OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2024: Question 24

6 marks · Hard difficulty · Extended Response

Determine the structure of unknown organic compound J using elemental analysis, mass spectrometry, IR spectroscopy, and proton NMR spectroscopy data.

Practise this question

Question

The question presents elemental analysis data for compound J showing percentage composition by mass for C (80.60%), H (7.46%), and O (11.94%), alongside a mass spectrum, an infrared spectrum, and a proton NMR spectrum with relative peak areas of 1, 5, 1, and 3. Students are asked to determine the structure of compound J showing all their reasoning in 6 marks.
Question text

Analysis of an unknown organic compound J produces the following results.

Elemental analysis by mass of compound J

C, 80.60%; H, 7.46%; O, 11.94%

Mass spectrum of compound J

Relative

intensity

25 50 75 100 125

m/z

IR spectrum of compound J

ransmittance

(%) 50

4000 3000 2000 1500 1000 500

Wavenumber / cm−1

Proton NMR spectrum of compound J

10 8 6 4 2 0

Chemical shift, δ/ppm

The numbers by the peaks are the relative peak areas.

Determine the structure of compound J, showing all your reasoning. [6]

Extra answer space if required.

Mark scheme

Show the mark scheme The mark scheme outlines a level-based response grid from Level 1 to Level 3 for 6 marks, detailing indicative scientific points including empirical formula calculation leading to C9H10O, mass spectrum and IR interpretations identifying a carbonyl group and a benzene ring, and NMR chemical shifts corresponding to specific proton environments in C6H5CHCH3CHO.

Question Answer Marks Guidance

Please refer to the marking instructions on page 4 of this 6 LOOK ON THE SPECTRA for labelled peaks and mark as

mark scheme for guidance on how to mark this question. SEEN

Level 3 (5–6 marks) Indicative scientific points:

Structure is C6H5CHCH3CHO 1. Empirical (and Molecular) Formulae

AND 80.60 7.46 11.94

• C : H : O = : :

Analyses data from all 3 scientific points 12.0 1.0 16.0

= 6.72 : 7.46 : 0.746

There is a well-developed line of reasoning which is clear and

= 9 : 10 : 1

logically structured. The information presented is relevant and

• Empirical formula = C9H10O

substantiated.

Level 2 (3–4 marks) 2. Mass spectrum and IR

Structure with most key features including O atom(s) Mass spectrum

AND • uses m/z = 134 to give molecular formula: C9H10O

Analyses data from at least 2 of the scientific points • Any possible fragments:

- m/z = 105 C H CHCH +

65 3

There is a line of reasoning presented with some structure. +

- m/z = 77 C6H5

The information presented is relevant and supported by some +

- m/z = 29 CHO

evidence.

IR

• C=O from ~1700 cm–1

Level 1 (1–2 marks)

• Likely to be aldehyde or ketone

Attempts analysis from at least 2 of the scientific points –1

• C=C (arenes) ~1500 cm

ALLOW Data Sheet ranges

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

reasoning. The information is in the most part relevant. 3. 1H NMR

• = 1.4 ppm, doublet, 3H CH3CH-

0 marks

No response or no response worthy of credit. • = 3.8 ppm, quintet, 1H next to 4 adjacent H

• = 7.3 ppm, singlet, 5H C6H5-

• = 9.0 ppm, doublet, 1H -CHCHO

ALLOW approximate values for chemical shifts

Structure

ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

32 ALLOW correct Kekulé representation of benzene

Key features

• Benzene ring

• C=O

• CH3

Correct structure

CH3

CH

H432/02 Mark Scheme CHO June 2024

• (C6H5CHCH3CHO)

Aspects of the communication statement being met might

typically include:

• Structures given are feasible and unambiguous

• Easy to follow layout on empirical formula calculation

• Empirical formula is shown to be same as molecular

• IR peaks linked clearly to bond it refers to not just

functional groups

• Positive charge given on MS fragments

• MS fragments plausible for the molecular formula

determined

• Clear information for each NMR peak

• No additional irrelevant/incorrect information given

BLANK PAGE

How to answer it

Elucidation of Organic Compound J

OCR A-Level Chemistry • 6 Mark Extended Response

What this question tests

This synthesis question assesses your ability to combine analytical data from four distinct techniques—combustion/elemental analysis, mass spectrometry, infrared (IR) spectroscopy, and proton (¹H) NMR spectroscopy—to logically deduce and justify an unknown organic molecular structure.

