Organic Chemistry · Structure Determination: Mass Spectrometry and Infrared Spectroscopy
Mass Spectrometry of Some Common Functional Groups
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In 30 seconds
Different functional groups fragment in characteristic, predictable ways — and those patterns are the fastest way to identify a functional group from a spectrum. An alcohol loses water (M−18) and cleaves next to the C–O bond; a ketone fragments to an Acylium ion R–C≡O+, formed by α-cleavage of a carbonyl Full entry → and undergoes the McLafferty rearrangement γ-H transfer through a six-membered ring, then cleavage to an alkene + enol ion Full entry →; an alkyl chloride or bromide announces itself with a chlorine (3:1) or bromine (1:1) Isotope doublet M and M+2 peaks from Cl or Br isotopes Full entry →. This topic teaches you to recognize those signatures: what the functional group does to the molecular ion, which bonds break most easily, and what the diagnostic fragment masses are.
Think of it as learning the "accents" of each functional group. Once you can hear the difference between an alcohol's M−18 and a ketone's M−43, spectra stop being random bar charts and become conversations about structure.
Why this matters
Spectrum interpretation is the chemist's version of reading a fingerprint. In research, a quick look at the mass spectrum tells you whether your product is the alcohol you wanted or the alkene you accidentally dehydrated it to. In industry and forensics, functional-group fragmentation patterns are the basis of library searching: a computer matches your unknown spectrum against thousands of reference spectra by these same characteristic peaks. On exams, the difference between a correct and incorrect structure usually comes down to one diagnostic fragment — a peak at m/z = 31 (primary alcohol), m/z = 43 (methyl ketone), or a 1:1 doublet (bromide).
The college version
Core Concepts
Alkanes and alkyl chains
Alkanes give weak or absent molecular ions and a series of fragment peaks 14 mass units apart (CnH2n+1+): 29, 43, 57, 71, 85… Each 14 Da step is one CH2. The most abundant alkyl fragment is usually the most stable carbocation, so branching shows up as a strong peak at the branch point (e.g., a tert-butyl group gives an intense m/z = 57, (CH3)3C+). The base peak of a straight-chain alkane is often m/z = 43 or 57.
Alcohols: water loss and α-cleavage
Alcohols have a weak or invisible molecular ion. Two diagnostic processes dominate:
- Dehydration (M−18) Loss of water from an alcohol Full entry →: loss of water gives an alkene fragment ion at M−18.
- α-Cleavage Breaking the bond between the functional group and the adjacent carbon Full entry →: the bond between the carbinol carbon (C–OH) and the adjacent carbon breaks, giving the resonance-stabilized oxocarbenium ion. For a primary alcohol RCH2OH, the ion CH2=OH+ appears at m/z = 31 — one of the most reliable diagnostics in mass spectrometry. Secondary alcohols shift the peak (e.g., RCH=OH+, m/z = 45 for the simplest case), and tertiary alcohols show the alkyl cation instead.
Ketones and aldehydes: acylium ions and McLafferty
Carbonyl compounds fragment by two famous pathways:
- α-Cleavage gives the acylium ion R–C≡O+: for a methyl ketone, CH3CO+ at m/z = 43. This is the same nominal mass as the propyl cation, so check the context.
- McLafferty rearrangement requires a γ-Hydrogen Hydrogen on the carbon three bonds from the carbonyl Full entry → (on the carbon three bonds from the carbonyl). The molecule transfers that hydrogen to the carbonyl oxygen through a six-membered-ring transition state, then breaks the α/γ bond, ejecting a neutral alkene. For 2-hexanone, this gives the enol radical cation at m/z = 58.
Carboxylic acids and esters
Carboxylic acids lose OH (M−17) and water (M−18), and show McLafferty peaks. Esters fragment at the acyl–O and O–alkyl bonds; methyl esters characteristically lose CH3O• (M−31) and show a McLafferty peak from the alkyl chain.
Amines
The nitrogen rule applies (odd mass for one N). Amines undergo α-cleavage to give iminium ions: a primary amine RCH2NH2 gives CH2=NH2+ at m/z = 30, exactly analogous to the alcohol's 31.
Alkyl halides
Chlorine (3:1) and bromine (1:1) isotope doublets at M and M+2 are immediate. Fragmentation typically loses the halogen radical: M−35/37 for Cl, M−79/81 for Br, leaving the alkyl cation. The molecular ion of an alkyl chloride or bromide is usually visible because the halogen's mass keeps it prominent.
