OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2023: Question 23
6 marks · Hard difficulty · Extended Response
Identify an unknown organic compound using 2,4-DNP test, elemental analysis, mass spectrometry, IR spectrum, and proton NMR spectrum.
Practise this questionQuestion
Question text
An unknown organic compound is analysed.
The results are shown below.
Addition of 2,4-DNP
No visible change
Elemental analysis by mass
C, 66.63%; H, 11.18%; O, 22.19%
Mass spectrum
Molecular ion peak at m/z = 144.0
IR spectrum
transmittance
(%) 50
4000 3000 2000 1500 1000 500
wavenumber / cm–1
Proton NMR spectrum
The numbers by each peak are the relative peak areas.
54 3 2 1 0
chemical shift, δ/ppm
Use the information to identify the organic compound.
Show all your reasoning.
… [6]
Additional answer space if required.
Mark scheme
Show the mark scheme
Question Answer Marks AO Guidance
element
23 Please refer to the marking instructions on page 4 of this 6 AO1.2 Mark spectra page as SEEN
mark scheme for guidance on how to mark this question. × 2 Indicative scientific points:
AO3.1
Level 3 (5–6 marks) × 2 1. Empirical Formulae
Structure is either CH3CH2COOCH2C(CH3)3 OR AO3.2 66.63 11.18 22.19
× 2 • C : H : O = 12.0 : 1.0 : 16.0
(CH3)3CCH2COOCH2CH3
AND = 5.55 : 11.18 : 1.39
Most of the data analysed. = 4 : 8 : 1
• Empirical formula = C4H8O
There is a well-developed line of reasoning which is clear
and logically structured. The information presented is 2. Molecular Formulae
relevant and substantiated. • uses m/z = 144.0 to determine molecular
formula as C8H16O2
Level 2 (3–4 marks)
Structure is an ester of C8H16O2 with some key features 3. Functional group
present From IR,
AND • → C=O from ~1740 cm–1
Analyses some of the data from at least 3 of the scientific
IGNORE references to C–O peaks
points.
No reaction with 2,4-DNP
There is a line of reasoning presented with some structure.
• → no carbonyl/no ketone and aldehyde
The information presented is relevant and supported by
some evidence. • Likely to be an ester
4. 1H NMR analysis
Level 1 (1–2 marks)
Attempts analysis from at least 2 of the scientific points. • = 0.9 ppm, singlet, 9H –C(CH3)3
• = 1.2 ppm, triplet, 3H CH3CH2–
There is an attempt at a logical structure with a line of • = 2.2 ppm, quartet, 2H CH3CH2CO
reasoning. The information is in the most part relevant.
• = 4.1 ppm, singlet, 2H –OCH2–
0 marks
No response or no response worthy of credit. ALLOW approximate values for chemical shifts.
34 element
Structure
ALLOW any combination of skeletal OR structural
OR displayed formula as long as unambiguous
Key features consistent with chemical shift data
and relative peak areas
• O-CH2
• C(CH3)3
• CH3CH2C=O
Correct Structure
• CH3CH2COOCH2C(CH3)3
How to answer it
Organic Structure Determination (6-Mark Synoptic Question)
This is a classic Level-of-Response synoptic organic analysis question. It tests your ability to methodically combine data from four distinct analytical techniques: combustion/elemental analysis (for empirical formula), mass spectrometry (for molecular formula), infrared (IR) spectroscopy (for functional groups), and proton (¹H) NMR spectroscopy (for carbon skeleton connectivity and molecular fragment integration).
Question 23: Complete Structural Identification
Full Mark Scheme & Analytical Breakdown
✅ Correct Answer
IUPAC / Structural Formula:
CH₃CH₂COOCH₂C(CH₃)₃ (or (CH₃)₃CCH₂COOCH₂CH₃ )
Name: 2,2-dimethylpropyl propanoate (or neopentyl propanoate)
💡 Key Knowledge: Analytical Toolkit
- 2,4-DNP: No reaction means NO aldehydes or ketones. Combined with oxygen present, points to an ester, carboxylic acid, or alcohol.
- IR Spectrum: Sharp peak around 1740 cm⁻¹ confirms a C=O ester stretch.
- Mass Spec: Molecular ion peak (M⁺) at m/z = 144.0 gives the exact relative molecular mass.
📐 Step-by-Step Calculation: Formulae
Step 1: Empirical Formula
- C: 66.63 / 12.0 = 5.55
- H: 11.18 / 1.0 = 11.18
- O: 22.19 / 16.0 = 1.39
- Divide by smallest (1.39): C = 4, H = 8, O = 1
- Empirical Formula: C₄H₈O (Mass = 72.0)
Step 2: Molecular Formula
- M⁺ / Empirical Mass = 144.0 / 72.0 = 2
- Molecular Formula: C₈H₁₆O₂
🧠 Exam Technique & NMR Fragment Analysis
Extract structural fragments systematically using ¹H NMR chemical shifts, splitting patterns, and relative peak areas (integration ratio 3 : 2 : 2 : 9):
- δ = 0.9 ppm (singlet, 9H): -C(CH₃)₃ (tertiary butyl group)
- δ = 1.2 ppm (triplet, 3H): CH₃CH₂- adjacent to CH₂
- δ = 2.2 ppm (quartet, 2H): CH₃CH₂CO- adjacent to carbonyl
- δ = 4.1 ppm (singlet, 2H): -OCH₂- attached to oxygen next to a quaternary carbon (no adjacent protons)
❌ Common Errors & Where Students Lose Marks
- Ignoring 2,4-DNP: Failing to state that a negative 2,4-DNP test eliminates aldehydes and ketones, leading candidates to incorrectly propose carbonyl isomers.
- Misinterpreting NMR Multiplicity: Confusing the 9H singlet for three separate methyl groups instead of recognising a neopentyl or tert-butyl structural environment.
- Arithmetic Slips in Empirical Formula: Rounding percentage values incorrectly before dividing by relative atomic masses. Always keep unrounded intermediate values in your calculator.
• Level 3 (5–6 marks): Correct structure provided with clear, logical validation covering elemental analysis, mass spec, IR, and full NMR assignments.
• Level 2 (3–4 marks): Correct ester family identified with partial structural data correctly analysed.
• Level 1 (1–2 marks): Correct calculation of empirical/molecular formula or successful analysis of 2-3 individual data points.
Topics
Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.3 Analysis
Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.