How to Distinguish Isotopic Peaks from Fragment Peaks

Suppose an electron-impact mass spectrum contains two prominent peaks at \(m/z=78\) and \(m/z=80\). Are these two different fragments of a molecule, or is the second peak simply a heavier version of the first?

If the peak at \(m/z=80\) is approximately one-third the intensity of the peak at \(m/z=78\), it is probably an isotopic peak caused by chlorine, not a fragment peak.

That is the central distinction:

• Isotopic peaks come from ions containing different isotopes of the same element. The ion has essentially the same chemical composition, but a slightly different mass.
• Fragment peaks come from molecular ions breaking apart. The resulting ions contain fewer atoms than the original molecule.

There is an important complication, though: fragment ions can also have isotopic peaks. So you need to identify both the isotope pattern and the fragmentation pathway before assigning a peak.

01Understand what each type of peak represents

Start by picturing a molecule as a collection of atoms joined by chemical bonds.

When a molecule enters a mass spectrometer, it can become a positively charged ion. That ion may remain intact, or some of its chemical bonds may break.

There are therefore two different ways to produce ions with different masses.

Isotopic peaks: the same ion with different isotopes

Consider chlorine.

Naturally occurring chlorine contains two main isotopes:

• Chlorine-35, \(^{35}\mathrm{Cl}\), makes up approximately 75.8% of natural chlorine.
• Chlorine-37, \(^{37}\mathrm{Cl}\), makes up approximately 24.2%.

Both isotopes are chlorine. They have the same number of protons, but different numbers of neutrons.

Imagine a molecule containing one chlorine atom. Some molecules will contain \(^{35}\mathrm{Cl}\), while others will contain \(^{37}\mathrm{Cl}\).

When these molecules form positive ions, the two groups produce ions with different masses.

Replacing \(^{35}\mathrm{Cl}\) with \(^{37}\mathrm{Cl}\) increases the mass by two units.

Consequently, the mass spectrum contains two molecular ion peaks separated by two \(m/z\) units.

The lighter isotope produces the larger peak because it is more abundant.

Fragment peaks: the molecule breaks apart

Now consider a different process.

An ionised molecule may have enough energy to break one of its chemical bonds. One part retains the positive charge and becomes a fragment ion. The other part leaves as a neutral particle.

For example:

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

The mass spectrometer detects \(A^+\), but not the neutral fragment \(B^\bullet\).

The resulting positive ion has a lower mass than the original molecular ion.

This is why fragment peaks generally appear at lower \(m/z\) values than their parent molecular ion.

A useful way to picture the difference

Think of building a model from two types of Lego bricks.

• An isotopic change is like replacing one brick with a slightly heavier version of the same brick. The overall model stays intact.
• Fragmentation is like removing part of the model. You are now weighing a smaller structure.

The analogy has limits: isotopes are atoms of the same element with different neutron numbers, and fragmentation involves actual chemical bond breaking. But it captures the key distinction.

An important detail: a fragment ion can itself contain different isotopes.

For example, a fragment containing chlorine might produce peaks at \(m/z=49\) and \(m/z=51\). Both are fragment ions, but the peak at 51 is also the isotopic partner of the peak at 49.

So the labels are not always mutually exclusive. One describes how the ion was formed; the other describes its isotopic relationship to another ion.

02How the mass spectrometer produces these peaks

To understand the evidence, it helps to follow the molecule through the instrument.

In electron-impact mass spectrometry, the experiment proceeds in a particular order.

Step 1: The sample is vaporised.

The sample enters the instrument and is converted into gaseous molecules under vacuum conditions.

Step 2: The molecules are ionised.

A beam of high-energy electrons collides with the gaseous molecules.

Some molecules lose an electron and become positive molecular ions.

\[ \mathrm{M\rightarrow M^{+\bullet}+e^-} \]

Here, \(M\) represents the original neutral molecule, and \(M^{+\bullet}\) represents its positive molecular ion.

