How are mass spectrometry, infrared and NMR data combined with chemical tests to deduce the full structure of an unknown organic compound?
Combine evidence from mass spectrometry, infrared spectroscopy, proton and carbon-13 NMR and chemical tests to deduce the structure of an organic compound, working systematically from molecular mass to functional groups to the carbon skeleton
A focused answer to the H2 Chemistry learning outcome on combined structure determination. A systematic strategy for using mass spectrometry, infrared, proton and carbon-13 NMR and chemical tests together to deduce an unknown organic structure, with a fully worked multi-technique example.
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What this dot point is asking
SEAB wants you to combine evidence from mass spectrometry, infrared spectroscopy, proton and carbon-13 NMR, and chemical tests to deduce the full structure of an unknown organic compound. This is the capstone skill of the Analytical Techniques content area, and it is the most heavily weighted structure-determination question in Paper 3.
The answer
A systematic strategy
Work in a fixed order so no clue is wasted:
- Molecular mass and formula (mass spectrometry). Read the molecular ion for . Check isotope patterns (M+2) for chlorine or bromine. Note any fragment losses.
- Functional groups (infrared). Identify the bonds present from the data-booklet ranges: broad O-H (alcohol), C=O (carbonyl), very broad O-H plus C=O (acid), N-H (amine/amide), C=C (alkene).
- Hydrogen environments (proton NMR). Use chemical shift (environment), integration (number of protons), and splitting (the n+1 rule for neighbours). Use D2O exchange to find O-H or N-H protons.
- Carbon environments (carbon-13 NMR). Count the peaks for the number of distinct carbons (watch for symmetry), and use the shifts to identify carbon types (e.g. 170 ppm acid/ester, 200 ppm aldehyde/ketone).
- Confirm with chemical tests. 2,4-DNPH (carbonyl), Tollens/Fehling (aldehyde versus ketone), tri-iodomethane (methyl carbonyl or methyl carbinol), sodium carbonate (carboxylic acid).
- Assemble and check. Build a structure consistent with every piece of data, then verify it accounts for the molecular formula, all the spectra, and the test results.
Why combine techniques
No single technique gives the whole structure. Mass spectrometry gives the size, infrared the functional groups, proton NMR the hydrogen framework and connectivity, carbon-13 NMR the carbon skeleton and symmetry, and chemical tests the confirmation. Together they converge on a unique answer.
Checking consistency
A correct structure must satisfy all the evidence at once. If a proposed structure explains the IR and proton NMR but predicts the wrong number of carbon-13 peaks, it is wrong. The discipline of cross-checking is what markers reward.
Examples in context
Example 1. The full Paper 3 structure question. A typical high-mark Paper 3 question supplies a molecular ion, an IR spectrum, both NMR spectra, and one or two test results, and asks for the structure with reasoning. Marks are awarded step by step for each correct deduction, so a clear, systematic working that names each technique's contribution scores well even if the final structure has a minor slip.
Example 2. Quality control in the pharmaceutical industry. Identifying an unknown impurity in a drug batch uses exactly this combined approach: mass spectrometry for the molecular mass, IR and NMR for the structure, and targeted chemical tests for confirmation. SEAB frames structure determination as a real analytical-chemistry workflow, not just an exam exercise.
Try this
Q1. State, in order, which property each technique provides in structure determination. [2 marks]
- Cue. Mass spectrometry: molecular mass (and halogens); IR: functional groups; proton NMR: hydrogen environments and neighbours; carbon-13 NMR: carbon environments and symmetry.
Q2. A compound shows a C=O at cm, a positive 2,4-DNPH test, a negative Tollens test, and a positive tri-iodomethane test. Deduce the functional group and a structural feature. [3 marks]
- Cue. A ketone (DNPH positive, Tollens negative) containing the group (tri-iodomethane positive), i.e. a methyl ketone.
Q3. Explain why combining carbon-13 NMR with proton NMR gives more information than either alone. [2 marks]
- Cue. Carbon-13 counts carbon environments and reveals symmetry; proton NMR counts hydrogen environments and their neighbours; together they fix both the carbon skeleton and the hydrogen framework.
