How to Compare Mass Spectra of Two Structural Isomers

Two compounds can have the same molecular formula, the same relative molecular mass, and still produce different mass spectra. That is exactly what makes mass spectrometry useful for distinguishing structural isomers.

Suppose two compounds both have the molecular formula \(\ce{C4H10O}\). One is butan-1-ol, and the other is ethoxyethane. Their molecular ions have the same mass-to-charge ratio, but their bonds are arranged differently. When the molecules fragment, they can produce different patterns of ions.

So, how do you compare their mass spectra and decide which structure is more likely?

The key is to compare three kinds of evidence:

1. **Molecular ion peaks:** Do the spectra support the same molecular mass and molecular formula?
2. **Fragment peaks:** Which stable fragments form, and what do their mass-to-charge ratios suggest about the original structure?
3. **Isotope patterns:** Do characteristic groups of peaks reveal the presence of particular elements, such as chlorine or bromine?

No single fragment peak automatically proves a structure. The strongest identification comes from combining several pieces of evidence and checking whether they agree with the proposed molecule.

## A method for comparing two structural isomers

Structural isomers have the same molecular formula but different arrangements of atoms. If you need to revise how to identify them, see [Structural Isomerism in Organic Compounds for HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-structural-isomerism-in-organic-compounds/).

Mass spectrometry gives you a way to investigate those different arrangements. A mass spectrum plots the relative abundance of detected positive ions against their mass-to-charge ratio, written as \(m/z\).

For the usual singly charged ions encountered in introductory organic mass spectrometry, \(m/z\) is numerically equal to the ion's mass in atomic mass units. A peak at \(m/z=43\), for example, represents an ion with a mass-to-charge ratio of 43.

[IMAGE: An annotated mass spectrum with mass-to-charge ratio m/z on the horizontal axis and relative abundance on the vertical axis. Label a molecular ion peak, a base peak, several smaller fragment peaks, and show that peak height represents relative abundance rather than the mass of an ion.]

Work through the spectra in a consistent order. This prevents you from jumping to a structural conclusion just because one familiar-looking peak appears.

### Step 1: Compare the molecular ion peaks

The molecular ion, represented by \(\mathrm{M}^{+\bullet}\), forms when a molecule loses an electron during electron ionisation. It retains the atoms of the original molecule but carries a positive charge and an unpaired electron.

Its \(m/z\) value can indicate the molecular mass of the original compound.

For two structural isomers with the same molecular formula, you would generally expect their molecular ion peaks to occur at the same \(m/z\), provided both molecular ions are detectable.

For example, butan-1-ol and ethoxyethane both have the formula \(\ce{C4H10O}\).

Their nominal molecular mass is:

\[
\begin{aligned}
M_r &= (4\times12)+(10\times1)+(1\times16)\\
&=74
\end{aligned}
\]

Both therefore have a molecular ion at \(m/z=74\), if that ion survives long enough to be detected.

This gives you an important starting point, but it does not distinguish the isomers. A matching molecular ion mass is evidence that two compounds may have the same molecular mass, not proof that they have the same structure.

### Step 2: Compare the fragment peaks

The molecular ion often has enough excess energy to break apart. These smaller charged particles are called fragment ions.

The fragments are particularly useful when comparing structural isomers because different arrangements of atoms favour different fragmentation pathways.

Think about snapping two differently constructed objects made from the same number of identical building blocks. Both objects contain the same materials, but their shapes and connections affect which pieces separate easily. Molecular fragmentation follows a more precise set of chemical rules, but the basic idea is similar.

A fragment's \(m/z\) value helps you work out its possible composition. You then ask whether that fragment could reasonably form from the proposed structure.

For example:

- \(m/z=15\) may correspond to \(\ce{CH3+}\).
- \(m/z=29\) may correspond to \(\ce{C2H5+}\).
- \(m/z=31\) may correspond to \(\ce{CH3O+}\).
- \(m/z=43\) may correspond to \(\ce{C3H7+}\).

