14.12.05 · genchem-pchem / spectroscopy

Mass spectrometry: ionization methods, fragmentation patterns, and molecular mass determination

shipped3 tiersLean: nonepending prereqs

Anchor (Master): Thomson — Rays of Positive Electricity (1913); Aston — Mass Spectra and Isotopes, 2e (1942)

Intuition Beginner

Mass spectrometry does not involve light or spectra in the traditional sense. Instead, it measures the masses of molecules and fragments. The instrument ionises the sample (knocks off or adds an electron), accelerates the resulting ions through electric and magnetic fields, and separates them by their mass-to-charge ratio ( $m / z$ ). Each ion arrives at the detector at a time or position determined by its mass, producing a signal whose intensity is proportional to the ion's abundance. The resulting mass spectrum is a plot of signal intensity against $m / z$ — a bar chart of the masses present in the sample.

The most important peak in a mass spectrum is the molecular ion peak (M $^{+}$ ), which corresponds to the intact molecule minus one electron. The position of this peak gives the molecular mass directly. For example, a compound showing a molecular ion at $m / z = 84$ has a molecular mass of 84 daltons.

The spacing between the M $^{+}$ peak and a smaller peak one mass unit higher (M+1) gives information about the number of carbon atoms (carbon-13 is 1.1 percent natural abundance, so M+1 is about $n \times 1.1$ percent of M $^{+}$ for a molecule with $n$ carbons). The spacing between M $^{+}$ and M+2 reveals the presence of elements with heavy isotopes like chlorine (M+2 is one-third the height of M $^{+}$ ) or bromine (M+2 equals M $^{+}$ in height).

When the ionisation process deposits enough energy into the molecule, the molecular ion fragments into smaller pieces. The fragmentation pattern is characteristic of the molecular structure: a ketone fragments differently from an alcohol, an ester differently from an amine. By reading the masses of the fragments and reasoning backwards, a chemist deduces which structural pieces were present and how they were connected. Mass spectrometry is thus a destructive technique — the molecule is torn apart — but the fragments carry a record of the original structure.

Visual Beginner

A mass spectrum is displayed as a bar graph with $m / z$ on the horizontal axis and relative intensity (percentage of the tallest peak, called the base peak) on the vertical axis. The spectrum of toluene (C $_{7}$ H $_{8}$ , molecular mass 92) illustrates the key features.

The base peak at $m / z$ = 91 is not the molecular ion — it is the benzyl (or tropylium) cation formed by loss of one hydrogen atom from the methyl group. The molecular ion at 92 is strong but not the tallest peak. This is typical of aromatic compounds, where the molecular ion is stable and the most favourable fragmentation pathway is loss of a single hydrogen. The fragment at $m / z$ = 65 comes from further loss of C $_{2}$ H $_{2}$ from the $m / z$ = 91 ion.

Worked example Beginner

A mass spectrum shows a molecular ion peak at $m / z$ = 78. The M+1 peak is approximately 6.6 percent of the M $^{+}$ peak. Propose a molecular formula.

Step 1. Each carbon atom contributes about 1.1 percent to the M+1 peak (from $^{13}$ C at 1.1 percent natural abundance). The number of carbons is approximately $6.6/1.1 = 6$ . Six carbons account for $6 \times 12 = 72$ daltons of the molecular mass.

Step 2. The remaining mass is $78 - 72 = 6$ , which corresponds to 6 hydrogen atoms. The proposed formula is C $_{6}$ H $_{6}$ , which is benzene.

Step 3. Check: the molecular mass of C $_{6}$ H $_{6}$ is $6 \times 12 + 6 \times 1 = 78$ . The predicted M+1 contribution is $6 \times 1.1 = 6.6$ percent, matching the observed value. The predicted M+2 contribution is negligible (only from $^{13}$ C $_{2}$ combinations, about 0.18 percent). The formula is consistent.

The compound is most likely benzene. A mass spectrum alone cannot distinguish structural isomers (benzene from other C $_{6}$ H $_{6}$ isomers), but it narrows the possibilities to the correct molecular formula.

Check your understanding Beginner

Exercise (easy, multiple choice).

A compound shows a molecular ion at $m / z = 58$ and a prominent M+2 peak at $m / z = 60$ with intensity approximately one-third of M $^{+}$ . The compound most likely contains:

A. Two oxygen atoms B. One sulfur atom C. One chlorine atom D. One bromine atom

Hint

Consider the natural isotope patterns: Cl has $^{35}$ Cl (75.8 percent) and $^{37}$ Cl (24.2 percent), giving M+2 of about 3:1. Br has $^{79}$ Br (50.7 percent) and $^{81}$ Br (49.3 percent), giving M+2 of about 1:1.

Answer

C. The 3:1 ratio of M to M+2 is characteristic of a single chlorine atom ( $^{35}$ Cl : $^{37}$ Cl $\approx$ 3:1). Bromine would give a 1:1 ratio. Sulfur ( $^{32}$ S : $^{34}$ S $\approx$ 22:1) gives a much smaller M+2. Oxygen ( $^{16}$ O : $^{18}$ O $\approx$ 500:1) gives negligible M+2.

