Welcome to Mass Spectrometry for A2 Chemistry!
Welcome to one of the most powerful analytical toolkits in organic chemistry. Imagine you are given an unlabelled vial containing a clear liquid. How do you find out what it is? You can't just look at individual molecules with a microscope—they are far too small! Instead, chemists use mass spectrometry.
Think of a mass spectrometer as an ultra-precise molecular weighing scale that also smashes molecules into recognizable pieces, like dropping a Lego model onto a table to see which blocks come apart. By looking at the weights of the whole molecule and its broken fragments, you can piece together the exact identity and structure of the compound.
Don't worry if mass spectra look like a confusing forest of lines at first glance. By following a few simple rules, you will be decoding them with confidence in no time!
1. The Molecular Ion Peak: Finding the Relative Molecular Mass
When an organic sample is introduced into a mass spectrometer, it is vaporised and bombarded with high-energy electrons. This knocks an electron off the molecule to form a positive ion known as the molecular ion (or parent ion), written as \([\text{M}]^{+\bullet}\) or simply \(\text{M}^+\).
\(\text{M(g)} + \text{e}^- \rightarrow [\text{M}]^{+\bullet}\text{(g)} + 2\text{e}^-\)
Because an electron has a negligible mass, the mass of this positive ion is essentially identical to the relative molecular mass (\(M_r\)) of the original molecule.
Key Features of the Molecular Ion Peak
• Location on the spectrum: The molecular ion peak is usually the peak with the highest mass-to-charge ratio (\(m/z\)) value, located towards the far right-hand side of the spectrum (ignoring small isotope peaks).
• Value: The \(m/z\) value of the \(\text{M}^+\) peak gives the relative molecular mass (\(M_r\)) of the compound directly, since the charge \(z\) is almost always \(+1\).
• The \((M+1)\) peak: You might notice a tiny peak exactly one mass unit to the right of the molecular ion peak. This is due to the natural abundance of the carbon-13 isotope (\(^{13}\text{C}\)), which makes up roughly \(1.1\%\) of all carbon atoms in nature.
Quick Key Takeaway: Look at the furthest right major peak—its \(m/z\) value gives you the compound's relative molecular mass (\(M_r\)).
2. High-Resolution Mass Spectrometry
In standard (low-resolution) mass spectrometry, masses are measured to the nearest whole integer (nominal mass). However, modern high-resolution mass spectrometry measures \(m/z\) values to four or five decimal places.
Why is High Resolution Necessary?
Different molecular formulas can share the exact same integer mass, but because individual isotopes have precise atomic masses that deviate slightly from whole numbers (due to nuclear binding energy), their accurate masses are distinct.
Example: Consider three compounds that all have a nominal \(M_r\) of \(60\):
• Propan-1-ol (\(\text{C}_3\text{H}_8\text{O}\)): Accurate mass = \(3(12.0000) + 8(1.0078) + 1(15.9949) = 60.0573\)
• Ethanoic acid (\(\text{C}_2\text{H}_4\text{O}_2\)): Accurate mass = \(2(12.0000) + 4(1.0078) + 2(15.9949) = 60.0210\)
• Methylethylamine (\(\text{C}_3\text{H}_9\text{N}\)): Accurate mass = \(3(12.0000) + 9(1.0078) + 1(14.0031) = 60.0733\)
A standard low-resolution spectrometer would show a peak at \(m/z = 60\) for all three. A high-resolution instrument allows you to determine the unique, unambiguous molecular formula instantly!
Quick Key Takeaway: High-resolution mass spectrometry measures masses to several decimal places, allowing chemists to distinguish between compounds with identical integer masses and find the exact molecular formula.
3. Identifying Halogens: Isotopic Patterns
Halogenated organic compounds produce very distinctive patterns in their molecular ion region because chlorine and bromine exist naturally as multiple abundant isotopes.
A. Compounds Containing Chlorine
Chlorine naturally exists as two main isotopes: \(^{35}\text{Cl}\) (approx. \(75\%\) abundance) and \(^{37}\text{Cl}\) (approx. \(25\%\) abundance). This gives an isotopic ratio of approximately \(3:1\).
