OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 21

6 marks · Hard difficulty · Extended Response

Deduce the structure of an unknown organic compound using elemental analysis, mass spectrometry, infrared spectroscopy, and proton NMR data.

Practise this question

Question

An exam question showing analytical data for an unknown organic compound: elemental analysis by mass (C: 73.17%, H: 7.32%, O: 19.51%), a molecular ion peak at m/z = 164.0, an infrared spectrum showing absorption peaks including a broad O-H stretch around 3000 cm^-1 and a C=O stretch around 1720 cm^-1, and a 1H NMR spectrum in D2O with peaks at approximately 1.6 ppm (doublet, relative area 3), 2.3 ppm (singlet, relative area 3), 2.7 ppm (quartet, relative area 1), and 7.1-7.5 ppm (multiplet, relative area 4). The question asks to use the results to suggest one possible structure for the unknown compound, showing all reasoning, worth 6 marks.
Question text

Analysis of an unknown organic compound produced the following results.

Elemental analysis by mass

C: 73.17%; H: 7.32%; O: 19.51%

Mass spectrum

Molecular ion peak at m/z = 164.0

Infrared spectrum

Transmittance

(%) 50

4000 3000 2000 1500 1000 500

Wavenumber / cm–1

1H NMR spectrum in D O

87 6 5 4 3 2 1 0

Chemical shift,d/ppm

The numbers by the peaks are the relative peak areas.

Use the results to suggest one possible structure for the unknown compound.

Show all your reasoning. [6]

Additional answer space if required

Mark scheme

Show the mark scheme A mark scheme outlining a 6-mark level-based response. Level 3 (5-6 marks) requires the correct structure CH3C6H4CH(CH3)COOH with most data analyzed and a well-developed line of reasoning. Guidance details the calculation of the empirical formula (C5H6O), the use of m/z = 164 to find the molecular formula (C10H12O2), and key spectral features for an aromatic carboxylic acid with specific methyl and CH environments.

AO

Question Answer Marks Guidance

element

Please refer to the marking instructions on page 4 of 6 AO1.2 Indicative scientific points:

this mark scheme for guidance on how to mark this × 2

question. AO3.1 Empirical and Molecular Formulae

× 2 73.17 7.32 19.51

AO3.2 • C : H : O = 12.0 : 1.0 : 16.0

Level 3 (5–6 marks)

Structure is CH3C6H4CH(CH3)COOH × 2 = 6.10 : 7.32 : 1.22

AND = 5 : 6 : 1

Most of the data analysed.

• Empirical formula = C5H6O

There is a well-developed line of reasoning which is • uses m/z = 164.0 to determine molecular formula as

clear and logically structured. The information C10H12O2

presented is relevant and substantiated.

Level 2 (3–4 marks)

A viable aromatic structure of C10H12O2 that contains

Structure

C=O AND most key features consistent with spectral

data ALLOW any combination of skeletal OR structural OR

AND displayed formula as long as unambiguous

Some of the spectral data analysed

Key features of an aromatic structure consistent with

There is a line of reasoning presented with some spectral data

structure. The information presented is relevant and • COOH group

supported by some evidence. • 4 aromatic H atoms

• single H atom that would give a quartet

• CH3 group that would give a doublet

• CH3 group that would give a singlet

29 AO

element

Correct Structure

Level 1 (1–2 marks) • CH3C6H4CH(CH3)COOH

Correct determination of empirical formula and/or ALLOW 2-, 3- OR 4- substitution of ring

molecular formula. i.e.

OR CH3

H3C

Analyses some of the IR and NMR data. H

H

OR

Analyses most of the NMR data. C COOH OR C COOH

CH3 CH

There is an attempt at a logical structure with a line 3

of reasoning. The information is in the most part

relevant. H

OR H3C C COOH

0 marks

No response or no response worthy of credit. CH3

Spectral analysis

1H NMR

• δ = 1.6 ppm, doublet, 3H CH3–CH–

• δ = 2.3 ppm, singlet, 3H Ar-CH3

• δ = 2.7 ppm, quartet, 1H CO–CH–CH3

OR Ar–CH–CH3 / C6H5–CH–CH3

• δ = 7.1–7.5 ppm, multiplet, 4H C6H4–

ALLOW approximate values for chemical shifts.

