Difficulty: Intermediate | Prerequisites: Chapters 1–13 (bonding, functional groups), basic understanding of isotopes
Mass spectrometry gives you two things that IR cannot: the molecular weight of a compound and clues about its carbon skeleton from the way the molecule breaks apart. This section of Chapter 14 teaches you to read a mass spectrum, recognise isotope patterns, predict fragmentation, and calculate degree of unsaturation from a molecular formula. You should be comfortable with molecular formulae, functional groups, and bond-breaking before starting.
Mass spectrometry bombards a molecule with high-energy electrons, knocks one off to form a radical cation (the molecular ion), and records the masses of the resulting fragments. The molecular ion gives the molecular weight, the base peak is the most abundant fragment, and isotope patterns reveal the presence of Cl, Br, or I. Degree of unsaturation tells you how many rings or double bonds a molecular formula implies.
Molecular ion (M⁺•)
The radical cation formed when the molecule loses one electron to the ionising beam. Its m/z value equals the molecular weight of the compound. Think of it as the intact molecule with one electron missing.
Base peak
The tallest peak in the mass spectrum, set to 100% relative intensity. It corresponds to the most abundant ion. The base peak is not necessarily the molecular ion; it is simply whichever fragment forms most easily.
Fragmentation
The process by which the unstable molecular ion breaks apart into smaller pieces. Bonds break to produce a cation (detected) and a radical or neutral molecule (not detected). Fragmentation patterns reveal structural features.
m/z (mass-to-charge ratio)
The x-axis of a mass spectrum. For singly charged ions (the usual case in introductory organic chemistry), m/z equals the mass of the ion.
M+1 peak
A small peak one mass unit above the molecular ion, caused by the natural abundance of ¹³C (about 1.1% per carbon atom). Useful for estimating the number of carbons.
M+2 peak
A peak two mass units above the molecular ion. Significant when Cl, Br, or S is present. For chlorine, the M+2 peak is roughly one-third the intensity of M⁺. For bromine, M and M+2 are nearly equal.
Degree of unsaturation (index of hydrogen deficiency, IHD)
The number of rings plus double bonds in a molecular formula. Each ring or double bond removes two hydrogens from the saturated formula. A degree of unsaturation of 4 is characteristic of a benzene ring.
How the molecular ion forms
A high-energy electron beam knocks one electron out of the molecule, producing a radical cation (M⁺•).
The molecular ion is produced by loss of one electron, not a pair. Loss of a pair would produce a dication, which is a different species entirely.
The mass of M⁺• equals the molecular weight of the original molecule.
The molecular ion is often unstable and fragments further.
The base peak
The base peak is always the most abundant ion in the spectrum (tallest peak, defined as 100% relative intensity).
The base peak may or may not be the molecular ion. In many compounds, a fragment ion is more abundant.
The base peak is not necessarily the lowest m/z or the cation. Those are common wrong answers on exams.
Isotope patterns for halogen identification
Isotope peaks are one of the most reliable ways to spot halogens in a mass spectrum.
Halogen | Key Isotopes | M : M+2 Ratio | What to Look For |
|---|---|---|---|
Chlorine (³⁵Cl/³⁷Cl) | 75.8% / 24.2% | 3 : 1 | M+2 peak about one-third the height of M⁺ |
Bromine (⁹⁹Br/⁸¹Br) | 50.7% / 49.3% | 1 : 1 | M and M+2 nearly equal height |
Iodine (¹²⁷I) | 100% one isotope | No M+2 pattern | Loss of 127 (M–127) is the giveaway |
When you see two peaks of roughly equal height separated by 2 mass units near the molecular ion, think bromine. When the M+2 is about a third of M, think chlorine.
How fragmentation works
The molecular ion breaks at the weakest or most stabilised bond. The cation fragment is detected; the radical or neutral piece is not. What you see in the spectrum is the mass of the charged fragment.
Common neutral losses and what they mean
Loss (M – x) | x = | Indicates |
|---|---|---|
M – 15 | 15 (CH₃) | A methyl group was attached |
M – 18 | 18 (H₂O) | An alcohol (dehydration) |
M – 29 | 29 (CHO or C₂H₅) | An aldehyde or an ethyl group |
M – 31 | 31 (CH₂OH) | A primary alcohol |
M – 45 | 45 (OC₂H₅) | An ethyl ester or ethoxy group |
M – 127 | 127 (I) | An iodine atom |
Alpha cleavage
The bond next to a heteroatom (O, N) or carbonyl breaks preferentially, because the resulting cation is stabilised by the lone pair on the heteroatom.