These are possible assignments, not unique identifications. Different ions can have the same nominal mass, so you must consider the molecular formula, chemical stability, and other peaks before assigning a fragment.

### Step 3: Check the isotope pattern

Most elements occur naturally as mixtures of isotopes. These isotopes have different masses, so they can produce additional peaks in a mass spectrum.

Two particularly useful examples are chlorine and bromine.

Chlorine occurs mainly as two isotopes:

- Chlorine-35, with a natural abundance of approximately 75%.
- Chlorine-37, with a natural abundance of approximately 25%.

A molecule containing one chlorine atom can therefore produce molecular ion peaks separated by two mass units, with an approximate intensity ratio of \(3:1\).

Bromine has two major isotopes, bromine-79 and bromine-81, with approximately equal natural abundances. A molecule containing one bromine atom can produce two molecular ion peaks separated by two mass units, with an approximate ratio of \(1:1\).

[IMAGE: Three clearly labelled illustrative mass-spectrum comparisons: a molecule containing no chlorine or bromine, a molecule containing one chlorine atom with M and M+2 peaks in an approximately 3:1 ratio, and a molecule containing one bromine atom with M and M+2 peaks in an approximately 1:1 ratio. Use separate horizontal axes and relative-abundance scales.]

An isotope pattern can help establish which elements are present. It does not normally tell you how the atoms are connected, so it is supporting evidence rather than a complete structural identification.

01What happens inside a mass spectrometer?

Understanding the instrument helps explain why the three types of evidence are useful.

In a simplified electron-ionisation (EI) mass spectrometer, the sample follows a sequence of steps:

1. Vaporisation: The sample enters the instrument as a gas.
2. Ionisation: A beam of high-energy electrons strikes the molecules, removing electrons and forming positive molecular ions.
3. Fragmentation: Some molecular ions break into smaller charged and neutral particles.
4. Acceleration: An electric field accelerates the positive ions.
5. Separation: A mass analyser separates ions according to their mass-to-charge ratios.
6. Detection: A detector records the ions that arrive, and the instrument produces a mass spectrum.

The important distinction is that the detector only records charged particles. If a molecule breaks into one positive ion and one neutral particle, only the positive ion contributes to the spectrum.

The mass analyser also does not directly identify the structure of an ion. It measures its mass-to-charge ratio.

The diagram describes one classic magnetic-sector design. Modern mass spectrometers can use different types of analysers, but the underlying measurement is still based on separating ions according to \(m/z\).

Why do structural isomers produce different spectra?

Consider two alcohols with the molecular formula \(\ce{C4H10O}\):

• Butan-1-ol: \(\ce{CH3CH2CH2CH2OH}\)
• Butan-2-ol: \(\ce{CH3CH(OH)CH2CH3}\)

Both contain four carbon atoms, ten hydrogen atoms, and one oxygen atom. Both have a molecular ion at \(m/z=74\), assuming that the molecular ion is detected.

However, the oxygen atom is attached to a different carbon position.

That changes which bonds can break to produce stable positive ions.

For butan-1-ol, breaking the bond beside the carbon attached to OH can produce the ion \(\ce{CH2OH+}\).

Its nominal mass is:

\[ m/z=12+(3\times1)+16=31 \]

For butan-2-ol, a corresponding cleavage can produce \(\ce{CH3CHOH+}\).

Its nominal mass is:

\[ m/z=(2\times12)+5+16=45 \]

The two compounds can therefore produce different dominant fragment peaks, even though their molecular ions have the same mass.

This type of bond breaking next to a functional group is commonly called alpha cleavage.

The key is to examine where the functional group sits and which fragment ions its position can produce.

02Worked examples: comparing actual structural possibilities

The following spectra use simplified, illustrative relative-abundance data. They are constructed to demonstrate the interpretation method, rather than presented as measurements from a particular experiment.

