Fragmentation of the Molecular Ion

Introduction

If a mass spectrum showed only the molecular ion, it would tell you the mass of a molecule but little else. In practice, electron impact spectra contain dozens of peaks at lower m/z values. These arise because the molecular ion breaks apart — it fragments — before reaching the detector. Far from being a nuisance, fragmentation is a gift: the pieces reveal how the atoms are joined, rather as the fragments of a broken vase reveal its original shape.

Core explanation

Why fragmentation happens. The bombarding electrons in EI typically carry 70 eV, while a C–C bond needs only about 3.6 eV (roughly 350 kJ mol⁻¹) to break. Removing one electron needs around 10 eV, so the molecular ion is left with a large surplus of internal energy. The ion also has one fewer bonding electron, so some bonds are weakened. Within microseconds, many molecular ions break one of their bonds.

What forms when a bond breaks. The molecular ion is a radical cation : it has both a positive charge and an unpaired electron. When one covalent bond breaks, the two resulting pieces share out the charge and the unpaired electron. One piece becomes a cation (charged, all electrons paired) and the other a radical (neutral, with an unpaired electron):

M⁺• → X⁺ + Y•

Only charged fragments are detected. The mass spectrometer can accelerate, separate and detect only ions. The neutral radical Y• is invisible: it drifts off and is pumped away. So each fragmentation produces one peak, at the m/z of X⁺.

Either piece can carry the charge. A molecular ion A–B⁺• can split in two ways:

A–B⁺• → A⁺ + B• or A–B⁺• → A• + B⁺

Both happen to some extent, so both A⁺ and B⁺ usually appear, but in different amounts depending on the stability of each cation.

Example: propane. Propane, CH₃CH₂CH₃, has M = 44. Breaking a C–C bond gives:

- CH₃CH₂CH₃⁺• → CH₃CH₂⁺ + CH₃• (peak at m/z 29) - CH₃CH₂CH₃⁺• → CH₃⁺ + CH₃CH₂• (peak at m/z 15)

The spectrum shows a large peak at 29 and a small one at 15.

Losing a neutral molecule. Sometimes a radical cation expels a small stable molecule, such as H₂O or CO, instead of a radical. The product is still a radical cation:

M⁺• → [M−18]⁺• + H₂O

Because a stable molecule is lost, such fragmentations are often favourable.

Further fragmentation. Fragment ions with enough energy can break again, producing a cascade of smaller ions. This is why long chains give families of peaks.

Step-by-step reasoning

To predict fragment peaks for a simple molecule:

1. Draw the structure and find Mr. 2. Choose a single bond to break. 3. Place the positive charge on one piece and the unpaired electron on the other. 4. Calculate the mass of the charged piece; that is the m/z of the peak. 5. Repeat with the charge on the other piece, then with other bonds.

Visual explanation

Draw propane with a positive charge and a dot over the central carbon to represent M⁺•. Draw a single-headed "fishhook" arrow from the C–C bond towards each carbon, showing the two electrons of the bond separating one each. The left CH₃ becomes a radical and the CH₃CH₂ group carries the positive charge.

Real-world analogy

Imagine a tired parent carrying a baby and a shopping bag who suddenly sets one down. Whichever is put down becomes "neutral" and stays behind; whatever is still being carried is the "charge". The observer at the door only sees the parent arrive with whatever was kept — the charged fragment.

Real-world example

Mass-spectral libraries used by forensic laboratories hold the fragmentation patterns of hundreds of thousands of compounds. When a suspected drug sample is analysed, its fragment pattern is matched against the library, and a close match identifies the substance with high confidence.

Why?

Why is one fragment detected but its partner not? Detection depends on charge. The cation is accelerated towards the detector; the radical, being neutral, feels no force from the electric field and is not directed anywhere. Only mass differences between peaks reveal what neutral species were lost.

Common misconception

"When M⁺ fragments, both pieces appear in the spectrum." Only the charged piece gives a peak. The neutral radical is lost, and its mass can be inferred only from the gap between the parent ion and the fragment ion.

Worked example

Question: Pentane, CH₃CH₂CH₂CH₂CH₃ (Mr 72), shows peaks at m/z 57, 43 and 29. Identify each fragment ion and the neutral species lost.

Reasoning: 72 − 57 = 15, loss of CH₃•. 72 − 43 = 29, loss of C₂H₅•. 72 − 29 = 43, loss of C₃H₇•.

Answer: m/z 57 is C₄H₉⁺, m/z 43 is C₃H₇⁺ and m/z 29 is C₂H₅⁺.

Quick check

1. Write an equation for the fragmentation of the ethane molecular ion into a methyl cation and a methyl radical. Answer: CH₃CH₃⁺• → CH₃⁺ + CH₃•, giving a peak at m/z 15.

Exam focus

Examiners often ask for an equation for the formation of a given fragment. Show the molecular ion as a radical cation on the left and a cation plus a radical on the right; charge and electrons must balance. Do not show the radical as charged, and do not forget the positive sign on the ion.

Advanced insight

Fragmentation can be driven by the charge site or the radical site. In α-cleavage next to a heteroatom, the unpaired electron on oxygen or nitrogen pairs with one electron from a neighbouring bond, breaking it homolytically. Rearrangements are also possible: in the McLafferty rearrangement, a hydrogen atom migrates through a six-membered ring before a neutral alkene is expelled.

Summary

Electron impact leaves molecular ions with excess energy, so many fragment. A radical cation splits into a cation and a radical (M⁺• → X⁺ + Y•), or loses a small neutral molecule. Only the charged fragment is detected. The mass difference between M and a fragment reveals the neutral piece lost, and fragment patterns reveal how the molecule is built.

Practice questions

1. Explain why the neutral fragment formed during fragmentation does not appear in a mass spectrum. Answer: It has no charge, so it is not accelerated or deflected and never reaches the detector. 2. Write an equation to show the formation of the ion at m/z 29 from the molecular ion of butane. Answer: CH₃CH₂CH₂CH₃⁺• → CH₃CH₂⁺ + CH₃CH₂• 3. A compound of Mr 60 shows a peak at m/z 45. What mass of neutral fragment was lost, and what might it be? Answer: 15, most likely a methyl radical, CH₃•. 4. An alcohol of Mr 74 gives a peak at m/z 56. Suggest what was lost and whether the fragment is a cation or a radical cation. Answer: 74 − 56 = 18, loss of H₂O, a neutral molecule, so the fragment is still a radical cation, [M−18]⁺•.