A-Level Chemistry: Structure Determination

Structure determination questions are among the most distinctive in A-level Chemistry papers: you are given a molecular formula and a set of data, and you have to put the evidence together into one structure. This material trains that way of thinking with problems in which every piece of data, including peak positions, integration values and splitting, is written out in the question, so you can practise the reasoning without needing a spectrum picture.

It starts with high-resolution mass spectrometry. Using precise atomic masses, you decide which of several formulas with the same nominal mass matches a measured molecular ion, a calculation that relies on four-decimal-place arithmetic. You then turn to nuclear magnetic resonance. In ¹H NMR you use chemical shift values to identify the environment of each set of protons (for example δ 3.7–4.1 for hydrogen on a carbon next to an ester oxygen and δ 9–10 for an aldehyde hydrogen), use integration traces to find the relative numbers of protons in each environment, and apply the n+1 rule to predict and interpret doublets, triplets and quartets. In ¹³C NMR you count carbon environments, using symmetry to explain why a molecule shows fewer peaks than it has carbon atoms. You also explain why tetramethylsilane is used as the standard and why samples are dissolved in deuterated solvents or CCl₄.

The combined problems bring the techniques together with infrared data: a carbonyl absorption near 1700–1750 cm⁻¹, the very broad O–H absorption of a carboxylic acid between 2500 and 3300 cm⁻¹, or the absence of an O–H band. From these you identify compounds such as ethyl ethanoate, propanone and propanoic acid, tell propanal from propanone, and separate isomers that a single technique cannot distinguish.

The final part covers chromatography beyond simple paper and thin-layer work: how column chromatography separates a mixture by the balance between solubility in the moving phase and retention by the stationary phase, how gas chromatography uses retention times compared with standards, and why coupling gas chromatography to a mass spectrometer lets each separated component be identified.

Three formats are offered. The 12-question quiz consists of data-based problems with worked explanations. The flashcards collect the n+1 rule, integration, the role of TMS and deuterated solvents, key chemical shift and infrared ranges, retention time and high-resolution mass spectrometry. The written work has eight longer problems to answer by hand, each with a reference answer showing how the evidence leads to the structure.

Chemical shift and infrared values used here are typical values of the kind given in exam-board data sheets; always use the data sheet provided in your own exam. The content is based on the A-level chemistry content on organic analysis, NMR spectroscopy and chromatography, as set out for example in the AQA A-level specification. It is revision practice written by Zestly, not an exam board resource.

  • Use precise atomic masses and a high-resolution molecular ion peak to choose a molecular formula
  • Use ¹H NMR chemical shifts, integration and the n+1 rule to deduce part structures
  • Predict and interpret the number of peaks in ¹³C NMR spectra, using symmetry
  • Explain the use of TMS as a standard and of deuterated solvents or CCl₄
  • Combine infrared, NMR and molecular formula data to identify an organic compound
  • Explain separation by column and gas chromatography, retention times and GC–MS

Practice material written by Zestly, based on the A-level chemistry content on organic analysis, NMR spectroscopy and chromatography (for example AQA A-level Chemistry 7405, sections 3.3.6, 3.3.15 and 3.3.16). Chemical shift and wavenumber values are typical data-sheet values.

Sample question

Which of the following statements correctly describe the number of peaks expected in the ¹³C NMR spectra of these compounds?

See the answer

Butanone (CH₃COCH₂CH₃) has 4 peaks, Propan-2-ol (CH₃CH(OH)CH₃) has 2 peaks

Butanone has 4 unique carbon environments (CH₃, C=O, CH₂, CH₃). Propan-2-ol has 2 unique carbon environments due to symmetry (the two CH₃ groups are equivalent).

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