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Identifying an unknown compound
Combining percentage composition, mass spectrometry, infrared, carbon-13 NMR and proton NMR, plus chemical tests, to work out the structure of an unknown organic compound, with a full worked example.
A drug-checking service at a music festival receives a pill that the owner bought as one substance. The chemists' answer may decide whether someone ends up in hospital, so it can't rest on a single test. No one technique gives a whole structure, but together they leave only one possibility. That is exactly the kind of reasoning the HSC asks for: several pieces of evidence, each ruling out options, until only one structure fits them all.
A strategy
| Evidence | What it tells you |
|---|---|
| Percentage composition | empirical formula |
| Mass spectrum | molar mass from M⁺ (so the molecular formula), Cl/Br from M + 2, and fragments |
| Infrared spectrum | functional groups: O–H, C=O, N–H, C=C |
| ¹³C NMR | number of carbon environments and their types |
| ¹H NMR | number of H environments, their types, how many H in each, and neighbours |
| Chemical tests | quick confirmation: fizzing with carbonate, decolourising bromine water, dichromate colour change |
Once you know the molecular formula, it helps to work out the degree of unsaturation: each ring or double bond removes two hydrogens from the saturated formula CnH2n+2. For C₃H₆O₂, the saturated compound would have 8 H; it has 6, so there is one double bond (or ring). If the IR shows C=O, that accounts for it.
A worked example
Compound X contains 48.6% C, 8.1% H and 43.2% O by mass. Its spectra are shown above. Identify X.
Tips for exam questions
- Use every piece of data, and say what each shows. Markers award marks for linking each piece of evidence to a feature of the structure, not just for the final name.
- Quote values with their source. "1715 cm⁻¹, which lies in the C=O range of 1680–1750 cm⁻¹ on the data sheet."
- Draw the structure and label which atoms give which NMR signal.
- Check against isomers. Explain briefly why the most likely alternatives don't fit.
- Look for patterns: a 3H triplet + 2H quartet means an ethyl group; a 6H doublet + 1H septet means an isopropyl group; a 3H singlet near 2.1 ppm means CH₃ next to C=O; a 3H singlet near 3.3–3.7 ppm means CH₃ on oxygen.
- Solve five unknowns. For each, write down which spectrum was decisive before you choose.
- When the formula is C₃H₈O, decide between propan-1-ol, propan-2-ol and methoxyethane. Which spectrum separates the two alcohols most quickly?
- When the formula is C₃H₆O₂, look at the ¹H NMR first. How does it separate the acid from the two esters?
- Pick a wrong answer on purpose and read the feedback. Use it to explain which feature of the data rules that isomer out.
Test yourself
Which technique gives the molar mass of an unknown compound most directly?
The molecular ion peak gives the relative molecular mass.
A compound with formula C₄H₈O has a strong IR absorption at 1715 cm⁻¹, no absorption near 2720 cm⁻¹, and a 3H singlet at 2.1 ppm in its ¹H NMR spectrum along with a 2H quartet and a 3H triplet. It is
C=O with no aldehyde H means a ketone. The CH₃ singlet next to C=O plus an ethyl group gives CH₃COCH₂CH₃.
A compound is 85.6% carbon and 14.4% hydrogen by mass and its molecular ion is at m/z 56. Its molecular formula is
C 7.13, H 14.3, so CH₂ (mass 14); 56 ÷ 14 = 4, so C₄H₈.
The ¹H NMR spectrum of an ester C₃H₆O₂ shows just two singlets, each 3H. The ester is
CH₃COOCH₃: neither CH₃ has an H on the adjacent atom, so both are singlets.
Name what each technique tells you.
Which technique gives each piece of evidence?
Percentage composition gives the formula. The molecular ion peak gives the . A 3H triplet and a 2H quartet in a ¹H spectrum show an group. A singlet near 12 ppm suggests a .
An unknown compound Y contains 66.7% C, 11.1% H and 22.2% O. Its mass spectrum has a molecular ion at m/z 72 and a base peak at m/z 43. The IR spectrum has a strong absorption at 1715 cm⁻¹ and no broad absorption above 3000 cm⁻¹. The ¹³C NMR spectrum has four signals, one at 209 ppm. The ¹H NMR spectrum shows a 3H singlet at 2.1 ppm, a 2H quartet at 2.4 ppm and a 3H triplet at 1.0 ppm. Determine the structure of Y, explaining how each piece of data supports your answer.
Formula. C $66.7 \div 12.01 = 5.55$; H $11.1 \div 1.008 = 11.0$; O $22.2 \div 16.00 = 1.39$. Dividing by 1.39 gives C 4 : H 8 : O 1, so the empirical formula is C₄H₈O (mass 72). M⁺ = 72, so the molecular formula is also C₄H₈O, with one degree of unsaturation.
IR. The strong absorption at 1715 cm⁻¹ is in the C=O range (1680–1750 cm⁻¹). There is no broad O–H, so Y is not an alcohol or acid: it is an aldehyde or ketone. The C=O accounts for the unsaturation.
¹³C NMR. Four signals means four different carbons, so no symmetry. The 209 ppm signal is a ketone or aldehyde C=O.
¹H NMR. There is no signal at 9–10 ppm, so there is no aldehyde H: Y is a ketone. The 3H singlet at 2.1 ppm is a CH₃ bonded to the C=O carbon (no neighbouring H). The 2H quartet (2.4 ppm, next to C=O) and the 3H triplet (1.0 ppm) form an ethyl group. The ratio 3 : 2 : 3 accounts for all 8 H.
Mass spectrum. The base peak at 43 is 72 − 29: loss of C₂H₅ leaves CH₃CO⁺, which fits a CH₃–C=O group.
Structure: Y is butanone, CH₃COCH₂CH₃. Butanal is ruled out by the lack of an aldehyde H.