03A reliable decision process for HSC questions

When comparing two spectra in an assessment, use the following sequence.

A useful way to organise your reasoning is to work backwards from each major fragment.

For example, suppose a compound has a molecular ion at \(m/z=86\), and a strong fragment at \(m/z=57\).

The mass difference is:

\[ 86-57=29 \]

This suggests that a neutral species with a nominal mass of 29 has been lost.

You could investigate whether the proposed structure allows that loss while leaving behind a reasonably stable positive ion.

However, a difference of 29 does not automatically prove that \(\ce{C2H5}\) was lost. Other possible neutral compositions can also have a nominal mass of 29.

The proposed molecular structure must support the assignment.

Common fragment masses and their possible identities

These examples are useful starting points when interpreting organic mass spectra.

These are possible assignments, not a list of exclusive structural fingerprints.

For example, the peak at \(m/z=43\) is especially easy to misinterpret. It could represent an alkyl ion or an oxygen-containing acylium ion.

The molecular structure and other peaks help determine which assignment is more reasonable.

04Important limitations when comparing spectra

Before attempting the harder questions, keep four limitations in mind.

1. The molecular ion may be weak or absent.

Some molecular ions fragment readily. A weak molecular ion does not mean that the molecule has a low molecular mass.

If the molecular formula is supplied, calculate the expected molecular mass independently.

2. The base peak is not necessarily the molecular ion.

The base peak is simply the most abundant ion in the spectrum. It is assigned a relative abundance of 100.

A smaller fragment can be more abundant than the molecular ion if it forms readily and is relatively stable.

3. Different ions can have the same nominal mass.

A peak at \(m/z=43\) does not automatically identify \(\ce{CH3CO+}\).

Without additional information, several ion compositions may be possible. High-resolution mass spectrometry can distinguish ions with the same nominal mass but different exact masses, although that extra measurement is not required for the basic HSC method.

4. A mass spectrum is evidence, not an infallible structural identification.

Different compounds can produce overlapping fragmentation patterns. Rearrangements can also produce ions that are not formed by simply breaking one bond.

When the evidence is inconclusive, another analytical technique, such as infrared spectroscopy or nuclear magnetic resonance spectroscopy, may be needed.

05Questions and solutions

The following questions progress from identifying isotope patterns to interpreting competing fragmentation pathways. All numerical spectra are illustrative teaching data.

Question 1

Two structural isomers have the molecular formula \(\ce{C3H7Cl}\):

• 1-chloropropane
• 2-chloropropane

An unknown sample produces two molecular ion peaks:

(a) Explain what the two peaks indicate about the elements present in the molecule.

(b) Can the spectrum distinguish between 1-chloropropane and 2-chloropropane? Explain your reasoning.

Solution 1

(a) The isotope pattern strongly indicates the presence of one chlorine atom.

Chlorine has two naturally occurring isotopes:

• \({}^{35}\mathrm{Cl}\), approximately 75% abundance.
• \({}^{37}\mathrm{Cl}\), approximately 25% abundance.

The isotopes differ in mass by two units.

Therefore, a molecule containing one chlorine atom can produce two molecular ion peaks separated by two mass units.

The expected approximate ratio is:

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

The measured ratio is:

\[ \frac{100}{32}=3.125 \]

This is close to the expected ratio of approximately 3:1.

The peaks at \(m/z=78\) and \(m/z=80\) are therefore consistent with the two chlorine isotopes.

(b) No. The isotope pattern cannot distinguish the two structural isomers.

Both 1-chloropropane and 2-chloropropane contain exactly one chlorine atom.

They also have the same molecular formula and molecular mass.

Consequently, both produce the same expected chlorine isotope pattern.

To distinguish them, you would need to compare their fragment ions. The different positions of the chlorine atom can influence which bonds break and which charged fragments form.

The important distinction is that isotope patterns help identify elements, while fragmentation patterns provide evidence about molecular structure.

