Why Neutral Fragments Are Not Detected in Mass Spectrometry
Mass spectrometers detect charged fragments, not neutral ones. Learn how charge retention, neutral-loss calculations, and isotope patterns help explain fragmentation and interpret mass spectra.
A molecule can break into two fragments in a mass spectrometer, yet the resulting mass spectrum might show only one of them. Where did the other fragment go?
The answer is charge. A standard electron-ionisation (EI) mass spectrometer detects charged particles, called ions. It does not directly detect neutral fragments because its mass analyser uses electric and magnetic fields to control the paths of charged particles. A neutral fragment is not steered towards the ion detector, so it produces no mass-spectrum peak.
For example, consider this fragmentation:
\[ \mathrm{AB}^{+\bullet}\rightarrow \mathrm{A}^{+}+\mathrm{B}^{\bullet} \]
The original molecule has a positive charge. When it breaks apart, fragment A retains that charge, while fragment B is neutral.
The instrument detects \(\mathrm{A}^{+}\), but not \(\mathrm{B}^{\bullet}\). Importantly, the neutral fragment still has mass. You can sometimes calculate its mass from the difference between the molecular ion and the detected fragment ion, even though the instrument never records it directly.
01What does a mass spectrometer actually detect?
To understand why neutral fragments are missing, follow a molecule through a simplified electron-ionisation mass spectrometer.
1. The molecule enters the ionisation chamber
The sample is first vaporised so that its molecules can enter the instrument as a gas.
A beam of high-energy electrons passes through the ionisation chamber. When one of these electrons collides with a molecule, it can remove an electron from that molecule.
This produces a positively charged molecular ion:
\[ \ce{M + e^- -> M^{+\bullet} + 2e^-} \]
Here:
- \(\mathrm{M}\) represents the original neutral molecule.
- \(\mathrm{M}^{+\bullet}\) represents the molecular ion.
- The \(+\) indicates a positive charge.
- The dot indicates an unpaired electron.
The molecular ion initially contains all the atoms of the original molecule. It may remain intact, or it may break apart into smaller fragments.
2. The ions are accelerated and separated
After ionisation, the positive ions are accelerated by an electric field towards the mass analyser.
In a simplified magnetic-sector instrument, the ions then enter a region containing a magnetic field. The field bends their paths.
The amount of bending depends on the ion’s mass-to-charge ratio, written as:
\[ \frac{m}{z} \]
Where:
- \(m\) is the mass of the ion, measured in atomic mass units (u).
- \(z\) is the positive charge number of the ion.
For example, a singly charged ion has \(z=1\). Its numerical \(m/z\) value is therefore equal to its mass in u.
Different ions follow different paths through the magnetic field. This allows the instrument to separate them according to their mass-to-charge ratios.
The detector is positioned to receive ions travelling along the appropriate path.
3. What happens to a neutral fragment?
Now consider what happens if the molecular ion breaks apart:
\[ \mathrm{AB}^{+\bullet}\rightarrow \mathrm{A}^{+}+\mathrm{B}^{\bullet} \]
Fragment A is positively charged, so it responds to the electric and magnetic fields.
Fragment B has no net electrical charge. It therefore does not experience the charge-dependent forces that steer ions through the analyser.
The underlying relationships are:
\[ F_E=qE \]
\[ F_B=qvB\sin\theta \]
Where:
- \(F_E\) is the magnitude of the electric force, in newtons (N).
- \(F_B\) is the magnitude of the magnetic force, in newtons.
- \(q\) is the magnitude of the particle’s charge, in coulombs (C).
- \(E\) is electric field strength, in N C\(^{-1}\).
- \(v\) is the particle’s speed, in m s\(^{-1}\).
- \(B\) is magnetic flux density, in tesla (T).
- \(\theta\) is the angle between the particle’s velocity and the magnetic field.
For a neutral particle, \(q=0\). Both charge-dependent forces are therefore zero.
The neutral fragment continues along its original direction rather than following the curved ion path. It misses the ion detector and does not produce a mass-spectrum signal.
There is another way to express this: when \(z=0\), the expression \(m/z\) is undefined. A neutral particle cannot be assigned a normal mass-to-charge ratio by this analyser.
The important distinction is that a neutral fragment has mass, but no measurable \(m/z\) in the standard ion-based mass spectrum.
Why is the analyser designed this way?
