Edexcel A-Level Chemistry Paper 2, June 2018: Question 8

10 marks · Medium difficulty · Open Response

Assess the reactions of phenol including bromination, nitration mechanisms, percentage yield calculation, and structural isomerism of 4-nitrophenol.

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Question

Question text

8 Phenol is a feedstock in the production of many organic molecules.

(a) Phenol reacts with bromine water.

(i) Complete the equation for the reaction of phenol with excess bromine water,

using the skeletal formula of the organic product.

(2)

+ … Br2 →

(ii) Compare and contrast the bromination of phenol with the bromination of benzene.

(3)

(b) Phenol can be nitrated to produce 4-nitrophenol.

18 NO2

(i) The mechanisms of the nitration of phenol and of benzene are similar. Complete*P52293A01824*

the diagram, using curly arrows, to show a possible mechanism for the reaction

between the electrophile, NO+, and phenol to produce 4-nitrophenol.

(3)

NO+

(ii) What is the mass, in grams, of 4-nitrophenol produced from 0.94g of phenol if

the yield of this isomer is 15%?

(1)

A 0.14

B 0.21

C 0.68

D 1.39

(iii) Draw two structural isomers of 4-nitrophenol which have a benzene ring.

(1)

(Total for Question 8 = 10 marks)

Mark scheme

Show the mark scheme

Question

Additional Guidance Mark

Number

8(a)(i) Ignore state symbols even if incorrect (2)

Structure of 2,4,6-tribromophenol (1) Do not award C6H3OBr3

Balanced equation (1) M2 dependent on M1

Question

Acceptable Answer Additional Guidance Mark

Number

8(a)(ii) An answer that makes reference to the following: Ignore comments of ease of reaction (3)

Similarity

Both electrophilic substitution (1) Should be stated clearly as a similarity

Any two from:

Contrast

No need of a halogen carrier with phenol (1) Accept reverse argument

Allow Fe/FeBr3/AlBr3 with benzene

Do not award just ‘catalyst’

oxygen’s lone pair of electrons interacts with the Allow reference to OH group

benzene ring of delocalised electrons so Allow ‘bromine’ for ‘electrophilic’

electrophilic attack more likely (1) Do not award for nucleophilic attack

Tri-substitution of phenol compared to mono for Allow “multiple-” for “tri-”

benzene (1)

Bromination of phenol requires bromine in

aqueous solution but benzene requires liquid

bromine (1)

Bromination of phenol requires room temperature

but benzene requires heating (under reflux) /

reflux (1)

Question

Acceptable Answer Additional Guidance Mark

Number

8(b)(i) An answer that makes reference to (3)

Electron pair movement from ring to Allow arrow that starts from anywhere within the

electrophile (1) hexagon but it must go to the nitrogen of the ion

Formula of intermediate ion (1) ‘Horseshoe’ to cover at least three carbon atoms,

facing the tetrahedral carbon and part of the + sign

to be inside the ‘horseshoe’

Do not award ‘+’ charge on the tetrahedral carbon

Do not award dotted bonds unless part of a 3D

structure

Curly arrow from C-H bond to reform Curly arrow to go from the bond to anywhere inside

delocalised ring and correct final the ring

structure with H+ also formed (1)

Accept the drawing of HSO − to remove the H from

the ring as long as H2SO4 is given as the product

instead of H+

Exemplar mechanism

Do not penalise attachment of OH/NO2 to benzene

ring

Penalise incorrect product: 1 mark

Question

Acceptable Answer Mark

Number

8(b)(ii) The only correct answer is B (1)

A is incorrect because this is 15% of the mass of the starting material

C is incorrect because this is the percentage of the starting mass over the max mass of product

D is incorrect because this is 100% yield and not 15%

Question

Additional Guidance Mark

Number

8(b)(iii) Ignore connectivity of OH/NO2 (1)

and

(Total for Question 8 = 10 marks)

How to answer it

Phenol Reactions and Electrophilic Substitution

📚 What this question tests

This question assesses your understanding of the activating effect of the hydroxyl (-OH) group on a benzene ring, comparison of reactivity between phenol and benzene, electrophilic substitution mechanisms using curly arrows, percentage yield calculations, and identification of structural isomers.

Part (a)(i): Reaction of Phenol with Bromine Water

Complete the equation for the reaction of phenol with excess bromine water

✅ Correct Answer

Balanced equation using skeletal formulas:

C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr

Note: Organic product must be drawn as the skeletal structure of 2,4,6-tribromophenol.

❌ Common Errors

Writing molecular formulas (e.g., C₆H₂Br₃OH ) instead of drawing the requested skeletal structure.
Failing to balance the stoichiometry (forgetting the 3 in front of Br₂ and HBr ). M2 is strictly dependent on M1.