Some molecular ions remain intact. Others break apart into smaller positive ions and neutral particles.

Step 3: The ions are accelerated.

An electric field accelerates the positive ions into the mass analyser.

Step 4: The ions are separated according to mass-to-charge ratio.

The mass analyser separates ions according to their mass-to-charge ratio, written as \(m/z\).

Step 5: The detector records the ions.

The detector measures the number of ions arriving at each \(m/z\) value.

The instrument produces a mass spectrum:

• The horizontal axis represents \(m/z\).
• The vertical axis represents relative abundance or relative intensity.

Here is the important experimental limitation: the mass spectrometer does not label a peak as an isotope or a fragment.

It measures the \(m/z\) values and relative intensities of ions. You must interpret those measurements using isotope abundances and chemically sensible fragmentation pathways.

03Recognise the characteristic isotope patterns

Certain elements produce particularly recognisable isotope patterns.

These patterns are among the most useful clues when distinguishing isotopic peaks from fragment peaks.

Chlorine: approximately a 3:1 ratio

As discussed earlier, natural chlorine contains approximately:

The ratio of their abundances is:

\[ \frac{75.8}{24.2}\approx 3.13 \]

Therefore, a molecule containing one chlorine atom will generally produce two molecular ion peaks separated by two \(m/z\) units, with an approximate intensity ratio of:

\[ M:M+2=3:1 \]

For example, if the lighter molecular ion peak has a relative intensity of 100, the heavier isotope peak should have an intensity of approximately:

\[ 100\times\frac{24.2}{75.8}=31.9 \]

So the expected intensities are approximately 100 and 32.

This is a strong indication that the ion contains one chlorine atom.

Bromine: approximately a 1:1 ratio

Bromine has two naturally abundant isotopes:

• \(^{79}\mathrm{Br}\), approximately 50.7%.
• \(^{81}\mathrm{Br}\), approximately 49.3%.

Their abundances are almost equal.

Consequently, a molecule containing one bromine atom generally produces two molecular ion peaks separated by two \(m/z\) units, with approximately equal intensities.

\[ M:M+2\approx 1:1 \]

For example:

This pattern suggests one bromine atom rather than one chlorine atom.

Before reading further, predict what would happen if a molecule contained two bromine atoms instead of one.

You would expect three peaks:

\[ M,\quad M+2,\quad M+4 \]

Their approximate intensity ratio would be:

\[ 1:2:1 \]

This occurs because there are three possible combinations of the two bromine isotopes:

• Two \(^{79}\mathrm{Br}\) atoms.
• One \(^{79}\mathrm{Br}\) atom and one \(^{81}\mathrm{Br}\) atom.
• Two \(^{81}\mathrm{Br}\) atoms.

Carbon: the M+1 peak

Carbon provides another useful isotope clue.

Approximately 1.1% of naturally occurring carbon atoms are \(^{13}\mathrm{C}\), while almost all the remaining carbon is \(^{12}\mathrm{C}\).

Replacing one \(^{12}\mathrm{C}\) atom with \(^{13}\mathrm{C}\) increases the mass by one unit.

This produces an \(M+1\) isotopic peak.

The more carbon atoms a molecule contains, the more likely it is that one will be \(^{13}\mathrm{C}\).

A useful approximation is:

\[ \text{M+1 intensity}\approx 1.1n\% \]

where \(n\) is the number of carbon atoms, relative to an \(M\) peak of 100.

For example, a molecule containing six carbon atoms may have an \(M+1\) peak of approximately:

\[ 6(1.1)=6.6 \]

This is a useful estimate, although other naturally occurring isotopes can also contribute.

Not every M+2 peak means chlorine or bromine

Be careful here.

A peak two mass units above another peak is not automatically evidence of chlorine or bromine.

For example:

• Sulfur contains a naturally occurring \(^{34}\mathrm{S}\) isotope, which can contribute to an M+2 peak.
• Oxygen contains \(^{18}\mathrm{O}\), which can also produce a small M+2 contribution.

