AQA A-Level Chemistry Paper 3, 2017: Question 29
1 mark · Medium difficulty · Multiple Choice
Determine the number of peaks in the 13C NMR spectrum of benzo[b]thiophene.
Practise this questionQuestion
Question text
How many peaks does this compound have in its 13C spectrum?
[1 mark]
A 5
B 6
C 7
D 8
Mark scheme
Show the mark scheme
29 D
How to answer it
Determining ¹³C NMR Chemical Environments in Fused Heterocycles
This question assesses your ability to deduce the number of unique carbon environments (and thus the number of separate signals in a ¹³C NMR spectrum) for a fused bicyclic aromatic compound (benzo[b]thiophene). It tests molecular symmetry recognition, identifying bridgehead/quaternary carbons, and understanding the electronic influence of heteroatoms (S).
Question 29: Peak Count in ¹³C NMR Spectrum
Multiple Choice Analysis • [1 Mark]
✅ Correct Answer
D — 8
💡 Key Knowledge
- Each unique (non-equivalent) carbon environment gives rise to exactly one signal in a ¹³C NMR spectrum.
- Symmetry reduces the number of peaks; an asymmetric molecule shows as many peaks as there are carbon atoms.
- Heteroatoms (such as sulfur) alter electron density unevenly across fused rings, completely destroying potential planes of symmetry.
- Bridgehead carbons (carbons shared by both rings) are quaternary and count as distinct carbon environments.
📐 Step-by-Step Carbon Environment Audit
Total formula of benzo[b]thiophene is C₈H₆S. Let us systematically verify every carbon:
| Carbon Position | Type | Environment & Chemical Connectivity | Equivalence |
|---|---|---|---|
| C2 | =CH– | Directly bonded to S atom in 5-membered ring | Unique (1) |
| C3 | =CH– | Bonded to C2 and the bridgehead carbon C3a | Unique (2) |
| C3a | >C= (junction) | Bridgehead carbon adjacent to C3 (not directly bonded to S) | Unique (3) |
| C7a | >C= (junction) | Bridgehead carbon directly bonded to electronegative S | Unique (4) |
| C4 | =CH– | Benzene ring CH, ortho to C3a | Unique (5) |
| C5 | =CH– | Benzene ring CH, meta to C3a | Unique (6) |
| C6 | =CH– | Benzene ring CH, meta to C7a | Unique (7) |
| C7 | =CH– | Benzene ring CH, ortho to C7a (close to S) | Unique (8) |
Conclusion: Because there is no line or plane of symmetry in benzo[b]thiophene, all 8 carbon atoms are in unique chemical environments → 8 peaks (Option D).
🧠 Exam Technique: Symmetry Check
- Step 1: Count total carbons first. Here: 6 in the benzene ring + 2 in the thiophene ring = 8 carbons.
- Step 2: Look for a plane of symmetry. Draw a line straight through the middle horizontally:
• Top has a sulfur atom ( –S– ).
• Bottom has a double bond ( –CH=CH– ).
Since S ≠ CH, there is no horizontal symmetry. - Step 3: Draw a vertical line:
• Left side is a 5-membered ring.
• Right side is a 6-membered benzene ring.
Since 5 ≠ 6, there is no vertical symmetry. - No symmetry means: Number of peaks = Total number of carbons = 8 .
❌ Common Student Traps
- Assuming benzene ring symmetry (Picking 5 or 6): Students often falsely assume the 6-membered ring carbons exist in pairs (like typical 1,2-disubstituted benzenes). Because the two attachment points (C3a and C7a) are different, the top and bottom of the benzene ring are NOT equivalent.
- Forgetting bridgehead carbons: Overlooking the two ring-junction quaternary carbons, counting only CH carbons (giving 6 peaks).
- Confusing ¹H and ¹³C NMR: Counting proton environments instead of carbon environments. (Note: ¹H spectrum would give 6 peaks, because there are 6 hydrogens!).
Topics
Organic Chemistry · 3.3.15 Nuclear Magnetic Resonance Spectroscopy
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.