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9701 · 37.3

Carbon-13 NMR spectroscopy — practice questions

Practice and worked examples for 9701 Carbon-13 NMR spectroscopy. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Predict the number of signals in the ¹³C NMR spectrum of 2-methylbutane.

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First, draw the structure of 2-methylbutane: (CH₃)₂CHCH₂CH₃.

  1. Identify potential symmetry. There is no plane of symmetry that makes all carbons equivalent.
  2. Let's label the carbons: C1 and C1' are the two methyl groups on C2. C2 is the CH group. C3 is the CH₂ group. C4 is the terminal CH₃ group.
  3. The two methyl groups attached to the CH carbon (C1 and C1') are chemically equivalent because of free rotation around the C-C bond. They will produce one signal.
  4. The CH carbon (C2) is in a unique environment.
  5. The CH₂ carbon (C3) is in a unique environment.
  6. The terminal CH₃ carbon (C4) is in a unique environment.

Therefore, there are four distinct carbon environments. The spectrum will show 4 signals.

Worked example 2

An unknown compound with molecular formula C₃H₆O₂ produces a ¹³C NMR spectrum with three peaks at δ = 174.5, 51.5, and 25.2 ppm. Deduce the structure of the compound.

Show solution outline
  1. Analyse the data: The formula is C₃H₆O₂. The spectrum has 3 peaks, meaning there are 3 distinct carbon environments. Since there are 3 carbons in the formula, all carbons are non-equivalent.
  2. Consult the Data Booklet for chemical shifts:
    • The peak at δ = 174.5 ppm is in the range for a carbonyl carbon of a carboxylic acid or ester (160-185 ppm). This suggests a -COO- group.
    • The peak at δ = 51.5 ppm is in the range for a carbon singly bonded to an oxygen atom (e.g., in an alcohol or ether, 50-90 ppm). This could be the C in a C-O-R or an O-CH₃ group.
    • The peak at δ = 25.2 ppm is in the alkyl region (5-40 ppm), suggesting a CH₃ or CH₂ group not directly bonded to an oxygen.
  3. Propose structures and evaluate: Let's consider isomers of C₃H₆O₂.
    • Propanoic acid (CH₃CH₂COOH): This has 3 non-equivalent carbons. The COOH carbon would be ~175-185 ppm. The CH₂ would be ~30 ppm. The CH₃ would be ~10 ppm. This is a possibility.
    • Methyl ethanoate (CH₃COOCH₃): This has 3 non-equivalent carbons. The C=O carbon would be ~170 ppm. The O-CH₃ carbon would be ~50-60 ppm. The CH₃-C=O carbon would be ~20 ppm.
    • Ethyl methanoate (HCOOCH₂CH₃): This has 3 non-equivalent carbons. The H-C=O carbon would be ~160 ppm. The O-CH₂ carbon would be ~60 ppm. The CH₃ carbon would be ~15 ppm.
  4. Compare and conclude: The observed shifts (174.5, 51.5, 25.2 ppm) are an excellent match for methyl ethanoate:
    • δ = 174.5 ppm corresponds to the ester C=O carbon.
    • δ = 51.5 ppm corresponds to the O-CH₃ carbon.
    • δ = 25.2 ppm corresponds to the CH₃-C=O carbon.

The structure is methyl ethanoate.

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