The intensity and the rest of the isotope pattern matter.

Here is a useful comparison:

These ratios are approximate and assume natural isotopic abundances.

04How to recognise a fragment peak

Isotope patterns tell you whether peaks belong to the same ion with different isotopes.

But how do you identify an actual fragment?

The most useful technique is to calculate the mass difference between the molecular ion and the proposed fragment ion.

For a singly charged positive molecular ion and fragment ion:

\[ \Delta(m/z)= (m/z)_{\text{parent}}-(m/z)_{\text{fragment}} \]

This difference represents the nominal mass of the neutral particle lost during fragmentation.

You can then ask whether that mass corresponds to a chemically plausible group of atoms.

For example:

• A mass difference of 15 could correspond to a \(\mathrm{CH_3^\bullet}\) radical.
• A mass difference of 16 could correspond to an oxygen atom.
• A mass difference of 17 could correspond to an \(\mathrm{OH^\bullet}\) radical.
• A mass difference of 35 could correspond to a \(^{35}\mathrm{Cl}\) atom.

These are possible assignments, not automatic conclusions. The proposed fragment must also make chemical sense.

05A reliable decision process for exam questions

When you are given an unfamiliar mass spectrum, use the following sequence.

Step 1: Identify the likely molecular ion.

Start with the highest-mass credible ion cluster, rather than automatically choosing the tallest peak.

The molecular ion represents the intact molecule after ionisation.

Remember that the molecular ion is not necessarily the base peak. The base peak is simply the most intense peak in the spectrum.

Step 2: Look for isotope partners.

Check for peaks separated by one or two \(m/z\) units.

Compare their relative intensities with the expected isotope patterns.

In particular:

• Approximately 3:1 at a separation of two suggests one chlorine atom.
• Approximately 1:1 at a separation of two suggests one bromine atom.
• A small M+1 peak can indicate the presence of \(^{13}\mathrm{C}\).

Step 3: Look for lower-mass peaks.

Calculate the difference between the molecular ion and the proposed fragment.

Ask whether the mass lost corresponds to a chemically plausible neutral particle.

Step 4: Check whether the fragment retains any isotopes of interest.

A fragment containing chlorine may produce its own 3:1 isotope pair.

A fragment containing bromine may produce its own approximately 1:1 isotope pair.

If the fragment no longer contains the halogen, you should not expect that halogen’s characteristic isotope pattern in that fragment.

Step 5: State what the evidence supports, not more than it proves.

A characteristic isotope pattern can provide strong evidence for the presence of an element.

A plausible fragmentation pathway can provide evidence about which atoms remain together.

However, neither observation automatically establishes the complete molecular structure.

06Common mistakes when interpreting mass spectra

What the spectrum cannot tell you by itself

A standard mass spectrum has some limitations.

The molecular ion may be weak or even absent if it fragments readily. This makes identifying the original molecular mass more difficult.

Different ions can also have the same nominal \(m/z\) value. A conventional low-resolution spectrum may not distinguish them.

More advanced techniques can help:

• High-resolution mass spectrometry measures mass more precisely, helping distinguish ions with similar nominal masses.
• Tandem mass spectrometry (MS/MS) selects a particular parent ion, causes it to fragment, and measures the resulting daughter ions. This provides stronger evidence about which fragments came from which parent.

These are extensions of the basic interpretation method. The central HSC-style reasoning remains isotope patterns, mass differences, and chemically plausible fragmentation.

If you want to revise the underlying isotope-abundance calculations, see tutorgum’s Relative Atomic Mass from Isotopic Composition.

07Questions and solutions

The following questions progress from straightforward isotope identification to more demanding problems involving overlapping isotope patterns and fragmentation.

Question 1

An electron-impact mass spectrum has an established molecular ion peak at \(m/z=100\), with a relative intensity of 100.

