2.10.2Unit 2. Energetics, Group Chemistry, Halogenoalkanes and Alcohols
Interpreting Mass Spectra — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 15 of 53 Unit quiz
Part 1 of 6Reading a Mass Spectrum
Mass spectrometry helps to identify organic compounds from the masses of ions formed in the instrument.
Axes
- The horizontal axis shows the mass-to-charge ratio, m/z, and the vertical axis shows the relative abundance of each ion.
Molecular ion
- The peak with the highest m/z, apart from small isotope peaks, is the molecular ion, and its m/z gives the relative molecular mass.
Formation
- The molecule loses one electron to become M⁺•, a radical cation, and this is the heaviest ion detected.
Base peak
- The tallest peak is the base peak and is assigned 100% abundance, and it may not be the molecular ion.
Part 2 of 6Fragmentation
Fragmentation
Process
- The molecular ion can break up into a positive fragment and a neutral radical, and only the positive fragment is detected.
Equation
- M⁺• → X⁺ + Y•, where the radical is not seen in the spectrum.
Common fragments
- The ions CH₃⁺ at 15, C₂H₅⁺ at 29, C₃H₇⁺ at 43 and C₄H₉⁺ at 57 are frequently seen.
Other fragments
- CH₃CO⁺ at 43 suggests a carbonyl group, and CH₂OH⁺ at 31 suggests a primary alcohol.
Stable ions
- Fragments that form more stable cations, such as secondary or tertiary carbocations, give taller peaks.
| m/z | Likely fragment |
|---|---|
| 15 | CH₃⁺ |
| 29 | C₂H₅⁺ |
| 43 | C₃H₇⁺ or CH₃CO⁺ |
| 57 | C₄H₉⁺ |
Part 3 of 6Working Out a Structure
1
Identify the molecular ion peak and note the relative molecular mass.
2
Subtract the m/z of each major peak from the molecular ion to find the mass lost.
3
Match each lost mass to a likely group, for example a loss of 15 means CH₃ and a loss of 17 means OH.
4
Match each peak to a possible fragment ion and write the structure of the ion.
5
Combine this with other data, such as infrared spectra, to decide on the final structure.
Part 4 of 6Halogen Isotope Patterns
Halogen Isotope Patterns
Chlorine
- Chlorine has isotopes ³⁵Cl and ³⁷Cl in about a 3:1 ratio, so a chloro compound shows peaks at M and M+2 in a 3:1 ratio.
Bromine
- Bromine has ⁷⁹Br and ⁸¹Br in about a 1:1 ratio, so a bromo compound shows peaks at M and M+2 of nearly equal height.
Use
- These patterns show that a halogen is present and which one it is.
Small M+1 peak
- A very small M+1 peak arises from molecules containing carbon-13 and is not a separate fragment.
Part 5 of 6Worked Example
Question
- The mass spectrum of a bromoalkane shows peaks at m/z 108 and 110 with almost equal heights.
- Identify the molecular formula.
- Equal heights at M and M+2 show the presence of one bromine atom, ⁷⁹Br and ⁸¹Br.
- The mass left over for the alkyl group is 108 − 79 = 29.
- A mass of 29 corresponds to C₂H₅, since 2 × 12 + 5 = 29.
Answer
- C₂H₅Br (bromoethane);
- the peaks are C₂H₅⁷⁹Br⁺ at 108 and C₂H₅⁸¹Br⁺ at 110.
Part 6 of 6Exam Checklist & Tips
2.10.2 – Key Points for the Exam
- m/z The mass-to-charge ratio of an ion detected in the mass spectrometer.
- Molecular ion The ion formed when a molecule loses one electron, whose m/z equals the relative molecular mass.
- Fragmentation The breaking up of the molecular ion into smaller pieces.
- Base peak The most abundant ion in the spectrum, given a height of 100%.
- Isotope pattern The set of peaks arising from the different isotopes of an element.
Common Mistakes
- Treating the tallest peak as the molecular ion in every case.
- Saying neutral fragments are detected.
- Forgetting that the charge of the ion is usually 1+, so m/z equals the mass.
- Using M+2 peak heights as evidence of two molecules rather than isotopes.
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