OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 23

10 marks · Hard difficulty · Structured Questions

Deduce the structures of organic compounds K, L, M, and N using infrared, mass, and NMR spectroscopy, alongside elemental analysis data.

Practise this question

Question

Question 23 contains two parts. Part (a) states that compound K (C4H10O) is refluxed with concentrated sulfuric acid to produce a mixture of compounds L and M. An infrared spectrum for K displays a broad absorption band between 3200 and 3600 cm⁻¹. Carbon-13 NMR spectra show that compound L has 4 peaks and compound M has 2 peaks. Students are asked to identify the functional group, explain their reasoning, and draw structures for K, L, and M. Part (b) is an extended-response question marked with an asterisk, giving the elemental composition of compound N as C 62.07%, H 10.34%, O 27.59%. Its mass spectrum has a molecular ion peak at m/z = 116 and major fragment peaks at m/z = 43 and 71. The 1H NMR spectrum exhibits four signals: a triplet at 1.1 ppm (area 3), a doublet at 1.4 ppm (area 6), a quartet at 2.4 ppm (area 2), and a heptet at 4.4 ppm (area 1). Students must deduce the complete structure of compound N.
Question text

23 This question is about spectral analysis.

(a) Compound K, C4H10O, is refluxed with H2SO4 to form a mixture of compound L and

compound M.

The infrared spectrum of compound K is shown.

Transmittance

(%)

4000 3000 2000 1500 1000 500

Wavenumber / cm−1

Compound L and compound M are analysed using 13C NMR spectroscopy.

• The 13C NMR spectrum of compound L has 4 peaks.

• The 13C NMR spectrum of compound M has 2 peaks.

Suggest the functional group in compound K, showing your reasoning, and suggest possible

structures for compound K, compound L and compound M.

Functional group …

Reasoning …

Possible structure

K

L

M

[4]

(b)* An unknown organic compound N is analysed and produces the following results.

Elemental analysis by mass

C, 62.07%; H, 10.34%; O, 27.59%

Mass spectrum

Relative

intensity

10 20 30 40 50 60 70 80 90 100 110 120

m/z

Proton NMR spectrum

Expansion of multiplet

65 4 3 2 1 0

Chemical shift, δ/ppm

The numbers by the peaks are the relative peak areas.

Determine the structure of the compound N, showing all your reasoning.

… [6]

Extra answer space if required.

Mark scheme

Show the mark scheme Mark scheme for Question 23. Part (a) awards 4 marks: identifying the functional group as alcohol due to the O-H absorption at 3200-3600 cm⁻¹, and drawing structures: K is butan-2-ol, L is but-1-ene (4 carbon environments), and M is but-2-ene (2 carbon environments, cis or trans). Part (b) is marked using a 3-level best-fit grid for 6 marks. Indicative content covers: 1) Empirical formula calculation yielding C3H6O. 2) Using the molecular ion peak at m/z = 116 to determine molecular formula C6H12O2 and identifying fragments such as m/z = 43 or 71. 3) Analyzing splitting and chemical shifts of 1H NMR signals to identify ethyl and isopropyl environments. 4) Deducing the final ester structure as either isopropyl propanoate or ethyl 2-methylpropanoate.

Question Answer Mark Guidance

23 (a) Functional group and reasoning for compound K 4 DO NOT ALLOW any other functional group e.g. phenol,

carboxylic acid

alcohol IGNORE ‘hydroxyl’ for ‘alcohol’ BUT ALLOW for O-H

AND ALLOW any stated value in range 3100 to 3600 (cm-1)

OH (consistent with spectrum provided)

AND IGNORE reference to C-O

peak at 3200 – 3600 (cm–1) ✓ If no or incomplete answer, check spectrum for

annotations – credit can be given from a correctly labelled

peak (values are not needed)

Possible structures of K, L and M ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

ALLOW one mark if L and M both correct but in the wrong

K = ✓ boxes

L = ✓

M = OR ✓

23 (b)* Please refer to the marking instructions on page 4 of this 6 Mark spectra page as SEEN

mark scheme for guidance on how to mark this question. Indicative scientific points:

Level 3 (5–6 marks) 1. Empirical Formula

Structure is either CH3CH2COOCH(CH3)2 OR 62.07 10.34 27.59

(CH3)2CHCOOCH2CH3 • C : H : O = 12.0 : 1.0 : 16.0

AND = 5.17 : 10.34 : 1.72

Most of the data analysed = 3 : 6 : 1

• Empirical formula = C3H6O

There is a well-developed line of reasoning which is clear and • ALLOW Alternative method using Mr of 116

logically structured. The information presented is relevant and e.g (0.6270 x116) ÷12 = 6

substantiated.

