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Spectroscopy · Mass Analysis

Mass Spectrometry

Mass spectrometry (MS) measures the mass-to-charge ratio (m/z) of ionised molecules and their fragments. From a single mass spectrum you can determine molecular weight, identify the molecular formula (with high-resolution MS), and deduce structural features from characteristic fragmentation patterns. This lab teaches the EI (electron ionisation) mass spectrum that students see most often: the molecular ion, fragment ions, isotope patterns, and the diagnostic fragmentations (alpha cleavage, McLafferty rearrangement, tropylium formation) that distinguish compound classes.

Theory — Mass Spectrometry

1. What does mass spec measure?

A mass spectrometer measures the mass-to-charge ratio (m/z) of charged species. For most organic chemistry, the charge is +1 (singly ionised cations), so m/z is essentially the mass of the ion in atomic mass units (Daltons, Da). The instrument has three parts: an ion source (where neutral molecules are converted to cations), a mass analyser (which separates ions by m/z), and a detector. The output is a spectrum: m/z on the x-axis, relative intensity (0-100%) on the y-axis. The most intense peak is called the base peak (set to 100% by convention).

2. Ionisation methods

The choice of ionisation method affects which ions you see:

MethodEnergyWhat you getBest for
EI (electron ionisation)70 eV (high)Lots of fragments + sometimes a weak molecular ionSmall to medium organic molecules; the "classical" mass spec
CI (chemical ionisation)~10 eV (soft)Mainly [M+H]⁺ or [M-H]⁻; little fragmentationWhen EI loses the molecular ion; gives MW directly
ESI (electrospray)Very softIons from solution; multiply charged for large moleculesPolar molecules, salts, biomolecules, proteins
MALDIPulsed laserMostly [M+H]⁺ ionsLarge molecules: proteins, polymers, lipids
APCISoftVolatile molecules from LC eluentLC-MS for small/medium polar molecules

Most undergraduate organic chemistry uses EI mass spectra from GC-MS instruments. EI provides rich fragmentation patterns that reveal structure; the rest of this lab focuses on EI.

3. The molecular ion (M⁺)

In EI, the high-energy electron beam knocks one electron out of the molecule, producing the radical cation M⁺· (the molecular ion, with the same mass as the neutral molecule). This ion appears at the highest m/z value in the spectrum and tells you the molecular weight directly. Many EI molecular ions are unstable and fragment quickly; the M⁺ peak may be small or absent.

EI ionisation M + e⁻ (70 eV) → M⁺· + 2 e⁻
M⁺· (excited; if stable, observed; if not, fragments to give A⁺ + B·)
The M⁺ peak in the spectrum has m/z = molecular weight of the compound.

4. Isotope peaks

Most elements have several stable isotopes. The most abundant isotope gives the main peak; less abundant isotopes give peaks 1, 2, or more mass units higher (M+1, M+2, etc.).

ElementMajor isotopeMinor isotopesPattern
C¹²C (98.9%)¹³C (1.1%)~1.1% per C in M+1 peak
H¹H (99.99%)²H (deuterium, trace)Negligible
N¹⁴N (99.6%)¹⁵N (0.4%)Negligible
O¹⁶O (99.8%)¹⁷O (0.04%), ¹⁸O (0.2%)Small M+2 contribution
Cl³⁵Cl (75.8%)³⁷Cl (24.2%)3:1 ratio (M:M+2) — very distinctive
Br⁷⁹Br (50.7%)⁸¹Br (49.3%)1:1 ratio (M:M+2) — very distinctive
I¹²⁷I (100%)None significantNo isotope pattern
S³²S (95%)³⁴S (4.2%)~4% M+2 contribution

The chlorine and bromine patterns are diagnostic. A 3:1 ratio of peaks separated by 2 mass units = chlorine. A 1:1 ratio of peaks separated by 2 mass units = bromine. These patterns let you identify halogen-containing compounds at a glance.

