Edexcel A-Level Chemistry Paper 3, June 2019: Question 10

Organic Structure Determination & Combustion Analysis

What this question tests

This multi-step synoptic organic question evaluates your ability to combine quantitative combustion data, mass spectrometry, and spectroscopic techniques (C-13 NMR) to deduce the empirical and molecular formula of an unknown organic compound containing a benzene ring, along with drawing constitutional isomers and analysing carbon environments.

Question 10(a)(i)

Combustion Apparatus Analysis

✅ Correct Answer (3 Marks)

The mass of solid X will increase more than expected (1 mark).
Because solid X will also absorb the water/moisture present in the undried oxygen gas (1 mark).
The mass of solid Y will stay the same (1 mark).
Because the water/moisture in the oxygen gas has already been completely absorbed by solid X before reaching Y (1 mark).

❌ Common Errors

A frequent error is assuming that because the oxygen isn't dry, carbon dioxide will also be affected, leading students to incorrectly state that solid Y's mass would increase. Remember, solid X (drying agent/water absorber) comes first in the train!

Question 10(ii)

Empirical Formula Calculation

📐 Step-by-Step Calculation

Find moles of CO₂:
Mass = 5.28 g. M(CO₂) = 12.0 + (16.0 × 2) = 44.0 g mol⁻¹.
Moles = 5.28 / 44.0 = 0.12 mol (hence moles of C = 0.12 mol).

Find moles of H₂O and H:
Mass = 0.72 g. M(H₂O) = (1.0 × 2) + 16.0 = 18.0 g mol⁻¹.
Moles of H₂O = 0.72 / 18.0 = 0.04 mol.
Moles of H = 0.04 × 2 = 0.08 mol .

Find moles of N:
Mass is given directly as 0.56 g.
Moles of N = 0.56 / 14.0 = 0.04 mol .

Find mass and moles of O:
Total mass of compound D = 3.36 g.
Mass of C = 0.12 × 12.0 = 1.44 g
Mass of H = 0.08 × 1.0 = 0.08 g
Mass of N = 0.56 g
Mass of O = 3.36 - (1.44 + 0.08 + 0.56) = 3.36 - 2.08 = 1.28 g .
Moles of O = 1.28 / 16.0 = 0.08 mol .

Determine Simplest Whole Number Ratio:C : H : N : O
C : 0.12
H : 0.08
N : 0.04
O : 0.08
Divide by smallest (0.04):
C = 3, H = 2, N = 1, O = 2.
Empirical Formula: C₃H₂NO₂ (5 marks total)

Examiner Tip: Always clearly label each element's mass and mole calculation steps. Explicitly show the division by the smallest number of moles to secure the final empirical formula mark.

Question 10(b)

Deducing Molecular Formula

💡 Key Knowledge

The molecular ion peak (M⁺) is located at the furthest peak to the right on the mass spectrum, representing the relative molecular mass (Mᵣ) of the compound.

Mᵣ from spectrum = 168
Mᵣ of empirical formula unit (C₃H₂NO₂) = (3×12) + (2×1) + 14 + (2×16) = 36 + 2 + 14 + 32 = 84.
Ratio = 168 / 84 = 2.

✅ Correct Answer

Molecular Formula: C₆H₄N₂O₄ (2 marks)

Justification: The molecular ion peak is at m/z = 168, which is twice the mass of the empirical formula (84).

Question 10(c)(i) & (ii)

Benzene Ring Fragments and Isomers

✅ (i) Fragment at m/z = 76 (1 Mark)

Species: C₆H₄⁺ (or allow H₄C₆⁺ )

Do not award just C₆H₁ .

🧠 (ii) Three Isomers containing a Benzene Ring (2 Marks)

Compound D contains a benzene ring and has the molecular formula C₆H₄N₂O₄ (which corresponds to dinitrobenzene, C₆H₄(NO₂)₂).

The three possible positional isomers are:

1,2-dinitrobenzene (ortho)
1,3-dinitrobenzene (meta)
1,4-dinitrobenzene (para)

Question 10(c)(iii)

C-13 NMR Spectroscopic Identification

✅ Correct Answer & Justification (3 Marks)

Identification: Compound D is identified as 1,3-dinitrobenzene.
Reasoning: Its structure yields exactly 4 different carbon environments, which matches the spectroscopic data stating the ¹³C NMR spectrum has four peaks.
Comparative breakdown for isomers:
- 1,3-dinitrobenzene = 4 carbon environments
- 1,2-dinitrobenzene = 3 carbon environments
- 1,4-dinitrobenzene = 2 carbon environments

Top-Level Examiner Note: To gain full marks, you must explicitly link the number of peaks in the ¹³C NMR spectrum (four peaks) to the unique carbon environments present in your selected isomer (1,3-dinitrobenzene).