OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 12
1 mark · Medium difficulty · Multiple Choice
Identify the compound that produces the given 13C NMR spectrum from a multiple-choice list of organic molecules.
Practise this questionQuestion
Question text
12 A compound produces the 13C NMR spectrum below.
80 70 60 50 40 30 20 10 0
chemical shift, d/ppm
Which compound could have produced this spectrum?
A Propane
B 2-Methylbutane
C 2-Methylpropan-1-ol
D 2-Methylpropan-2-ol
Your answer
[1]
Mark scheme
Show the mark scheme
12 C 1 AO2.5
How to answer it
Interpreting Carbon-13 NMR Spectra
What this question tests
This question assesses your ability to interpret ¹³C NMR spectra by correlating the number of peaks (carbon environments) and chemical shift values (δ / ppm) with the structural isomers provided in multiple-choice options.
Determining the Structure from ¹³C NMR
✅ Correct Answer: C
The correct option is C (2-Methylpropan-1-ol). The spectrum shows exactly 3 distinct peaks, matching the 3 unique carbon environments in 2-methylpropan-1-ol, including one peak around 72 ppm corresponding to a carbon bonded to an oxygen atom (-CH₂-OH).
💡 Key Knowledge
- Number of peaks: Each peak corresponds to a carbon environment in a non-symmetrical molecule. Here we see 3 peaks, meaning 3 carbon environments.
- Chemical shift (δ): Look at the Data Booklet values. A peak at ~72 ppm indicates a carbon attached to an electronegative oxygen atom (C-O single bond).
- Carbon environments in option C: (CH₃)₂CH-CH₂-OH gives 3 environments (the two methyl groups are equivalent by symmetry).
🧠 Exam Technique
Eliminate options systematically:
- Propane (A): Only 2 carbon environments.
- 2-Methylbutane (B): 4 carbon environments, and no oxygen to push shift to 70+ ppm.
- 2-Methylpropan-2-ol (D): Only 2 carbon environments (the three methyl groups are equivalent, plus the central quaternary carbon).
❌ Common Errors
Students often fall into these traps:
- Confusing ¹³C NMR with ¹H NMR (looking for splitting patterns or integration traces, which do not exist in standard decoupled ¹³C spectra).
- Failing to account for molecular symmetry (e.g., assuming all 4 carbons in 2-methylpropan-2-ol give separate peaks).
- Ignoring the chemical shift region around 70 ppm which immediately rules out non-alcohol hydrocarbons.
Topics
Module 6: Organic chemistry and analysis · 6.3 Analysis
Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.