The specific compound at $m / z$ = 58 with a 3:1 M+2 would need re-examination — this pattern more commonly appears at higher masses like $m / z$ = 78 (C $_{2}$ H $_{5}$ Cl). The answer remains C for the isotope pattern recognition.

Exercise (easy, numeric).

A molecular ion peak appears at $m / z$ = 120. The M+1 peak is 7.7 percent of M $^{+}$ . Estimate the number of carbon atoms in the molecule.

Hint

Each carbon contributes about 1.1 percent to the M+1 peak.

Answer

Number of carbons $\approx 7.7/1.1 = 7$ . Seven carbons account for $7 \times 12 = 84$ daltons, leaving $120 - 84 = 36$ for the remaining elements. The carbon count from M+1 is an approximation — elements like nitrogen ( $^{15}$ N at 0.37 percent) also contribute to the M+1 peak, so accurate mass measurement is needed for definitive formula assignment.

Checking candidates: C $_{8}$ H $_{8}$ O gives $96 + 8 + 16 = 120$ with M+1 of $8 \times 1.1 = 8.8$ percent. C $_{9}$ H $_{12}$ gives $108 + 12 = 120$ with M+1 of $9 \times 1.1 = 9.9$ percent. The observed 7.7 percent is closest to 7 carbons, suggesting additional M+1 contributions from heteroatoms rather than pure hydrocarbons.

Formal definition Intermediate+

Mass spectrometry measures the mass-to-charge ratio ( $m / z$ ) of gas-phase ions. A mass spectrometer has three components: an ion source that converts neutral molecules into gas-phase ions, a mass analyser that separates ions by $m / z$ , and a detector that records the ion abundance.

Definition (mass-to-charge ratio). The $m / z$ value of an ion is its mass in daltons (atomic mass units, u) divided by its charge number $z$ (in units of the elementary charge $e$ ). For singly charged ions ( $z = 1$ ), $m / z$ equals the ion mass in daltons. For doubly charged ions ( $z = 2$ ), $m / z$ is half the ion mass. Most small-molecule mass spectra use $z = 1$ exclusively; electrospray ionisation of proteins produces ions with $z = 10$ to $50$ .

Definition (electron ionisation, EI). The sample is vaporised and bombarded with 70 eV electrons from a heated filament. The electron impact removes one electron from the molecule:

M + e^{-} \to M^{+∙} + 2 e^{-}

The resulting molecular ion M $^{+∙}$ (a radical cation) contains 70 eV of energy — far more than typical bond energies (3-5 eV) — so it fragments readily. EI is the standard ionisation method for small molecules (up to about 1000 Da) and produces reproducible fragmentation patterns that can be searched against reference libraries.

Definition (nitrogen rule). A molecule that contains only C, H, O, and halogens has an even nominal molecular mass if it contains zero or an even number of nitrogen atoms, and an odd nominal molecular mass if it contains an odd number of nitrogen atoms. This follows because C (12), O (16), H (1), and halogens (35, 37 for Cl; 79, 81 for Br) all have even mass or come in even-odd pairs, while nitrogen (14, even mass) contributes zero parity per atom but requires a change in the total hydrogen count that shifts the parity.

Counterexamples to common slips

Molecular ion not always present. Under EI conditions, some molecules fragment so readily that the M $^{+}$ peak is absent or very weak. Alcohols, for instance, readily lose water or a hydrogen atom, and the M $^{+}$ may be invisible. Using softer ionisation methods (CI, ESI) preserves the molecular ion.
Odd-electron versus even-electron ions. The molecular ion M $^{+∙}$ is an odd-electron species (radical cation). Most fragment ions are even-electron cations produced by loss of a neutral radical. Odd-electron fragment ions (from rearrangement reactions like the McLafferty rearrangement) are diagnostic because they indicate a specific fragmentation mechanism.
Isotope peaks are not impurities. The M+1, M+2, etc. peaks are from the natural isotopic abundances of the elements in the molecule, not from impurities. A compound with one chlorine atom always shows M+2 at one-third of M $^{+}$ ; a compound with one bromine always shows M+2 equal to M $^{+}$ . These isotope patterns are structural information.

Key result Intermediate+

Theorem (isotopic distribution for a binary isotope system). For an element with two isotopes of masses $m_{1}$ and $m_{2}$ ( $m_{2} = m_{1} + 1$ or $m_{1} + 2$ ) and natural abundances $p$ and $1 - p$ , a molecule containing $n$ atoms of this element has an isotopic peak pattern given by the binomial distribution $(p + q)^{n}$ where $q = 1 - p$ . The peak at M+ $k$ has relative intensity $(k n) p^{n - k} q^{k}$ .

