OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 15

Identifying Peak Numbers in ¹³C NMR Spectra

📋 What this question tests

This question evaluates your ability to determine the number of distinct carbon chemical environments in organic molecules by analysing molecular symmetry in aliphatic, aromatic, and alicyclic compounds:

Recognising lines and planes of symmetry in straight-chain diols (hexane-3,4-diol).
Understanding symmetry in disubstituted benzene rings (specifically 1,4-disubstituted aromatic systems).
Correctly identifying non-equivalent CH₂ groups in cyclic alkanes (cyclohexane-1,3-diol).
Navigating multiple-choice format traps (Type K questions: combinations of statements 1, 2, and 3).

Question 15 Analysis

Multiple Choice (1 Mark)

✅ Correct Answer: D (Only 1)

Only Compound 1 possesses exactly three peaks in its ¹³C NMR spectrum. Compound 2 exhibits 2 peaks, and Compound 3 exhibits 4 peaks.

Mark Scheme Allocation:
• Correct option letter D = 1 mark.
• Any other response = 0 marks.

Detailed Compound Breakdown

In ¹³C NMR spectroscopy, each separate carbon environment gives rise to exactly one peak. Symmetrical carbons are in identical electronic environments and contribute to the same peak.

3 Peaks Compound 1: Hexane-3,4-diol

Structure: CH₃–CH₂–CH(OH)–CH(OH)–CH₂–CH₃

This molecule is a symmetrical aliphatic diol with an internal plane of symmetry located directly between C3 and C4:

Environment 1: The two terminal methyl carbons, C1 and C6 ( –CH₃ ) → 1 peak
Environment 2: The two adjacent methylene carbons, C2 and C5 ( –CH₂– ) → 1 peak
Environment 3: The two hydroxyl-bearing methine carbons, C3 and C4 ( –CH(OH)– ) → 1 peak

Total peaks = 1 + 1 + 1 = 3 peaks. Compound 1 satisfies the condition.

2 Peaks Compound 2: Benzene-1,4-diol (Hydroquinone)

Structure: A benzene ring with –OH groups at positions 1 and 4 (para-substitution).

Benzene-1,4-diol has two perpendicular planes of symmetry:

Plane A: Passes straight through C1, C4, and both –OH groups.
Plane B: Bisects the C2–C3 bond and the C5–C6 bond perpendicular to Plane A.

Due to this high symmetry:

Environment 1: The two ring carbons attached to –OH groups (C1 and C4) → 1 peak (~150 ppm)
Environment 2: All four aromatic CH carbons (C2, C3, C5, and C6) are chemically and electronically equivalent → 1 peak (~115 ppm)

Total peaks = 1 + 1 = 2 peaks. Compound 2 does NOT have three peaks.

4 Peaks Compound 3: Cyclohexane-1,3-diol

Structure: A six-membered saturated ring with –OH groups at carbons 1 and 3.

A plane of symmetry passes through C2 (the carbon between the two –OH groups) and C5 (the carbon directly opposite C2):

Environment 1: Carbon 2 ( –CH₂– sandwiched between two –CH(OH)– carbons) → 1 peak
Environment 2: Carbons 1 and 3 ( –CH(OH)– carbons, equivalent by symmetry) → 1 peak
Environment 3: Carbons 4 and 6 ( –CH₂– carbons adjacent to C3 and C1) → 1 peak
Environment 4: Carbon 5 ( –CH₂– opposite C2, furthest from –OH groups) → 1 peak

Total peaks = 1 + 1 + 1 + 1 = 4 peaks. Compound 3 does NOT have three peaks.

💡 Key Knowledge

¹³C NMR Peak Count: Equals the number of non-equivalent carbon environments in the molecule.
Symmetry Simplification: Whenever a molecule has a line, plane, or axis of symmetry, carbons related by symmetry map to the exact same resonance frequency.
1,4-Disubstituted Arenes: If the two substituents are identical (e.g. 1,4-diol), there are only 2 peaks. If the substituents are different (e.g. 4-nitrophenol), there are 4 peaks.

🧠 Exam Technique

Draw planes of symmetry: In paper exams, draw a dashed pencil line straight through the molecule to visually mirror identical halves.
Label each unique carbon: Write letters (a, b, c...) at each vertex. If two carbons have identical connectivity in both directions, give them the same letter.
Elimination tactic: As soon as you establish Compound 1 has 3 peaks, eliminate C. When Compound 2 only has 2 peaks, eliminate A and B. You arrive at D with 100% confidence.

❌ Common Errors

The Benzene-1,4-diol Trap: Mistakenly thinking C2/C6 are different from C3/C5 in hydroquinone. Because both substituents are identical, all four CH carbons are equivalent, giving 2 peaks (not 3!). Choosing B is the most common student error.
Lumping all ring CH₂ groups together: Assuming all –CH₂– carbons in cyclohexane-1,3-diol are identical. C2, C4/C6, and C5 are at different distances from the electronegative oxygen atoms and therefore constitute three separate environments.