AQA A-Level Chemistry Paper 2, June 2025: Question 5

11 marks · Medium difficulty · State/Explain/Numerical

Write the overall equation for the formation of TNT, calculate the mass of TNT formed given the percentage yield, and outline the mechanism for the nitration of methylbenzene.

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Question text

05 This question is about making the explosive 2-methyl-1,3,5-trinitrobenzene, commonly

known as TNT.

TNT can be made by reacting methylbenzene with concentrated nitric acid in the

presence of concentrated sulfuric acid.

05.1 Give the overall equation for the reaction between methylbenzene and nitric acid.

[2 marks]

05.2 The yield of TNT in this reaction is 42.5%.

Calculate the mass, in g, of TNT (Mr = 227.0) obtained from

1050 cm3 of methylbenzene.

Methylbenzene has density = 0.867 g cm–3 and M = 92.0

[4 marks]

Mass of TNT g

05.3 In the mechanism for this reaction, concentrated nitric acid reacts with

concentrated sulfuric acid to produce the nitronium ion O N+

*16Give an equation for the reaction to produce the nitronium ion.*

State the role of the nitronium ion in the reaction with methylbenzene.

Outline the mechanism for the reaction of the nitronium ion with methylbenzene to

give 2-methyl-1-nitrobenzene.

[5 marks]

Equation

Role of nitronium ion

Mechanism

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

M1 for structure of TNT

M2 for the rest of the equation

05.1

(2 x AO1)

M1 Mass 1050 cm3 methyl benzene = 1050 × 0.867 = 910.35 g

910.35

M1 M2 n C6H5CH3 = 92 = 9.895 mol

M2 n C6H5CH3 = 92 4

05.2

M3 n C6H2CH3(NO2)3 = 9.895 × 0.425 = 4.205 (4 x AO2)

M3 n C6H2CH3(NO2)3 = M2 × 0.425

mol

M4 Mass TNT = M3 × 227 = 955 g (3sf) allow 954.6 to 2 or more sf M4 Mass TNT = 4.205 × 227 = 955 g (3sf) allow

954.6 or more sf

– A-LEVEL CHEMISTRY – 7405/2 –

M1 HNO + 2 H SO ⟶ O N+ + 2 HSO – + H O+ M1 HNO + H SO ⟶ O N+ + HSO –+ H O

32 4 2 4 3 3 2 4 2 4 2

OR via two equations 25

M2 Electrophile

05.3 (5 x

AO1)

M3 Positive must be on N and arrow from inside hexagon to

N or + on N

M4 Intermediate structure with horseshoe not beyond C2–C6

M5 Arrow from C–H bond back into the hexagon

How to answer it

Synthesis of TNT & Electrophilic Substitution

WHAT THIS QUESTION TESTS

This question assesses your mastery of aromatic chemistry (Arenes) and quantitative chemistry calculations, specifically:

Writing stoichiometric equations for multi-step nitration reactions yielding 2,4,6-trinitrotoluene (TNT).
Multi-step mole calculations integrating liquid volume, density (ρ = m / V), percentage yield, and relative formula mass (M_r).
The generation of the active nitronium ion (NO₂⁺) electrophile using conc. HNO₃ and conc. H₂SO₄.
Accurate drawing and curly arrow notation for the electrophilic aromatic substitution mechanism on a substituted benzene ring.

Question 05.1

Overall Equation for the Synthesis of TNT

2 Marks • AO1

✅ Correct Answer

C₆H₅CH₃ + 3HNO₃ → C₆H₂(NO₂)₃CH₃ + 3H₂O

Structural representation: A benzene ring with a -CH₃ group at C1, and -NO₂ groups at positions 2, 4, and 6 (also called 2-methyl-1,3,5-trinitrobenzene in the question's IUPAC naming convention).

Mark Scheme Breakdown:
• [M1]: Correct structure of TNT (showing three -NO₂ groups substituted alternately around the ring relative to -CH₃).
• [M2]: Balanced equation with 3HNO₃ and 3H₂O .