Proof. Each of the $n$ atoms independently contributes either the light isotope (probability $p$ ) or the heavy isotope (probability $q = 1 - p$ ). The probability of exactly $k$ heavy isotopes among $n$ atoms is given by the binomial distribution $(k n) p^{n - k} q^{k}$ . Each heavy isotope adds one (for $^{13}$ C, $^{2}$ H, $^{15}$ N, $^{17}$ O) or two (for $^{18}$ O, $^{37}$ Cl, $^{81}$ Br, $^{34}$ S) mass units to the molecular mass, so the peak at M+ $k$ (or M+ $2 k$ for elements with $+ 2$ isotopes) has intensity proportional to $(k n) p^{n - k} q^{k}$ . For multiple elements, the overall isotopic pattern is the convolution of the individual binomial distributions. $□$

Bridge. The isotopic distribution is the primary tool for determining molecular formulae from mass spectra. For a compound containing $n$ carbon atoms, the M+1 peak is approximately $1.1 n$ percent of M $^{+}$ (from $^{13}$ C at 1.1 percent abundance). For a compound containing one chlorine, M+2 is 32.5 percent of M $^{+}$ (from $^{37}$ Cl at 24.2 percent abundance). These numerical fingerprints allow a chemist to read the elemental composition from the isotope pattern without knowing the structure.

Worked example at intermediate level

Predict the M, M+1, M+2, and M+4 peak intensities for CH $_{2}$ Cl $_{2}$ (dichloromethane, molecular mass 84).

Dichloromethane has 1 carbon, 2 hydrogen, and 2 chlorine atoms. The isotope pattern is dominated by chlorine: $^{35}$ Cl has 75.8 percent abundance and $^{37}$ Cl has 24.2 percent.

For the two chlorine atoms, the binomial distribution gives:

Both $^{35}$ Cl: $(0.758)^{2} = 0.575$ (M peak, $m / z$ = 84)
One $^{35}$ Cl + one $^{37}$ Cl: $2 \times 0.758 \times 0.242 = 0.367$ (M+2 peak, $m / z$ = 86)
Both $^{37}$ Cl: $(0.242)^{2} = 0.059$ (M+4 peak, $m / z$ = 88)

Normalising to the M peak as 100 percent: M = 100, M+2 = 63.8, M+4 = 10.2.

The M+1 contribution from $^{13}$ C ( $1.1$ percent) and $^{2}$ H ( $0.015$ percent $\times 2 = 0.03$ percent) adds about 1.1 percent to the M+2 peak (which already has 63.8 percent from $^{37}$ Cl, so the carbon contribution is negligible in comparison).

The resulting pattern — M : M+2 : M+4 in the approximate ratio 9 : 6 : 1 — is diagnostic of two chlorine atoms and allows immediate visual identification of a dichloro compound in a mass spectrum.

Exercises Intermediate+

Exercise 2 (easy, symbolic).

State the nitrogen rule: for a molecule containing C, H, N, O, and halogens, what is the relationship between the nominal molecular mass and the number of nitrogen atoms?

Hint

Consider the parity (even/odd) of the nominal atomic masses: C = 12 (even), H = 1 (odd), N = 14 (even), O = 16 (even).

Answer

The nitrogen rule states: a molecule containing only C, H, N, O, and halogens has an odd nominal molecular mass if it contains an odd number of nitrogen atoms, and an even nominal molecular mass if it contains an even number (including zero) of nitrogen atoms.

Proof by parity: C (12), N (14), O (16) all have even mass. H (1) is odd, F (19) is odd, Cl (35, 37) are both odd, Br (79, 81) are both odd. A neutral molecule with formula C $_{a}$ H $_{b}$ N $_{c}$ O $_{d}$ X $_{e}$ has mass $12 a + b + 14 c + 16 d + (odd) \times e$ . The parity is the parity of $(b + e)$ . For a neutral molecule, the total number of bonds (and hence hydrogen + halogen count) is constrained: each C makes 4 bonds, each N makes 3, each O makes 2. The sum $4 a + 3 c + 2 d$ gives the total bonding capacity. Each H or halogen uses one bond. So $b + e = 4 a + 3 c + 2 d - 2 \times (number of C-C, C-N, C-O, etc. bonds)$ . The parity of $b + e$ equals the parity of $c$ (nitrogen count). Therefore odd $c$ gives odd mass, even $c$ gives even mass.

Exercise 3 (medium, symbolic).

Predict the M, M+2, and M+4 peak intensities for C $_{6}$ H $_{5}$ Br (bromobenzene, molecular mass 156/158).

Hint

Bromine has two isotopes: $^{79}$ Br (50.69 percent) and $^{81}$ Br (49.31 percent). A compound with one bromine shows M and M+2 of approximately equal intensity.

Answer

The molecular ion appears as a doublet at $m / z$ = 156 ( $^{79}$ Br) and $m / z$ = 158 ( $^{81}$ Br) with relative intensities:

M ( $^{79}$ Br): 50.69 percent
M+2 ( $^{81}$ Br): 49.31 percent

Normalising to M = 100: M+2 = $49.31/50.69 \times 100 = 97.3$ percent.

The M+1 contribution from $^{13}$ C ( $6 \times 1.1 = 6.6$ percent) is superimposed on these peaks: the peak at $m / z$ = 157 includes the $^{13}$ C contribution to the $^{79}$ Br molecular ion, and the peak at $m / z$ = 159 includes the $^{13}$ C contribution to the $^{81}$ Br molecular ion.