Question 2

Two ketones have the molecular formula \(\ce{C5H10O}\):

• Pentan-2-one: \(\ce{CH3COCH2CH2CH3}\)
• Pentan-3-one: \(\ce{CH3CH2COCH2CH3}\)

Their molecular ions both occur at \(m/z=86\).

Two simplified spectra produce the following data:

(a) Identify which spectrum is more consistent with pentan-2-one.

(b) Identify which spectrum is more consistent with pentan-3-one.

(c) Explain why the symmetrical structure of pentan-3-one favours the formation of its major fragment.

Solution 2

(a) Spectrum A is more consistent with pentan-2-one. Spectrum B is more consistent with pentan-3-one.

Start with the strongest peak in Spectrum A, which occurs at \(m/z=43\).

Pentan-2-one contains the structure:

\[ \ce{CH3COCH2CH2CH3} \]

Breaking the bond between the carbonyl carbon and the propyl group can leave the charged fragment:

\[ \ce{CH3CO+} \]

This is an acetyl ion.

Its nominal mass is:

\[ m/z=(2\times12)+3+16=43 \]

The strong peak at \(m/z=43\) therefore supports pentan-2-one.

The additional peak at \(m/z=71\) is also consistent with loss of a methyl radical from the molecular ion:

\[ 86-15=71 \]

(b) Spectrum B is more consistent with pentan-3-one.

Pentan-3-one has the structure:

\[ \ce{CH3CH2COCH2CH3} \]

Breaking either carbon-carbon bond beside the carbonyl group can produce:

\[ \ce{CH3CH2CO+} \]

This is a propionyl ion with the formula \(\ce{C3H5O+}\).

Its nominal mass is:

\[ m/z=(3\times12)+5+16=57 \]

This explains the dominant peak at \(m/z=57\).

(c) Pentan-3-one has two equivalent ethyl groups attached to its carbonyl group.

This symmetry is important.

Cleavage on either side of the carbonyl produces the same type of oxygen-containing positive fragment.

The two equivalent pathways therefore favour the same major fragment at \(m/z=57\).

Pentan-2-one is not symmetrical in the same way. Its methyl and propyl groups can produce different fragments when bonds beside the carbonyl group break.

This is why the two ketones can have different dominant peaks despite having the same molecular formula.

The tempting shortcut is to assign \(m/z=57\) to any compound containing four carbons. But the proposed ion composition and the arrangement of atoms matter. In this case, the ketone structure supports an oxygen-containing acylium ion rather than simply an alkyl ion.

Question 3

An unknown organic compound has the molecular formula \(\ce{C4H8O}\).

Two possible structures are:

• Butanal: \(\ce{CH3CH2CH2CHO}\)
• Butan-2-one: \(\ce{CH3COCH2CH3}\)

Its electron-ionisation mass spectrum contains:

A student argues that the compound must be butan-2-one because it has a dominant peak at \(m/z=43\).

Evaluate this conclusion. Explain whether the available data uniquely establish the structure.

Solution 3

The student’s conclusion is not justified. The spectrum does not uniquely establish butan-2-one.

First, both candidate compounds have the same molecular formula.

Their molecular mass is:

\[ \begin{aligned} M_r&=(4\times12)+(8\times1)+16\\\ &=72 \end{aligned} \]

The peak at \(m/z=72\) is consistent with either molecular ion.

Next, consider the dominant peak at \(m/z=43\).

Butan-2-one can produce:

\[ \ce{CH3CO+} \]

This is an acetyl ion with a nominal mass of 43.

However, butanal can also produce a fragment with a nominal mass of 43, such as:

\[ \ce{C3H7+} \]

Therefore, the same peak position can be consistent with different fragmentation pathways.

The peak at \(m/z=29\) is also ambiguous.

It could represent:

• \(\ce{CHO+}\), which is consistent with fragmentation involving an aldehyde group.
• \(\ce{C2H5+}\), an ethyl ion.

