pi bond) and the pi electrons results in partial positive charge

on chlorine (atom)

Allow polarises the chlorine forming a δ+ Cl

Note M2 must have idea that the partial charge is

induced / brought about / produced so a diagram

alone does not score M2

Question

Answer Additional Guidance Mark

Number

7(c)(ii) An answer that makes reference to the following (3)

points:

• curly arrow from one double bond to chlorine (1) Do not award if arrow shown to partial negative / negative Cl

or just beyond

• correct dipole on chlorine molecule (1)

• curly arrow from Cl-Cl bond to Cl (1)

• carbocation intermediate (1) Do not award primary carbocation

(1) Do not award Cl / Clδ– , but allow TE into BP6

• lone pair on chloride ion

• curly arrow from chloride ion to carbocation (1) If lone pair evident, curly arrow must originate from it

and

correct final product

6 bullet points scores 3 marks, 4/5 bullet points

scores 2 marks, 2/3 bullet points scores 1 mark, 0/1

bullet points scores 0 marks

Note – correct dipole discussed in (c)(i) can be awarded as part

of M1 if not shown on the diagram

Question

Answer Additional Guidance Mark

Number

7(c)(iii) An answer that makes reference to the following point: (1)

• addition (polymerisation) Do not award condensation

Question

Answer Additional Guidance Mark

Number

7(d)(i) An explanation that makes reference to the following points: (2)

• both compounds have hydrogen bonds (1) Allow can hydrogen bond to each other

Allow H-bond for hydrogen bond

Allow correct diagram for M1

• so the intermolecular forces / hydrogen bonds they (1) Allow ‘they break strong H bonds but then form

form together will be similar in strength (to the strong H bonds together’

intermolecular forces / hydrogen bonds they have Allow the new hydrogen bonds are stronger than the

before mixing) ones between water molecules / ethane-1-2-diol

Question

Answer Additional Guidance Mark

Number

7(d)(ii) Example of calculation (2)

• calculation of mass and moles of polyester (1) 4.25 × 10–3 ÷ 8400 = 5.0595 × 10–7 (mol)

(1) 5.0595 × 10–7 × 6.02 × 1023 = 3.0458 × 1017 (molecules)

• calculation of number of molecules of polyester

Ignore SF

Allow TE from M1

If incorrect units given in M1 and M2, penalise only once

Correct answer with or without working scores 2 marks

Question

Answer Mark

Number

7(d)(iii) The only correct answer is B (hydrolysis) (1)

A is not correct as the reaction as dehydration is the removal of water

C is not correct as the reaction as neutralisation is a reaction between an acid and a base

D is not correct as the reaction as redox is oxidation and reduction

(Total for Question 7 = 17 marks)

How to answer it

Polymers, Electrophilic Addition & Mole Calculations

What this question tests

This exam question tests your knowledge of organic nomenclature (IUPAC naming), ideal gas law calculations (PV = nRT), electrophilic addition mechanisms involving non-polar molecules (induced dipoles), intermolecular forces (hydrogen bonding in diols), Avogadro constant calculations, and the functional group chemistry of polyesters (hydrolysis).

Part (a) - IUPAC Naming

Naming Chloroprene

✅ Correct Answer

D: 2-chlorobuta-1,3-diene

❌ Common Errors

Students often misnumber the carbon chain. Remember to give functional groups the lowest possible numbers. Starting from the right end gives carbon 1 at the terminal alkene closest to the chlorine substituent.

Marks available: 1 mark

Part (b) - Ideal Gas Equation Calculation

Calculating Gas Volume

📐 Step-by-Step Calculation

Molar mass of chloroprene (C₄H₅Cl): (4 × 12.0) + (5 × 1.0) + 35.5 = 88.5 g mol⁻¹
Moles of chloroprene: n = mass / M = 10.0 / 88.5 = 0.11299 mol
Convert units: Temperature T = 80.0 + 273.15 = 353.15 K (or 353 K). Pressure P = 205 kPa = 205 × 10³ Pa (or 205000 Pa).
Rearrange ideal gas equation (V = nRT / P): V = (0.11299 × 8.31 × 353) / 205000 = 1.6169 × 10⁻³ m³
Convert m³ to cm³ and round: Multiply by 1,000,000 to get 1616.9 cm³ . Rounding to 2 or 3 significant figures gives 1600 cm³ or 1620 cm³ .