2. Molecular Formula and Mass Spectrum

Level 2 (3–4 marks) • uses m/z = 116.0 to determine molecular formula as

Structure has a molecular formula of C6H12O2 with at least two C6H12O2

key features present • Any possible fragments, for example:

AND - m/z = 43 (CH ) CH+

Analyses some of the data from the NMR - m/z = 29 CH CH +

There is a line of reasoning presented with some structure. The 3. 1H NMR analysis

information presented is relevant and supported by some

• = 1.1 ppm, triplet, 3H CH3CH2–

evidence.

• = 1.4 ppm, doublet, 6 H –CH(CH3)2

Level 1 (1–2 marks) • = 2.4 ppm, quartet, 2H CH3CH2CO

Attempts analysis from at least 2 of the scientific points. • = 4.4 ppm, heptet, 1H –OCH(CH3)2

There is an attempt at a logical structure with a line of reasoning. ALLOW approximate values for chemical shifts.

The information is in the most part relevant. ALLOW multiplet for heptet

0 marks 4. Structure

No response or no response worthy of credit. ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

Key Features

• Carbonyl -C=O

• C-O bond

• Ethyl group -CH2CH3

• Dimethyl group -C(CH3)2

Correct structure

O

CH CH C CH3

O C CH3

H

Justified from NMR chemical shift analysis

OR

Justified from MS fragment as (CH ) CHCO+ at 71

Aspects of the communication statement being met might

typically includes;

• Structures given are feasible and unambiguous

• Justifying the correct structure from the evidence

• Easy to follow layout on empirical formula calculation

• Plausible MS fragments (ideally with a positive charge

shown)

• Clear information for each NMR peak, including use of

correct terminology for splitting patterns

• No additional irrelevant/incorrect information given

How to answer it

Structure Determination by Spectral Analysis

What this question tests

This question evaluates your ability to combine multiple analytical techniques: IR spectroscopy (functional group identification), dehydration of alcohols, ¹³C NMR symmetry, empirical formula calculations, mass spectrometry (molecular ion and fragmentation), and detailed ¹H NMR multiplet splitting & chemical shift analysis to deduce unknown organic structures.

Part (a) — 4 Marks

Dehydration of Alcohol K & ¹³C NMR Symmetry Analysis

Identifying functional group, reasoning, and structures K, L, and M

✅ Model Answers & Mark Scheme

  • Functional group: Alcohol (DO NOT write phenol or carboxylic acid; 'hydroxyl' is ignored)
  • Reasoning: O–H absorption / broad peak at 3200–3600 cm⁻¹ (or any value in the range 3100–3600 cm⁻¹).
  • Structure K: Butan-2-ol, CH₃CH(OH)CH₂CH₃
  • Structure L (4 ¹³C peaks): But-1-ene, CH₂=CHCH₂CH₃
  • Structure M (2 ¹³C peaks): But-2-ene, CH₃CH=CHCH₃ (either cis- or trans- / (Z)- or (E)-)
Mark distribution:
• 1 mark: Alcohol + O–H peak at 3200–3600 cm⁻¹
• 1 mark: Structure K (butan-2-ol)
• 1 mark: Structure L (but-1-ene)
• 1 mark: Structure M (but-2-ene)

💡 Key Knowledge

  • Acid-catalysed dehydration: Refluxing an alcohol with concentrated H₂SO₄ produces an alkene via elimination of H₂O.
  • Unsymmetrical elimination: Butan-2-ol can eliminate H from C1 to give but-1-ene, or from C3 to give but-2-ene.
  • ¹³C NMR and Symmetry:
    • But-1-ene: 4 chemically unique carbon environments → 4 peaks.
    • But-2-ene: Symmetrical molecule with only 2 unique carbon environments ( CH₃ and =CH– ) → 2 peaks.

🧠 Exam Technique: How to deduce K, L, and M

  • Formula C₄H₁₀O with general formula CnH2n+2O indicates a saturated, non-cyclic alcohol or ether.
  • IR shows a characteristic broad trough at ~3350 cm⁻¹, confirming an alcohol O–H bond.
  • Reflux with H₂SO₄ eliminates water to give isomers with molecular formula C₄H₈ .
  • If K were butan-1-ol, elimination would yield only but-1-ene (ignoring rearrangement). Since a mixture of L and M is formed, K must be butan-2-ol.

