Why a Molecular Ion Can Be Weak or Missing

A mass spectrum can have a huge peak at m/z 43 and only a tiny peak at m/z 74. Yet the neutral molecule can still have a molecular mass of 74. In fact, some molecules produce such unstable molecular ions that the molecular-ion peak is barely visible or disappears altogether.

The main reason is fragmentation. During electron-impact ionisation, a molecule loses an electron and becomes a positively charged molecular ion. However, the ion can also gain enough internal energy to break apart before reaching the detector.

This creates an important distinction: the most abundant ion in a mass spectrum is not necessarily the molecular ion. The tallest peak is called the base peak, and it often represents a particularly stable fragment.

Let’s work through why this happens, how isotope patterns can help, and what you can conclude when the molecular-ion peak is missing.

01Why doesn’t every molecule produce a strong molecular-ion peak?

Imagine removing one brick from a carefully balanced structure. The structure might remain standing, or the change might cause several other parts to collapse.

Electron-impact ionisation creates a similar situation. Removing an electron changes the electronic structure of a molecule. Depending on the molecule and the energy involved, the resulting ion may remain intact or fragment into smaller ions.

Before looking at the chemistry, consider how the instrument actually makes its measurements.

02How electron-impact mass spectrometry works

The measurement problem is straightforward: how can we determine the mass of something too small to place on a balance?

A mass spectrometer solves this by converting molecules into charged particles and separating those particles according to their mass-to-charge ratio.

In a classic electron-impact magnetic-sector mass spectrometer, the process follows this sequence:

1. Sample introduction: The sample is vaporised and introduced into an evacuated ionisation chamber. The vacuum reduces collisions with other particles.
2. Ionisation: A beam of high-energy electrons passes through the sample vapour. Collisions between the electrons and molecules produce positive ions.
3. Acceleration: An electric field accelerates the positive ions through narrow openings.
4. Mass separation: The ions enter a magnetic field, which bends their paths. Their trajectories depend on their mass-to-charge ratios.
5. Detection: Ions reaching the detector produce a measurable signal. The instrument records their abundance at different m/z values.

The instrument does not directly detect neutral molecules. It detects ions.

The resulting mass spectrum plots the mass-to-charge ratio, m/z, against relative ion abundance.

Here:

• \(m\) is the mass of the ion.
• \(z\) is its charge number.
• Relative intensity describes the abundance of an ion compared with the most abundant ion.

The most abundant ion is assigned a relative intensity of 100. This is the base peak.

For the standard HSC interpretation, we generally assume singly charged positive ions, meaning \(z=1\). In that case, the numerical m/z value is approximately equal to the ion’s nominal mass in atomic mass units.

What should the instrument show?

Suppose a molecule has a molecular mass of 74.

If it loses one electron but remains intact, it forms a molecular ion with an m/z value of approximately 74.

We represent this process as:

\[ \mathrm{M(g)+e^- \rightarrow M^{+\bullet}(g)+2e^-} \]

Here, \(M^{+\bullet}\) is the molecular ion. The positive charge indicates that an electron has been removed, while the dot represents an unpaired electron.

The original prediction is that intact molecular ions should produce a peak at m/z 74.

However, experimental mass spectra often contain many additional peaks at lower m/z values. Sometimes the molecular-ion peak is extremely small.

This observation tells us that ionisation is not simply a process of removing an electron while leaving everything else unchanged.

03How fragmentation makes the molecular-ion peak weak

When a molecule is ionised, some of the energy from the collision can remain as internal energy in the molecular ion.

If the ion has sufficient internal energy, bonds can break or atoms can rearrange.

The molecular ion may then undergo fragmentation:

\[ \mathrm{M^{+\bullet} \rightarrow A^+ + B^\bullet} \]

Here:

• \(M^{+\bullet}\) is the original molecular ion.
• \(A^+\) is a positively charged fragment ion.
• \(B^\bullet\) is a neutral radical.

Notice that only \(A^+\) is detected by the mass spectrometer. The neutral radical does not produce a peak.

This is the crucial point: if the molecular ion fragments before reaching the analyser, the instrument detects the smaller fragment ion instead of the original molecular ion.

Why are some fragment ions more abundant than others?

Not all fragmentation pathways are equally favourable.