❌ Common Calculation Traps

Forgetting to convert kPa into Pa (multiplying by 10³).
Forgetting to convert Celsius into Kelvin (+273).
Forgetting the final unit conversion from m³ to cm³ (missing multiplication by 10⁶).

Marks available: 5 marks

Part (c)(i) - Electrophile Theory

Polarisation of a Symmetrical Molecule

💡 Key Knowledge

Even though Cl₂ is non-polar, when it approaches the electron-rich C=C double bond (which has a high electron density of pi electrons), the high negative charge repels the bonding electrons in Cl-Cl. This induces a temporary dipole, creating a partial positive charge (δ⁺) on the chlorine atom closest to the double bond.

🧠 Exam Technique

Annotating a quick diagram showing the C=C double bond close to a Cl-Cl molecule with a δ⁺ on the nearest chlorine atom is an effective way to secure both marks cleanly.

Marks available: 2 marks

Part (c)(ii) - Reaction Mechanism

Mechanism for Step 1

✅ Correct Answer / Features Required

Curly arrow starting from the C=C double bond pointing to the δ⁺ chlorine atom.
Correct dipole label (δ⁺-δ⁻) on the Cl₂ molecule.
Curly arrow showing the Cl-Cl bond breaking, with the electron pair moving to the distal Cl atom to form a chloride ion (Cl⁻).
Formation of a correct secondary carbocation intermediate (do not draw a primary carbocation).
A lone pair of electrons on the chloride ion, with a curly arrow originating from this lone pair pointing directly to the positively charged carbon of the carbocation.
Final neutral addition product drawn accurately.

❌ Common Errors

A frequent error is drawing the first curly arrow pointing to the Cl-Cl bond instead of specifically to the δ⁺ chlorine atom. Ensure curly arrows start precisely from bonds or lone pairs.

Marks available: 3 marks

Part (c)(iii) - Polymerisation Type

Type of Polymerisation

✅ Correct Answer

Addition (polymerisation)

❌ Common Errors

Do not write condensation. Neoprene is formed via the addition of unsaturated alkene monomers opening their double bonds without the elimination of small molecules like water or HCl.

Marks available: 1 mark

Part (d)(i) - Intermolecular Forces

Solubility of Ethane-1,2-diol

💡 Key Knowledge

Ethane-1,2-diol contains two hydroxyl (-OH) groups. These groups can form hydrogen bonds with water molecules. The energy released when new hydrogen bonds form between the diol and water molecules compensates for breaking the existing hydrogen bonds in pure water and pure diol, rendering it fully soluble.

Marks available: 2 marks

Part (d)(ii) - Avogadro Calculation

Number of Polymer Molecules

📐 Step-by-Step Calculation

Convert mass to grams: 4.25 mg = 4.25 × 10⁻³ g
Calculate moles of polyester: Moles = mass / molar mass = (4.25 × 10⁻³ g) / 8400 g mol⁻¹ = 5.0595 × 10⁻⁷ mol
Calculate number of molecules: Multiply moles by the Avogadro constant (L):
Number = (5.0595 × 10⁻⁷) × (6.02 × 10²³) = 3.05 × 10¹⁷ molecules

❌ Common Calculation Traps

Watch out for the mass unit prefix! The question gives milligrams (mg), which must be converted to grams (g) using ×10⁻³ before dividing by the molar mass.

Marks available: 2 marks

Part (d)(iii) - Functional Group Reactions

Identifying Chemical Degradation

✅ Correct Answer

B: hydrolysis

💡 Key Knowledge

Polyesters contain ester linkages (-COO-). Aqueous sodium hydroxide supplies OH⁻ ions which attack the electrophilic carbonyl carbon of the ester group, breaking down the polymer chain. This base-hydrolysis (saponification) breaks the ester bonds, causing the laboratory coat to form a hole.

Marks available: 1 mark

Topics

Physical Chemistry · Organic Chemistry · Core Practicals · Topic 5: Formulae, Equations and Amounts of Substance · Topic 6: Organic Chemistry I · Topic 17: Organic Chemistry II · Topic 2: Bonding and Structure

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.