Halogen isotope pattern recognition Two peaks 2 mass units apart, ratio 3:1 (larger : smaller) → one Cl
Two peaks 2 mass units apart, ratio 1:1 → one Br
Three peaks 2 mass units apart, ratio 9:6:1 → two Cl
Three peaks 2 mass units apart, ratio 1:2:1 → two Br
If you see "M and M+2 about equal" in the spectrum, the compound contains Br. If you see "M significantly larger than M+2 in 3:1 ratio", the compound contains Cl.

5. Fragmentation patterns — alpha cleavage

The most common fragmentation in EI is alpha cleavage: cleavage of a C-C bond NEXT to a heteroatom or a π bond. The result: the heteroatom keeps its lone pair on the cation side (so the cation is stabilised by resonance with the heteroatom), and a free radical is lost.

For acetone (CH₃COCH₃, MW 58), the EI spectrum shows: M⁺ at m/z 58 (small); alpha cleavage gives [CH₃CO]⁺ (acetyl cation, m/z 43, base peak); further loss of CO gives [CH₃]⁺ (m/z 15). The pattern 58 / 43 / 15 with 43 as base peak is classic ketone alpha cleavage.

6. McLafferty rearrangement

For carbonyl compounds with a γ-hydrogen (a hydrogen on the third carbon from the carbonyl), a six-membered cyclic transition state allows transfer of the γ-H to the carbonyl O, with simultaneous cleavage of the α-β bond. This gives an enol cation (preserves the carbonyl ion) and a neutral alkene.

McLafferty rearrangement (for carbonyls with γ-H) R-CH₂-CH₂-CHR\'-C(=O)-R\'\' → [CH₂=CR\'-C(=OH)-R\'\']⁺ + R-CH=CH₂
The enol cation has m/z = [original M] - [neutral alkene mass]
Diagnostic for esters, ketones, aldehydes, carboxylic acids with γ-H. The McLafferty fragment is typically a strong peak.

7. Tropylium and other aromatic fragmentations

Aromatic compounds with benzylic CH₂ groups give a very stable cation: the tropylium cation (C⁷H⁷⁺, m/z 91), a 7-membered aromatic ring. Toluene\'s mass spectrum has m/z 91 as the base peak (!) because the initial benzyl cation ([PhCH₂]⁺, m/z 91) rearranges to the tropylium cation (also m/z 91) which is much more stable. Almost any benzyl-type cleavage product gives m/z 91 as a major peak. m/z 77 (phenyl cation, [C₆H₅]⁺) and m/z 65 (cyclopentadienyl cation from tropylium loss of acetylene) are also common.

8. Common fragment ion masses

m/zFragmentIndicator of
15[CH₃]⁺Methyl group present
17[OH]⁺Hydroxyl loss (alcohols)
18[H₂O]⁺ or M-18Water loss (alcohols, acids)
28[CO]⁺, [N₂]⁺, or [C₂H₄]⁺Carbonyl, nitrogen, ethylene
29[CHO]⁺ or [C₂H₅]⁺Aldehyde C-H of carbonyl, or ethyl
43[CH₃CO]⁺ or [C₃H⁷]⁺Acetyl cation (ketones), propyl
57[C₄H⁹]⁺ or [CH₃CH₂CO]⁺Butyl, propanoyl
77[C₆H₅]⁺Phenyl cation
91[C⁷H⁷]⁺Tropylium / benzyl — aromatic with -CH₂-
105[C₆H₅CO]⁺Benzoyl cation (aromatic ketone)

9. Real vs simulated spectra in this lab

The instrument bench in this lab uses a mix of real reference spectra (annotated as REAL) and simulated spectra (annotated as SIMULATED). Real spectra are computer renderings of published data from the NIST WebBook, SDBS, or commercial databases; simulated spectra are constructed from known fragmentation patterns and isotope ratios. Each spectrum in the lab is labeled at the top so you always know which you\'re looking at. Both modes preserve the diagnostic pattern (M⁺, base peak, isotope pattern) accurately.

