Section 1 of 6Producing the Spectrum
The pattern of peaks acts as a fingerprint for the molecule.
Ionisation
- A high-energy electron removes an electron from a molecule: M(g) + e⁻ → M⁺•(g) + 2e⁻.
Molecular ion
- The ion M⁺• is called the molecular ion, and it still contains all the atoms of the molecule.
Fragmentation
- The molecular ion breaks into smaller pieces, giving a positive fragment ion and a neutral radical.
Detection
- Only positively charged fragments are detected, and the neutral fragments are not seen.
Spectrum axes
- The spectrum plots abundance against the mass-to-charge ratio (m/z), with a charge of 1+ for most ions.
Section 2 of 6Reading a Spectrum
Reading a Spectrum
Molecular ion peak
- The peak with the highest m/z gives the relative molecular mass of the compound.
Base peak
- The tallest peak is the base peak and is the most abundant fragment ion.
Fragment peaks
- Peaks at lower m/z come from fragment ions, and the differences between peaks show which groups were lost.
Common fragments
- m/z 15 is CH₃⁺, 29 is C₂H₅⁺, 43 is C₃H₇⁺ or CH₃CO⁺, and 57 is C₄H₉⁺.
| m/z | Likely fragment |
|---|---|
| 15 | CH₃⁺ |
| 29 | C₂H₅⁺ or CHO⁺ |
| 43 | C₃H₇⁺ or CH₃CO⁺ |
| 45 | COOH⁺ or C₂H₅O⁺ |
| 57 | C₄H₉⁺ |
Section 3 of 6Interpreting Fragmentation
1
Identify the molecular ion peak and read the relative molecular mass of the compound.
2
Work out the mass lost between the molecular ion and each major fragment ion.
3
A difference of 15 corresponds to loss of a CH₃ group, and a difference of 29 to loss of a C₂H₅ group.
4
Match each fragment mass with a possible structure from the data table.
5
Propose a structure consistent with the fragments and with the molecular mass.
Section 4 of 6Practical Considerations
Pure samples
- Mixtures are usually separated first because a combined spectrum would be too hard to interpret.
Tiny samples
- Only a very small amount of sample is needed, which is useful when little is available.
Isotopes
- The relative molecular mass from the M⁺• peak is based on the most abundant isotopes.
Radical not seen
- A neutral radical such as •CH₃ produced in fragmentation is not detected as it carries no charge.
Section 5 of 6Worked Example
Question
- The mass spectrum of propanone, CH₃COCH₃, has peaks at m/z 58, 43 and 15.
- Identify the species responsible for each peak.
- m/z 58 is the molecular ion, since the Mr of C₃H₆O = 3 × 12 + 6 × 1 + 16 = 58.
- The loss from 58 to 43 is 15, which is the mass of a CH₃ radical, giving CH₃CO⁺.
- m/z 43 is therefore [CH₃CO]⁺, which is the base peak.
- m/z 15 is CH₃⁺, from the other C–C bond cleavage.
Answer 58 = [CH₃COCH₃]⁺•, 43 = [CH₃CO]⁺, 15 = [CH₃]⁺
Section 6 of 6Exam Checklist & Tips
4.7.2 – Key Points for the Exam
- Molecular ion The ion formed when a molecule loses one electron, with m/z equal to its relative molecular mass.
- Fragmentation Breaking of the molecular ion into smaller pieces.
- m/z The mass-to-charge ratio of an ion.
- Base peak The most abundant ion in the spectrum, shown as the tallest peak.
Common Mistakes
- Confusing the base peak with the molecular ion peak; the molecular ion is the one with the largest m/z.
- Forgetting to include the positive charge when writing fragment ions.
- Assuming the neutral fragment appears as a peak.
- Reading the relative molecular mass from the tallest peak.