OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 8

1 mark · Medium difficulty · Multiple Choice

Identify which isomer of C6H12O2 produces the smallest number of peaks in its 13C NMR spectrum.

Practise this question

Question

Multiple-choice question asking which isomer of C6H12O2 produces the smallest number of peaks in its 13C NMR spectrum. Four options A, B, C, and D show skeletal formulas of carboxylic acid and ester isomers: A is 2,3-dimethylbutanoic acid, B is methyl 3-methylbutanoate, C is 3,3-dimethylbutanoic acid, and D is 2,2-dimethylbutanoic acid. A blank box for the answer and a mark allocation of [1] are shown at the bottom.
Question text

8 Which isomer of C H O produces the smallest number of peaks in its 13C NMR spectrum?

6 12 2

O

A

OH

O

B

O

O

C

OH

O

D

OH

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme table indicates the correct answer is C, with a mark allocation of 1.

8 C 1 2.2

How to answer it

Carbon-13 NMR Isomer Analysis

📌 What this question tests

This question tests your ability to interpret structural isomerism alongside Carbon-13 (¹³C) NMR spectroscopy. You must determine the number of distinct carbon environments in different structural isomers of C₆H₁₂O₂ to find the one with the highest degree of molecular symmetry (yielding the smallest number of peaks).

Question 8: Multiple Choice Analysis

Determining the Isomer with the Fewest ¹³C NMR Peaks

✅ Correct Answer: C

Isomer C (3,3-dimethylbutanoic acid) produces the smallest number of peaks in its ¹³C NMR spectrum due to its high degree of molecular symmetry around the quaternary carbon centre.

💡 Key Knowledge: ¹³C NMR Environments

  • Each unique carbon environment in a molecule corresponds to one peak in a ¹³C NMR spectrum.
  • Lines of symmetry within a molecule reduce the number of distinct carbon environments.
  • Quaternary carbons (bonded to four other carbons) and methyl groups ( -CH₃ ) often create predictable symmetry patterns.

🧠 Exam Technique

  • Systematically count the carbon environments for every given structure rather than guessing.
  • Look for branching, especially gem-dimethyl groups ( -C(CH₃)₂ ), which frequently introduce planes of symmetry.
  • Check functional group carbons (e.g., the carbonyl carbon in -COOH or esters) as they always count as a distinct environment.

❌ Common Errors

  • Confusing ¹³C NMR (counting carbon environments) with ¹H NMR (counting proton environments and splitting patterns).
  • Forgetting to count the carbonyl carbon atom in carboxylic acids or esters.
  • Assuming identical alkyl chains automatically have the same chemical environment if they are attached to different structural frameworks.
Examiner Insight: Top-level responses quickly evaluated the symmetry of structure C. By recognizing that the two methyl groups attached to the C-3 position are equivalent due to free rotation, and comparing the carbon environments methodically, students successfully eliminated asymmetric distractors like A, B, and D.

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), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.