10. The systematic MS analysis approach

  1. Find M⁺ (highest m/z peak). This gives the molecular weight. Check the M+1 and M+2 peaks for isotope clues (especially Cl/Br pattern).
  2. Identify the base peak (most intense peak). This is the most stable fragment; it tells you what kind of cleavage dominates.
  3. Calculate mass differences from M⁺. Common losses: 15 (CH₃), 17 (OH), 18 (H₂O), 28 (CO), 29 (CHO), 43 (CH₃CO).
  4. Check for diagnostic patterns: m/z 91 (tropylium = aromatic with CH₂); m/z 77 (phenyl); halogen 3:1 or 1:1 patterns; McLafferty fragments at predicted positions.
  5. Combine with IR + NMR + integration formula. MS gives mass; IR gives functional groups; NMR gives structure detail. The three together identify most unknowns.

Instructions

This lab\'s Simulation section has four parts. Complete them in order.

1
Section I — Fragmentation Library. Eight fragment-identification cases. For each: identify the m/z position; identify the fragment ion; identify the parent compound class.
2
Section II — MS Instrument Bench. Six unknown samples. For each: select the bottle, click "Inject & Ionize", view the mass spectrum, identify the compound class, and verify using the fragment labels (revealed after you answer).
3
Section III — Spectrum Interpretation. Eight harder puzzles: McLafferty rearrangements, halogen isotope patterns, M⁺ identification, ortho-effect, alpha cleavage prediction.
4
Section IV — SDS & Microscale. Read SDS extracts for four MS-related materials (16 questions: EI source filaments, ESI methanol/formic acid solutions, GC-MS column phases, sample-handling solvents). Then six microscale technique scenarios (when to use EI vs ESI, GC-MS vs LC-MS, sample prep for direct injection vs dissolved, etc.).
5
Prepare your lab notebook. Use the Example Report as your template.

Prerequisite: Familiarity with basic organic functional groups; ideally completion of the Functional Group Tests, Aldehydes & Ketones, and IR Spectroscopy labs first. Mass spec is most useful when combined with IR and NMR for full structural identification.

Simulation

Four interactive parts. Use the ↺ Reset Simulation button at any time to clear all answers and start over.

MS Workbench Section I — Fragmentation Library

Eight fragment-identification cases. For each: (a) m/z position type; (b) which fragment ion; (c) the compound class indicated.

Score: 0 / 24 (3 questions × 8 cases)

Six unknown samples. Select a sample, click Inject & Ionize, then identify the compound class from the spectrum. Fragment labels appear after you answer.

Score: 0 / 6

Eight harder puzzles: McLafferty rearrangements, halogen isotope patterns, M⁺ identification, alpha cleavage prediction.

Score: 0 / 8

Round 1 — SDS interpretation

Four common MS materials and reagents. Each has 4 questions.

SDS score: 0 / 16

Round 2 — Microscale technique scenarios

Six scenarios. Identify the appropriate MS technique or sample prep.

Microscale score: 0 / 6

Team Questions

Discuss with your team before answering.

Question 1 — Molecular ion. What is the molecular ion (M⁺) and what does it tell you about your sample?
Question 2 — Cl vs Br. A mass spectrum shows two peaks at m/z 78 and 80 in roughly equal intensity (1:1). What halogen is present?
Question 3 — Tropylium. What does a strong peak at m/z 91 most often indicate, and why is this fragment so stable?
Question 4 — Acetone fragmentation. Acetone (M=58) shows a base peak at m/z 43. What fragment is at m/z 43, and what mechanism produces it?
Question 5 — McLafferty. What structural feature is required for a compound to undergo a McLafferty rearrangement?
Question 6 — Limitations. Name two important things mass spectrometry CANNOT tell you directly about a molecule.

Example Lab Notebook Entry

Use the format below as a template.

Mass Spectrometry — Lab Notebook Entry

Submitted by: [Student Name]

Course: Organic Chemistry I · Section: 201-A · Date: May 8, 2026

Objective

To learn how electron ionisation (EI) mass spectrometry produces molecular and fragment ions; to identify the molecular ion (M⁺) and base peak; to recognise diagnostic fragmentation patterns including alpha cleavage, McLafferty rearrangement, and tropylium formation; to recognise halogen isotope patterns (Cl 3:1, Br 1:1); to interpret mass spectra of unknown compounds using systematic analysis.