IR:

• peak at 2300–3700 (cm–1) is O–H

• peak at ~1720 (cm–1) is C=O

• unknown is a carboxylic acid

ALLOW ranges from Data Sheet

IGNORE references to C–O peaks

Total 6

How to answer it

Compound Structure Determination from Spectra and Elemental Analysis

Question 21 • 6 Marks • Level of Response (Levels 1–3)

What this question tests

This synthesis question tests your ability to combine multiple analytical techniques to deduce an unknown organic structure. You must calculate empirical and molecular formulae using elemental mass percentages and mass spectrometry ( m/z ), interpret infrared (IR) functional group absorption regions, and analyze high-resolution 1H NMR splitting patterns and integration values to assemble a fully connected aromatic structure.

Complete Solution & Mark Breakdown

Question 21 (6 Marks Total)

✅ Correct Answer / Final Structure

CH₃C₆H₄CH(CH₃)COOH (or any valid 2-, 3-, or 4-substituted positional isomer on the benzene ring).

Awarded Level 3 (5–6 marks) for a fully correct structure with a clear, logical, and fully substantiated line of reasoning covering all analytical data.

💡 Key Knowledge Required

  • Converting elemental mass percentages into empirical formulae via molar mass division.
  • Relating molecular ion peak ( m/z = 164.0 ) to molecular formula multipliers.
  • Recognizing broad IR absorptions (O–H in acids) and sharp carbonyl peaks (C=O).
  • Applying the n + 1 rule for 1H NMR peak splitting and using integration values for proton environments.

Step-by-Step Analytical Breakdown

📐 Step 1: Empirical and Molecular Formula Calculation

Calculations:

  • C: 73.17 / 12.0 = 6.10 → 5
  • H: 7.32 / 1.0 = 7.32 → 6
  • O: 19.51 / 16.0 = 1.22 → 1

Empirical formula = C₅H₆O (Relative mass = 5(12) + 6(1) + 1(16) = 82.0)

Molecular ion peak gives Mᵣ = 164.0 . Since 164.0 / 82.0 = 2, multiply the empirical formula by 2:

Molecular Formula = C₁₀H₁₂O₂

💡 Step 2: Infrared (IR) Spectrum Analysis

  • Broad absorption band between 2300–3700 cm⁻¹ indicates an O–H stretch in a carboxylic acid.
  • Sharp absorption peak around 1720 cm⁻¹ indicates a C=O stretch.
  • Conclusion: The compound contains a –COOH (carboxylic acid) group.

💡 Step 3: 1H NMR Spectrum Analysis

  • δ = 1.6 ppm (doublet, 3H): A –CH₃ group attached to a carbon with 1 adjacent proton ( –CH(CH₃)– ).
  • δ = 2.3 ppm (singlet, 3H): An aromatic ring methyl group ( Ar–CH₃ ) with no adjacent protons.
  • δ = 2.7 ppm (quartet, 1H): A CH proton adjacent to a methyl group ( Ar–CH(CH₃)– ).
  • δ = 7.1–7.5 ppm (multiplet, 4H): A 1,4-disubstituted (or other disubstituted) benzene ring ( C₆H₄ ) with 4 aromatic protons.

Examiner Insights & Pitfalls

🧠 Exam Technique for Top Marks

To secure Level 3 (5–6 marks), structure your answer methodically: start with formula calculations, move to functional group identification from IR, analyze every NMR signal (chemical shift, splitting, and integration), and finish by assembling the pieces into a coherent structure. Explicitly linking the proton counts to molecular fragments prevents lost marks.

❌ Common Errors & Calculation Traps

  • Rounding too early: Rounding elemental ratios prematurely leads to incorrect empirical formulae. Always keep decimal places until the final simplification step.
  • Ignoring D₂O shake data: The NMR is noted as being run in D₂O, which exchanges acidic protons (like the carboxylic acid –COOH proton), explaining why the COOH proton peak does not appear on the spectrum. Students often waste time looking for it.
  • Ambiguous structures: Ensure skeletal or structural formulas clearly show linkages, especially how the alkyl branches attach to the benzene ring and the carboxylic acid group.

Topics

Module 6: Organic chemistry and analysis · Module 2: Foundations in chemistry · 6.1 Aromatic compounds, carbonyls and acids · 6.3 Analysis · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.