OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 20

17 marks · Medium difficulty · Structured Questions

Explain the evidence for benzene's delocalised model, compare the reactivities of benzene and phenol, and outline the synthesis, mechanism, directing effects, and reduction of substituted aromatic compounds.

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Question

Question 20 spans several sub-questions on aromatic chemistry: (a) Comparing Kekulé and delocalised models of benzene to suggest two pieces of evidence for the delocalised model; (b) Reaction observations of benzene versus phenol with bromine, requiring an explanation of their relative reactivities and a balanced equation for the formation of 2,4,6-tribromophenol; (c) Multi-step syntheses of 1-methyl-2-nitrobenzene (compound G), including identifying reagents/halogen carriers for Friedel-Crafts alkylation, outlining the electrophilic substitution mechanism for nitration with H2SO4 catalyst, explaining directing groups (ortho/para vs meta) in alternative synthesis routes, and completing the equation/reagents for reduction of compound G to an amine.
Question text

20 This question is about aromatic compounds.

(a) The diagrams show the Kekulé and delocalised models of benzene.

Kekulé Delocalised

One piece of evidence for the delocalised model is that benzene does not decolourise aqueous

bromine.

Suggest two other pieces of evidence which support the delocalised model of benzene.

… [2]

(b) A student investigates the reactions of benzene and phenol with excess bromine.

The student’s observations are given in Table 20.1.

Table 20.1

Compound Benzene Phenol

Observation with decolourises and forms a

no change

excess bromine white precipitate

(i) Explain the difference in the relative reactivities of benzene and phenol with bromine.

… [3]

(ii) The white precipitate has a relative molecular mass of 330.7.

Construct an equation for the reaction of phenol with excess bromine shown in Table 20.1.

Show the structure of the organic species.

[2]

(c) Compound G is an organic compound used in the production of pesticides.

Compound G can be synthesised from benzene in the two-stage process shown.

Synthesis 1

CH3 CH3

Stage 1 Stage 2 NO2

Benzene Methylbenzene Compound G

(i) Suggest a reagent and halogen carrier that could be used in Stage 1 of Synthesis 1.

… [1]

(ii) Stage 2 of Synthesis 1 is the conversion of methylbenzene to compound G.

Outline the mechanism for the reaction in Stage 2.

Show the role of H2SO4 as a catalyst.

[5]

(iii) A student suggested an alternative two-stage synthesis of compound G from benzene.

Synthesis 2

CH3

Stage 1 NO2 Stage 2 NO2

Benzene Nitrobenzene Compound G

When Synthesis 2 is attempted, the synthesis is not successful and a different organic

compound from compound G is produced.

• Suggest why Synthesis 1 is successful but Synthesis 2 is not successful.

• Suggest the structure of the different organic compound that is formed in Synthesis 2.

[2]

(iv) In the production of a pesticide, compound G is reduced to amine H.

Suggest suitable reagents and complete the equation for this reduction.

Reagents …

Equation:

CH3 CH3

NO2 NH2

+ … [H] + …

Compound G Amine H

[2]

Mark scheme

Show the mark scheme Mark scheme for Question 20: (a) 2 marks for C-C bond lengths being equal/intermediate and enthalpy of hydrogenation being less exothermic than expected; (b)(i) 3 marks for lone pair on oxygen delocalised into ring, higher electron density, and greater susceptibility to electrophilic attack/polarisation of Br2; (b)(ii) 2 marks for structure of 2,4,6-tribromophenol and balanced equation with 3 Br2 and 3 HBr; (c)(i) 1 mark for CH3Cl and AlCl3; (c)(ii) 5 marks for catalyst generation/regeneration equations, curly arrow from benzene ring to NO2+, correct positively charged intermediate, and curly arrow from C-H bond regenerating the aromatic ring; (c)(iii) 2 marks for explaining -CH3 is 2-directing while -NO2 is 3-directing, and drawing 1-methyl-3-nitrobenzene; (c)(iv) 2 marks for Sn and HCl reagents, balanced equation with 6 [H] and 2 H2O.