A second peak occurs at \(m/z=102\), with a relative intensity of 32.

Assuming a pure sample, singly charged ions, and no overlapping signals, explain whether the peak at \(m/z=102\) is more likely to be an isotope peak or a fragment peak. Identify the element that its intensity pattern most strongly suggests.

Solution 1

The peak at \(m/z=102\) is most likely the chlorine-37 isotope peak of the molecular ion, rather than a fragment.

The mass difference is:

\[ 102-100=2 \]

Chlorine has two naturally occurring isotopes, with approximate abundances of 75.8% and 24.2%.

The expected intensity ratio is:

\[ 75.8:24.2 \]

Dividing both values by 24.2 gives approximately:

\[ 3.13:1 \]

If the lighter peak has an intensity of 100, the expected heavier peak intensity is:

\[ 100\times\frac{24.2}{75.8}=31.9 \]

The observed intensity is 32, which agrees closely.

A fragment produced by breaking the molecular ion would normally have a lower \(m/z\) value, not a higher one.

Therefore, the peak at 102 is strong evidence for one chlorine atom in the molecular ion.

Question 2

An unknown organic compound has the molecular formula \(\mathrm{C_4H_9Cl}\).

Its electron-impact mass spectrum contains the following peaks:

Assume all detected ions are singly charged.

(a) Explain the relationship between the peaks at 92 and 94.

(b) Identify a plausible formula for the fragment ion at \(m/z=57\).

(c) Identify the relationship between the peaks at 49 and 51.

(d) Calculate the neutral mass lost to produce the fragment at \(m/z=49\), and suggest a plausible neutral fragment.

Solution 2

(a) The peaks at 92 and 94 are the molecular ion’s chlorine isotope pair.

The molecular ion containing chlorine-35 has a nominal mass of:

\[ 4(12)+9(1)+35=92 \]

Replacing chlorine-35 with chlorine-37 gives:

\[ 4(12)+9(1)+37=94 \]

The intensity ratio is:

\[ 100:32 \]

This is approximately the expected 3:1 ratio for one chlorine atom.

Therefore, the peak at 94 is an isotopic peak, not a fragment.

(b) The peak at \(m/z=57\) can be assigned to \(\mathrm{C_4H_9^+}\).

Its nominal mass is:

\[ 4(12)+9(1)=57 \]

The molecular ion can produce this fragment by losing a chlorine radical.

For the lighter isotopic molecular ion:

\[ 92-57=35 \]

This corresponds to chlorine-35.

The fragment has lost the chlorine atom, so it does not show the characteristic chlorine isotope pair.

(c) The peaks at 49 and 51 are isotopic partners of a chlorine-containing fragment ion.

Their intensity ratio is:

\[ 24:8=3:1 \]

Their mass difference is:

\[ 51-49=2 \]

A plausible assignment is:

\[ m/z=49:\quad \mathrm{CH_2{}^{35}Cl^+} \]

\[ m/z=51:\quad \mathrm{CH_2{}^{37}Cl^+} \]

These ions are fragments of the original molecule, and the peak at 51 is the heavier isotope version of the peak at 49.

(d) The neutral mass lost is 43, corresponding to a plausible \(\mathrm{C_3H_7^\bullet}\) radical.

Using the lighter molecular ion and fragment:

\[ \Delta(m/z)=92-49 \]

\[ \Delta(m/z)=43 \]

The nominal mass of \(\mathrm{C_3H_7^\bullet}\) is:

\[ 3(12)+7(1)=43 \]

The proposed fragmentation is therefore:

\[ \mathrm{C_4H_9Cl^{+\bullet} \rightarrow CH_2Cl^+ + C_3H_7^\bullet} \]

The heavier isotope version also gives a mass difference of 43:

\[ 94-51=43 \]

This supports the proposed assignment.

The important point is that the fragment at 49 retains chlorine, whereas the fragment at 57 does not.