Complete Question Analysis & Solution Guide

Question 24: Determine the structure of compound J

✅ Correct Answer

Structure: C₆H₅CH(CH₃)CHO (2-phenylpropanal)

Key structural fragments identified:

  • Benzene ring ( C₆H₅- )
  • Aldehyde group ( -CHO )
  • Methyl group attached to a CH group ( -CH(CH₃)- )

💡 Key Knowledge

  • Elemental Analysis: Converts % mass to empirical formula via molar masses.
  • Mass Spec: The molecular ion peak ( m/z ) gives the relative molecular mass (Mr). Fragment ions provide structural clues.
  • IR Spec: Sharp absorption at ~1700 cm⁻¹ indicates a C=O bond (carbonyl).
  • ¹H NMR: Chemical shifts, integration traces (peak areas), and splitting patterns (n+1 rule) reveal hydrogen environments and connectivity.

🧠 Exam Technique (Level 3 Strategy)

To secure a Level 3 (5–6 marks), you must systematically present evidence from all 3 scientific analysis areas (Elemental/Molecular Formula, Mass Spec/IR, and ¹H NMR) with a clear, logical chain of reasoning leading to the final structure.

❌ Common Errors

  • Failing to link IR absorption numbers to specific bonds (e.g., just writing "functional group present" without naming C=O).
  • Omitting positive charge symbols ( + ) on mass spec fragment ions.
  • Misinterpreting the ¹H NMR splitting patterns (e.g., missing that a quintet means 4 adjacent protons).

📐 Step-by-Step Calculation: Empirical & Molecular Formula

  1. Divide percentage masses by relative atomic masses (Ar):
    C: 80.60 / 12.0 = 6.72
    H: 7.46 / 1.0 = 7.46
    O: 11.94 / 16.0 = 0.746
  2. Find the simplest whole number ratio (divide by the smallest value, 0.746):
    C: 6.72 / 0.746 = 9
    H: 7.46 / 0.746 = 10
    O: 0.746 / 0.746 = 1
  3. Empirical Formula: C₉H₁₀O
  4. Molecular Formula Confirmation: The mass spectrum shows a molecular ion peak ( M⁺ ) at m/z = 134 . Calculating the Mr of C₉H₁₀O gives (9×12) + (10×1) + 16 = 134. Therefore, the empirical formula is identical to the molecular formula.

🔬 Spectral Breakdown & Reasoning

  • IR Spectrum: Strong absorption peak at ~1700 cm⁻¹ confirms the presence of a carbonyl group ( C=O ), typical of an aldehyde or ketone. Additional peaks around 1500 cm⁻¹ indicate a benzene ring ( C=C arenes).
  • Mass Spectrum Fragments: m/z = 105 ( C₆H₅CHCH₃⁺ ), m/z = 77 ( C₆H₅⁺ ), and m/z = 29 ( CHO⁺ ).
  • ¹H NMR Analysis:
    • δ = 9.0 ppm (doublet, 1H) corresponds to an aldehyde proton ( -CHO ), split into a doublet by 1 adjacent proton.
    • δ = 7.3 ppm (singlet, 5H) confirms a monosubstituted benzene ring ( C₆H₅- ).
    • δ = 3.8 ppm (quintet, 1H) represents the CH group positioned adjacent to 4 other protons.
    • δ = 1.4 ppm (doublet, 3H) indicates a methyl group ( -CH₃ ) adjacent to 1 proton.
Mark Scheme Thresholds:
• Level 3 (5–6 marks): Correct structure ( C₆H₅CH(CH₃)CHO ) + analysis from all 3 scientific points + coherent logical reasoning.
• Level 2 (3–4 marks): Structure with most key features + analysis from at least 2 scientific points.
• Level 1 (1–2 marks): Attempts analysis from at least 2 scientific points with partial structure development.

Topics

Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · 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 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.