Instrument bench results (Section II)

SampleCompoundM⁺Base peakKey diagnostic
1Methanol3231Loss of H from M⁺ (alcohol pattern)
2Acetone5843 (CH₃CO⁺)Alpha cleavage of ketone
3Hexane8657 (C₄H⁹⁺)Alkane series: 57, 43, 29 (loss of CH₂ units)
42-Butanone7243 (CH₃CO⁺)Alpha cleavage + McLafferty (m/z 58)
5Chlorobenzene112/114 (3:1)112 (M⁺)Cl isotope pattern (3:1)
6Toluene9291 (C⁷H⁷⁺ tropylium)Tropylium formation from benzyl cation

Fragmentation summary (key m/z values memorised)

m/zFragmentIndicator
15[CH₃]⁺Methyl loss
18M-18 (water)Alcohol or carboxylic acid
29[CHO]⁺ or [C₂H₅]⁺Aldehyde C-H or ethyl
43[CH₃CO]⁺Methyl ketone (alpha cleavage)
57[C₄H⁹]⁺Butyl, branched alkyl
77[C₆H₅]⁺Phenyl cation
91[C⁷H⁷]⁺Tropylium / aromatic with -CH₂-
105[C₆H₅CO]⁺Aromatic ketone (PhCO⁺)

Discussion

Mass spectrometry by electron ionisation gives two pieces of information from one spectrum: (1) molecular weight from M⁺, and (2) structural information from the fragmentation pattern. The molecular ion is the highest m/z peak (its abundance varies; some compounds have weak M⁺). Below it, fragment ions appear at lower m/z. The most intense peak is the base peak, normalised to 100%. Other peaks are reported as percentage of the base peak.

Alpha cleavage is the most common fragmentation: a C-C bond next to a heteroatom or pi system breaks, giving a stabilised cation on the heteroatom side. For ketones, this produces an acylium cation (R-C≡O⁺). Acetone (CH₃COCH₃) gives the acetyl cation (CH₃CO⁺) at m/z 43 as the base peak. Higher ketones give characteristic acylium fragments at predictable m/z values: 43 (acetyl), 57 (propanoyl), 71 (butanoyl), etc.

McLafferty rearrangement is a six-membered cyclic transition state that transfers a gamma-hydrogen to a carbonyl oxygen and cleaves the alpha-beta bond. The result is an enol cation (still containing the carbonyl) and a neutral alkene. McLafferty fragments are diagnostic of carbonyls with at least three carbons in the alpha chain. 2-butanone has a McLafferty fragment at m/z 58 (the enol of acetone, CH₂=C(OH)CH₃) plus the standard alpha cleavage at m/z 43.

Halogen isotope patterns are the most useful spectral feature for identifying Cl and Br. Chlorine\'s ³⁵Cl/³⁷Cl ratio is 3:1, so a chlorinated compound shows two peaks separated by 2 mass units in 3:1 ratio. Bromine\'s ⁷⁹Br/⁸¹Br ratio is roughly 1:1, so a brominated compound shows two peaks of nearly equal height separated by 2 mass units. Multiple halogens give characteristic patterns (e.g., two Cl gives three peaks 9:6:1).

Tropylium formation in aromatic compounds with benzylic CH₂ groups gives the diagnostic m/z 91 peak (often the base peak). The initial benzyl cation [PhCH₂]⁺ rearranges to the seven-membered tropylium cation, which is aromatic (6 pi e⁻) and exceptionally stable. Toluene, ethylbenzene, propylbenzene all show m/z 91 as the base peak. Other aromatic fragments include m/z 77 (phenyl cation) and m/z 65 (cyclopentadienyl from tropylium loss of acetylene).

The systematic approach to interpreting an MS: (1) identify M⁺ and check its isotope pattern for Cl/Br; (2) identify the base peak and what mechanism produces it; (3) calculate mass differences from M⁺ to identify common losses (15 = CH₃, 17 = OH, 18 = H₂O, 28 = CO, 43 = CH₃CO); (4) match the pattern to a compound class. Mass spec works best in combination with IR (functional groups) and NMR (connectivity); each gives different information, and the three together identify most undergraduate unknowns.