Question 3

An unknown compound produces the following molecular ion and fragment-ion peaks in its electron-impact mass spectrum.

Assume:

• All ions have a charge of \(+1\).
• The molecular ion cluster and the lower-mass pair belong to the same compound.
• The lower-mass pair represents one fragment-ion family.
• Chlorine has approximate isotope abundances of 75% and 25%.

(a) Determine how many chlorine atoms are likely present in the original molecule.

(b) Determine how many chlorine atoms remain in the fragment ion.

(c) Calculate the nominal mass of the neutral particle lost during the fragmentation.

(d) Explain why the peak at \(m/z=79\) should not be interpreted as an independent fragmentation pathway from the peak at \(m/z=77\).

Solution 3

(a) The original molecule most likely contains two chlorine atoms.

For two chlorine atoms, there are three possible isotopic combinations:

• Two chlorine-35 atoms.
• One chlorine-35 atom and one chlorine-37 atom.
• Two chlorine-37 atoms.

Using the approximate 75:25 abundance ratio, the expected probabilities are:

\[ P(M)=0.75^2=0.5625 \]

\[ P(M+2)=2(0.75)(0.25)=0.375 \]

\[ P(M+4)=0.25^2=0.0625 \]

Therefore:

\[ M:M+2:M+4=9:6:1 \]

Normalising the first peak to 100 gives:

\[ 100:66.7:11.1 \]

The observed ratio is:

\[ 100:64:10 \]

This is a close match, allowing for approximate abundances and illustrative peak intensities.

The molecular ion therefore most likely contains two chlorine atoms.

(b) The fragment ion most likely retains one chlorine atom.

The lower-mass peaks have an intensity ratio of:

\[ 30:10=3:1 \]

They are also separated by two mass units:

\[ 79-77=2 \]

This is consistent with a fragment containing one chlorine atom.

The original molecular ion contains two chlorine atoms, while the fragment contains one.

(c) The nominal mass of the neutral particle lost is 35.

Using the lighter isotopic peaks:

\[ \Delta(m/z)=112-77 \]

\[ \Delta(m/z)=35 \]

A neutral chlorine-35 radical has a nominal mass of 35.

The proposed fragmentation therefore involves the loss of one chlorine atom:

\[ \mathrm{Parent^{+\bullet}\rightarrow Fragment^+ + Cl^\bullet} \]

(d) The peak at 79 is the chlorine-37 isotope partner of the fragment at 77.

The two peaks are separated by two mass units and have an approximate 3:1 intensity ratio.

These are exactly the expected characteristics of a fragment-ion family containing one chlorine atom.

Treating the peak at 79 as a separate fragmentation event ignores the isotope pattern.

The spectrum therefore provides evidence for a parent molecule containing two chlorine atoms and a fragment that retains one of them.

Question 4

A singly charged electron-impact mass spectrum contains the following peaks:

Assume:

• The molecular ion is at \(m/z=150\).
• The compound contains only carbon, hydrogen, and oxygen.
• The M+1 contribution is dominated by \(^{13}\mathrm{C}\).
• The lower-mass peaks at 135 and 136 belong to the same fragment-ion family.
• The fragment is formed directly from the molecular ion.

(a) Estimate the number of carbon atoms in the original molecule.

(b) Explain why the peak at \(m/z=135\) is a fragment peak rather than an isotopic peak of the molecular ion.

(c) Determine whether the peak at \(m/z=136\) is more likely to be a separate fragment or an isotope peak of the fragment at 135.

(d) Use the intensity ratio of the lower-mass peaks to estimate how many carbon atoms remain in the fragment. Suggest a plausible neutral loss.

Solution 4

(a) The original molecule contains approximately seven carbon atoms.

Each carbon atom contributes approximately 1.1% to the M+1 peak through the presence of \(^{13}\mathrm{C}\).

