Edexcel A-Level Chemistry Paper 3, June 2025: Question 2

14 marks · Medium difficulty · Open Response

Answer questions on the reactions, mechanism, bonding, isomerism, and water solubility of prop-2-en-1-ol and related unsaturated compounds.

Practise this question

Question

Question text

2 This question is about compounds that contain both an alkene and an alcohol

functional group.

(a) Prop-2-en-1-ol, present in garlic, has the structure shown.

H CH2OH

C C

H H

(i) Describe a test, and the positive result, that confirms the presence of a

C C group.

(2)

Test

Result

(ii) Prop-2-en-1-ol reacts with hydrogen bromide to form 3-bromopropan-1-ol as

one of the products.

Draw a mechanism for this reaction.

Include curly arrows, and relevant lone pairs and dipoles.

(3)

H CH2OH

C C

H H

(b) Prop-2-en-1-ol can polymerise to form an addition polymer.

Write a balanced equation for this polymerisation.

(2)

C C bonds, as shown.

H H

4 C C C

*P77834A0436*HH

propadiene

Explain why the CH2 groups at each end of the molecule are at right angles to

each other.

Use your knowledge of how orbitals overlap to form π-bonds.

(2)

(d) The compound bombykol is a pheromone secreted by female silk moths.

Like prop-2-en-1-ol, it can also be classified as both an alkene and an alcohol.

H H H

C C C OH

H H

bombykol prop-2-en-1-ol

Bombykol can exist as a number of geometric isomers.

(i) Give the number of different geometric isomers of bombykol.

(1)

… *P77834A0536*

(ii) Give the reason why prop-2-en-1-ol does not exhibit geometric isomerism.

(1)

(iii) Explain the expected solubilities in water of bombykol and prop-2-en-1-ol.

(3)

(Total for Question 2 = 14 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

2(a)(i) An answer that makes reference to the following points: Only award M2 if linked to the correct test (2)

Test

• add bromine water / bromine

(1) Allow Br2 or Br2(aq)

Result Do not award if Br− or bromide used

• which decolourises / (turns) colourless

(1) Allow if near miss on reagent e.g. Br

Allow initial colours of orange, yellow, brown or

or brown-red

Ignore references to heating

Do not award incorrect initial colour e.g. red to

Test colourless

• add acidified KMnO4 / Do not award clear

acidified potassium manganate(VII) / (1)

acidified potassium permanganate

Allow initial colours of purple or pink

If formula is given it must be correct

Result Do not award incorrect initial colour e.g. red to

• which decolourises / (turns) colourless colourless

(1) Do not award incorrect initial colour e.g. red to

colourless

Do not award clear

Allow if near miss on reagent e.g. omission of

acidified

or name / formula recognisable but incorrect

Ignore references to heating

Question

Answer Additional Guidance Mark

Number

2(a)(ii) An answer that makes reference to the following points: Do not award partial charge on C=C (3)

• dipole on HBr

• curly arrow from C=C bond to H (δ+)

• curly arrow from H−Br bond to Br or just beyond Allow structural, skeletal or hybrid formulae

• correct intermediate Penalise half arrows once only

Penalise additional arrows once only

• lone pair and negative charge on Br– Penalise use of Br2 or other incorrect reactant once

only

• curly arrow from lone pair (if shown) on Br– to C+

All 6 points scores 3

4 or 5 points scores 2

2 or 3 points scores 1

Question

Answer Additional Guidance Mark

Number

2(b) An answer that makes reference to the (2)

following points:

• correct structure of monomer and n on Allow placement of ‘n’ anywhere vertically if it is in front of

both sides of equation (1) the reactant and after the product

• correct structure of repeating unit with

extension bonds (1)

Allow structural or skeletal formulae throughout

Do not award molecular formulae

Examples

n CH2CHCH2OH [ CH2CHCH2OH ] n

Vertical connectivity must be to C of CH2OH; penalise once

only

Ignore conditions / catalysts on the arrow even if incorrect

Ignore brackets

Question

Answer Additional Guidance Mark

Number

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

• pi bonds form by sideways overlap of (p) orbitals

(1) Do not award ‘head-on’ overlap

Ignore references to ‘s’ orbitals

May be shown on a (labelled) diagram

• so two (p) orbitals are at right angles / perpendicular

(to each other) / π bonds are at right angles / (1) Allow ‘one orbital on central C is Py, the other is

perpendicular (to each other) Pz’ (any pair of x, y and z orbitals will suffice)

May be shown on a (labelled) diagram

Ignore repulsion between electrons / bonds etc

Question

Answer Additional Guidance Mark

Number

2(d)(i) An answer that makes reference to the following point: (1)

