Deduce a simple structure from proton NMR evidence
Combine signal count, chemical shifts, relative areas and splitting to evaluate possible molecular structures.
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
Which NMR feature gives relative proton numbers?
What you will learn
- Combine four kinds of proton NMR evidence
- Match each signal to a proton environment
- Check a proposed structure against the full spectrum
Use each feature for a different constraint
The number of signals gives the number of proton environments; chemical shifts suggest those environments; integrals give relative proton counts; splitting indicates equivalent neighbouring protons on adjacent carbons.
Use the molecular formula to set absolute atom counts and eliminate impossible candidates. A candidate structure must explain all signals, not just one convenient peak.
Build and test a candidate
Assign a tentative group to each signal, use integral ratios to place proton counts, then use splitting to connect neighbouring groups. Check carbon and hydrogen counts against the formula.
If more than one candidate satisfies the supplied evidence, state the ambiguity or identify which additional spectrum could help. A simple spectrum does not always determine a unique structure by itself.
Pause and explain
What must a proposed structure explain?
Worked example
A compound C₃H₆O has a 3H triplet near 1.0 ppm, a 2H quartet near 2.4 ppm and a 1H signal near 9.8 ppm. Which simple structure fits?
Show the worked solution
- The 1H signal is in the aldehyde region.
- The 2H group next to CH₃ gives a quartet; the CH₃ next to CH₂ gives a triplet.
- Check that the atom count and all signals fit.
Answer Propanal, CH₃CH₂CHO, fits: the CH₃ and CH₂ form an ethyl group, and the remaining proton is an aldehyde proton. The formula totals C₃H₆O.
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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 C₂H₆O spectrum has a 3H triplet in the alkyl region, a 2H quartet next to oxygen and a variable 1H signal. Give a structure consistent with the formula and pattern.
Give me a hint
Use the 3H and 2H carbon-bound groups, then account for the remaining hydrogen.
Compare my reasoning
- The triplet and quartet indicate adjacent CH₃ and CH₂ groups.
- The CH₂ is next to oxygen.
- The formula leaves one hydrogen for an O–H group.
Ethanol, CH₃CH₂OH, fits: the CH₃ gives a triplet, CH₂–O gives a quartet, and the variable exchangeable O–H signal accounts for the sixth hydrogen overall.
Look for these in your work
- Formula considered
- Neighbour pattern used
- Exchangeable proton assigned cautiously
A candidate has a 6H integral for two methyl groups but the spectrum shows two different 3H signals. What structural feature would explain this better than a symmetric molecule?
Give me a hint
Different 3H environments indicate the methyl groups are not equivalent.
Compare my reasoning
- The two methyl groups experience different surroundings.
- The molecule lacks symmetry exchanging them.
- Choose a candidate with two distinct methyl environments.
An asymmetric candidate with two non-equivalent methyl groups is more consistent than a structure that makes them symmetry-equivalent.
Look for these in your work
- Integral interpreted
- Equivalence distinguished
- Structural constraint stated
Why should the n + 1 pattern be checked against the proposed carbon skeleton?
Give me a hint
Adjacent-proton counts depend on which carbon atoms are bonded together.
Compare my reasoning
- Splitting suggests neighbouring proton counts.
- The candidate determines which groups are adjacent.
- A route that fails this adjacency check is inconsistent.
The candidate must have the adjacent carbon-bound proton arrangement implied by the splitting; otherwise it cannot explain the observed multiplet.
Look for these in your work
- Neighbour rule used
- Connectivity checked
- Consistency evaluated
Common mistakes
- Using one peak to identify the whole structure
- Treating integration and splitting as the same information
- Ignoring the formula
- Forcing a unique answer when evidence is incomplete
What can you explain now?
Which NMR feature gives relative proton numbers?
Compare with the explanation
Integrated peak areas
The integral measures relative protons in each environment.
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Make it stick
How do signal count, shift, area and splitting work together to test a candidate 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.