Alcohols: cleavage of the C–C bond adjacent to the C–OH gives a cation stabilised by oxygen. For 2-methyl-2-butanol (MW 88), the dominant cleavage loses an ethyl radical (29) to give m/z = 59, or loses a methyl radical (15) to give m/z = 73. Water loss (M – 18 = 70) is also prominent. Loss of 16 (an oxygen atom alone) is not a typical fragmentation and would not be a prominent peak.
Amines: cleavage next to nitrogen is favoured. For ethylamine (CH₃CH₂NH₂, MW 45), alpha cleavage loses a methyl radical (15) to give m/z = 30 as the base peak (the CH₂=NH₂⁺ cation).
Branching and the base peak
More branched compounds tend to fragment at the branch point because the resulting carbocation is more substituted and therefore more stable. For neopentane-type structures, loss of a methyl group to form a tertiary cation (m/z = 57 for C₄H₉⁺ or m/z = 43 for C₃H₇⁺) is typical.
Identifying a compound from its mass spectrum
Look at three things:
The molecular ion (M⁺) gives the molecular weight. An odd M⁺ with no nitrogen suggests an odd number of nitrogens is present (nitrogen rule).
The fragmentation pattern tells you which pieces broke off.
Isotope patterns near M⁺ tell you whether Cl, Br, or I is present.
The formula
For a molecular formula C_c H_h N_n O_o X_x (where X = any halogen):
Degree of unsaturation = (2c + 2 + n – h – x) / 2
Oxygen does not appear in the formula. Halogens count the same as hydrogen (they replace one H each). Nitrogen adds one to the numerator.
What each value means
0 = fully saturated, no rings or double bonds.
1 = one ring or one double bond.
2 = two of the above in any combination.
4 = a benzene ring (three double bonds + one ring), or equivalent.
Worked examples from the practice set
C₅H₅Br₂NO: DoU = (2×5 + 2 + 1 – 5 – 2) / 2 = 6/2 = 3.
C₆H₅Br: DoU = (2×6 + 2 + 0 – 5 – 1) / 2 = 8/2 = 4.
C₈H₁₀BrIO: DoU = (2×8 + 2 + 0 – 10 – 2) / 2 = 6/2 = 3.
C₉H₁₁NO: DoU = (2×9 + 2 + 1 – 11 – 0) / 2 = 10/2 = 5.
Compounds II (C₆H₅Br, DoU = 4) and IV (C₉H₁₁NO, DoU = 5) do not match. Compounds I and III both have DoU = 3, so they share the same degree of unsaturation.
Mass spectrometry is the workhorse behind drug testing, forensic identification, and environmental monitoring. When a forensic lab identifies an unknown substance, the fragmentation pattern is compared against a database of known spectra, much the way a fingerprint is matched. Pharmaceutical companies use MS to confirm that a synthesised drug has the correct molecular weight and structure.
Students often think the molecular ion is formed by losing a pair of electrons. It is formed by losing one electron, producing a radical cation. Losing a pair would give a dication, which is a different process.
Students confuse the base peak with the molecular ion. The base peak is simply the most abundant ion. It is the tallest peak in the spectrum, regardless of its m/z value.
Students forget that iodine has only one stable isotope, so there is no M+2 pattern for iodine. Instead, look for loss of 127.
Students neglect oxygen when calculating degree of unsaturation. Oxygen does not appear in the DoU formula at all.
⚠️ Know the M : M+2 ratios for Cl (3:1) and Br (1:1). Expect to be shown mass spectra and asked to identify which halogen is present.
⚠️ Be able to predict the base peak from a given structure. Think about which bond cleavage gives the most stable cation.
⚠️ Common neutral losses (15, 18, 29, 127) are tested frequently. Know what each number implies.
⚠️ Degree of unsaturation calculations appear in multiple-choice and as part of structure-determination problems. Practise the formula until it is automatic.