🎯 Mark Breakdown: 2 marks total (1 mark for correct skeletal structure of 2,4,6-tribromophenol; 1 mark for the balanced equation).

Part (a)(ii): Comparing Phenol and Benzene Bromination

Compare and contrast the bromination of phenol with the bromination of benzene

💡 Key Knowledge

The lone pair of electrons on the oxygen atom of the -OH group in phenol is partially delocalized into the pi-system of the benzene ring. This increases electron density, polarizing approaching electrophiles more effectively.

🧠 Exam Technique

You must give one similarity and two distinct contrasts. Bullet points make your answer clear and ensure the examiner can easily award marks.

✅ Acceptable Marking Points (Select 1 Similarity + 2 Contrasts)

Similarity: Both reactions are electrophilic substitutions.
Contrast 1: Phenol does not require a halogen carrier catalyst (unlike benzene, which needs FeBr₃ or AlBr₃ ).
Contrast 2: Phenol undergoes tri-substitution (2,4,6-tribromophenol), whereas benzene undergoes mono-substitution.
Contrast 3: Phenol reacts with aqueous bromine water, whereas benzene requires liquid bromine.
Contrast 4: Phenol reacts at room temperature, whereas benzene requires heating (under reflux).

🎯 Mark Breakdown: 3 marks total (1 mark for the similarity, 2 marks for any two valid contrasts).

Part (b)(i): Nitration Mechanism

Mechanism for the reaction between the electrophile (NO₂⁺) and phenol to produce 4-nitrophenol

🧠 Mechanism Requirements

Arrow 1: Curly arrow starting from inside the benzene ring (delocalized ring) pointing directly to the nitrogen atom of the NO₂⁺ ion.
Intermediate: A horseshoe-shaped intermediate with a positive charge located centrally inside the broken ring. The partial ring must cover at least 3 carbon atoms, and the plus sign must face the tetrahedral carbon.
Arrow 2: Curly arrow starting from the C-H bond (of the carbon bonded to the incoming nitro group) pointing back into the ring to restore delocalization, accompanied by the release of H⁺ .

❌ Common Errors

Starting the first curly arrow from outside the ring or directly from a carbon atom rather than the delocalized pi-system.
Drawing the positive charge floating outside the horseshoe or directly attached to the tetrahedral carbon atom.
Omitting the release of H⁺ at the final step.

🎯 Mark Breakdown: 3 marks total (1 mark for electron pair movement to electrophile; 1 mark for intermediate ion structure; 1 mark for C-H bond breaking arrow and correct final product).

Part (b)(ii): Percentage Yield Calculation

Calculate the mass of 4-nitrophenol produced from 0.94 g of phenol at 15% yield

📐 Step-by-Step Calculation

Molar mass of phenol (C₆H₅OH): (6 × 12.0) + (6 × 1.0) + 16.0 = 94.0 g mol⁻¹
Moles of phenol used: 0.94 g / 94.0 g mol⁻¹ = 0.010 mol
Theoretical moles of 4-nitrophenol: 1:1 molar ratio, so theoretical yield = 0.010 mol
Molar mass of 4-nitrophenol (C₆H₅NO₃): (6 × 12.0) + (5 × 1.0) + 14.0 + (3 × 16.0) = 139.0 g mol⁻¹
Theoretical maximum mass: 0.010 mol × 139.0 g mol⁻¹ = 1.39 g
Actual mass at 15% yield: 1.39 g × 0.15 = 0.2085 g → 0.21 g

✅ Correct Option

B: 0.21

Why other options are incorrect:

A (0.14): Calculated as 15% of the starting mass of phenol (0.94 × 0.15).
C (0.68): Incorrect ratio calculation.
D (1.39): Represents 100% theoretical yield, ignoring the 15% yield constraint.

🎯 Mark Breakdown: 1 mark for selecting option B.

Part (b)(iii): Structural Isomers

Draw two structural isomers of 4-nitrophenol which have a benzene ring

✅ Correct Answer

Any two distinct isomers of nitrophenol containing the benzene ring:

2-nitrophenol (ortho-nitrophenol): The -NO₂ group is adjacent to the -OH group (carbon-2).
3-nitrophenol (meta-nitrophenol): The -NO₂ group is in the meta position relative to the -OH group (carbon-3).

🧠 Exam Technique

Ensure the skeleton or benzene ring is clearly drawn with the -OH and -NO₂ substituents placed correctly at either positions 2 and 1 or 3 and 1.

🎯 Mark Breakdown: 1 mark total for drawing any two correct structural isomers with a benzene ring.

Topics

Organic Chemistry · Physical Chemistry · Topic 18: Organic Chemistry III · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.