Therefore:

\[ n_C\approx \frac{\text{M+1 intensity}}{1.1} \]

Substituting the observed intensity:

\[ n_C\approx\frac{7.7}{1.1} \]

\[ \boxed{n_C\approx7} \]

The original molecule therefore contains approximately seven carbon atoms.

(b) The peak at \(m/z=135\) is a fragment ion because it is 15 mass units lighter than the molecular ion.

Calculate the difference:

\[ 150-135=15 \]

An isotope substitution makes an ion heavier, not 15 units lighter.

A neutral particle with nominal mass 15 could be a methyl radical, \(\mathrm{CH_3^\bullet}\).

A plausible fragmentation pathway is therefore:

\[ \mathrm{M^{+\bullet}\rightarrow Fragment^+ + CH_3^\bullet} \]

The peak at 135 is consequently consistent with a fragment formed by loss of a methyl radical.

(c) The peak at \(m/z=136\) is most likely the M+1 isotope peak of the fragment at 135.

The mass difference is:

\[ 136-135=1 \]

This is consistent with replacing one \(^{12}\mathrm{C}\) atom in the fragment with \(^{13}\mathrm{C}\).

We can test this by comparing their intensities.

(d) The fragment most likely retains six carbon atoms.

The relative M+1 intensity of the fragment is:

\[ \frac{2.6}{40}\times100=6.5\% \]

Using the approximate 1.1% contribution per carbon:

\[ n_C\approx\frac{6.5}{1.1} \]

\[ n_C\approx5.9 \]

Therefore, the fragment contains approximately six carbon atoms.

The original molecule contained approximately seven carbon atoms, while the fragment contains approximately six.

This agrees with the proposed loss of one carbon-containing neutral particle, specifically a methyl radical:

\[ \mathrm{CH_3^\bullet} \]

The interesting point is that the peak at 136 is not an isotope peak of the original molecular ion at 150. It is an isotope peak of the smaller fragment at 135.

You must always identify which ion family an isotopic peak belongs to.

Question 5

An unknown organic compound is analysed using electron-impact mass spectrometry.

The non-halogen portion of its molecular ion has a nominal mass of 50.

The following peaks are observed:

Assume:

• All ions are singly charged.
• The compound contains only one chlorine atom and one bromine atom.
• Use approximate isotope abundances of 75:25 for chlorine and 50:50 for bromine.
• The peak clusters do not contain contributions from unrelated compounds or overlapping ion species.
• The lower-mass clusters are produced by fragmentation of the molecular ion.

Answer the following:

(a) Explain why the molecular ion produces three peaks at 164, 166, and 168 rather than just two.

(b) Determine which isotopic combinations produce the peaks at 164, 166, and 168.

(c) Explain why the peak at 166 cannot be interpreted as the isotope peak of chlorine alone.

(d) Identify the likely relationship between the peaks at 129 and 131. What does their intensity pattern tell you about the fragment?

(e) Suggest a plausible relationship between the peaks at 79 and 81 and the original molecule.

(f) Construct a consistent explanation of the entire spectrum. State what you can infer about the halogen atoms and fragmentation pathways, and identify one important conclusion you cannot draw from the data.

Solution 5

(a) The molecular ion produces three peaks because the molecule contains both chlorine and bromine, and each element has two naturally occurring isotopes.

This is more complicated than a molecule containing only chlorine or only bromine.

The non-halogen portion has a nominal mass of 50.

The lightest molecular ion contains chlorine-35 and bromine-79:

\[ 50+35+79=164 \]

However, either halogen can independently be replaced by its heavier isotope.

This creates three possible mass values because two different isotope substitutions can produce the same total mass increase of two units.

(b) The three peaks represent four possible isotopic combinations, with two combinations contributing to the middle peak.

The combinations are:

The peak at 166 therefore contains contributions from two different isotopic combinations.

This is the important detail that makes the question more difficult.

We can calculate the expected relative probabilities using the approximate isotope abundances.