Consider two possible ways for a molecular ion to break apart:

• Pathway A produces a relatively unstable positive fragment.
• Pathway B produces a more stable positive fragment.

Pathway B may be favoured because the resulting ion is better able to accommodate its positive charge.

For example, some fragmentation pathways produce:

• Resonance-stabilised ions, where the positive charge is distributed across several atoms.
• Benzylic or allylic carbocations, where the positive charge is stabilised by neighbouring pi electrons.
• Relatively stable tertiary carbocations.

These ions can be produced in significant quantities, giving them large peaks in the mass spectrum.

However, ion stability alone does not determine peak intensity. The probability of forming the ion, competing fragmentation pathways, and the ion’s ability to survive and reach the detector also matter.

A strong fragment peak therefore tells you that a particular ion is abundant. It does not automatically tell you the exact structure of the original molecule.

Weak molecular ion versus missing molecular ion

There are two related but different situations.

A missing molecular-ion peak does not mean the original molecule has no molecular mass.

It means the molecular ion has not been detected under those experimental conditions.

There can also be other explanations. The sample might decompose during heating, fail to vaporise properly, or produce an ion outside the instrument’s scan range. Instrument sensitivity and sample concentration also affect which peaks are visible.

04Isotope patterns can help identify a weak molecular ion

If the molecular-ion peak is weak, isotope patterns provide another useful source of evidence.

Remember that isotopes are atoms of the same element with different numbers of neutrons.

For example, most naturally occurring carbon is carbon-12, but a small proportion is carbon-13. If a molecule contains several carbon atoms, some of its molecular ions will contain one carbon-13 atom instead of carbon-12.

These isotopic versions of the same ion have slightly different masses.

The M+1 peak

A molecular ion containing one carbon-13 atom has a mass approximately one unit greater than the corresponding ion containing only carbon-12 atoms.

This produces an M+1 peak.

The natural abundance of carbon-13 is approximately 1.1%, so a useful estimate is:

\[ \frac{I_{M+1}}{I_M} \approx 1.1n \]

where:

• \(I_{M+1}\) is the relative intensity of the M+1 peak.
• \(I_M\) is the relative intensity of the molecular-ion peak.
• \(n\) is the number of carbon atoms.

The expression gives the approximate percentage of M+1 intensity relative to M.

For example, a molecule containing four carbon atoms has an expected M+1 peak of approximately 4.4% of the molecular-ion peak.

This is an estimate, not an exact rule. Other isotopes, overlapping peaks, and measurement limitations can affect the observed ratio.

The M+2 peak: chlorine and bromine

Chlorine and bromine produce particularly recognisable isotope patterns.

Chlorine has two important naturally occurring isotopes:

• Chlorine-35, with an abundance of approximately 75%.
• Chlorine-37, with an abundance of approximately 25%.

For an ion containing one chlorine atom, the molecular-ion peak and its M+2 isotope peak have an approximate intensity ratio of 3:1.

Bromine behaves differently:

• Bromine-79 has an abundance of approximately 50%.
• Bromine-81 has an abundance of approximately 50%.

An ion containing one bromine atom therefore produces two peaks separated by two mass units, with approximately equal intensities.

There is an important limitation here.

An isotope pattern belongs to the ion being detected, not automatically to the original molecule.

For example, a fragment ion containing one chlorine atom will also produce an M+2 peak with an approximate 3:1 ratio. You cannot assume that every chlorine isotope pattern represents the molecular ion.

If you want to practise distinguishing these possibilities, see How to Distinguish Isotopic Peaks from Fragment Peaks.

07What if the molecular-ion peak is completely missing?

So far, we have considered molecules that produce at least a small detectable molecular-ion peak.

But what happens when no molecular-ion peak appears at all?

Suppose an EI spectrum contains:

The peaks at 77 and 79 have an intensity ratio of 3:1.

This supports the presence of one chlorine atom in the ion at m/z 77.

However, it does not establish that the neutral molecule has a molecular mass of 77.

The ion could be a fragment containing chlorine. The original molecule could have a considerably greater molecular mass, with the intact molecular ion fragmenting before detection.

This is an important limitation of interpreting mass spectra: a recognisable isotope pattern can identify an element in an ion without identifying whether that ion is the molecular ion.

