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A Level · Cambridge (CIE) · 9701

Deduce a simple structure from proton NMR evidence

Combine signal count, chemical shifts, relative areas and splitting to evaluate possible molecular structures.

Section 37 · Analytical techniques

Analytical techniques · Cambridge International AS & A Level Chemistry 9701 · 37.4

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?

Put the idea to work

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
  1. The 1H signal is in the aldehyde region.
  2. The 2H group next to CH₃ gives a quartet; the CH₃ next to CH₂ gives a triplet.
  3. 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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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.

1 · Guided

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
  1. The triplet and quartet indicate adjacent CH₃ and CH₂ groups.
  2. The CH₂ is next to oxygen.
  3. 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
2 · Independent

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
  1. The two methyl groups experience different surroundings.
  2. The molecule lacks symmetry exchanging them.
  3. 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
3 · Transfer

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
  1. Splitting suggests neighbouring proton counts.
  2. The candidate determines which groups are adjacent.
  3. 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
Recall without your notes

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.

After trying it yourself, choose your next review. This is your self-assessment.

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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.

Keep learning Chemistry

Course scope and assessment

Four hundred and forty-two original lessons map the published learning outcomes across AS Chemistry Topics 1–22 and A Level Topics 23–37, with guided, independent and transfer practice and explained answers. Every numbered subject-content topic and listed outcome tag now has a mapped lesson sequence. Topic 32 maps all 12 hydroxy-compound outcome tags in 11 lessons; Topic 33 maps all 15 carboxylic-acid and derivative outcome tags in 16 lessons; Topic 34 maps all 22 nitrogen-compound outcome tags in 20 lessons; Topic 35 maps all 12 polymerisation outcome tags in 16 lessons; Topic 36 maps all four organic-synthesis outcome tags in 14 lessons; Topic 37 maps all 22 chromatography and NMR outcome tags in 18 lessons. Cambridge's 2028–2030 Version 1 syllabus states that there are no significant changes which affect teaching from the preceding edition; the routes are aligned to that edition. This is 37 of the 37 numbered subject-content topics. Full cumulative exam assessment, Paper 3 laboratory competence, Paper 5 investigation practice and human subject-teacher review remain outstanding. Written practice is self-checked and does not certify exam readiness.

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