Question Answer Mark Guidance

20 (a) (In the delocalised model): 2 IGNORE less susceptible to electrophilic attack (given in

(C–C) bond length is between single (C–C) and question)

double bond (C=C)

OR all (C–C) bond lengths are the same ✓

DO NOT ALLOW enthalpy of hydration OR ΔhydH

(Enthalpy change of) hydrogenation is less exothermic (than ALLOW ‘less negative’ OR ‘more positive’ OR ‘less (heat)

expected) ✓ energy released’ for ‘less exothermic’ IGNORE higher/lower

ALLOW comparison of values that show delocalised is less

exothermic e.g. Kekulé is -360 delocalised is -208

20 (b) (i) (In phenol) a (lone) pair of electrons on O is (partially) 3 ALLOW the electron pair in a p-orbital on O atom becomes

delocalised/donated into the ring / π-system ✓ part of the ring / π-system

ALLOW diagram to show movement of lone pair into ring

ALLOW lone pair of electrons on O is (partially)

drawn/attracted/pulled/ into ring / π-system

ALLOW lone pair on O

DO NOT ALLOW (two) lone pairs are (partially) delocalised

into ring / π-system

Electron density increases/is higher (than benzene) ✓ Responses must be comparative for 2nd and 3rd marking point

ORA IGNORE activating

IGNORE charge density

IGNORE electronegativity

(phenol is) more susceptible to electrophilic attack

OR ALLOW Br+ for electrophile

(phenol can) attract/accept electrophile/Br2 more IGNORE Br for electrophile

OR ALLOW Benzene can’t polarise electrophile/Br2 but phenol

(phenol can) polarise electrophile/Br2 more ✓ can (polarise electrophile/Br2) ORA

ORA IGNORE phenol reacts more readily/easily with electrophiles

Question Answer 23 Mark Guidance

20 (b) (ii) 2

ALLOW Kekulé structure

DO NOT ALLOW substitution at any three positions on ring,

must be 2, 4, 6.

IGNORE molecular formula even if incorrect

2,4,6-tribromophenol product ✓ DO NOT ALLOW ECF for organic product with incorrect

number of substituted Br

Rest of equation and balancing ✓ (Second mark is dependent on getting first mark)

20 (c) (i) CH3Cl AND AlCl3 ✓ 1 ALLOW names if no formulae given

ALLOW CH3Br for reagent

ALLOW AlBr3 OR FeCl3 OR FeBr3 for the Lewis acid

ALLOW suitable non-specification alternative Lewis acids e.g.

zeolite, BF3

20 (c) (ii) Role of H2SO4 catalyst 2 marks 5 ANNOTATIONS MUST BE USED

Forming electrophile ALLOW for forming an electrophile

HNO + H SO → NO + + HSO – + H O ✓ HNO + 2H SO → NO + + 2HSO – + H O+

32 4 2 4 2 3 2 4 2 4 3

Reforming catalyst ALLOW for forming an electrophile

H+ + HSO – → H SO ✓ HNO + H SO → H NO + + HSO –

42 4 3 2 4 2 3 4

AND then H NO + → NO + + H O

23 2 2

NOTE: curly arrows can be straight, snake-like etc.

Electrophilic attack 1 mark but NOT double headed or half headed arrows

Curly arrow from π-bond to NO + ✓ ALLOW use of +NO OR NO +

22 2

1st curly arrow must:

• start from, OR close to circle or benzene ring

AND

• go to anywhere on NO +

Correct intermediate 1 mark DO NOT ALLOW mark for intermediate if methyl group is

missing, or substitution is in wrong position

IGNORE connectivity (mark is for correct substitution position

and position of π-ring)

✓ DO NOT ALLOW following intermediates:

π -ring should cover approximately 4

of the 6 sides of the benzene ring

structure

OR AND

the correct orientation, i.e. gap

towards C with NO2 and H

ALLOW + sign anywhere inside the ‘hexagon’ of intermediate

2nd curly arrow must:

Reforming benzene ring 1 mark

• start from, OR be traced back to any part of the C – H

Curly arrow from C–H bond to reform -ring bond

AND H+ as product ✓ AND

• go inside the ‘hexagon’ of the intermediate

ALLOW mechanism using Kekulé structures with wheland

intermediate

+ H+

20 (c) (iii) (Synthesis 1 shows that the) –CH3 is 2, (4)-directing 2 ALLOW CH3 is 2, (4)-directing AND –NO2 / nitro group is not

AND (2, 4-directing)

(Synthesis 2 is not successful because) –NO2 / nitro group is ALLOW –NO2 / nitro group is 3-directing AND -CH3 is not

3-directing ✓

IGNORE CH3 and NO2 have different directing effects (the

directing effect of at least one of the groups must be specified)