Using a different ionisation method

Scientists can investigate a missing molecular-ion peak by changing the ionisation method.

Electron-impact ionisation is considered a relatively energetic ionisation method. Other methods, such as chemical ionisation and electrospray ionisation, can often produce less fragmentation.

These are examples of softer ionisation methods.

Instead of simply removing an electron, some methods transfer a proton to the molecule.

For example:

\[ \mathrm{M+H^+\rightarrow[M+H]^+} \]

The resulting ion is called a protonated molecule.

Its m/z value is approximately one unit greater than the neutral molecular mass, assuming a singly charged ion.

Consider a neutral molecule with a nominal molecular mass of 150.

Its expected ions under two different methods are:

If the EI spectrum has no peak at 150 but the chemical-ionisation spectrum has a strong peak at 151, the second experiment provides evidence for the intact molecule’s mass.

The peak at 151 is not the EI molecular ion. It is a different type of ion formed through protonation.

This revised experiment addresses a limitation of the original measurement. EI shows which ions survive the relatively energetic ionisation process, but it may not preserve enough intact molecular ions to determine the molecular mass directly.

A softer ionisation method can provide an additional measurement while reducing fragmentation.

There are still limitations. Different ionisation methods have different sample requirements, and electrospray ionisation can also produce adducts or multiply charged ions. You must identify the ion type before assigning a molecular mass.

08A practical method for interpreting a weak or missing molecular ion

When you encounter an unfamiliar mass spectrum, use the following reasoning sequence.

A few common mistakes are worth avoiding:

• Do not assume the base peak represents the molecular mass.
• Do not assume that the highest-mass peak is automatically the molecular ion.
• Do not identify an M+2 peak as a fragment without considering isotope patterns.
• Do not assume that a chlorine or bromine isotope pattern must belong to the whole molecule.
• Do not claim that a molecule has no molecular mass simply because its molecular-ion peak is missing.
• Do not use a protonated molecule, \(\mathrm{[M+H]^+}\), as though it were the EI molecular ion, \(\mathrm{M^{+\bullet}}\).

09Questions and solutions

Question 1

A pure volatile compound has the molecular formula \(\mathrm{C_5H_{12}}\).

Its EI mass spectrum contains the following peaks:

The instrument has a reporting threshold of 0.3 relative-intensity units.

Assume singly charged ions and that carbon-13 is the only significant contributor to the M+1 peak.

Identify the molecular-ion peak and base peak. Explain why the M+1 peak may not be detected.

Solution 1

The molecular-ion peak is at m/z 72, while the base peak is at m/z 43.

The molecular mass can be calculated from the formula:

\[ M_r=(5\times12)+(12\times1)=72 \]

Therefore, the molecular ion appears at m/z 72.

The peak at m/z 43 has the greatest relative intensity, so it is the base peak. It represents a fragment ion rather than the intact molecule.

To estimate the M+1 peak:

\[ \frac{I_{M+1}}{I_M}\approx5\times1.1\% \]

\[ =5.5\% \]

The molecular-ion peak has an intensity of 4, giving:

\[ I_{M+1}=4\times0.055 \]

\[ =0.22 \]

The predicted intensity is 0.22, which is below the instrument’s reporting threshold of 0.3.

Therefore, the M+1 peak may not appear even though the molecular-ion peak is present.

Question 2

An EI mass spectrum contains a possible molecular-ion peak at m/z 86 with a relative intensity of 20.

A smaller peak appears at m/z 87 with a relative intensity of 0.9.

Assume that the peak at 87 is the M+1 isotope peak, that carbon-13 is the dominant contributor, and that there is no overlap between peaks.

Estimate the number of carbon atoms in the molecule.

Explain why this estimate supports the proposed molecular-ion assignment but does not independently establish the molecular formula.

Solution 2

The molecule is estimated to contain approximately four carbon atoms.

First, calculate the relative intensity of the M+1 peak:

\[ \frac{I_{M+1}}{I_M} =\frac{0.9}{20} \]

\[ =0.045=4.5\% \]

Each carbon atom contributes approximately 1.1% to the M+1 peak.

Therefore:

\[ n_C\approx\frac{4.5}{1.1} \]

\[ n_C\approx4.1 \]

The estimated number of carbon atoms is approximately four.

