How do we determine the structure of an unknown molecule?
Mass spectrometry, infrared spectroscopy, proton and carbon-13 NMR, and chromatographic separation in structure determination.
A focused answer to WJEC A-Level Chemistry Unit 4, covering mass spectrometry and fragmentation, infrared spectroscopy, proton and carbon-13 NMR, and chromatographic separation in determining the structure of organic molecules.
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What this dot point is asking
WJEC wants you to use mass spectrometry, infrared spectroscopy, proton and carbon-13 NMR, and chromatography together to determine the structure of an unknown organic compound.
The answer
Mass spectrometry
Infrared spectroscopy
NMR
Proton () NMR gives the number of different proton environments (number of peaks), the relative numbers in each (integration), and neighbouring protons (splitting, by the rule). Carbon-13 NMR gives the number of different carbon environments.
Chromatography
Chromatography separates a mixture between a stationary and a mobile phase; components are identified by their retention factor () in TLC or retention time in gas chromatography, often coupled to mass spectrometry (GC-MS) for identification.
A strategy for an unknown
Markers reward a logical order of reasoning. Work from the mass spectrum first to fix the molecular mass and likely formula, then use infrared to identify functional groups, then use NMR to count and place the carbon and hydrogen environments. Each technique narrows the possibilities until only one structure remains. For example, a molecular ion at with a broad infrared band and two proton environments points to ethanol rather than its isomer methoxymethane, which would show only one proton environment.
NMR splitting and the n+1 rule
In high-resolution proton NMR, a peak is split into a number of lines equal to one more than the number of hydrogens on the adjacent carbon (the rule). A next to a appears as a triplet (), and the appears as a quartet (). This pattern reveals which groups are next to each other, which is why the ethyl group gives its characteristic triplet-and-quartet signature.
Examples in context
Drug testing by GC-MS. Forensic and anti-doping labs separate complex mixtures by gas chromatography then identify each component by its mass spectrum, the combined power of the two techniques. MRI from NMR. Medical magnetic resonance imaging is built on the same nuclear magnetic resonance physics used to read proton NMR spectra in the lab.
Try this
Q1. State what the molecular ion peak in a mass spectrum tells you. [1 mark]
- Cue. The relative molecular mass of the compound.
Q2. State the approximate infrared absorption of a bond. [1 mark]
- Cue. Around cm.
Q3. State what the number of peaks in a proton NMR spectrum represents. [1 mark]
- Cue. The number of different proton (hydrogen) environments.
Q4. State the splitting pattern expected for the protons in an ethyl group. [1 mark]
- Cue. A quartet (, from the three hydrogens on the adjacent ).
Q5. A mass spectrum loses from the molecular ion. Suggest the group lost. [1 mark]
- Cue. A methyl group, .
Exam-style practice questions
Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
WJEC 20194 marksThe mass spectrum of a compound shows a molecular ion peak at m/z = 74 and a strong fragment at m/z = 57. Suggest the identity of the fragment lost and what this indicates about the structure. The compound is a carboxylic acid.Show worked answer β
The loss is , which corresponds to the loss of a hydroxyl group (mass ).
Loss of from the molecular ion is characteristic of a carboxylic acid, leaving an acylium ion .
A molecular ion of matches propanoic acid (, ), and the fragment at is .
Markers reward the mass difference of , identifying it as , and linking the fragmentation to the carboxylic acid (propanoic acid).
WJEC 20214 marksAn organic compound has the molecular formula C3H6O. Its infrared spectrum shows a strong absorption at about 1715 cm-1 but no broad absorption around 3200 to 3550 cm-1. Its proton NMR shows a single peak. Deduce the structure and justify your answer.Show worked answer β
The strong absorption near cm-1 is a stretch, so the compound contains a carbonyl group.
The absence of a broad to cm-1 absorption rules out , so it is not an alcohol or carboxylic acid.
For with a carbonyl and a single NMR peak, all six hydrogens must be equivalent, which fits propanone (), where the two methyl groups are identical.
Markers reward the from cm-1, ruling out , the single NMR peak meaning equivalent hydrogens, and the conclusion of propanone.
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Sources & how we know this
- WJEC A-level Chemistry specification β WJEC (2015)