The near-equal M and M+2 peaks (1:1 ratio) are the unmistakable signature of a single bromine atom. A compound with two bromine atoms would show M, M+2, M+4 in approximately 1:2:1 ratio.

Exercise 4 (medium, symbolic).

The McLafferty rearrangement is a fragmentation mechanism in which a hydrogen atom is transferred from a gamma-carbon to a carbonyl oxygen through a six-membered transition state, producing an enol radical cation and an olefin. Write the mechanism for 2-pentanone (CH $_{3}$ COCH $_{2}$ CH $_{2}$ CH $_{3}$ ) and predict the $m / z$ of the charged fragment.

Hint

The McLafferty rearrangement requires a gamma-hydrogen relative to the carbonyl. The hydrogen transfers to the oxygen, and the bond between the alpha and beta carbons breaks, giving a charged enol and a neutral alkene.

Answer

2-Pentanone: CH $_{3}$ -C(=O)-CH $_{2}$ -CH $_{2}$ -CH $_{3}$ . The carbonyl carbon is the alpha position, CH $_{2}$ is beta, CH $_{2}$ is gamma (with two H atoms), and CH $_{3}$ is delta.

Mechanism: The gamma-hydrogen transfers to the carbonyl oxygen through a six-membered cyclic transition state. Simultaneously, the bond between the alpha-beta carbons breaks.

Products:

Charged fragment (enol radical cation): CH $_{2}$ =C(OH)CH $_{3}$ at $m / z$ = 58 ( $14 + 13 + 16 + 15 = 58$ , specifically: C $_{2}$ H $_{3}$ O $^{+}$ + CH $_{3}$ = actually CH $_{2}$ =C(OH)(CH $_{3}$ ) is C $_{3}$ H $_{6}$ O = 58)
Neutral fragment (alkene): CH $_{2}$ =CH $_{2}$ (ethene, $m / z$ = 28, not detected)

The McLafferty rearrangement peak at $m / z$ = 58 is diagnostic of methyl ketones with at least three carbons in the alkyl chain. The alpha-cleavage (loss of an alkyl group from the carbonyl) gives a complementary peak at $m / z$ = 43 (CH $_{3}$ CO $^{+}$ ). The presence of both peaks — 58 from McLafferty and 43 from alpha-cleavage — is characteristic of methyl ketones.

Exercise 5 (medium, symbolic).

A compound with molecular mass 104 shows fragment peaks at $m / z$ = 89, 76, 61, and 43. The M+2 peak is approximately 4.4 percent of M $^{+}$ . Propose a molecular formula and identify the likely fragments.

Hint

The M+2 of 4.4 percent suggests about 4 carbons (4 $\times$ 1.1 = 4.4 percent from $^{13}$ C $_{2}$ , but M+2 from carbon is actually about $n (n - 1) /2 \times (0.011)^{2} \times 100$ , which is small). Actually, M+2 = 4.4 percent could also indicate one sulfur atom ( $^{34}$ S at 4.2 percent abundance). Mass 104 minus S (32) = 72, consistent with C $_{5}$ H $_{12}$ (72 + 12 = 84, no). Consider C $_{4}$ H $_{8}$ O $_{2}$ = $48 + 8 + 32 = 88$ (no). Try oxygen: M+2 from $^{18}$ O is negligible.

Answer

The M+2 of 4.4 percent is a strong indicator of one sulfur atom ( $^{34}$ S has 4.21 percent natural abundance). Subtracting S (32) leaves 72, consistent with C $_{5}$ H $_{12}$ ( $60 + 12 = 72$ ) but that gives formula C $_{5}$ H $_{12}$ S with mass $60 + 12 + 32 = 104$ . Check: C $_{5}$ H $_{12}$ S (pentanethiol or methylbutanethiol).

Fragmentation:

$m / z$ = 89: loss of 15 (CH $_{3}$ $^{∙}$ ) $\to$ C $_{4}$ H $_{9}$ S $^{+}$ (alpha-cleavage at the sulfur)
$m / z$ = 76: loss of 28 (C $_{2}$ H $_{4}$ or CO) from M $^{+}$
$m / z$ = 61: loss of 43 (C $_{3}$ H $_{7}$ $^{∙}$ ) $\to$ CH $_{3}$ S $^{+}$ or C $_{2}$ H $_{5}$ S $^{+}$
$m / z$ = 43: C $_{3}$ H $_{7}^{+}$ (propyl cation from alpha-cleavage)

The nitrogen rule confirms: mass 104 is even, and the formula C $_{5}$ H $_{12}$ S contains zero nitrogen atoms (even number), consistent. The dominant fragmentation at $m / z$ = 43 (C $_{3}$ H $_{7}^{+}$ ) and 61 (CH $_{2}$ =SH $^{+}$ or CH $_{3}$ S $^{+}$ ) suggests 1-pentanethiol, where alpha-cleavage at the C-S bond gives complementary fragments.

Exercise 6 (hard, symbolic).

Derive the time-of-flight (TOF) relation $t = L m / (2 z V)$ for an ion of mass $m$ and charge $z e$ accelerated through potential $V$ in a field-free drift tube of length $L$ . Calculate the mass resolution $m /Δ m$ achievable if the flight time can be measured to a precision of $Δ t$ , and show that the resolution increases with the square root of $m$ .

