Explain mass-spectrometric ionisation and fragmentation

Electron impact removes one electron: CH₄ + e⁻ → CH₄⁺• + 2e⁻.

GCE A-Level H3 Chemistry 9813-2027

On this page

Learning objectives

Explain mass-spectrometric ionisation and fragmentation

Form the molecular radical cation

Electron impact removes one electron: CH₄ + e⁻ → CH₄⁺• + 2e⁻.

The parent ion is both positively charged and odd-electron.

Follow the particles through the instrument: the sample is converted to gas-phase positive ions, those ions may fragment, and only charged species are separated and detected. Neutral fragments do not produce their own peaks.

Separate ionisation from cleavage

Ionisation creates M⁺•; fragmentation is subsequent bond cleavage.

A spectrum may contain intact M⁺• and smaller charged fragments.

Electron ionisation removes an electron from a molecule to form an odd-electron molecular ion, M⁺•. Further bond cleavage can produce an even-electron positive fragment and a neutral radical while conserving atoms and charge.

Track detected charge

Ethanol M⁺• can cleave to CH₂OH⁺ m/z 31 plus CH₃•.

Only charged products are accelerated/detected; neutral radicals are absent.

Write electron ionisation as M(g) + e⁻ → M⁺•(g) + 2e⁻. The incoming electron is not added to the molecule; one molecular electron is removed and two electrons leave the collision.

Fragmentation is not random atom loss. A proposed channel must start from the supplied connectivity, identify the broken bond, place the positive charge on one product and balance the complementary neutral species.

Balance an EI equation

CH₃Cl + e⁻ → CH₃Cl⁺• + 2e⁻ conserves atoms and charge.

Do not write CH₃Cl⁻ or omit the second outgoing electron.

For ethanol, first form C₂H₆O⁺• at m/z 46. Cleavage beside oxygen can give CH₂OH⁺ at m/z 31 and CH₃• as the neutral radical; the charged fragment reaches the detector.

Try this

Write a balanced electron-ionisation equation for chloromethane, CH₃Cl, and give the m/z of the ion formed when the molecule contains ³⁵Cl.

Check your answer

CH₃Cl(g) + e⁻ → CH₃Cl⁺•(g) + 2e⁻. The atoms balance, and so does charge: −1 on the left and +1 − 2 = −1 on the right. The product is a radical cation, not CH₃Cl⁻, because an electron is removed rather than added. With ³⁵Cl, m/z = (12 + 3 + 35)/1 = 50.

Assign charge retention

Propanone⁺• → CH₃CO⁺ m/z 43 + CH₃• places charge on the acylium fragment.

The neutral methyl radical has mass 15 but makes no m/z 15 peak; only CH₃⁺ would make that peak.

Audit a channel in four steps: parent formula, broken bond, charged formula with m/z, and neutral partner. If any atoms or charge disappear, the proposed fragmentation is incomplete.

Try this

Propanone, CH₃COCH₃, forms M⁺• at m/z 58. Breaking one C–C bond beside the carbonyl group gives a peak at m/z 43. Write the fragmentation equation, label the charged and neutral products, and state which one is detected.

Check your answer

CH₃COCH₃⁺• → CH₃CO⁺ + CH₃•. The acylium ion CH₃CO⁺ (24 + 3 + 16 = 43) carries the positive charge and gives the m/z 43 peak. The methyl radical CH₃• (mass 15) is neutral, so it gives no peak. Atoms balance (C₃H₆O = C₂H₃O + CH₃), charge balances (+1 = +1 + 0) and the masses add up: 43 + 15 = 58.

Compare competing cleavage

Butan-2-one can yield CH₃CO⁺ m/z 43 or C₂H₅CO⁺ m/z 57.

Relative stability/cleavage propensity changes abundance; both assignments conserve the parent atoms.

Compare two possible cleavages of an unsymmetrical molecule rather than assuming the largest fragment must be charged. The spectrum provides evidence about which charged ions are sufficiently formed and stable to detect.

Try this

Butan-2-one, CH₃COCH₂CH₃, gives M⁺• at m/z 72 and fragments at m/z 43 and 57. Write the two cleavage equations for the C–C bonds beside the carbonyl group and check that each balances atoms, charge and mass.

Check your answer

CH₃COC₂H₅⁺• → CH₃CO⁺ (m/z 43) + C₂H₅• (29), and 43 + 29 = 72. CH₃COC₂H₅⁺• → C₂H₅CO⁺ (m/z 57) + CH₃• (15), and 57 + 15 = 72. Each channel breaks a bond next to C=O, conserves C₄H₈O and leaves the charge on an acylium ion, while the complementary neutral radical gives no peak.

Common mistake: every-fragment detection

Neutral loss can be chemically essential yet invisible to the detector.

A peak represents an ion, not every product of bond cleavage.

The molecular ion is not the original neutral molecule with an electron attached. Electron ionisation makes a positive radical cation by removing an electron.

Try this

Correct this statement: ‘When CH₃COCH₃⁺• breaks into CH₃CO⁺ and CH₃•, peaks appear at both m/z 43 and m/z 15.’

Check your answer

Only the charged product is accelerated, separated and detected. In this channel CH₃CO⁺ gives the m/z 43 peak, while CH₃• is neutral and gives no peak. A peak at m/z 15 needs CH₃⁺, which forms in a different channel, CH₃COCH₃⁺• → CH₃⁺ + CH₃CO•, where the charge stays on the methyl fragment.

Cross-check a fixed spectrum

Ethanol peaks m/z 46 and 31 can be M⁺• and CH₂OH⁺ respectively.

Next: m/z.

A full mechanism answer writes the ionisation equation, labels M⁺•, shows the broken bond, balances the neutral radical and explains why only the positive product gives the stated peak.

Try this

The ethanol spectrum, CH₃CH₂OH, has peaks at m/z 46 and 31. Write the ionisation equation that produces the m/z 46 ion and the fragmentation equation that produces the m/z 31 ion, balancing atoms and charge.

Check your answer

CH₃CH₂OH + e⁻ → CH₃CH₂OH⁺• + 2e⁻ forms M⁺• at 24 + 6 + 16 = 46. Breaking the C–C bond next to the oxygen-bearing carbon then gives CH₃CH₂OH⁺• → CH₂OH⁺ + CH₃•. CH₂OH⁺ has m/z 12 + 3 + 16 = 31 and is detected; CH₃• (15) is neutral and undetected, and 31 + 15 = 46.