OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 10

1 mark · Medium difficulty · Multiple Choice

Identify which given compound shows 4 peaks in its carbon-13 NMR spectrum.

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Question

Multiple choice question 10 asking which compound shows 4 peaks in its carbon-13 NMR spectrum, with four options labelled A to D showing different substituted cyclohexane structures. Option A is 1,1-dimethylcyclohexane, option B is 1,2-dimethylcyclohexane, option C is 1,3-dimethylcyclohexane, and option D is 1,4-dimethylcyclohexane.
Question text

10 Which compound shows 4 peaks in its carbon-13 NMR spectrum?

A

B

C

D

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is B for question 10, worth 1 mark.

10 B 1 AO2.5

How to answer it

Identifying Carbon-13 NMR Peaks in Cyclic Compounds

What this question tests

This question assesses your understanding of carbon-13 (C-13) NMR spectroscopy, specifically how molecular symmetry relates to the number of non-equivalent carbon environments in organic molecules. You must be able to visualise 3D molecular symmetry planes to count unique carbon environments accurately.

Question 10: Identifying the 4-Peak Carbon-13 NMR Spectrum

Full Mark Scheme Solution & Breakdown

✅ Correct Answer

The correct option is B (1,2-dimethylcyclohexane stereoisomers/arrangement shown).

Awarded 1 mark for selecting B.

💡 Key Knowledge

  • Each unique carbon environment in a molecule produces one distinct peak in a C-13 NMR spectrum.
  • Lines of symmetry within a molecule reduce the number of unique carbon environments because equivalent carbons absorb at the exact same chemical shift.
  • Cyclic structures require careful tracking of ring carbons, substituent positions (e.g., ortho, meta, para relationships), and side-chain carbons.

🧠 Exam Technique

  • Draw or visualize a vertical or horizontal line of symmetry directly down/across the molecule.
  • Label each unique carbon environment starting from one side and working around the ring to avoid double-counting.
  • Check methyl or branching carbons separately from ring backbone carbons.

❌ Common Errors

  • Counting total carbons instead of environments: Students often count how many carbons are present in total rather than grouping them by chemical equivalence.
  • Ignoring stereochemistry and symmetry planes: Assuming all substituted cycloalkanes are symmetrical without checking the relative positions of substituents.

🔍 Step-by-Step Analysis of Compound B (The Correct Answer)

Compound B features a cyclohexane ring with two methyl groups on adjacent carbons (1,2-dimethylcyclohexane). Due to the lack of a simple mirror plane bisecting the two methyl-bearing carbons symmetrically in this specific cis/trans orientation, let's count the unique carbon environments:

  • 2 CH(CH₃) carbons: The two ring carbons bonded to the methyl groups are equivalent by symmetry. (1 peak)
  • 2 CH₃ carbons: The two methyl substituent groups attached to the ring are equivalent. (1 peak)
  • Remaining ring carbons: The CH₂ groups in the rest of the cyclohexane ring split into distinct environments due to their unsymmetrical positioning relative to the two methyl groups, giving 2 further unique CH₂ environments. (2 peaks)
  • Total environments = 1 + 1 + 2 = 4 peaks.

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.