Hint

The ion gains kinetic energy $z e V = m v^{2} /2$ , giving velocity $v = 2 z e V / m$ . The flight time is $t = L / v$ . Differentiate to find $d t / d m$ and convert to $m /Δ m$ .

Answer

An ion of mass $m$ and charge $z e$ accelerated through potential $V$ gains kinetic energy $K E = z e V = m v^{2} /2$ . The velocity is

v = \frac{2 z e V}{m} .

The time to traverse the drift tube of length $L$ is

t = \frac{L}{v} = L \frac{m}{2 z e V} .

Differentiating with respect to $m$ :

\frac{d t}{d m} = \frac{L}{2 2 z e V} \cdot m^{- 1/2} = \frac{t}{2 m} .

The mass resolution is

\frac{m}{Δ m} = \frac{m}{2 m \cdot Δ t / t} \cdot \frac{t}{1} = \frac{t}{2Δ t} .

Substituting $t = L m / (2 z e V)$ :

\frac{m}{Δ m} = \frac{L}{2Δ t} \frac{m}{2 z e V} .

The resolution is proportional to $m$ — heavier ions spend more time in the drift tube and are resolved better. For a practical TOF instrument with $L = 1$ m, $V = 2000$ V, and $Δ t = 1$ ns:

$t = 1 \times 100/ (2 \times 2000) \approx 1 \times 0.025 = 0.158$ ms for $m / z = 100$ .

Resolution $= 0.158 \times 1 0^{- 3} / (2 \times 1 0^{- 9}) = 79, 000$ . Modern reflectron TOF instruments achieve resolutions of $m /Δ m > 40, 000$ by compensating for the initial kinetic-energy spread of the ions.

Ionisation methods: from EI to ESI and MALDI Master

Electron ionisation (EI) at 70 eV is the standard for small-molecule mass spectrometry because it produces reproducible, library-searchable fragmentation patterns. The 70 eV energy is a historical convention: above about 50 eV the fragmentation pattern becomes nearly independent of electron energy, so 70 eV provides stable, reproducible spectra that can be compared across instruments and laboratories. The NIST mass spectral library contains over 300 000 EI spectra searchable by fragmentation pattern.

However, EI has two limitations. First, the sample must be vaporised before ionisation, which excludes thermally labile and non-volatile compounds (most biomolecules, polymers, and pharmaceuticals). Second, the 70 eV energy causes extensive fragmentation, and for many compounds the molecular ion is absent. Chemical ionisation (CI) addresses the second limitation by using a reagent gas (methane, isobutane, or ammonia) that is ionised by electron impact and then transfers a proton to the analyte in a soft proton-transfer reaction. The resulting [M+H] $^{+}$ quasi-molecular ion is much less energetic than M $^{+∙}$ from EI, so fragmentation is minimal and the molecular mass is easily identified. CI sacrifices the rich fragmentation information of EI in exchange for a reliable molecular ion.

Two ionisation methods developed in the late 1980s transformed mass spectrometry from a small-molecule technique into a universal analytical tool. Electrospray ionisation (ESI), developed by Fenn ^[Fenn2002], dissolves the analyte in a volatile solvent and sprays it through a fine needle at high voltage. The charged droplets evaporate, Coulombic repulsion breaks them into smaller droplets, and eventually individual gas-phase ions are released. ESI produces multiply charged ions from large biomolecules — a 50 kDa protein with 30 protons attached appears at $m / z \approx 1667$ , well within the range of a standard quadrupole analyser. This charge-state distribution is itself informative: the spacing between adjacent charge states gives the molecular mass independently of calibration. Fenn shared the 2002 Nobel Prize in Chemistry for ESI.

Matrix-assisted laser desorption/ionisation (MALDI), developed by Karas and Hillenkamp ^[Karas1988] and recognised by Tanaka's share of the 2002 Nobel ^[Tanaka2002], embeds the analyte in a solid matrix (such as sinapinic acid or dihydroxybenzoic acid) that strongly absorbs UV laser light. A short laser pulse vaporises and ionises the matrix, which transfers charge to the analyte in a soft process that produces predominantly singly charged ions. MALDI combined with time-of-flight (TOF) analysis gives a mass range limited only by the detector, enabling analysis of proteins, DNA, polymers, and even entire viruses with masses above 1 MDa. The development of ESI and MALDI opened mass spectrometry to biology and medicine, creating the fields of proteomics, metabolomics, and lipidomics.

Mass analysers: quadrupole, TOF, ion trap, Orbitrap, and FT-ICR Master

The mass analyser separates ions by $m / z$ and delivers them to the detector. Five analyser types dominate modern mass spectrometry.

The quadrupole analyser uses four parallel rods with a combination of DC and RF voltages. Ions oscillate in the transverse electric field, and only ions of a specific $m / z$ have stable trajectories that pass through to the detector; all others hit the rods. Scanning the voltages sweeps the transmitted $m / z$ across the mass range. The quadrupole is compact, inexpensive, fast-scanning, and tolerant of high pressure — it is the workhorse analyser in GC-MS and LC-MS instruments. Its mass resolution is modest ( $m /Δ m \approx 1000$ -4000), sufficient to separate unit-mass differences but not isobaric species.

