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Weigh the pieces.

Six original Cambridge-style questions on the mass spectra of organic compounds.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

M, fragments, M+1, M+2.

01
Calculation
[2 marks]

In the mass spectrum of a hydrocarbon the M+ peak at m/z 86 has a relative abundance of 20.0 and the M+1 peak 1.32. Calculate the number of carbon atoms and give the molecular formula.

  • n = 100 × 1.32 / (1.1 × 20.0) = 132 / 22.0 = 6.00, so 6 carbon atoms. ✓
  • 6 × 12 = 72; 86 − 72 = 14 H, so C6H14. ✓
02
Recall
[2 marks]

Write an equation for the formation of the molecular ion of propanone, and an equation for its fragmentation to give the peak at m/z 43.

  • CH3COCH3 + e− → [CH3COCH3]+ + 2e− (m/z 58). ✓
  • [CH3COCH3]+ → CH3CO+ + CH3•; CH3CO+ is detected at m/z 43. ✓
03
Analysis
[3 marks]

A compound contains carbon, hydrogen and one halogen. Its spectrum shows molecular ion peaks at m/z 78 and 80 with heights in the ratio 3 : 1, and a large peak at 43. Identify the halogen and suggest a structure.

  • 3 : 1 pair two units apart: one chlorine atom (35Cl and 37Cl). ✓
  • 78 − 35 = 43, so the rest of the molecule is C3H7; the peak at 43 is C3H7+. ✓
  • C3H7Cl: 1-chloropropane or 2-chloropropane. ✓
04
Calculation
[2 marks]

The M+ peak of an ester has height 24.0 and the M+1 peak 2.11. Mr = 144. Find the number of carbon atoms and suggest a molecular formula with two oxygen atoms.

  • n = 100 × 2.11 / (1.1 × 24.0) = 211 / 26.4 = 7.99, so 8 carbon atoms. ✓
  • 8 × 12 + 2 × 16 = 128; 144 − 128 = 16 H, so C8H16O2. ✓
05
Explain
[2 marks]

Explain why the mass spectrum of 1-bromopropane shows two peaks of almost equal height at m/z 122 and 124.

  • Bromine exists as 79Br and 81Br in roughly equal amounts. ✓
  • C3H779Br+ has m/z 43 + 79 = 122 and C3H781Br+ has 43 + 81 = 124, so the two molecular ions are equally common. ✓
06
Analysis
[3 marks]

Compound Q, C3H8O, is an alcohol. Its mass spectrum has M+ at 60 and a large peak at m/z 31 but only a small peak at 45. Suggest the structure of Q and explain the peak at 31.

  • The alcohol is propan-1-ol or propan-2-ol. ✓
  • m/z 31 is CH2OH+, formed when the C1–C2 bond breaks; this needs an OH on a CH2 at the end of the chain. ✓
  • So Q is propan-1-ol, CH3CH2CH2OH. Propan-2-ol would give a large peak at 45, CH3CHOH+, from loss of CH3. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.