Combine carbon-13 and proton NMR evidence
Use carbon and proton environments together to test a structure against signal counts, shifts and splitting.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 37. Use the supplied spectra, shift table and chromatography conditions. Interpret peaks as evidence and state when a short trace cannot uniquely identify a substance. Written practice is self-checked
Before you begin
- Recognise functional groups, molecular structures, polarity, intermolecular interactions and the distinction between equivalent and different chemical environments.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does combining ¹³C and ¹H NMR add to structure deduction?
What you will learn
- Combine ¹³C and ¹H signal counts
- Use proton areas and splitting to test connectivity
- State when more than one structure remains possible
Let each spectrum answer a different question
Carbon-13 NMR counts distinct carbon environments and uses shifts to indicate carbon types. Proton NMR gives proton environments, typical shifts, relative numbers from areas and adjacent-proton information from splitting.
Use the molecular formula to constrain the total atoms. A candidate should explain carbon environments and every proton signal without leaving atoms unassigned.
Test structure hypotheses step by step
Start with a carbonyl, aromatic, alkene or alkyl clue in the carbon spectrum. Use proton shifts and integral ratios to assign groups, then splitting to connect neighbouring groups along the carbon skeleton.
If a signal appears exchangeable, use a D₂O comparison if supplied. If multiple structures satisfy the data, name the additional evidence needed rather than presenting one as certain.
Pause and explain
If a candidate fits the carbon-13 signal count but fails the proton integral ratio, what should you conclude?
Worked example
A C₃H₆O compound has two carbon-13 environments, including one carbonyl carbon, and one proton signal with a 6H integral near 2.2 ppm. Which simple structure fits?
Show the worked solution
- Two carbon environments fit a carbonyl carbon and two equivalent methyl carbons.
- One 6H proton environment fits two equivalent methyl groups.
- Check the shift and total formula.
Answer Propanone, CH₃COCH₃, fits: its two methyl groups are equivalent, giving one 6H proton environment, while its carbonyl carbon and equivalent methyl carbons give two carbon-13 environments.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A C₂H₄O₂ candidate has two carbon environments, one carboxyl carbon, and proton area ratio 3:1. Give a plausible structure and explain the evidence briefly.
Give me a hint
Use a methyl group plus one exchangeable acid proton.
Compare my reasoning
- The carboxyl carbon gives one carbon environment.
- A methyl group gives a 3H signal.
- A carboxylic-acid O–H can supply the 1H environment.
Ethanoic acid, CH₃COOH, is consistent: two carbon environments (methyl and carboxyl), with 3H in CH₃ and 1H in O–H. The acid proton shift is variable.
Look for these in your work
- Formula fits
- Carbon environments fit
- Proton ratio explained
A structure has three carbon environments but a carbon-13 spectrum shows four signals. What does this mismatch imply?
Give me a hint
Each distinct carbon environment should correspond to one signal in a simple spectrum.
Compare my reasoning
- The candidate predicts three environments.
- The observation indicates four.
- Recheck the proposed structure or data interpretation.
The candidate is inconsistent with the stated spectrum; either the structure, equivalence assessment or signal assignment must be revised.
Look for these in your work
- Predicted count compared
- Mismatch identified
- Next check named
Why is it useful to compare ¹H splitting with a proposed carbon skeleton after assigning shifts and integrals?
Give me a hint
Splitting provides a neighbour relationship that can connect groups.
Compare my reasoning
- Shift suggests proton environment.
- Area gives relative proton count.
- Splitting tests adjacency in the proposed structure.
Splitting checks whether the proposed groups are adjacent in the way required by the n + 1 pattern, so it tests connectivity beyond shifts and proton counts.
Look for these in your work
- Three evidence types distinguished
- Connectivity tested
- Role explained
Common mistakes
- Using carbon-13 peak count as a proton count
- Ignoring missing proton integration
- Applying the n + 1 rule without checking adjacent atoms
- Claiming a unique structure from incomplete spectra
What can you explain now?
What does combining ¹³C and ¹H NMR add to structure deduction?
Compare with the explanation
Independent constraints on carbon and proton environments
The two spectra describe different atoms and relationships.
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Make it stick
How do ¹³C environments, ¹H shifts, integrals and splitting jointly test a proposed structure?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.