3.6.1Organic Chemistry
Mass Spectrometry — Edexcel AS Chemistry Revision Notes & Topical Questions
5 min read 6 parts Lesson 31 of 42
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Part 1 of 6Producing the Spectrum
A sample passes through the instrument in a fixed order of stages.
1
A small vaporised sample is bombarded with a beam of high-energy electrons, which knock one electron out of some molecules.
2
Each molecule that loses an electron becomes a positive molecular ion written M⁺•, carrying one unpaired electron.
3
The positive ions are accelerated by an electric field so that they all travel towards the detector region.
4
Ions are deflected, and those with lower mass-to-charge ratio or higher charge are deflected more strongly.
5
The detector records how many ions arrive for each m/z value, and the results are plotted as a spectrum of abundance against m/z.
Part 2 of 6Reading the Peaks
Each feature of a spectrum tells you something different about the sample.
Molecular ion peak
- Molecular ion peak The peak with the highest m/z (ignoring isotope peaks) gives the relative molecular mass of the compound.
Base peak
- Base peak The tallest peak in the spectrum, showing the most abundant ion, which is given a height of 100%.
m/z value
- m/z value The mass of an ion divided by its charge;
- for ions with a single positive charge it equals the ion's mass.
[M+1] peak
- [M+1] peak A very small peak one unit above M⁺, caused by the natural 1.1% of carbon-13 atoms in the molecule.
Isotope patterns
- Isotope patterns Elements such as chlorine and bromine give pairs of peaks because they exist as two common isotopes.
| Feature | What it shows |
|---|---|
| M⁺ peak | Relative molecular mass |
| [M+1] height | More carbon atoms give a taller peak |
| Peak pairs | Presence of Cl (3 : 1) or Br (1 : 1) |
Part 3 of 6Fragmentation
Molecular ions are unstable and break apart into smaller pieces.
Fragmentation
Positive fragments
- Positive fragments Only the fragment that keeps the positive charge is detected;
- the neutral radical or molecule lost is not seen.
Common losses
- Common losses Peaks at 18, 28 and 44 below M⁺ often show loss of H₂O, CO and CO₂ respectively.
Alkane fragments
- Alkane fragments Breaking C–C bonds gives ions such as CH₃⁺ at 15, C₂H₅⁺ at 29 and C₃H₇⁺ at 43.
Alcohols
- Alcohols A peak at M − 18 shows loss of water, and CH₂OH⁺ at m/z 31 is typical of a primary alcohol.
Part 4 of 6Using Spectra for Structure
Propan-1-ol
- Propan-1-ol M⁺ at 60, with fragments at 59 (loss of H), 42 (loss of H₂O), 31 (CH₂OH⁺) and 29 (C₂H₅⁺).
Halogenoalkanes
- Halogenoalkanes Bromomethane shows M⁺ peaks at 94 and 96 of equal height because ⁷⁹Br and ⁸¹Br are about equally abundant.
Isotope abundance
- Isotope abundance Peak heights give the percentage of each isotope, so the relative atomic mass can be calculated.
Matching to data
- Matching to data Comparing fragment peaks with a table of known fragments lets you propose which groups are present.
Part 5 of 6Worked Example
Question
- A mass spectrum of boron shows 19.9% of ¹⁰B and 80.1% of ¹¹B.
- Calculate the relative atomic mass of boron to 3 significant figures.
- Multiply each mass number by its abundance as a fraction: 10 × 0.199 = 1.99.
- Do the same for the second isotope: 11 × 0.801 = 8.811.
- Add the two contributions1.99 + 8.811 = 10.801.
Answer Relative atomic mass of boron = 10.8
Part 6 of 6Exam Checklist & Tips
3.6.1 – Key Points for the Exam
- Molecular ion
- The ion formed when a molecule loses one electron, M⁺•;
- its m/z gives the relative molecular mass.
- The ion formed when a molecule loses one electron, M⁺•;
- Base peak The most intense peak in a mass spectrum, from the most abundant ion.
- m/z The mass of an ion divided by its charge.
- Fragmentation Breaking of the molecular ion into smaller charged and uncharged pieces.
- [M+1] peak A small peak caused by molecules containing a carbon-13 atom.
Common Mistakes
- Choosing the tallest peak as the molecular ion;
- the base peak is only the most abundant ion, not necessarily the heaviest.
- Treating the [M+1] peak as the molecular ion because it has a higher m/z value.
- Forgetting that only positively charged fragments are detected, so the neutral piece lost has to be worked out by subtraction.
- Writing the fragment without its positive charge, for example CH₃ instead of CH₃⁺.
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