For \(m/z=164\):

\[ P(164)=0.75(0.50)=0.375 \]

For \(m/z=166\), there are two possible combinations:

\[ P(166)=0.25(0.50)+0.75(0.50) \]

\[ P(166)=0.125+0.375=0.500 \]

For \(m/z=168\):

\[ P(168)=0.25(0.50)=0.125 \]

Therefore:

\[ 164:166:168 \]

\[ =0.375:0.500:0.125 \]

Simplifying:

\[ \boxed{3:4:1} \]

The observed intensities are:

\[ 75:100:25 \]

which simplify to:

\[ 3:4:1 \]

The measured spectrum therefore agrees with the expected isotope pattern for a molecule containing one chlorine atom and one bromine atom.

(c) The peak at 166 cannot be attributed to chlorine alone because bromine also produces a two-unit mass increase.

A student might see a peak two units above 164 and immediately assign it to chlorine-37.

That interpretation is incomplete.

There are actually two ways to produce an ion at 166:

• Replace chlorine-35 with chlorine-37 while retaining bromine-79.
• Replace bromine-79 with bromine-81 while retaining chlorine-35.

Both substitutions increase the mass by two units.

The detector records both contributions at the same nominal \(m/z\) value.

This is why the middle peak is more intense than either of the other two peaks.

It is not simply one chlorine isotope peak. It represents two overlapping isotopic contributions.

(d) The peaks at 129 and 131 are fragment-ion isotopic partners, and their approximately 1:1 ratio suggests that the fragment retains bromine but not chlorine.

The two peaks have a mass difference of:

\[ 131-129=2 \]

Their intensities are approximately equal:

\[ 100:98\approx1:1 \]

This is the characteristic pattern of a fragment containing one bromine atom.

The original molecular ion has a nominal mass of 164 when it contains chlorine-35 and bromine-79.

The fragment at 129 has a nominal mass of 129.

The difference is:

\[ 164-129=35 \]

This is consistent with the loss of a chlorine-35 radical.

The fragment therefore retains the non-halogen framework and bromine, but loses chlorine.

Its nominal composition can be represented as:

\[ X{}^{79}\mathrm{Br^+} \]

where \(X\) represents the non-halogen framework with nominal mass 50.

Replacing bromine-79 with bromine-81 produces the isotope partner:

\[ X{}^{81}\mathrm{Br^+} \]

at \(m/z=131\).

(e) The peaks at 79 and 81 are consistent with bromine-containing fragment ions.

The mass difference is:

\[ 81-79=2 \]

Their intensities are approximately equal.

This is consistent with the two naturally occurring bromine isotopes.

A plausible assignment is:

\[ ^{79}\mathrm{Br^+} \]

and

\[ ^{81}\mathrm{Br^+} \]

These ions could be produced by further fragmentation of the bromine-containing molecular fragments, with the non-halogen framework separating from the bromine.

The original molecule therefore provides a plausible source of both the 129/131 fragment family and the 79/81 bromine-ion family.

(f) The complete spectrum supports a molecule containing one chlorine atom and one bromine atom, with fragmentation that can remove chlorine while retaining bromine.

The evidence fits together as follows:

The molecular ion cluster shows that the original molecule contains both halogens.

The lower-mass fragment cluster at 129 and 131 suggests that chlorine can be lost while bromine remains attached to the non-halogen framework.

The peaks at 79 and 81 provide further evidence for bromine-containing ions.

However, the spectrum does not establish the exact arrangement of atoms or the complete molecular structure. Different compounds can produce similar fragment ions.

That final distinction matters: mass spectrometry provides evidence about molecular mass, isotope composition, and fragmentation, but a single spectrum does not necessarily prove a unique structure.

The next step in a full structural identification would be to combine these mass spectral assignments with evidence from techniques such as infrared spectroscopy or nuclear magnetic resonance spectroscopy. Those techniques provide additional information about functional groups and the arrangement of atoms, helping distinguish structures that have similar mass spectra.