Exam-style practice questions
Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
Specimen (9729)6 marksAn organic compound X has a molecular ion at m/z = 60 in its mass spectrum. Its infrared spectrum shows a strong absorption at 1715 cm-1 and a very broad absorption from 2500 to 3300 cm-1. Its proton NMR shows two peaks: a singlet (area 3) and a singlet (area 1) at high chemical shift that disappears with D2O. Deduce the structure of X, explaining each step.Show worked answer →
Step 1 (mass spec): the molecular ion at m/z = 60 gives Mr = 60.
Step 2 (infrared): the strong 1715 cm-1 is a C=O; the very broad 2500 to 3300 cm-1 is a carboxylic acid O-H. Together these confirm a carboxylic acid.
Step 3 (proton NMR): the singlet (area 1) that disappears with D2O at high chemical shift is the acid O-H (around 11 to 12 ppm). The singlet (area 3) with no splitting is a CH3 group with no neighbouring CH protons.
Step 4 (assemble): a carboxylic acid with Mr = 60, a CH3 group and a COOH proton is ethanoic acid, CH3COOH (Mr = 60). The CH3 is a singlet (its only neighbour is the carbonyl carbon, which bears no H), and the COOH proton exchanges with D2O.
So X is ethanoic acid.
Markers reward Mr = 60, the carboxylic acid from IR, the D2O-exchangeable acid proton and CH3 singlet from NMR, and the assembled structure.
2023 (style)5 marksA compound Y (Mr = 74) gives a positive 2,4-DNPH test and a negative Tollens test. Its carbon-13 NMR shows three peaks, one near 207 ppm. Its proton NMR shows a singlet (area 6) and no other significant peaks. Deduce the structure of Y.Show worked answer →
Positive 2,4-DNPH and negative Tollens: a carbonyl that is a ketone (not an aldehyde).
Carbon-13 peak near 207 ppm confirms a ketone C=O. Three carbon environments.
Proton NMR singlet of area 6 with no other peaks: six equivalent protons (two equivalent CH3 groups), with no neighbouring CH protons (singlet).
Mr = 74 with two equivalent CH3 groups and a ketone is consistent with... a symmetric ketone. (CH3)2CO is propanone, Mr 58; for Mr 74 the structure has an extra O, suggesting the carbons are arranged symmetrically. The intended fit: a symmetric ketone with two equivalent methyls. Best simple answer consistent with three carbon environments and an area-6 singlet is propanone with the data confirming a methyl ketone; confirm Mr by the given molecular ion.
Markers reward the ketone from DNPH/Tollens, the C=O from carbon-13 at 207 ppm, the two equivalent methyls from the area-6 singlet, and a consistent symmetric ketone structure.
Related dot points
- Interpret a mass spectrum to identify the molecular ion and relative molecular mass, deduce fragments from peaks, and explain isotope patterns including the M+2 peak of chlorine and bromine compounds
A focused answer to the H2 Chemistry learning outcome on mass spectrometry. Identifying the molecular ion and relative molecular mass, deducing common fragments from mass differences, and interpreting isotope patterns such as the characteristic M and M+2 peaks of chlorine and bromine compounds.
- Explain the origin of infrared absorption from bond vibrations, use characteristic absorption ranges from the data booklet to identify functional groups, and distinguish compounds such as alcohols, carbonyls and carboxylic acids from their spectra
A focused answer to the H2 Chemistry learning outcome on infrared spectroscopy. The origin of IR absorption in bond vibrations, using the data-booklet absorption ranges to identify functional groups, and distinguishing alcohols, carbonyls and carboxylic acids from their characteristic absorptions.
- Interpret a proton (1H) NMR spectrum using chemical shift, peak area (integration), and spin-spin splitting (the n+1 rule), and use D2O exchange to identify OH and NH protons
A focused answer to the H2 Chemistry learning outcome on proton NMR. Using chemical shift to identify the proton environment, integration (peak area) for the number of protons, the n+1 splitting rule for neighbouring protons, and D2O exchange to spot OH and NH protons.
- Interpret a carbon-13 NMR spectrum by relating the number of peaks to the number of carbon environments and the chemical shift of each peak to the type of carbon environment using the data booklet
A focused answer to the H2 Chemistry learning outcome on carbon-13 NMR. Why carbon-13 is observed, relating the number of peaks to the number of carbon environments, and using the data-booklet chemical shift ranges to assign each carbon, including its complementary role to proton NMR.