Aromatic rings
Benzene rings give an unusually strong molecular ion (the aromatic ring is stable) and a characteristic loss of H• (M−1). A m/z = 77 peak (C6H5+) and m/z = 91 (C7H7+, Tropylium ion C7H7+ (m/z = 91), a stable aromatic cation Full entry → from benzyl compounds) are classic fingerprints for phenyl and benzyl groups.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| m/z 31 (alcohol) | m/z 30 (amine) | CH2=OH+ = 31 for primary alcohols; CH2=NH2+ = 30 for primary amines — one mass unit apart |
| m/z 43 acylium ion | m/z 43 propyl cation | Both are 43; acylium (ketone) vs C3H7+ (alkane) — use M−18/M−28 context and other peaks |
| McLafferty rearrangement | α-cleavage | McLafferty needs a γ-H and transfers it through a 6-membered ring (rearrangement); α-cleavage is a direct bond break at the functional group |
| M−18 (water, alcohol) | M−17 (OH from carboxylic acid) | One mass unit difference; carboxylic acids also give M−18 |
| M−31 (methoxy loss, esters) | M−29 (ethyl or aldehyde H) | 31 = CH3O•/CH2OH•; 29 = C2H5•/CHO• |
| Chlorine 3:1 doublet | Bromine 1:1 doublet | Both at M and M+2; the ratio tells them apart (3:1 vs 1:1) |
| Strong M⁺• = stable molecule | Strong M⁺• = aromatic | Aromatic rings give strong M⁺•, but so do some small stable molecules; use M−1 and m/z 77/91 to confirm aromaticity |
| Molecular ion mass | Fragment mass | M⁺• is the highest cluster; fragments are lower. A "molecular ion" below the true M is a misassignment — check the nitrogen rule |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Every molecule is like a Lego castle that breaks apart in its own special way when it gets bumped. A castle with a water-tower (alcohol) always loses its water bucket first, so you see a "missing 18 bricks" clue. A castle with a garage door (ketone) snaps right next to the door and sometimes even rearranges its own bricks before breaking. And a castle built with chlorine bricks shows up as two nearly identical towers two bricks apart — because chlorine comes in two weights. When you see how the castle broke, you know what it was made of.
Worked example
Example 1: 1-hexanol vs 2-hexanone
These two C₆ compounds have different molecular masses — 1-hexanol (C₆H₁₄O) is 102 Da, 2-hexanone (C₆H₁₂O) is 100 Da — so their spectra differ immediately at the molecular ion, and each shows its own functional-group fingerprint.
1-Hexanol, CH3CH2CH2CH2CH2CH2OH (M = 102):
- Weak molecular ion at m/z = 102.
- M−18 = 84 (water loss) — the dehydration signature.
- m/z = 31: α-cleavage gives CH2=OH+. Check the mass: 12 + 2(1) + 16 + 1 = 31 ✓. This is the primary-alcohol fingerprint.
- A smaller m/z = 56 from dehydration followed by further fragmentation.
2-Hexanone, CH3COCH2CH2CH2CH3 (M = 100):
- Visible molecular ion at m/z = 100.
- m/z = 43: α-cleavage gives the acylium ion CH3CO+ (24 + 3 + 16 = 43 ✓) — usually the base peak of a methyl ketone.
- m/z = 58: McLafferty rearrangement. The γ-hydrogens sit on C4 (the carbon three bonds from the carbonyl). One transfers to the carbonyl oxygen through a six-membered-ring transition state, then the C2–C3 bond breaks. The enol radical cation CH3C(OH)=CH2+• (58 Da) is detected while neutral propene CH3CH=CH2 (42 Da) escapes: 58 + 42 = 100 = M ✓.
Reading the two spectra side by side: both contain C₆ chains, but the alcohol announces itself with M−18 and m/z 31, while the ketone announces itself with m/z 43 and m/z 58. One glance at the base peak region separates them.
Example 2: Bromide fingerprint
An unknown shows a 1:1 doublet at m/z = 122 and 124, and a fragment at m/z = 43. Identify the compound.