• One chlorine atom present: Produces two molecular ion peaks separated by \(2\) mass units: an \(\text{M}^+\) peak (containing \(^{35}\text{Cl}\)) and an \([\text{M}+2]^+\) peak (containing \(^{37}\text{Cl}\)) in a peak height ratio of \(3:1\).
• Two chlorine atoms present: Produces three peaks: \(\text{M}^+\), \([\text{M}+2]^+\), and \([\text{M}+4]^+\) in a height ratio of \(9:6:1\) (derived from expanding \((3 + 1)^2 = 9 + 6 + 1\)).
B. Compounds Containing Bromine
Bromine naturally exists as two isotopes in almost equal amounts: \(^{79}\text{Br}\) (approx. \(50.7\%\)) and \(^{81}\text{Br}\) (approx. \(49.3\%\)), giving a ratio of approximately \(1:1\).
• One bromine atom present: Produces an \(\text{M}^+\) peak and an \([\text{M}+2]^+\) peak of virtually equal height (\(1:1\)).
• Two bromine atoms present: Produces three peaks: \(\text{M}^+\), \([\text{M}+2]^+\), and \([\text{M}+4]^+\) in a height ratio of \(1:2:1\) (derived from expanding \((1 + 1)^2 = 1 + 2 + 1\)).
Summary of Halogen Peak Ratios
• \(1 \times \text{Cl} \implies \text{M} : [\text{M}+2] = 3:1\)
• \(2 \times \text{Cl} \implies \text{M} : [\text{M}+2] : [\text{M}+4] = 9:6:1\)
• \(1 \times \text{Br} \implies \text{M} : [\text{M}+2] = 1:1\)
• \(2 \times \text{Br} \implies \text{M} : [\text{M}+2] : [\text{M}+4] = 1:2:1\)
Quick Key Takeaway: Twin peaks of \(3:1\) ratio mean one chlorine atom is present; twin peaks of equal height (\(1:1\)) mean one bromine atom is present.
4. Fragmentation Patterns
The molecular ion \([\text{M}]^{+\bullet}\) is often unstable because it has had an electron stripped away and has excess vibrational energy. As a result, it breaks apart into smaller pieces in a process called fragmentation.
How Fragmentation Works
A molecular ion generally fragments into two parts: a positive ion and an uncharged neutral radical:
\([\text{M}]^{+\bullet} \rightarrow \text{X}^+ + \text{Y}^\bullet\)
Crucial Rule: The mass spectrometer can ONLY detect charged particles. The positive ion \(\text{X}^+\) will reach the detector and produce a peak on the spectrum. The neutral radical \(\text{Y}^\bullet\) is undetected and invisible!
The Base Peak
The tall peak with the greatest intensity on the spectrum is called the base peak. It represents the most stable and most abundant positive ion formed during fragmentation. The base peak is assigned an arbitrary relative abundance of \(100\%\), and all other peaks are scaled relative to it.
Common Stable Fragments and Their \(m/z\) Values
Memorising common fragment masses makes analysing spectra much quicker:
• \(m/z = 15 \implies [\text{CH}_3]^+\) (methyl cation)
• \(m/z = 17 \implies [\text{OH}]^+\) (hydroxyl cation)
• \(m/z = 29 \implies [\text{C}_2\text{H}_5]^+\) (ethyl cation) or \([\text{CHO}]^+\) (aldehyde fragment)
• \(m/z = 31 \implies [\text{CH}_2\text{OH}]^+\) (characteristic of primary alcohols)
• \(m/z = 43 \implies [\text{C}_3\text{H}_7]^+\) (propyl cation) or \([\text{CH}_3\text{CO}]^+\) (acylium / ethanoyl cation)
• \(m/z = 45 \implies [\text{COOH}]^+\) (carboxyl cation, found in carboxylic acids)
• \(m/z = 57 \implies [\text{C}_4\text{H}_9]^+\) (butyl cation)
• \(m/z = 77 \implies [\text{C}_6\text{H}_5]^+\) (phenyl cation, showing an aromatic benzene ring)
Common Neutral Losses
Sometimes it is easier to look at the difference between the molecular ion peak and a fragment peak:
• Loss of \(15\) \(\implies\) loss of a \({}^\bullet\text{CH}_3\) radical
• Loss of \(17\) \(\implies\) loss of an \({}^\bullet\text{OH}\) radical
• Loss of \(18\) \(\implies\) loss of a neutral \(\text{H}_2\text{O}\) molecule (common in alcohols)
• Loss of \(28\) \(\implies\) loss of \(\text{CO}\) or \(\text{C}_2\text{H}_4\)
• Loss of \(29\) \(\implies\) loss of a \({}^\bullet\text{C}_2\text{H}_5\) or \({}^\bullet\text{CHO}\) radical
Quick Key Takeaway: Mass spectra show positive fragment ions, NOT neutral radicals. Look at both the fragment peak values and the differences (mass losses) from the molecular ion peak.