❌ Common Errors to Avoid

  • Vague IR ranges: Quoting an incorrect range like "1600–1700" or confusing alcohol O–H (3200–3600 cm⁻¹) with carboxylic acid O–H (2500–3300 cm⁻¹).
  • Swapping L and M: Forgetting that higher symmetry reduces the number of peaks in ¹³C NMR. The symmetrical isomer (but-2-ene) has 2 peaks (M), while the unsymmetrical isomer (but-1-ene) has 4 peaks (L).
  • Wrong terminology: Calling the functional group "hydroxide" or "hydroxyl" instead of alcohol.
Part (b)* — 6 Marks (Level of Response)

Deducing the Structure of Unknown Organic Compound N

Elemental Analysis + Mass Spectrometry + ¹H NMR Spectral Analysis

📐 Step-by-Step Mathematical Deducations

Step 1: Determine the Empirical Formula

Element Percentage (%) Moles (divide by Ar) Divide by smallest (1.724) Simplest whole number ratio
C 62.07 62.07 / 12.0 = 5.1725 5.1725 / 1.724 = 3.00 3
H 10.34 10.34 / 1.0 = 10.340 10.340 / 1.724 = 6.00 6
O 27.59 27.59 / 16.0 = 1.7244 1.7244 / 1.724 = 1.00 1

Empirical formula = C₃H₆O (empirical mass = 3(12) + 6(1) + 16 = 58 g mol⁻¹)

Step 2: Determine the Molecular Formula from Mass Spectrum

  • Molecular ion peak (M⁺) is at m/z = 116 .
  • Ratio = 116 / 58 = 2.
  • Molecular formula = (C₃H₆O) × 2 = C₆H₁₂O₂.

💡 Analysis of ¹H NMR Spectrum

Chemical Shift (δ) Area (H) Splitting (n+1) Adjacent H (n) Deduction / Fragment
1.1 ppm 3H Triplet 2 adjacent H CH₃–CH₂– (part of an ethyl group)
1.4 ppm (or 1.2) 6H Doublet 1 adjacent H –CH(CH₃)₂ (two equivalent methyls next to CH)
2.4 ppm 2H Quartet 3 adjacent H –C(=O)CH₂–CH₃ (deshielded by carbonyl)
4.4 ppm (or 5.0) 1H Heptet (7 peaks) 6 adjacent H –O–CH(CH₃)₂ (strongly deshielded by ester oxygen)

✅ Structure Determination & MS Fragmentation

The fragments assemble into an ester:

CH₃–CH₂–COOCH(CH₃)₂
Isopropyl propanoate (propanoic acid, 1-methylethyl ester)

(Note: The mark scheme also permits the isomer (CH₃)₂CHCOOCH₂CH₃ [ethyl 2-methylpropanoate] if correctly justified by chemical shifts and MS fragments).

Key Mass Spec Fragment Confirmations:

  • m/z = 43 : (CH₃)₂CH⁺ or CH₃CO⁺ (base peak)
  • m/z = 57 : CH₃CH₂CO⁺
  • m/z = 71 : (CH₃)₂CHCO⁺ (if ethyl 2-methylpropanoate)
  • m/z = 29 : CH₃CH₂⁺

🧠 Level of Response (6-Mark) Strategy

  • Level 3 (5–6 marks): Fully correct structure with complete analysis of empirical formula, molecular ion, and all 4 NMR peaks (shifts, areas, and splitting) with logical and unambiguous reasoning.
  • Level 2 (3–4 marks): Molecular formula C₆H₁₂O₂ determined with at least two key functional fragments identified (e.g. ester carbonyl, ethyl, or isopropyl group) and partial NMR analysis.
  • Level 1 (1–2 marks): Only attempts two scientific points (e.g. empirical formula calculation + molecular ion identification).

❌ Common Examiner Traps

  • Forgetting charges on MS fragments: Writing CH₃CH₂ instead of CH₃CH₂⁺ or [CH₃CH₂]⁺ loses credit for fragmentation.
  • Confusing the ester oxygen side: The chemical shift at 4.4 ppm is strongly downfield, proving that the –CH(CH₃)₂ carbon is directly bonded to the electronegative –O– atom, NOT to the carbonyl –C(=O)– carbon (which typically appears at ~2.0–2.5 ppm).
  • Missing the coupling partner: Stating a peak is a triplet without specifying that it means it must be adjacent to 2 protons ( n+1 = 3 → n = 2 ).

Topics

Module 6: Organic chemistry and analysis · Module 4: Core organic chemistry · Module 2: Foundations in chemistry · 6.3 Analysis · 6.1 Aromatic compounds, carbonyls and acids · 4.2 Alcohols, haloalkanes and analysis · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.