The time-of-flight (TOF) analyser, described in Exercise 6, measures the transit time of ions through a field-free drift tube. All ions are accelerated to the same kinetic energy, so lighter ions travel faster and arrive first. The reflectron TOF uses an electrostatic mirror to correct for the initial kinetic-energy spread, improving resolution to $m /Δ m > 40, 000$ . TOF analysers have an unlimited mass range and very high acquisition speed (microseconds per spectrum), making them ideal for MALDI-TOF and for coupling with fast chromatographic separations.

The ion trap (quadrupole ion trap or Paul trap) confines ions in a three-dimensional RF quadrupole field. Trapped ions can be stored, fragmented by collision-induced dissociation (CID), and then ejected in order of $m / z$ for detection. The ion trap enables tandem mass spectrometry (MS/MS or MS $^{n}$ ): a precursor ion is isolated, fragmented, and the product ion spectrum is recorded. Sequential rounds of isolation and fragmentation (MS $^{3}$ , MS $^{4}$ , etc.) provide detailed structural information about complex molecules.

The Orbitrap analyser, invented by Makarov in 1999, traps ions in an electrostatic field between an inner spindle-shaped electrode and an outer barrel electrode. The ions orbit the central electrode and simultaneously oscillate along its axis at a frequency proportional to $z / m$ . A Fourier transform of the detected oscillation frequencies gives the mass spectrum with resolution up to $m /Δ m > 500, 000$ and mass accuracy below 1 ppm. The Orbitrap has become the dominant high-resolution analyser in proteomics and metabolomics because of its combination of high resolution, high mass accuracy, compact size, and relatively low cost compared with FT-ICR.

The Fourier-transform ion cyclotron resonance (FT-ICR) analyser traps ions in a strong magnetic field (typically 7-15 tesla) and measures their cyclotron frequency $ω_{c} = z B / m$ . The ions are excited to coherent cyclotron orbits by an RF pulse, and the image current on detector plates is Fourier-transformed to give the mass spectrum. FT-ICR achieves the highest mass resolution of any analyser ( $m /Δ m > 1 0^{6}$ ) and the highest mass accuracy (sub-ppm), but the required superconducting magnet makes it the most expensive mass spectrometer.

Fragmentation theory and the McLafferty rearrangement Master

The fragmentation of a molecular ion under EI conditions is governed by two principles: the stability of the charged fragment and the stability of the neutral fragment. Bonds break preferentially to produce the most stable cation and the most stable radical. This predictability is what makes mass spectral interpretation systematic.

Alpha-cleavage is the most common fragmentation mechanism. In a molecule containing a heteroatom (O, N, S), the radical site on the molecular ion (typically localised on the heteroatom by the ionisation process) induces cleavage of the adjacent C-C bond. The charge remains on the heteroatom-containing fragment. For 2-pentanone, alpha-cleavage at the carbonyl carbon gives CH $_{3}$ CO $^{+}$ at $m / z$ = 43 and the complementary radical CH $_{2}$ CH $_{2}$ CH $_{3}$ $^{∙}$ (neutral, not detected). Alpha-cleavage also operates at the C-N bond in amines, the C-O bond in ethers and alcohols, and the C-S bond in thiols.

The McLafferty rearrangement ^{[McLafferty1959]} is a characteristic fragmentation of carbonyl compounds that have a gamma-hydrogen. The mechanism involves a concerted six-membered transition state: the gamma-hydrogen transfers to the carbonyl oxygen while the alpha-beta bond breaks, producing an enol radical cation (charged, detected) and an alkene (neutral, not detected). The McLafferty peak is typically strong and appears at a predictable $m / z$ determined by the structure of the enol fragment. Methyl ketones always show a McLafferty peak at $m / z$ = 58 (for butanone) or $m / z$ = 72 (for 2-pentanone), etc. The rearrangement requires a specific structural motif (carbonyl with gamma-H), so the presence of the McLafferty peak is diagnostic of this structural element.

Retro-Diels-Alder fragmentation operates in cyclohexene derivatives: the molecular ion fragments into a diene and an alkene (or dienophile) in the reverse of the Diels-Alder reaction. The charged fragment carries the mass of either component. This fragmentation is diagnostic of cyclohexene and norbornene ring systems in natural products and is one of the few fragmentation mechanisms that produces odd-electron fragment ions from an odd-electron precursor.

Loss of small neutrals provides additional structural information. Loss of 15 (CH $_{3}$ ) indicates a methyl group; loss of 17 (OH) indicates an alcohol; loss of 18 (H $_{2}$ O) is characteristic of alcohols and carboxylic acids; loss of 28 (CO or C $_{2}$ H $_{4}$ ) is common in carbonyls and aromatics; loss of 29 (C $_{2}$ H $_{5}$ or CHO) indicates an ethyl group or an aldehyde; loss of 31 (OCH $_{3}$ ) indicates a methyl ester; loss of 44 (CO $_{2}$ ) is characteristic of carboxylic acids and esters; loss of 45 (OC $_{2}$ H $_{5}$ or CHS) indicates an ethyl ester or a thiol. The masses of the lost neutrals are a first-pass structural analysis tool: by listing the neutral losses from the molecular ion, a chemist identifies the functional groups present.

