OCR A-Level Chemistry Unified chemistry (03), June 2025: Question 1
8 marks · Medium difficulty · Structured Questions
Deduce subatomic particle numbers and isotopic molecular ion peak origins with relative abundances for bromine, and predict proton NMR peaks and splitting patterns for two dibromoalkanes.
Practise this questionQuestion
Question text
1 This question is about spectra.
(a) The mass spectrum for Br2 has peaks at m/z values of 79, 81, 158, 160 and 162.
The peaks at m/z 79 and 81 have virtually the same relative abundance.
(i) Complete the table to show the number of protons, neutrons and electrons in the particles that
cause the peaks at m/z 79 and 81.
Number of Number of Number of
m/z
protons neutrons electrons
[2]
(ii) Explain why there are peaks with m/z values of 158, 160 and 162 and predict their relative
abundances.
Explanation …
Relative abundances …
[3]
(b) This question looks at proton NMR spectra of dibromoalkanes.
Predict the following.
Compound Number of peaks CH2 splitting pattern(s)
BrCH2CH2Br
BrCH2CH2CH2Br
[3]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
1 (a) (i) 2
Number of Number of Number of
m/z
protons neutrons electrons
79 35 44 34
81 35 46 34
Protons AND Neutrons correct ✓
Electrons correct ✓
(a) (ii) Explanation 3 Mark explanation and Relative Abundances together
(m/z) 158: 79Br79Br OR 79Br OR 79 + 79
IGNORE charges
(m/z) 160: 79Br81Br OR 81Br79Br OR 79 + 81
m/z value needs linking with isotopes in Br2
(m/z) 162: 81Br81Br OR 81Br OR 81 + 81
3 correct ✓ ✓
2 correct ✓
Relative abundancies ALLOW 0.25 : 0.5 : 0.25
1 : 2 : 1 ✓
OR ALLOW peak at 160 is twice/double 158 and 162
25 : 50 : 25
(b) 3 For BrCH2CH2Br splitting pattern,
Compound Number of peaks CH2 splitting ALLOW ‘No splitting’ OR ‘None’
pattern(s)
For BrCH2CH2CH2Br splitting pattern
ALLOW
1 Singlet ✓ multiplet,
BrCH2CH2Br
quinruplet, quinlet, quinret, pentet, pentuplet, pentex
etc
AND
ALLOW splitting patterns shown as numbers
Triplet i.e. ‘1’ for singlet, ‘3’ for triplet, ‘5’ for quintet
2 ✓
BrCH2CH2CH2Br AND
ALLOW diagrams to show splitting pattern
Quintet ✓ e.g.
Mark number of peaks and splitting patterns independently
How to answer it
Mass Spectrometry of Br₂ & ¹H NMR of Dibromoalkanes
This question assesses your mastery of two foundational analytical techniques:
- Mass Spectrometry: Identifying subatomic particles in mass spectrometry ions (accounting for positive charge) and explaining molecular ion combinations and isotopic ratios in diatomic halogens.
- Proton (¹H) NMR Spectroscopy: Identifying molecular symmetry to determine the number of distinct proton environments, and applying the (n + 1) spin-spin coupling rule to predict splitting patterns.
Part (a)(i) — Subatomic Particles in Br⁺ Ions
Deducing protons, neutrons, and electrons for isotopic fragment ions [2 Marks]
✅ Completed Table
| m/z | Protons | Neutrons | Electrons |
|---|---|---|---|
| 79 | 35 | 44 | 34 |
| 81 | 35 | 46 | 34 |
• Protons AND Neutrons correct for both rows = [1 mark]
• Electrons correct (34 for both rows) = [1 mark]
❌ Common Errors & Pitfalls
- Writing 35 electrons: The most frequent mistake! Mass spectrometers only detect positive ions (Br⁺). A neutral Br atom has 35 electrons, but the particle detected at m/z 79 is ⁷⁹Br⁺, which has lost 1 electron: 35 − 1 = 34 electrons.
- Arithmetic slips: Neutrons = Mass number − Atomic number (79 − 35 = 44, and 81 − 35 = 46).