We can check this by predicting the M+1 intensity for four carbon atoms:

\[ 4\times1.1\%=4.4\% \]

\[ I_{M+1}=20\times0.044=0.88 \]

The predicted intensity of 0.88 is very close to the observed intensity of 0.9.

This supports the interpretation that m/z 86 is a molecular-ion peak belonging to a molecule with approximately four carbon atoms.

However, this does not establish the complete molecular formula. Different combinations of hydrogen, oxygen, nitrogen, and other elements can produce the same nominal molecular mass.

The isotope ratio provides supporting evidence, not a unique structural identification.

Question 3

An unknown compound produces the following EI mass spectrum:

No other peaks are detected above m/z 79.

Assume that the peaks are singly charged ions and that there are no overlapping signals.

A student argues that the molecular mass must be 77 because the peaks at 77 and 79 have a 3:1 ratio.

Evaluate this argument.

Explain what can be concluded from the isotope pattern and what additional information would be needed to establish the molecular mass.

Solution 3

The student’s conclusion is not justified. The peaks at m/z 77 and 79 support the presence of one chlorine atom in the detected ion, but they do not establish that this ion is the molecular ion.

First, calculate the intensity ratio:

\[ \frac{I_{77}}{I_{79}} =\frac{18}{6}=3 \]

This is consistent with the natural isotope abundance ratio of chlorine-35 to chlorine-37.

The ion at m/z 77 therefore likely contains one chlorine atom, with its chlorine-37 isotopologue appearing at m/z 79.

However, the ion could be a fragment produced from a larger molecule.

The absence of peaks above m/z 79 does not prove that the original molecule has a molecular mass of 77. Its molecular ion may have fragmented too readily to be detected.

The peak at m/z 41 is the base peak, indicating that it is the most abundant detected ion. It does not provide the original molecular mass either.

To establish the molecular mass, additional evidence would be needed. This could include an independently determined molecular formula or a softer ionisation method that produces a detectable protonated molecule.

The defensible conclusion is that the ion at m/z 77 likely contains one chlorine atom. Its identity as a molecular ion or fragment remains unresolved.

Question 4

A neutral compound is analysed using two different ionisation methods.

The EI mass spectrum contains no detectable peak at m/z 150. Its base peak appears at m/z 43.

A positive-mode electrospray ionisation (ESI) spectrum produces a dominant peak at m/z 151.

The instrument information confirms that the ESI peak represents a singly charged protonated molecule, \(\mathrm{[M+H]^+}\), with no sodium or solvent adducts.

Determine the nominal molecular mass of the neutral compound.

Explain why the molecular-ion peak can be missing from the EI spectrum and why the peak at m/z 151 should not be identified as the EI molecular ion.

Solution 4

The neutral compound has a nominal molecular mass of 150.

The ESI spectrum contains a singly charged protonated molecule:

\[ \mathrm{[M+H]^+} \]

The proton adds approximately one mass unit to the neutral molecule.

Therefore:

\[ M_r=151-1 \]

\[ \boxed{M_r=150} \]

The ESI result provides evidence for the intact neutral molecule even though the EI spectrum does not show a detectable molecular-ion peak.

In the EI experiment, the molecular ion may fragment rapidly after losing an electron. The resulting fragment ions can be much more abundant than the intact molecular ion.

The peak at m/z 43 is therefore consistent with an abundant fragment.

The peak at m/z 151 has a different origin. It represents a protonated molecule formed by electrospray ionisation, not the radical cation produced by electron-impact ionisation.

The two experiments provide complementary information about the same neutral compound.

However, the ESI result does not prove that fragmentation is the only reason the EI molecular ion is missing. Sample introduction, thermal decomposition, and instrument sensitivity may also contribute.

Question 5

A purified volatile compound is analysed using both electron-impact ionisation and proton-transfer chemical ionisation.

The following illustrative data are obtained.

EI spectrum

Proton-transfer chemical-ionisation spectrum

The following assumptions apply:

• Both experiments use singly charged positive ions.
• The chemical-ionisation spectrum contains only protonated molecules and their natural isotopic variants. There are no adducts.
• The approximate natural abundance ratio of chlorine-35 to chlorine-37 is 3:1.
• Both spectra come from the same purified compound.
• The detection threshold is 0.5 relative-intensity units.