ALLOW alternatives for ‘directing’ e.g. substitutes at

ALLOW –NO2 / nitro group is meta directing

ALLOW -CH3 is ortho, (para) directing

OR IGNORE 5-directing for -NO2

IGNORE 6-directing for -CH3

ALLOW Kekulé structure

✓

20 (c) (iv) Reagents 2 ALLOW names if no formulae given

Sn AND HCl ✓ IGNORE dilute for HCl

IGNORE H2 (with Sn and HCl)

Equation IGNORE NaOH if seen as a reagent to convert

nitro group into amine

e.g ‘Sn/(concentrated) HCl then NaOH’ scores the

mark

ALLOW suitable non-specification alternatives:

e.g. Zn AND HCl

H2O product AND balancing ✓

OR H2 AND Ni

How to answer it

Aromatic Chemistry: Benzene, Phenol, and Synthetic Pathways

Module 6: Organic Chemistry & Analysis

What this question tests:

This multi-step aromatic chemistry question assesses your understanding of the theoretical models of benzene, relative reactivities of arenes, reaction mechanisms, and synthetic planning:

  • Theoretical Models: Thermochemical and X-ray diffraction evidence supporting delocalised benzene over the Kekulé structure.
  • Activating Effects: Explaining why phenol reacts faster than benzene through lone pair p-orbital delocalisation and π-system electron density.
  • Electrophilic Substitution Mechanism: Detailed 5-mark nitration mechanism including inorganic generation and regeneration of the H₂SO₄ catalyst.
  • Directing Groups: Directing effects in substituted benzenes (2,4-directing alkyl vs 3-directing nitro) and synthetic sequencing.
  • Functional Group Conversions: Reduction of aromatic nitro groups to primary aromatic amines using Sn and concentrated HCl.
Part (a) • 2 Marks

Evidence for the Delocalised Model of Benzene

Contrasting the Kekulé triene model with experimental findings

✅ Accepted Evidence (Choose Two)

  • Bond Lengths: All six C–C bond lengths in benzene are identical (0.139 nm), which is intermediate between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm).
  • Enthalpy Change of Hydrogenation: The enthalpy of hydrogenation is less exothermic (less negative / -208 kJ mol⁻¹) than the expected value (-360 kJ mol⁻¹) predicted for cyclohexa-1,3,5-triene.

❌ Common Errors & Disallowed Terms

  • Repetition: Mentioning that "benzene does not decolourise bromine water" gains 0 marks as this was explicitly excluded in the question stem.
  • Vague Enthalpy Language: Do NOT say "enthalpy of hydration". Saying enthalpy is "higher" without specifying numerical value or saying "less heat released" is often rejected.
  • Stability alone: Writing "benzene is more stable" is an explanation, not experimental evidence.
Mark Scheme Note: 1 mark for identical/intermediate C–C bond lengths; 1 mark for hydrogenation enthalpy being less exothermic than theoretical values.
Part (b)(i) • 3 Marks

Relative Reactivity: Benzene vs Phenol with Bromine

Explaining why phenol reacts without a halogen carrier catalyst

✅ 3-Step Model Response

  1. Lone pair donation: A lone pair of electrons on the oxygen atom of the –OH group is partially delocalised/donated into the aromatic π-system.
  2. Increased electron density: This increases the electron density of the aromatic ring in phenol compared to benzene.
  3. Polarisation of electrophile: The higher electron density polarises Br₂ molecules more effectively (making phenol more susceptible to electrophilic attack).

🧠 Exam Technique & Keywords

  • Must specify that the lone pair is on the oxygen atom. Do not write "the OH group donates a lone pair".
  • Use comparative language for points 2 & 3: phenol has higher electron density than benzene and polarises bromine more.
  • Do not confuse "electronegativity" or "charge density" with "electron density".
Mark Breakdown: 1 mark: lone pair on O partially delocalised into π-system; 1 mark: higher electron density than benzene; 1 mark: polarises electrophile (Br₂) more / more susceptible to attack.
Part (b)(ii) • 2 Marks

Equation for the Bromination of Phenol

Reaction with excess bromine forming a white precipitate (Mᵣ = 330.7)

🔧 Mᵣ Calculation Check

Phenol reacts with excess bromine to undergo triple substitution at positions 2, 4, and 6:

  • Product: 2,4,6-tribromophenol (C₆H₃Br₃O)
  • Mᵣ = (6 × 12.0) + (3 × 1.0) + (3 × 79.9) + 16.0 = 72.0 + 3.0 + 239.7 + 16.0 = 330.7
  • Stoichiometric ratio: 1 Phenol : 3 Br₂ → 1 C₆H₂Br₃OH : 3 HBr

✅ Balanced Equation

C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr
Organic Structure Required: Draw a benzene ring with an –OH group at position 1 and –Br atoms bonded directly at positions 2, 4, and 6 (2,4,6-tribromophenol).