💡 Key Knowledge
Bromine has atomic number Z = 35. Naturally occurring bromine exists in two main stable isotopes, ⁷⁹Br and ⁸¹Br, in an approximate 50:50 (1:1) ratio. Peak detection at m/z 79 and 81 corresponds to singly charged monatomic ions: [⁷⁹Br]⁺ and [⁸¹Br]⁺.
Part (a)(ii) — Molecular Ion Peaks and Ratios for Br₂
Explaining peaks at m/z 158, 160, 162 and predicting relative abundances [3 Marks]
✅ Correct Answer
Explanation:
- m/z 158: [⁷⁹Br—⁷⁹Br]⁺ (or 79 + 79)
- m/z 160: [⁷⁹Br—⁸¹Br]⁺ and [⁸¹Br—⁷⁹Br]⁺ (or 79 + 81)
- m/z 162: [⁸¹Br—⁸¹Br]⁺ (or 81 + 81)
Relative abundances: 1 : 2 : 1 (or 25% : 50% : 25% / 0.25 : 0.5 : 0.25)
• All 3 isotope combinations identified = [2 marks] (2 correct = [1 mark])
• Correct abundance ratio (1:2:1 or description that 160 is double 158/162) = [1 mark]
Note: Marking guidance allows omitting charges on formulas in this question.
📐 Step-by-Step Probability Calculation
Since the abundances of ⁷⁹Br and ⁸¹Br are virtually equal (probability = 0.5 each):
- Peak 158: P(79, 79) = 0.5 × 0.5 = 0.25
- Peak 160: Can form two ways:
P(79, 81) + P(81, 79) = (0.5 × 0.5) + (0.5 × 0.5) = 0.50 - Peak 162: P(81, 81) = 0.5 × 0.5 = 0.25
Ratio = 0.25 : 0.50 : 0.25 = 1 : 2 : 1
🧠 Exam Technique & Comparison with Chlorine
- Bromine (Br₂): Isotopes are 1:1, producing molecular ion peaks in a 1 : 2 : 1 ratio.
- Chlorine (Cl₂): ³⁵Cl and ³⁷Cl are in a 3:1 ratio, producing peaks at 70, 72, 74 in a 9 : 6 : 1 ratio.
- Always link the m/z value directly to the exact isotopes involved.
Part (b) — ¹H NMR of Dibromoalkanes
Predicting proton environments and splitting patterns [3 Marks]
✅ Completed Table
| Compound | Number of peaks | CH₂ splitting pattern(s) |
|---|---|---|
| BrCH₂CH₂Br | 1 | Singlet |
| BrCH₂CH₂CH₂Br | 2 | Triplet AND Quintet |
• Number of peaks correct (1 AND 2) = [1 mark]
• BrCH₂CH₂Br splitting: Singlet (allow "no splitting" or "none") = [1 mark]
• BrCH₂CH₂CH₂Br splitting: Triplet AND Quintet (allow "multiplet" or "pentet") = [1 mark]
💡 Understanding the NMR Splitting
1. BrCH₂CH₂Br (1,2-dibromoethane):
- Symmetrical molecule: all four protons are in an identical chemical environment → 1 peak.
- Crucial Rule: Chemically equivalent protons do not couple/split each other! Hence, the signal is a singlet.
2. BrCH₂CH₂CH₂Br (1,3-dibromopropane):
- Symmetric across the central carbon:
• Environment 1: Two terminal -CH₂Br groups (4H total).
• Environment 2: One central -CH₂- group (2H total).
Total = 2 peaks. - Terminal -CH₂Br is adjacent to 2 protons → (2 + 1) = triplet.
- Central -CH₂- is adjacent to 4 protons (2 on each neighbouring CH₂) → (4 + 1) = quintet.
❌ Common Errors in Part (b)
- Calling BrCH₂CH₂Br a triplet: Students often mistakenly apply the (n + 1) rule to the adjacent CH₂ group, forgetting that protons on both carbons are equivalent. Equivalent protons show no splitting!
- Forgetting one splitting pattern in 1,3-dibromopropane: The prompt asks for CH₂ splitting pattern(s). Because there are two distinct CH₂ environments, you must state both the triplet and the quintet to secure the final mark.
Topics
Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 2.1 Atoms and reactions · 4.2 Alcohols, haloalkanes and analysis · 6.3 Analysis
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.