Answer the following:

(a) Determine the nominal molecular mass of the neutral compound.

(b) Determine how many chlorine atoms are present in the original molecule. Explain why the isotope pattern supports this number rather than three chlorine atoms.

(c) Explain what the peaks at m/z 77 and 79 indicate in the EI spectrum. How does their mass relationship to the neutral molecule support a possible fragmentation pathway?

(d) Explain why the peak at m/z 41 is not the molecular mass, and state what can and cannot be concluded about the original molecular structure.

Solution 5

The neutral molecule has a nominal molecular mass of 112 and contains two chlorine atoms. The EI spectrum shows a one-chlorine fragment at m/z 77, while the peak at m/z 41 is the most abundant detected fragment.

Part (a): Determine the nominal molecular mass.

The chemical-ionisation spectrum produces protonated molecules:

\[ \mathrm{[M+H]^+} \]

The first peak, at m/z 113, corresponds to the protonated molecule containing chlorine-35 isotopes.

Since protonation adds one mass unit:

\[ M_r=113-1 \]

\[ \boxed{M_r=112} \]

The neutral molecule therefore has a nominal molecular mass of 112.

Part (b): Determine the number of chlorine atoms.

The three peaks in the chemical-ionisation spectrum occur at:

• m/z 113
• m/z 115
• m/z 117

Their relative intensities are 90, 60, and 10.

Dividing each value by 10 gives:

\[ 90:60:10=9:6:1 \]

This is the expected isotope pattern for two chlorine atoms.

There are three possible combinations:

• Two chlorine-35 atoms produce the M peak.
• One chlorine-35 and one chlorine-37 atom produce the M+2 peak.
• Two chlorine-37 atoms produce the M+4 peak.

The approximate probabilities follow the ratio:

\[ 3^2:2(3)(1):1^2 \]

\[ \boxed{9:6:1} \]

It is also worth checking whether three chlorine atoms could explain the results.

For three chlorine atoms, the expected pattern is:

\[ 27:27:9:1 \]

This would produce four peaks, including an M+6 peak at m/z 119.

If the first peak had an intensity of 90, the expected M+6 intensity would be:

\[ I_{M+6}=\frac{90}{27} \]

\[ I_{M+6}\approx3.33 \]

This is well above the detection threshold of 0.5.

Since no peak is detected at m/z 119, the three-chlorine interpretation is inconsistent with the supplied data.

The isotope pattern therefore supports exactly two chlorine atoms in the original molecule.

Part (c): Interpret the peaks at m/z 77 and 79.

The intensity ratio is:

\[ \frac{18}{6}=3 \]

This supports the presence of one chlorine atom in the ion at m/z 77.

The peak at m/z 79 is its chlorine-37 isotopic counterpart.

The neutral molecule has a nominal molecular mass of 112, while the fragment ion has an m/z value of 77.

The difference is:

\[ 112-77=35 \]

A possible explanation is that the molecular ion loses a neutral chlorine-35 radical, leaving behind a positively charged fragment that still contains one chlorine atom.

This is consistent with the observed isotope pattern.

However, the nominal mass difference alone does not prove that exact fragmentation pathway. Other fragmentation and rearrangement pathways may produce ions with the same nominal mass.

Part (d): Explain the peak at m/z 41 and evaluate the structural evidence.

The peak at m/z 41 is the base peak because it has the highest relative intensity.

It represents a particularly abundant fragment ion, not the original molecule.

The chemical-ionisation spectrum independently establishes the neutral molecular mass as 112. Therefore, m/z 41 cannot represent the molecular mass.

The evidence supports the following conclusions:

• The neutral molecule has a nominal molecular mass of 112.
• The original molecule contains two chlorine atoms.
• The EI spectrum contains an abundant fragment at m/z 41.
• The EI spectrum contains a chlorine-containing fragment at m/z 77, supported by its isotope peak at m/z 79.

However, the evidence does not establish the exact structural arrangement of the atoms.

Different structural isomers can produce similar fragment ions, and a single peak rarely proves a unique molecular structure.

The next step is to test proposed fragment formulas against the established molecular formula. Every proposed fragment must contain atoms that could actually come from the original molecule.

For that next stage, see How to Check Whether a Fragment Fits a Molecular Formula.