True or false: The molecular ion is formed by loss of a pair of electrons.
False. It is formed by loss of one electron.
Fill in the blank: The base peak is the peak corresponding to the _______ ion.
Most abundant.
True or false: For a compound containing one chlorine atom, the M+2 peak is roughly equal in height to the M peak.
False. The M+2 peak is about one-third the height of M for chlorine. Equal heights indicate bromine.
Fill in the blank: A loss of 18 mass units from the molecular ion suggests the compound is an _______.
Alcohol (loss of water).
True or false: Oxygen must be included in the degree of unsaturation formula.
False. Oxygen does not appear in the formula.
Q: Which statement about the molecular ion is not true: (a) formed by loss of one electron, (b) mass equals the molecule's mass, (c) formed by loss of a pair of electrons, (d) often unstable and fragments?
A: (c). The molecular ion is formed by loss of one electron, not a pair.
Q: What is always true about the base peak?
A: It is the peak corresponding to the most abundant ion in the spectrum.
Q: For which compound will the M+2 peak be about one-third the intensity of M: CH₃CH₂CH₂Br, CH₃CH₂CH₂OH, CH₃CH₂CH₂Cl, or CH₃CH₂CH₂NH₂?
A: CH₃CH₂CH₂Cl. Chlorine's two isotopes (³⁵Cl and ³⁷Cl) give an M : M+2 ratio of approximately 3 : 1.
Q: Which of the following is not a prominent peak in the mass spectrum of 2-methyl-2-pentanol: M–15, M–18, M–29, or M–16?
A: M–16. Loss of a single oxygen atom is not a standard fragmentation pathway. Loss of CH₃ (15), H₂O (18), and C₂H₅ (29) are all common.
Q: Which compound will show a prominent peak at M–127: CH₃CH₂CH₂I, (CH₃CH₂)₃CCH₂Cl, (CH₃)₃CCH₂CH₂Br, or (CH₃)₂CHCH(CH₃)₂?
A: CH₃CH₂CH₂I. Loss of 127 corresponds to loss of an iodine atom.
Q: Which compound has a base peak at m/z = 43: hexane, 3-methylpentane, neopentane (2,2-dimethylpropane derivative), or 2,3-dimethylbutane?
A: (CH₃)₃CCH₂CH₃ (the neopentyl-type structure). Cleavage at the quaternary carbon loses an ethyl group and produces a tert-butyl cation, (CH₃)₃C⁺, at m/z = 57, or loses a tert-butyl radical and produces CH₂CH₃⁺. However, for m/z = 43 specifically, structures where a C₃H₇⁺ cation forms readily (such as branched C₆ hydrocarbons cleaving to give CH₃CO⁺ or (CH₃)₂CH⁺) should be considered. The answer from the practice set is (CH₃)₃CCH₂CH₃, where loss of the ethyl group gives a tert-butyl cation at m/z = 57, and the isopropyl cation at m/z = 43 forms through rearrangement.
Q: What is the expected base peak in the mass spectrum of 2-methyl-2-butanol (MW = 88)?
A: m/z = 59. Loss of the ethyl group (29) from alpha cleavage gives the most stable cation. However, m/z = 59 corresponds to M – 29. The tert-butyl alcohol also readily loses a methyl (M – 15 = 73) or water (M – 18 = 70). Among the answer choices, 59 is the base peak.
Q: What is the expected base peak for CH₃CH₂NH₂ (MW = 45)?
A: m/z = 30. Alpha cleavage next to nitrogen loses a methyl radical (15), giving CH₂=NH₂⁺ at m/z = 30.
Mass spectrometry ties into everything you have learnt about carbocation stability (Chapter 6–7). The fragments you see in a mass spectrum are the same stable cations you studied in substitution and elimination reactions: tertiary cations form preferentially, resonance-stabilised cations are favoured, and heteroatom-stabilised cations (from alpha cleavage) dominate when O or N is present.
Degree of unsaturation connects to aromaticity (Chapter 15–16). A DoU of 4 with a molecular formula that includes six carbons and only a few hydrogens is a strong hint that a benzene ring is present.
Together with IR spectroscopy (the companion study notes), mass spectrometry forms half of the structure-determination toolkit. The other half, NMR spectroscopy, is typically covered in the next chapter.
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