Connections Master

Electronic spectroscopy 14.12.04. Photoionisation (a specific ionisation method used in some mass spectrometers) is the mass-spectrometric analogue of electronic spectroscopy: a photon promotes an electron from a molecular orbital to the ionisation continuum, and the resulting ion is analysed by $m / z$ . Photoionisation efficiency curves (ion yield versus photon energy) provide the ionisation energy of the molecule and the energy thresholds for specific fragmentation pathways, connecting the electronic energy levels to the mass-spectrometric fragmentation pattern.
Vibrational spectroscopy 14.12.03. IR multiphoton dissociation (IRMPD) uses infrared photons to excite specific vibrational modes of a trapped ion, depositing enough energy to break bonds. By scanning the IR wavelength and monitoring the fragment yield, an IR spectrum of the mass-selected ion is obtained. This IR-MS combination provides vibrational spectroscopy of individual ionic species that cannot be isolated in sufficient quantity for conventional IR spectroscopy.
Quantum chemistry 14.04.01. Accurate mass measurement (to 1 ppm or better) requires knowledge of exact atomic masses, which depend on nuclear binding energies derived from mass-defect calculations. The mass defect — the difference between the nominal mass and the exact mass — is characteristic of the elemental composition and is used in formula assignment algorithms that combine accurate mass with isotopic pattern matching.
Statistical mechanics 14.07.02. The quasi-equilibrium theory (QET) of mass spectral fragmentation, developed by Rosenstock, Wahrhaftig, and Eyring in 1952, treats the fragmentation rates as unimolecular dissociation rates computed from the density of states of the molecular ion. The rate constant for each fragmentation pathway is $k (E) = (1/ h) \int_{0}^{E - E_{0}} ρ^{‡} (ϵ) d ϵ / ρ (E)$ , where $E_{0}$ is the activation energy, $ρ (E)$ is the density of states of the molecular ion, and $ρ^{‡}$ is the density of states of the transition state. This is a direct application of RRKM theory to mass spectrometry.
Organic chemistry: functional groups 15.02.01. The fragmentation patterns of organic molecules map onto the functional-group classification. Alpha-cleavage at carbonyls, alcohols, amines, and ethers produces diagnostic fragment ions whose masses identify the functional group. The McLafferty rearrangement identifies carbonyl compounds with gamma-hydrogens. The loss of specific neutrals (H $_{2}$ O, CO, CO $_{2}$ , CH $_{3}$ OH) identifies specific structural elements. Mass spectral interpretation is the mass-spectrometric analogue of the IR group-frequency approach.

Historical notes Master

Mass spectrometry began with J. J. Thomson's discovery of canal rays (positive ions) in 1912 and his development of the first mass spectrograph, published in his 1913 book Rays of Positive Electricity ^{[Thomson1913]}. Thomson's apparatus used electric and magnetic fields to deflect positive ions onto a photographic plate, producing parabolic curves whose positions depended on $m / z$ . Using this instrument, Thomson discovered that neon had two isotopes ( $^{20}$ Ne and $^{22}$ Ne), establishing the existence of isotopes and earning him the 1906 Nobel Prize in Physics (for the electron, not for isotopes).

Francis Aston refined Thomson's mass spectrograph into a precision instrument capable of measuring atomic masses to four decimal places. His 1919 mass spectrograph and the subsequent development of the mass spectrograph through the 1920s and 1930s ^[Aston1942] established the exact masses and abundances of the naturally occurring isotopes of all the elements. Aston's whole-number rule — that all atomic masses are approximately integers when measured in atomic mass units — and the discovery of the mass defect (the small deviation from integer values) provided one of the earliest experimental confirmations of Einstein's mass-energy equivalence $E = m c^{2}$ . Aston received the 1922 Nobel Prize in Chemistry.

The development of modern mass spectrometry proceeded through several transformative advances. The double-focusing mass spectrometer (Mattauch and Herzog, 1934) improved mass resolution by combining electric and magnetic sector focusing. The time-of-flight analyser (Cameron and Eggers, 1948; Stephens, 1946) enabled fast spectral acquisition. The quadrupole mass filter (Paul and Steinwedel, 1953; Nobel Prize 1989) provided a compact, scanning mass analyser that became the basis of GC-MS instruments. The ion trap (Paul, 1960) enabled tandem mass spectrometry in a single device.

The coupling of mass spectrometry with gas chromatography (GC-MS) in the late 1950s was a turning point. By separating a mixture into its components by GC and then identifying each component by its mass spectrum, GC-MS became the most powerful tool for analysing complex organic mixtures. The development of computerised spectral libraries in the 1970s made GC-MS identification essentially automatic, and GC-MS became the standard method for drug testing, environmental analysis, forensic science, and metabolite identification.