• four / 4

Question

Answer Additional Guidance Mark

Number

2(d)(ii) An answer that makes reference to the following point: Allow doesn’t have different groups at the (1)

end of the (one of the) C=C bonds

• has two hydrogen (atoms) on (first) carbon (atom) of Do not award molecules

double bond / has same group on (first) carbon (atom) of Allow ‘hydrogen groups’

double bond

Question

Answer Additional Guidance Mark

Number

2(d)(iii) An explanation that makes reference to the following points: (3)

• prop-2-en-1-ol is more soluble than bombykol (1) Allow prop-2-en-1-ol is soluble and

bombykol is insoluble

• both (have OH / alcohol group that) can form hydrogen

bonds (with water) (1)

• bombykol has a (very) long (hydro)carbon chain

or (1) Allow as bombykol has a (very) long

forms stronger London forces hydrophobic / non-polar hydrocarbon chain

(Total for Question 2 = 14 marks)

How to answer it

Alkenes, Polymerisation, Orbital Overlap & Stereoisomerism

📌 What this question tests

This question assesses core organic and bonding chemistry from across the specification:

Qualitative Analysis: Testing for C=C unsaturation using bromine water or acidified potassium manganate(VII).
Reaction Mechanisms: Electrophilic addition of hydrogen halides (HBr) to unsymmetrical alkenes, including intermediate carbocation stability and regioisomers.
Addition Polymerisation: Writing full balanced equations showing the monomer, repeating unit, and extension bonds.
Advanced Bonding: Explaining molecular geometry in cumulated dienes via the sideways overlap of perpendicular p-orbitals to form π-bonds.
Stereoisomerism & Solubility: Identifying conditions required for E/Z isomerism, calculating stereoisomer totals (2ⁿ), and weighing hydrogen bonding against long non-polar hydrophobic alkyl chains.

Part (a)(i) — Chemical Test for the C=C Bond

Identifying unsaturation in prop-2-en-1-ol (2 Marks)

✅ Correct Answer

Test: Add bromine water / bromine / Br₂(aq) [or acidified KMnO₄ / acidified potassium manganate(VII)].

Result: Decolourises / turns colourless (from orange/brown/yellow to colourless; or from purple/pink to colourless for KMnO₄).

❌ Common Errors & Pitfalls

Writing that the solution turns "clear" instead of "colourless" (loses the observation mark!).
Stating "bromide" or "Br⁻" instead of "bromine" or "Br₂".
Omitting "acidified" when quoting potassium manganate(VII).

Mark Scheme Breakdown:
• M1: Reagent: Bromine water / bromine / Br₂ / Br₂(aq) [1 mark]
• M2: Result: Decolourises / turns colourless (dependent on correct reagent) [1 mark]

Part (a)(ii) — Electrophilic Addition Mechanism

Formation of 3-bromopropan-1-ol from prop-2-en-1-ol (3 Marks)

✅ Mechanism Steps to Draw

Dipole: Show permanent dipole on HBr as H^δ+—Br^δ- .
First Curly Arrow: From the electron-rich C=C double bond pointing directly to H^δ+ .
Second Curly Arrow: From the H—Br covalent bond onto the Br atom.
Carbocation Intermediate: Because the question specifies 3-bromopropan-1-ol, the H must add to C2:
⁺CH₂—CH₂—CH₂OH (positive charge explicitly on the terminal carbon C3).
Bromide Ion: Draw :Br⁻ with its lone pair and negative charge.
Final Curly Arrow: From the lone pair of :Br⁻ to the positively charged carbon ( C⁺ ).

🧠 Exam Technique: Regiochemistry Alert!

Normally, electrophilic addition follows Markovnikov's rule to produce the major product via the more stable secondary carbocation (giving 2-bromopropan-1-ol).

However, the question specifically asks for the mechanism forming 3-bromopropan-1-ol! You must draw the primary carbocation intermediate ( ⁺CH₂—CH₂—CH₂OH ) to access full marks.

Mark Scheme Points (6 checkpoints):
(1) Dipole on HBr • (2) Arrow from C=C to H • (3) Arrow from H—Br bond to Br • (4) Correct primary carbocation intermediate • (5) Lone pair and negative charge on Br⁻ • (6) Arrow from lone pair on Br⁻ to C⁺.
Scoring: All 6 points = 3 marks; 4 or 5 points = 2 marks; 2 or 3 points = 1 mark.