Step 1 — halogen: 1:1 M:M+2 = one Br. Molecular mass 122 (nominal, using 79Br). Step 2 — formula: 122 − 79 = 43 → C3H7 (36 + 7 = 43). Formula C3H7Br. Step 3 — DU: 3 - (7+1)/2 + 1 = 0 — saturated, no rings. So: bromopropane. Step 4 — isomer: strong m/z = 43 (C3H7+) means the bromine was lost to give the propyl cation; both isomers give this. The 2-isomer would also show an enhanced M−15 (loss of CH₃). Additional evidence: the spectrum of 1-bromopropane shows m/z 43 (loss of Br) with the M⁺• doublet; the 2-isomer shows m/z 43 plus stronger m/z 107 (M−15). If only the doublet and 43 appear, 1-bromopropane is the better assignment.
Example 3: Benzyl group detection
A compound with molecular ion m/z = 106 (strong) and a major peak at m/z = 91. What structural feature is present?
Reasoning: a strong M⁺• suggests aromaticity; m/z = 91 is the tropylium ion C7H7+ (7(12) + 7 = 91 ✓), formed when a benzyl group (PhCH2R) loses the R fragment. M = 106 with a benzyl group: C6H5CH2 = 91, leaving 106 − 91 = 15 = CH3. So the compound is ethylbenzene (M = 106, C8H10, DU = 8 − 10/2 + 1 = 4 — benzene ring ✓), which fragments by benzylic cleavage to tropylium at 91.
Answer: a benzyl/aromatic system; specifically ethylbenzene.
Key takeaways
- Alkanes: fragment series 29, 43, 57, 71 (each +14 = CH₂); branching gives intense peaks at branch carbocations.
- Alcohols: weak M⁺•; M−18 (water loss); m/z = 31 (CH2=OH+, primary alcohols).
- Ketones/aldehydes: acylium ions (CH3CO+ = 43); McLafferty rearrangement needs a γ-H and gives an enol radical cation (e.g., 58 for 2-hexanone).
- Amines: odd nominal mass (1 N); m/z = 30 (CH2=NH2+, primary amines).
- Alkyl halides: Cl = 3:1 doublet (M, M+2); Br = 1:1 doublet; loss of X• gives M−35/37 or M−79/81.
- Carboxylic acids: M−17 (OH loss) and M−18 (water loss).
- Aromatics: strong M⁺•, M−1, and tropylium m/z = 91 for benzyl groups.
- Fragment ions are even-electron cations; the charge stays with the fragment that gives the most stable cation.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
What two diagnostic features identify a primary alcohol in a mass spectrum?
Show answer
A weak molecular ion with an M−18 peak (water loss) and an α-cleavage peak at m/z 31 (CH2=OH+).
What structural requirement must a ketone meet to undergo the McLafferty rearrangement?
Show answer
It must have a hydrogen on the γ-carbon (three bonds from the carbonyl), which transfers through a six-membered-ring transition state.
A spectrum shows a 3:1 doublet at M and M+2. What element is present?
Show answer
One chlorine atom: 35Cl and 37Cl in ~3:1 natural abundance.
What is the mass of the McLafferty fragment ion of 2-hexanone, and what neutral is lost?
Show answer
The enol radical cation CH3C(OH)=CH2+• at m/z 58, with propene (C3H6, 42 Da) lost as the neutral: 58 + 42 = 100 = M of 2-hexanone.
Why do branched alkanes show intense peaks at the branch point?
Show answer
Fragmentation favors the most stable carbocation; a branch point (e.g., tert-butyl) gives a tertiary cation (m/z 57), which is very stable and therefore abundant.
What fragment at m/z 91 suggests a benzyl group, and what is its structure?
Show answer
The tropylium ion C7H7+, a stable aromatic cation formed by cleavage of the benzylic C–C bond in PhCH2R.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- α-Cleavage
- Breaking the bond between the functional group and the adjacent carbon
- Acylium ion
- R–C≡O+, formed by α-cleavage of a carbonyl
- McLafferty rearrangement
- γ-H transfer through a six-membered ring, then cleavage to an alkene + enol ion
- γ-Hydrogen
- Hydrogen on the carbon three bonds from the carbonyl
- Dehydration (M−18)
- Loss of water from an alcohol
- Iminium ion
- R2C=N+R2 type cation from amine α-cleavage
- Tropylium ion
- C7H7+ (m/z = 91), a stable aromatic cation
- Isotope doublet
- M and M+2 peaks from Cl or Br isotopes
- Even-electron rule
- Fragment ions are cations with no unpaired electrons
- Fragment series
- Peaks spaced by 14 Da (CH2) in alkyl chains
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