Both assignments have a nominal mass of 29.

The strongest conclusion is that the spectrum is consistent with both candidate structures, but the supplied peak positions are insufficient to distinguish them conclusively.

To improve the identification, you could examine additional diagnostic fragment peaks, compare the full spectra with reference spectra, or use an independent technique.

For example, infrared spectroscopy can provide evidence for an aldehyde through its characteristic aldehydic C-H absorption, while NMR spectroscopy can provide information about the hydrogen and carbon environments.

The important lesson is that a strong peak is not automatically a unique structural fingerprint.

Question 4

Two structural isomers have the molecular formula \(\ce{C4H9Cl}\):

• 1-chlorobutane: \(\ce{CH3CH2CH2CH2Cl}\)
• 2-chlorobutane: \(\ce{CH3CHClCH2CH3}\)

An unknown sample produces the following peaks:

Assume that the peaks at 92 and 94 form the molecular ion isotope pair, and that the peaks at 63 and 65 form a fragment isotope pair.

(a) Explain why the peaks at 63 and 65 indicate that the fragment retains chlorine.

(b) Calculate the nominal mass of the neutral species lost when the molecular ion forms the fragment at \(m/z=63\).

(c) Explain why these results do not necessarily distinguish the two structural isomers.

Solution 4

(a) The fragment retains one chlorine atom.

The peaks at \(m/z=63\) and \(m/z=65\) are separated by two mass units.

Their relative abundance is:

\[ \frac{40}{13}\approx3.08 \]

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

A possible fragment composition is:

\[ \ce{C2H4^{35}Cl+} \]

Its nominal mass is:

\[ (2\times12)+4+35=63 \]

Replacing chlorine-35 with chlorine-37 increases the fragment mass to 65.

The two peaks therefore indicate that chlorine remains in the fragment.

(b) The neutral species lost has a nominal mass of 29.

The original molecular ion has \(m/z=92\).

The fragment ion has \(m/z=63\).

Therefore:

\[ \begin{aligned} \text{Neutral loss}&=92-63\\\ &=29 \end{aligned} \]

The complementary neutral fragment could be a species with a nominal mass of 29, such as \(\ce{C2H5^.}\).

The corresponding chlorine-37 pathway also gives:

\[ 94-65=29 \]

Both pathways are consistent with the same neutral loss.

(c) The data do not uniquely establish the chlorine position.

Both structural isomers contain one chlorine atom and have the same molecular formula.

Both can undergo carbon-carbon bond cleavage to produce fragments that retain chlorine.

The peak at \(m/z=57\) is also informative. It could represent \(\ce{C4H9+}\), formed when chlorine is lost from the molecular ion.

Because this fragment contains no chlorine, it does not have the corresponding chlorine-37 isotope partner.

However, this type of fragmentation does not uniquely reveal whether the chlorine was attached to carbon 1 or carbon 2.

The data support the presence of chlorine and provide information about which fragments retain it. More diagnostic fragmentation evidence or an independent structural technique would be needed to establish the chlorine position.

Question 5

Two pure compounds, P and Q, are known to be either butan-1-ol or ethoxyethane.

Both have the molecular formula \(\ce{C4H10O}\), and both have a nominal molecular mass of 74.

Their simplified electron-ionisation mass spectra contain the following peaks:

A student notices that both compounds have a strong peak at \(m/z=31\).

They confidently conclude that both compounds must contain a primary alcohol group.

Evaluate this reasoning.

(a) Identify compounds P and Q, assuming the only candidates are butan-1-ol and ethoxyethane.

(b) Explain why the peak at \(m/z=31\) cannot independently establish the presence of an alcohol.

(c) Explain how the other fragment peaks provide stronger evidence for the two different structures.

(d) Would the spectra, without the supplied candidate list, prove the identity of either compound? Explain.

Solution 5

(a) Compound P is butan-1-ol, and compound Q is ethoxyethane, within the stated two-candidate set.