The 2002 Nobel Prize in Chemistry was shared by Fenn (for ESI) and Tanaka (for soft laser desorption), recognising the transformation of mass spectrometry from a small-molecule technique into a universal tool for biomolecular analysis. ESI and MALDI enabled the measurement of proteins, DNA, and other macromolecules with mass accuracies of better than 0.01 percent, creating the field of proteomics. The subsequent development of the Orbitrap analyser (Makarov, 1999) and hybrid instruments (Q-TOF, LTQ-Orbitram) has pushed mass resolution and accuracy to levels that enable the identification of thousands of proteins from a single biological sample, making mass spectrometry the central analytical technology in modern biology.

Bibliography Master

@book{Atkins2022,
  author = {Atkins, P. and de Paula, J. and Keeler, J.},
  title = {Physical Chemistry},
  edition = {12},
  publisher = {Oxford University Press},
  year = {2022},
}

@book{Pavia2015,
  author = {Pavia, D. L. and Lampman, G. M. and Kriz, G. S. and Vyvyan, J. R.},
  title = {Introduction to Spectroscopy},
  edition = {5},
  publisher = {Cengage},
  year = {2015},
}

@book{Gross2017,
  author = {Gross, J. H.},
  title = {Mass Spectrometry: A Textbook},
  edition = {3},
  publisher = {Springer},
  year = {2017},
}

@book{McLafferty1993,
  author = {McLafferty, F. W. and Turecek, F.},
  title = {Interpretation of Mass Spectra},
  edition = {4},
  publisher = {University Science Books},
  year = {1993},
}

@book{Thomson1913,
  author = {Thomson, J. J.},
  title = {Rays of Positive Electricity and Their Application to Chemical Analyses},
  publisher = {Longmans, Green and Co.},
  year = {1913},
}

@book{Aston1942,
  author = {Aston, F. W.},
  title = {Mass Spectra and Isotopes},
  edition = {2},
  publisher = {Edward Arnold},
  year = {1942},
}

@article{McLafferty1959,
  author = {McLafferty, F. W.},
  title = {Mass Spectrometric Analysis: Molecular Rearrangements},
  journal = {Analytical Chemistry},
  volume = {31},
  year = {1959},
  pages = {82--87},
}

@article{Karas1988,
  author = {Karas, M. and Hillenkamp, F.},
  title = {Laser Desorption Ionization of Proteins with Molecular Masses Exceeding 10,000 Daltons},
  journal = {Analytical Chemistry},
  volume = {60},
  year = {1988},
  pages = {2299--2301},
}

@book{Fenn2002,
  author = {Fenn, J. B.},
  title = {Electrospray Ionization Mass Spectrometry: How It All Began (Nobel Lecture)},
  journal = {Angewandte Chemie International Edition},
  volume = {42},
  year = {2003},
  pages = {3871--3894},
}

@book{Tanaka2002,
  author = {Tanaka, K.},
  title = {The Origin of Macromolecule Ionization by Laser Irradiation (Nobel Lecture)},
  journal = {Angewandte Chemie International Edition},
  volume = {42},
  year = {2003},
  pages = {3860--3870},
}

Prerequisites

14.12.01 pending

Tier anchors

beginner: Atkins, Ch. 11.7; Pavia — Introduction to Spectroscopy, 5e (2015), Ch. 1
intermediate: Gross — Mass Spectrometry, 3e (2017), Ch. 1-2; McLafferty — Interpretation of Mass Spectra, 4e (1993)
master: Thomson — Rays of Positive Electricity (1913); Aston — Mass Spectra and Isotopes, 2e (1942)

References

Atkins, P. & de Paula, J. — Physical Chemistry, 12e (Oxford UP, 2022) · Ch. 11.7 Mass spectrometry
Pavia, D. L., Lampman, G. M., Kriz, G. S. & Vyvyan, J. R. — Introduction to Spectroscopy, 5e (Cengage, 2015) · Ch. 1 Mass spectrometry
Gross, J. H. — Mass Spectrometry: A Textbook, 3e (Springer, 2017) · Ch. 1-2 Introduction and ionisation methods
McLafferty, F. W. & Turecek, F. — Interpretation of Mass Spectra, 4e (University Science Books, 1993) · Ch. 1-5 Fragmentation theory and interpretation
Thomson1913 · Thomson, J. J. — Rays of Positive Electricity and Their Application to Chemical Analyses (Longmans, Green, 1913)
Aston1942 · Aston, F. W. — Mass Spectra and Isotopes, 2e (Edward Arnold, 1942); Nobel Prize 1922
McLafferty1959 · McLafferty, F. W., Anal. Chem. 31 (1959), 82-87 — correlation of mass spectra with molecular structure
Karas1988 · Karas, M. & Hillenkamp, F., Anal. Chem. 60 (1988), 2299-2301 — MALDI ionisation
Tanaka2002 · Tanaka, K., Nobel Lecture (2002) — soft laser desorption for mass spectrometry of biomolecules
Fenn2002 · Fenn, J. B., Nobel Lecture (2002) — electrospray ionisation for mass spectrometry of large biomolecules

Estimated time

beginner: 12m
intermediate: 30m
master: 45m