Conclusion

Mass spectrometry is the most direct way to determine molecular weight and a powerful tool for structural identification through fragmentation patterns. The key skills covered are: identifying M⁺, recognising halogen isotope patterns, predicting alpha cleavage and McLafferty fragments, and matching fragmentation patterns to compound classes. Combined with IR and NMR (which provide complementary functional group and connectivity information), mass spectrometry forms the structural identification toolkit used in research, pharmaceutical analysis, and quality control. The next labs in this Spectroscopy quartet (NMR and UV/Vis) build on the same systematic analysis logic.

References

1. Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed., Cengage, 2015, Ch 8.
2. Silverstein, R. M.; Webster, F. X.; Kiemle, D. J. Spectrometric Identification of Organic Compounds, 8th ed., Wiley, 2014, Ch 1.
3. NIST Mass Spectrometry Data Center, https://webbook.nist.gov/chemistry.
4. SDBS — Spectral Database for Organic Compounds, AIST Japan, https://sdbs.db.aist.go.jp.
5. McLafferty, F. W.; Turecek, F. Interpretation of Mass Spectra, 4th ed., University Science Books, 1993.

Practice Questions

Work through each before peeking at the hint.

Practice 1 — Molecular weight
An EI mass spectrum has the highest m/z peak at 78. There are no peaks above 78. The compound contains C and H only. What molecular formula is most likely, and what is the compound?
Hint: M⁺ = 78. Try formulas: C₆H₆ = 78. Yes — this is benzene. Other isomers? C₅H₂O is 84 (no). C₆H₆ = 78 fits cleanly. Confirms benzene as the most likely structure.
Practice 2 — Alcohol
An MS spectrum has M⁺ = 60 and a strong peak at m/z 45 (loss of 15). Another peak at m/z 31. What compound class is suggested?
Hint: M⁺ = 60 (could be propanol C₃H⁸O = 60, or acetic acid C₂H₄O₂ = 60). Loss of 15 = CH₃. Peak at m/z 31 = [CH₂OH]⁺ or [CHO]⁺. The 31 peak (CH₂OH⁺) is very characteristic of primary alcohols (alpha cleavage). So this is likely 1-propanol or 2-propanol. The pattern M-15 + m/z 31 suggests propanol (alcohol).
Practice 3 — Halogen pattern
An EI spectrum shows two peaks at m/z 122 and 124 in 1:1 ratio. What halogen is present, and what is the likely molecular formula?
Hint: 1:1 ratio of two peaks 2 mass units apart = bromine (one Br). M = 122 (lower peak); the compound contains one Br plus other elements totalling 122 - 79 = 43 mass units of non-Br part. Possible: C₃H⁷Br = 79+43 = 122. Yes, this is propyl bromide (1-bromopropane or 2-bromopropane).
Practice 4 — Tropylium
A compound has M⁺ = 106 and base peak at m/z 91. What is the compound likely to be?
Hint: Base peak at m/z 91 is the tropylium / benzyl cation. M-91 = 15, which is loss of CH₃. So we have an aromatic compound with -CH₂CH₃ (ethylbenzene) or -(CH₃)₂ on the ring (xylene). Both have MW 106. Ethylbenzene shows m/z 91 as base peak (alpha cleavage of CH₃ gives PhCH₂⁺ → tropylium). Xylenes also show m/z 91 (loss of one methyl, then tropylium rearrangement). Either fits the spectrum.
Practice 5 — McLafferty
2-Hexanone (M = 100) has a strong peak at m/z 58. What is this fragment, and how is it formed?
Hint: 100 - 58 = 42 = C₃H₆ (propene, neutral alkene). The McLafferty rearrangement transfers a gamma-H from C5 to the carbonyl O, and cleaves the alpha-beta C-C bond (between C3 and C4). This gives the enol of acetone (CH₂=C(OH)CH₃)⁺ at m/z 58 + propene (C₃H₆) as the neutral alkene. Diagnostic for carbonyls with gamma-H.
Practice 6 — Two halogens