Part (b) — Addition Polymerisation

Balanced equation for the formation of poly(prop-2-en-1-ol) (2 Marks)

✅ Balanced Equation

n CH₂=CH(CH₂OH) ⟶ ⸤ CH₂—CH(CH₂OH) ⸥_n

Or in displayed form: double bond opens up into a single C—C bond, brackets enclose the unit, extension bonds extend through the brackets, and n is placed before the reactant and after the bracket as a subscript.

❌ Common Errors

Forgetting the trailing extension bonds extending out through the brackets.
Leaving the double bond intact inside the bracket.
Incorrect vertical connectivity: ensure the C—C bond connects directly to the carbon of the —CH₂OH group, not to H or O.
Omitting ' n ' on either side of the equation.

Mark Scheme Breakdown:
• M1: Correct monomer structure and ' n ' on both sides of the equation [1 mark]
• M2: Correct repeating unit with extension bonds cutting through brackets [1 mark]

Part (c) — Orbital Overlap in Propadiene (Allene)

Explaining why the terminal CH₂ groups are perpendicular (2 Marks)

💡 Key Knowledge: Cumulated π-Bonds

A π (pi) bond is formed by the sideways overlap of adjacent p-orbitals.
The central carbon in propadiene ( CH₂=C=CH₂ ) is sp hybridised and has two unhybridised p-orbitals (e.g. p_y and p_z).
These two p-orbitals on the central carbon are mutually perpendicular (at right angles / 90°) to each other.
Therefore, the two π-bonds must form in perpendicular planes, forcing the two terminal CH₂ groups to lie at right angles to each other.

✅ Model Answer

1. π (pi) bonds are formed by the sideways overlap of p-orbitals.

2. The two p-orbitals on the central carbon atom are at right angles / perpendicular to each other, so the two π-bonds form at right angles to each other.

Mark Scheme Breakdown:
• M1: π bonds form by sideways overlap of (p) orbitals [1 mark]
• M2: The two (p) orbitals on central C / the two π bonds are at right angles / perpendicular to each other [1 mark]
Note: Do NOT award "head-on" overlap (that defines a σ bond).

Part (d) — Bombykol vs Prop-2-en-1-ol

Isomerism and Physical Properties (5 Marks Total)

(i) Number of Geometric Isomers of Bombykol (1 Mark)

✅ Correct Answer

4 (or four).

📐 Calculation / Logic

Bombykol contains 2 stereogenic C=C double bonds. Each double bond has two different groups attached to each carbon atom, allowing E/Z isomerism:

Total geometric isomers = 2ⁿ = 2² = 4

(Specifically: E,E; E,Z; Z,E; and Z,Z isomers).

• M1: 4 / four [1 mark]

(ii) Why Prop-2-en-1-ol Does NOT Exhibit Geometric Isomerism (1 Mark)

✅ Correct Answer

One of the double-bonded carbons (carbon-3 / the terminal C=C carbon) has two identical atoms (two hydrogens) attached to it.

🧠 Exam Technique

Always refer explicitly to the individual carbon atom. Stating "the molecule doesn't have different groups" is too vague and scores 0. You must specify: "two identical groups / two hydrogen atoms on the same carbon atom of the C=C bond."

• M1: Has two hydrogen atoms on the (first/same) carbon atom of the double bond [1 mark]

(iii) Explaining Water Solubilities (3 Marks)

✅ Full-Mark Explanation

Relative solubility: Prop-2-en-1-ol is soluble (or more soluble than bombykol), whereas bombykol is insoluble / poorly soluble.
Hydrogen bonding: Both compounds possess an —OH (hydroxyl) group capable of forming hydrogen bonds with water molecules.
Hydrophobic effect: Bombykol has a very long non-polar / hydrophobic hydrocarbon chain which disrupts the hydrogen-bonded network of water (and forms strong London dispersion forces between bombykol molecules), preventing it from dissolving.

❌ Common Errors

Failing to state clearly which molecule is more soluble (missing M1).
Forgetting to mention that both molecules can form hydrogen bonds with water due to their —OH groups (missing M2).
Not identifying the long carbon chain in bombykol as non-polar / hydrophobic (missing M3).

Mark Scheme Breakdown:
• M1: Prop-2-en-1-ol is more soluble than bombykol (or prop-2-en-1-ol is soluble and bombykol is insoluble) [1 mark]
• M2: Both have an OH / alcohol group that can form hydrogen bonds (with water) [1 mark]
• M3: Bombykol has a very long (hydrophobic/non-polar) hydrocarbon chain OR forms stronger London forces [1 mark]

Topics

Organic Chemistry · Physical Chemistry · Topic 6: Organic Chemistry I · Topic 2: Bonding and Structure

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.