AQA A-Level Chemistry Paper 3, June 2018: Question 16

1 mark · Medium difficulty · Multiple Choice

Determine the number of peaks in the 13C NMR spectrum of 1,4-dimethylbenzene.

Practise this question

Question

Question 16 asks: 'How many peaks are there in the 13C NMR spectrum of 1,4-dimethylbenzene?' followed by four multiple-choice options: A: 8, B: 4, C: 3, D: 2.
Question text

16 How many peaks are there in the 13C NMR spectrum of 1,4-dimethylbenzene?

[1 mark]

A 8

B 4

C 3

D 2

Mark scheme

Show the mark scheme Mark scheme table row showing question number 16 with the correct answer C.

16 C

How to answer it

Determining ¹³C NMR Peaks in Substituted Arenes

What this question tests

This question assesses your ability to determine molecular symmetry in substituted aromatic compounds and deduce the number of unique carbon environments (which directly equals the number of peaks in a ¹³C NMR spectrum).

Question 16

Multiple Choice — 1 Mark

✅ Correct Answer

C (3 peaks)

1,4-dimethylbenzene has high molecular symmetry containing two planes of symmetry, reducing its 8 total carbon atoms into exactly 3 unique chemical environments.

💡 Key Knowledge

  • Rule: In ¹³C NMR, the number of distinct peaks corresponds directly to the number of non-equivalent carbon environments.
  • Formula: 1,4-dimethylbenzene (also called p-xylene) has the formula C₈H₁₀.
  • Symmetry: Possesses two internal mirror planes (one passing through carbons 1 and 4, and another bisecting the C2–C3 and C5–C6 bonds).

📐 Step-by-Step Breakdown of Carbon Environments

Sketch the benzene ring with methyl groups at positions 1 and 4 (opposite each other):

  1. Environment 1 (Methyl carbons): The two methyl carbons ( –CH₃ ) attached at C1 and C4 are completely equivalent by symmetry (1 peak, typically δ ≈ 21 ppm).
  2. Environment 2 (Substituted aromatic carbons): Carbons 1 and 4 in the benzene ring (the quaternary ring carbons bonded to –CH₃ groups) are equivalent (1 peak, typically δ ≈ 135 ppm).
  3. Environment 3 (Unsubstituted aromatic carbons): Carbons 2, 3, 5, and 6 each have a C–H bond. Because of vertical and horizontal reflection symmetry, all four of these CH carbons are in identical chemical environments (1 peak, typically δ ≈ 129 ppm).

Total peaks = 1 + 1 + 1 = 3 peaks

❌ Common Errors & Traps

  • Choosing A (8 peaks): Forgetting symmetry completely and assuming every carbon atom in the molecule generates its own peak.
  • Choosing B (4 peaks): Mistakenly thinking only one plane of symmetry exists (e.g. counting the ring as 3 environments and the methyl group as 1, which happens in 1,2-dimethylbenzene).
  • Choosing D (2 peaks): Forgetting to count the methyl carbons ( –CH₃ ) entirely and only identifying the two environments on the aromatic ring.

🧠 Exam Technique

  • Draw lines of symmetry: In paper exams, quickly sketch the structure and draw the mirror planes through the molecule.
  • Label with letters: Label each identical carbon with the same letter (e.g., 'a' on both methyls, 'b' on C1/C4, 'c' on C2/C3/C5/C6). The highest letter reached gives your peak count.
  • Don't confuse ¹H with ¹³C: In ¹H NMR, 1,4-dimethylbenzene has only 2 peaks (methyl hydrogens and ring hydrogens). Ensure you read carefully whether the question asks for ¹H NMR or ¹³C NMR.
Mark Scheme Award: 1 mark for selecting option C.

Topics

Organic Chemistry · 3.3.15 Nuclear Magnetic Resonance Spectroscopy

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.