An EI spectrum has three peaks at m/z 84, 86, and 88 in approximately 9:6:1 ratio. How many chlorine atoms does the compound contain?
Hint: Three peaks 2 mass units apart in 9:6:1 ratio = TWO chlorines. (One Cl gives 3:1; two Cl gives 9:6:1; three Cl gives 27:27:9:1.) M = 84; with two Cl = 70 mass units, the rest is 84 - 70 = 14, which is CH₂. So the compound is CH₂Cl₂ (dichloromethane), MW 84.
Practice 7 — Aldehyde
An aldehyde has M⁺ = 72 and shows characteristic peaks at m/z 71 (M-1, very small), 43, and 29. What is m/z 29 from, and what is the aldehyde?
Hint: m/z 29 = [CHO]⁺ (the aldehyde C-H carbonyl cation, the diagnostic aldehyde fragment). m/z 43 = loss of 29, so M-29 = 72-29 = 43 = [C₃H⁷]⁺. Compound = C₄H⁸O = 72 = butanal (CH₃CH₂CH₂CHO). Alpha cleavage gives [CH₃CH₂CH₂]⁺ (m/z 43) + [CHO] (which then ionises to m/z 29).
Practice 8 — Combination
A compound shows M⁺ = 122 with no isotope pattern (small M+1 from C). Base peak at m/z 77, second strongest at m/z 105 (small). IR shows a strong C=O at 1685 and aromatic ring features. What is the compound?
Hint: M⁺ = 122. m/z 77 = [C₆H₅]⁺ (phenyl cation). m/z 105 = M - 17 = 122-17 = 105, possibly [C₆H₅CO]⁺ (benzoyl cation, PhCO⁺). 122-105 = 17 = OH loss; 122-77 = 45 = COOH loss. So the compound has a phenyl ring + carboxylic acid: PhCOOH = benzoic acid (MW 122). The IR confirms: aromatic ring + C=O 1685 (slightly lower because of conjugation). This is benzoic acid.
Practice 9 — Why M⁺ is small
For most ketones in EI MS, the M⁺ peak is small while the alpha-cleavage fragment is the base peak. Why is the alpha-cleavage product so much more abundant?
Hint: Alpha cleavage produces an acylium cation R-C≡O⁺, which is highly stabilised by resonance with the carbonyl oxygen lone pair. The molecular ion (M⁺·) is a radical cation — both an unpaired electron and a positive charge. Fragmentation to give a stable acylium + neutral alkyl radical is favourable both because of the cation stability and because radical formation has a low activation energy. In EI (high-energy ionisation, 70 eV), there\'s plenty of internal energy in M⁺· to drive fragmentation.
Practice 10 — Big picture
An unknown solid sample is sent for analysis. Mass spec gives M⁺ = 138, base peak at m/z 91, and strong peaks at 77 and 65. IR shows a sharp C=O at 1715 and aromatic features. What is the compound, and what does each technique contribute to the identification?
Hint: M⁺ = 138. m/z 91 base peak = tropylium (aromatic with CH₂). m/z 77 = phenyl. m/z 65 = cyclopentadienyl (often from tropylium losing acetylene). IR: C=O 1715 = aliphatic ester or ketone in conjugation. Loss from M⁺: 138-91 = 47 = C(=O)OH or similar acid/ester moiety. Plus C₆H₅-CH₂-COOR has ester. Most likely: methyl phenylacetate (CH₃OC(O)CH₂Ph), MW 150 (no). Try methyl benzoate PhCOOMe, MW 136 (close). Phenylacetic acid PhCH₂COOH, MW 136 (no). Reconsider: 138 + need C=O. (C₄H₆O₃ structures... wait). Most likely: phenylacetic acid is 136, not 138; ethyl benzoate is 150; could be phenylethanol C₆H₅CH₂CH₂OH = 122. Re-examining: 91 base + C=O suggests methyl phenylacetate or similar. The student should try several formulas matching M = 138 with aromatic + carbonyl. Real example: methyl benzoylformate MW 138, PhCOCOOH MW 150. Actually MW 138 with aromatic and carbonyl fits anisaldehyde MW 136 (close) or 4-methylbenzoic acid MW 136. The general lesson: combine MS (138) + IR (C=O + aromatic) + NMR (which would tell ortho/meta substitution) to fully identify. The point of this question is the LOGIC of combining techniques, not the specific compound.