Both compounds have the same molecular mass:

\[ M_r=74 \]

Their molecular ion peaks therefore provide no distinction.

The important evidence comes from the different fragment patterns.

(b) The peak at \(m/z=31\) does not uniquely identify an alcohol.

The tempting argument is that primary alcohols commonly produce the hydroxymethyl ion:

\[ \ce{CH2OH+} \]

This ion has a nominal mass of:

\[ 12+3+16=31 \]

However, the same nominal mass can be produced through fragmentation of other oxygen-containing structures.

Ethoxyethane can also produce a strong \(m/z=31\) ion during electron-ionisation fragmentation.

In particular, electron-impact studies of diethyl ether report a prominent ion at this mass. The precise fragmentation pathways and relative abundances depend on the ionisation conditions. 【turn187361search7】【turn187361search11】

Therefore, a peak at \(m/z=31\) is not exclusive to primary alcohols.

This is the critical flaw in the student’s reasoning.

(c) The remaining fragments provide the stronger structural evidence.

For compound P, the strongest peak occurs at \(m/z=56\).

Compare this with the molecular ion:

\[ \begin{aligned} \text{Mass difference}&=74-56\\\ &=18 \end{aligned} \]

A loss of 18 mass units is consistent with dehydration:

\[ \ce{H2O} \]

Butan-1-ol contains an OH group, so dehydration provides a plausible fragmentation pathway.

The strong peak at \(m/z=31\) also supports the presence of the terminal \(\ce{-CH2OH}\) group.

Together, these peaks strongly favour butan-1-ol.

Compound Q has a different pattern.

Its most prominent fragments occur at \(m/z=45\) and \(m/z=59\).

These can be assigned to oxygen-containing ions:

\[ \ce{C2H5O+}\qquad m/z=45 \]

\[ \ce{C3H7O+}\qquad m/z=59 \]

Ethoxyethane has an oxygen atom connecting two ethyl groups:

\[ \ce{CH3CH2-O-CH2CH3} \]

Cleavage around this central oxygen can produce oxygen-containing fragments.

The strong peaks at 45 and 59, together with the peak at 29, support the ether structure.

The very weak peak at 56 also provides less support for the dehydration pathway expected from an alcohol.

(d) No. Without the candidate list, the spectra would support these assignments but would not prove them uniquely.

The two spectra demonstrate why structural identification requires several pieces of evidence.

Both compounds have:

• The same molecular formula.
• The same nominal molecular mass.
• A molecular ion at \(m/z=74\), if detectable.
• A strong fragment at \(m/z=31\).

Their different fragmentation patterns provide the more useful evidence.

However, other structures could potentially produce overlapping peaks or similar fragmentation pathways.

A reliable identification would compare the full mass spectrum with reference data and, where necessary, use another analytical technique.

This is the difference between finding a structure that fits the evidence and proving that no other structure could fit it.

06The main takeaway: compare patterns, not isolated peaks

When comparing mass spectra of structural isomers, remember what each type of evidence actually tells you.

• The molecular ion helps establish molecular mass.
• Isotope patterns provide evidence about the elements present.
• Fragment ions reveal plausible parts of the original molecular structure.
• Relative abundances indicate which fragmentation pathways are favoured.
• The complete pattern helps determine which proposed structure best fits the evidence.

The biggest mistake is treating one peak as a guaranteed structural fingerprint.

A strong \(m/z=43\) peak, for example, might indicate an acetyl ion or a propyl ion. The difference matters because those ions suggest different structures.

Once you can interpret mass spectra this way, the next useful step is combining them with infrared spectroscopy and NMR spectroscopy. Mass spectrometry helps establish molecular mass and fragmentation, infrared spectroscopy identifies characteristic bonds and functional groups, and NMR provides information about the chemical environments of atoms. Together, these techniques give a much stronger basis for identifying an unknown organic compound.