5. Step-by-Step Guide to Solving Mass Spectra Problems
When presented with an unknown mass spectrum in an exam, follow these systematic steps:
Step 1: Check for Halogens
Look at the molecular ion region. Do you see a pair of peaks separated by \(2\) units with a \(3:1\) ratio (\(\text{Cl}\)) or a \(1:1\) ratio (\(\text{Br}\))? If yes, factor the halogen into your formula.
Step 2: Identify the Molecular Ion Peak (\(M_r\))
Find the highest non-isotope \(m/z\) peak to determine the relative molecular mass of the molecule.
Step 3: Identify Key Fragment Peaks
List the major peaks (especially the base peak) and identify the formula of the corresponding positive ion (e.g. \(m/z = 43 \implies [\text{CH}_3\text{CO}]^+\)). Remember to always include the positive charge!
Step 4: Check Mass Losses
Calculate the difference between the molecular ion and major fragment peaks to see what neutral species were lost.
Step 5: Distinguish Isomers
Use fragmentation pathways to distinguish between structural isomers.
Worked Example: Propanal vs. Propanone (both \(\text{C}_3\text{H}_6\text{O}\), \(M_r = 58\))
• Propanone (\(\text{CH}_3\text{COCH}_3\)): Cleavage of a \(\text{C}-\text{C}\) bond releases a \({}^\bullet\text{CH}_3\) radical (mass \(15\)), leaving behind an acylium ion \([\text{CH}_3\text{CO}]^+\) at \(m/z = 43\).
• Propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)): Cleavage releases a \({}^\bullet\text{CHO}\) radical (mass \(29\)) leaving an ethyl ion \([\text{C}_2\text{H}_5]^+\) at \(m/z = 29\), or releases a \({}^\bullet\text{C}_2\text{H}_5\) radical leaving \([\text{CHO}]^+\) at \(m/z = 29\).
• A strong peak at \(m/z = 43\) confirms propanone, whereas a dominant peak at \(m/z = 29\) confirms propanal.
6. Common Mistakes to Avoid
• Forgetting the positive charge: When asked to write the formula for a fragment ion responsible for a peak, writing \(\text{CH}_3\) or \(\text{C}_2\text{H}_5\) will lose marks. You MUST write \([\text{CH}_3]^+\) or \([\text{C}_2\text{H}_5]^+\).
• Confusing the base peak and the molecular ion peak: The base peak is the tallest peak (most abundant); the molecular ion peak is the peak furthest to the right (highest mass, excluding isotope peaks).
• Assigning neutral radicals to peaks: A peak at \(m/z = 15\) is caused by \([\text{CH}_3]^+\), never by the loss of \({}^\bullet\text{CH}_3\).
• Mixing up chlorine and bromine ratios: Remember that \(3:1\) is for chlorine (\(^{35}\text{Cl} : ^{37}\text{Cl}\)) and \(1:1\) is for bromine (\(^{79}\text{Br} : ^{81}\text{Br}\)).
Quick Review Summary
• Molecular Ion (\(\text{M}^+\)): Furthest right major peak; its \(m/z\) equals the \(M_r\).
• High Resolution: Gives accurate mass to multiple decimal places; identifies the exact molecular formula.
• Chlorine: Produces \(\text{M}^+\) and \([\text{M}+2]^+\) in a \(3:1\) ratio.
• Bromine: Produces \(\text{M}^+\) and \([\text{M}+2]^+\) in a \(1:1\) ratio.
• Fragmentation: \([\text{M}]^{+\bullet} \rightarrow \text{Ion}^+ + \text{Radical}^\bullet\). Only positive ions are detected.
• Base Peak: The most abundant peak (height = \(100\%\)).