The big idea: To work out the structure of an unknown organic compound, chemists combine three instruments. No single technique gives the whole answer, but together they pin down one structure.
| Technique | The question it answers |
|---|---|
| Mass spectrometry (MS) | how heavy is the molecule, and what breaks off? |
| Infrared (IR) | which functional group is present? |
| ¹H NMR | how many different hydrogen environments are there? |
The right-most peak (m/z = 46) is the molecular ion M⁺, so Mr = 46. The gap down to 31 is a loss of 15 (a CH₃ group); the peak at 31 is CH₂OH⁺.
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Key terms
- Molecular ion, M⁺ — the peak at the highest m/z; its value is the relative molecular mass, Mr.
- Fragment — a smaller piece left after a bond breaks; the mass lost (M⁺ − fragment) tells you which group came off.
- Functional group — the reactive part of the molecule that IR detects.
- Hydrogen environment — a set of H atoms in the same chemical position; each one gives one ¹H NMR peak.
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In a mass spectrometer the molecule is ionised and broken into pieces.
Reading the peaks
- The peak at the highest m/z is the molecular ion M⁺.
- Its value is the relative molecular mass, Mr.
- The other peaks are fragments.
- The mass lost between two peaks tells you which group broke off.
Common fragment losses: - Loss of 15 → a CH3 (methyl) group. - Loss of 17 → an OH group. - Loss of 29 → CHO (or C2H5). - A fragment at m/z = 43 is often CH3CO⁺ (a methyl ketone or ethanoyl group).
M⁺ = 58 gives Mr. The base peak at 43 is M⁺ minus 15 (loss of a CH₃ group), leaving the CH₃CO⁺ fragment — a tell-tale of a methyl ketone.
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Worked example — reading a mass spectrum
An unknown compound shows its highest-mass peak at m/z = 46 and a strong peak at m/z = 31. Deduce Mr and the group lost.
Solution
- Mr: the highest m/z peak is the molecular ion M⁺, so Mr = 46.
- Mass lost: 46 − 31 = 15, which is a CH3 group.
- So the molecule contains a CH3 group and the fragment at 31 (CH2OH⁺) suggests an –OH — consistent with ethanol, CH3CH2OH.
Final answer
Mr = 46; a CH3 group (mass 15) is lost.
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Infrared (IR) light makes bonds vibrate. Each bond absorbs at a characteristic wavenumber (cm⁻¹), so a peak in the IR spectrum identifies a functional group. You don't memorise the numbers — they are given in the data booklet.
| Bond | Found in | Wavenumber / cm⁻¹ | Appearance |
|---|---|---|---|
| O–H | alcohols | 3200–3600 | broad, strong |
| O–H | carboxylic acids | 2500–3000 | very broad |
| C=O | aldehydes, ketones, acids, esters | 1700–1750 | strong, sharp |
| C–O | alcohols, esters | 1000–1300 | strong |
| C=C | alkenes | 1620–1680 | medium |
| N–H | amines | 3300–3500 | medium |
The two IR peaks examiners love: - A broad peak at 3200–3600 cm⁻¹ → an O–H group (an alcohol). - A strong, sharp peak near 1700 cm⁻¹ → a C=O group (an aldehyde, ketone, acid or ester).
See both together (with the very broad 2500–3000 O–H) → a carboxylic acid.
¹H NMR counts hydrogen environments. Hydrogens in the same chemical position are equivalent and give one peak; the number of peaks = number of different H environments. (Reading the shift values is Higher Level; at SL you count the environments.)
| Molecule | Different H environments | Number of ¹H NMR peaks |
|---|---|---|
| CH3CH3 (ethane) | all 6 H are equivalent | 1 |
| CH3CH2OH (ethanol) | CH3, CH2, OH | 3 |
| CH3OCH3 (methoxymethane) | all 6 H equivalent | 1 |
| CH3COCH3 (propanone) | both CH3 equivalent | 1 |
Counting environments: To count environments, ask: are these H atoms in exactly the same position? In ethanol, CH3CH2OH, the three sets (CH3, CH2, OH) are all different → 3 peaks. By symmetry, the two CH3 groups in propanone are identical → just 1 peak.
How this is tested: Combined-spectra deduce questions are a signature Paper 2 style (and the integrated IR/¹H NMR/MS interpretation is flagged as new-syllabus content).
You are handed an MS, an IR and a ¹H NMR result and asked to deduce the structure. The markers want you to use each technique, then put them together.
Use every technique
MS
Mr from M⁺.
IR
The functional group. The data booklet supplies the IR table.
¹H NMR
The number of environments.
Combine
Put them together to deduce the structure.
MS → IR → ¹H NMR → combine
Common trap: Don't quote only one technique. A C=O peak alone cannot distinguish an aldehyde from a ketone — you must bring in the ¹H NMR environments and the MS fragments to decide.
IB-style question — deduce the structure (a)
Compound X has molecular formula C2H6O. Its mass spectrum shows M⁺ at m/z = 46 and a strong peak at m/z = 31. Its IR spectrum has a broad absorption at 3350 cm⁻¹. Its ¹H NMR spectrum shows three peaks. (a) Deduce the structure of X, using all three spectra. [3]
How to score the marks
- MS: M⁺ = 46 gives Mr = 46 (matches C2H6O); the loss of 15 (46 → 31) is a CH3 group, with the fragment at 31 being CH2OH⁺.
- IR: the broad peak at 3350 cm⁻¹ is an O–H group → X is an alcohol (not the ether).
- ¹H NMR: three peaks means three H environments — the CH3, CH2 and OH of ethanol. (The isomer CH3OCH3 would give only one peak and no O–H.)
Final answer
X is ethanol, CH3CH2OH: Mr 46 (MS), O–H alcohol (IR), three H environments (¹H NMR).
IB-style question — distinguish the isomers (b)
(b) Compound Y is an isomer of X with the same formula C2H6O but only one peak in its ¹H NMR and no broad IR absorption above 3000 cm⁻¹. Identify Y and explain how the spectra distinguish it from X. [2]
How to score the marks
- Identify: Y is methoxymethane (dimethyl ether), CH3OCH3 — the other C2H6O isomer.
- Explain: Y has no O–H, so there is no broad IR peak above 3000 cm⁻¹; and its six H atoms are all equivalent, giving just one ¹H NMR peak (whereas ethanol shows three).
Final answer
Y = CH3OCH3: no O–H (IR) and one H environment (¹H NMR), unlike ethanol's O–H and three environments.