❌ Fatal Trap

Students commonly forget the inorganic byproduct 3HBr or write mono-substituted bromophenol. Note that mark 2 (balancing) is dependent on getting the correct organic structure first!

Part (c)(i) • 1 Mark

Stage 1: Friedel-Crafts Alkylation

Synthesising methylbenzene from benzene

✅ Reagents Required

Haloalkane reagent: CH₃Cl (chloromethane) or CH₃Br (bromomethane)

Halogen carrier (Lewis acid): AlCl₃ , FeCl₃ , or FeBr₃

Example Pair: CH₃Cl AND AlCl₃

🧠 Examiner Tip

Ensure the halogen carrier matches the halogen in the haloalkane (e.g., use AlCl₃ with CH₃Cl, or FeBr₃ with CH₃Br) to avoid mixed halogen exchanges.

Part (c)(ii) • 5 Marks

Mechanism: Nitration of Methylbenzene (Stage 2)

Electrophilic aromatic substitution with H₂SO₄ acting as a catalyst

💡 Role of H₂SO₄ Catalyst (2 Marks)

1. Generation of Electrophile (NO₂⁺):

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

2. Regeneration of Catalyst:

H⁺ + HSO₄⁻ → H₂SO₄

🧠 Mechanism Steps & Drawing (3 Marks)

  • Arrow 1 (Electrophilic Attack): Curly arrow starts from inside or on the π-ring of methylbenzene and points directly to the positive nitrogen on NO₂⁺ .
  • Intermediate Structure: A partially delocalised carbocation ring. The horseshoe must encompass at least 4 carbon atoms with the open ends pointing towards carbon-2 (which carries both –CH₃ and –NO₂ / –H). The positive sign must be inside the ring.
  • Arrow 2 & Loss of Proton: Curly arrow starts from the C–H bond at carbon-2 into the broken ring to regenerate aromaticity, releasing an H⁺ ion.

❌ Critical Examiner Traps in Nitration Mechanisms

  • Arrow origin: The arrow MUST start from the π-electron ring, NOT from a specific carbon atom or the methyl group.
  • Horseshoe gap orientation: The open part of the broken π-system MUST face the carbon bonded to the incoming NO₂ and H. If the opening faces the wrong carbon, the intermediate mark is lost.
  • Omission of H⁺: You must show H⁺ being eliminated to achieve catalyst regeneration.
Part (c)(iii) • 2 Marks

Directing Effects and Synthesis Sequencing

Why nitrating first fails to produce Compound G (1-methyl-2-nitrobenzene)

✅ Model Explanation (Mark 1)

In Synthesis 1, the methyl group (–CH₃) is 2- and 4-directing (ortho/para-directing), which directs the incoming nitro group to position 2 to form compound G.

In Synthesis 2, the nitro group (–NO₂) is 3-directing (meta-directing), directing alkylation to position 3 instead.

✅ Alternative Product Formed (Mark 2)

The organic compound formed in Synthesis 2 is 1-methyl-3-nitrobenzene (3-nitrotoluene).

Structure: Benzene ring with a –NO₂ group at position 1 and a –CH₃ group at position 3 (separated by one unsubstituted ring carbon).
Examiner Insight: Simply stating "they have different directing effects" scores 0. You must specify which group is 2/4-directing and/or which group is 3-directing.
Part (c)(iv) • 2 Marks

Reduction of Nitro Group to Primary Amine

Conversion of Compound G to Amine H

✅ Reagents (1 Mark)

Sn (tin) AND conc. HCl

(Followed by NaOH in laboratory workup, though Sn + HCl alone scores the mark).

✅ Balanced Equation (1 Mark)

Compound G + 6[H] → Amine H + 2H₂O
  • [H] needed: 6 reducing equivalents are required per nitro group.
  • Byproduct: Exactly 2 molecules of H₂O are formed.

❌ What Lost Marks

  • Using NaBH₄ or LiAlH₄ : these reduce aldehydes/ketones/carboxylic acids, not aromatic nitro groups to amines.
  • Writing 3H₂ instead of 6[H] when the equation explicitly printed ...[H] .
  • Forgetting the stoichiometric balancing: balancing with 2H₂O requires exactly 6[H] .

Topics

Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.