❌ Common Errors

Forgetting water: Producing H₂ gas instead of H₂O as the co-product.
Incorrect stoichiometry: Forgetting the balancing coefficient 3 in front of HNO₃ and H₂O.
Connectivity errors: Connecting the nitro group through oxygen ( C–O–NO ) instead of nitrogen ( C–NO₂ or O₂N–C on the left side).

Question 05.2

Calculation: Mass of TNT Formed

4 Marks • AO2

📐 Step-by-Step Calculation

Step 1: Calculate mass of methylbenzene Mass = Volume × Density m = 1050 cm³ × 0.867 g cm⁻³ = 910.35 g [M1]

Step 2: Calculate moles of methylbenzene Moles = Mass / M_r n = 910.35 g / 92.0 g mol⁻¹ = 9.895 mol [M2]

Step 3: Apply percentage yield (42.5%) Mole ratio (methylbenzene : TNT) is 1 : 1 Theoretical moles of TNT = 9.895 mol Actual moles = 9.895 × (42.5 / 100) = 4.205 mol [M3]

Step 4: Calculate actual mass of TNT Mass = Actual Moles × M_r(TNT) Mass = 4.205 mol × 227.0 g mol⁻¹ = 954.6 g Final Answer: 955 g (quoted to 3 sig figs) [M4]

🧠 Exam Technique & Traps

Rounding errors: Never round intermediate values on your calculator; keep the full precision until the final step (954.6 g rounds neatly to 955 g).
Molar ratio check: Although 3 moles of HNO₃ are used, each mole of methylbenzene produces one mole of TNT (1:1 ratio).
Density trap: Students sometimes divide by density instead of multiplying. Always check units: cm³ × (g / cm³) = g.
Acceptable range: The mark scheme allows 954.6 g up to 955 g depending on rounding precision.

Question 05.3

Electrophile Formation & Mechanism

5 Marks • AO1

✅ Part 1: Equation for Electrophile Generation

HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺

(Alternative acceptable balanced equation: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O)

[M1] Correct balanced equation forming the nitronium ion (NO₂⁺).

✅ Part 2: Role of the Nitronium Ion

Electrophile (or electron pair acceptor).

[M2] Correct stated role.

💡 Part 3: Mechanism Details (3 Marks)

To form 2-methyl-1-nitrobenzene (substitution adjacent to -CH₃):

1. Attack of Electrophile [M3]:

Curly arrow must start from inside the delocalised π-electron ring of methylbenzene and point directly to the N atom (or positive charge on the N) of ⁺NO₂.

2. Arenium Carbocation Intermediate [M4]:

Ring must show a partially broken delocalised system represented as a horseshoe opening towards carbon-2 (the sp³ carbon where NO₂ and H are attached).
The horseshoe arc must span at most 5 ring carbons (C3 around to C1) and not extend beyond C2–C6.
A distinct positive charge (+) must reside inside the open horseshoe cavity.

3. Loss of Proton [M5]:

Curly arrow must originate cleanly from the C–H bond on carbon-2 and point directly back into the ring to restore the complete aromatic π-system.

❌ Critical Mechanism Pitfalls

Horseshoe shape too wide: Extending the horseshoe delocalisation over the tetrahedral C bonded to -H and -NO₂ loses M4 instantly.
Misplaced charge: Placing the ‘+’ on a specific carbon or outside the ring rather than within the horseshoe cavity.
Arrow origins: Starting the curly arrow from the ring's boundary carbon rather than from the electron-rich delocalised π-cloud.
Arrow terminus in step 3: Directing the arrow from the H atom itself rather than the C–H covalent bond line.

💡 Why Sulfuric Acid is the Catalyst

Concentrated sulfuric acid is a stronger acid than concentrated nitric acid. H₂SO₄ protonates HNO₃, which then loses water to generate the powerful NO₂⁺ electrophile. H₂SO₄ is subsequently regenerated when HSO₄⁻ accepts the proton lost in the final step of the substitution:

H⁺ + HSO₄⁻ → H₂SO₄

Topics

Organic Chemistry · Physical Chemistry · 3.3.10 Aromatic Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.