Imagine placing a detector directly behind the ionisation chamber. Particles might reach it, but the instrument would not separate ions according to their masses.
The mass analyser solves that problem by using fields to guide ions along mass-dependent paths before they reach the detector.
That design makes mass separation possible. However, it also means that neutral products are not guided into the ion detector.
This is the experimental limitation to remember: a mass spectrum shows the ions that reach the detector, not every particle produced when molecules break apart.
The magnetic-sector arrangement is a simplified model. Other mass analysers use different separation methods, but standard mass spectrometry still relies on ions rather than directly measuring neutral fragments.
02Why does only one fragment usually retain the charge?
When a singly charged molecular ion breaks into two pieces, its total charge must be conserved.
Consider:
\[ \mathrm{AB}^{+\bullet}\rightarrow \mathrm{A}^{+}+\mathrm{B}^{\bullet} \]
The original ion has a charge of \(+1\). The products have a combined charge of \(+1\).
The positive charge has remained on A, leaving B neutral.
However, the charge does not always have to remain on the same side of the molecule.
An alternative fragmentation pathway is possible:
\[ \mathrm{AB}^{+\bullet}\rightarrow \mathrm{A}^{\bullet}+\mathrm{B}^{+} \]
Now B is the detected fragment, while A is neutral.
Which pathway occurs depends partly on the stability of the resulting ions and the energy available during fragmentation.
This gives us an important distinction:
- One molecular ion undergoing a simple two-fragment cleavage normally produces one charged fragment and one neutral fragment.
- Different molecular ions from the same sample can undergo different fragmentation pathways.
- Consequently, both A+ and B+ can appear in the same mass spectrum, even though they are not produced together in one simple fragmentation event.
The mass spectrum is an accumulation of signals from many separate ion events.
03Worked example 1: Where did the chlorine go?
An illustrative positive-EI mass spectrum of 1-chloropropane, \(\ce{C3H7Cl}\), contains the following peaks:
| Peak | Relative intensity pattern |
|---|---|
| \(m/z=78\) | Larger molecular-ion peak |
| \(m/z=80\) | Approximately one-third the intensity of 78 |
| \(m/z=43\) | Strong fragment peak |
| \(m/z=45\) | No significant corresponding peak |
Assume that all ions are singly charged, use the nominal masses C = 12, H = 1, \({}^{35}\mathrm{Cl}=35\), and \({}^{37}\mathrm{Cl}=37\), and consider \(\ce{C3H7+}\) as the proposed assignment for the fragment at \(m/z=43\).
Explain why the fragment at 43 is detected, while the chlorine-containing fragment is not.
Step 1
The molecular ion containing chlorine-35 has a nominal mass of:
\[ \begin{aligned} m &= (3\times12)+7+35\\ &=78\ \mathrm{u} \end{aligned} \]
The corresponding ion containing chlorine-37 has a mass of:
\[ m=36+7+37=80\ \mathrm{u} \]
The peaks at 78 and 80 are separated by two mass units and have an approximate 3:1 intensity ratio.
This is the characteristic isotope pattern of a molecule containing one chlorine atom.
Step 2
The proposed fragment is \(\ce{C3H7+}\).
Its nominal mass is:
\[ m=(3\times12)+7=43\ \mathrm{u} \]
Because it carries a positive charge of \(+1\):
\[ \frac{m}{z}=\frac{43}{1}=43 \]
This explains the fragment peak at \(m/z=43\).
Step 3
For the molecular ion containing chlorine-35:
\[ 78-43=35\ \mathrm{u} \]
For the molecular ion containing chlorine-37:
\[ 80-43=37\ \mathrm{u} \]
The complementary fragment is therefore chlorine, with either isotope.
The proposed fragmentation pathways are:
\[ \begin{aligned} \mathrm{C_3H_7{}^{35}Cl}^{+\bullet} &\rightarrow \mathrm{C_3H_7}^{+} +{}^{35}\mathrm{Cl}^{\bullet} \\ \mathrm{C_3H_7{}^{37}Cl}^{+\bullet} &\rightarrow \mathrm{C_3H_7}^{+} +{}^{37}\mathrm{Cl}^{\bullet} \end{aligned} \]
Step 4
In these proposed pathways, the positive charge remains on \(\ce{C3H7+}\).
The chlorine fragments are neutral radicals. They are not accelerated and separated by the analyser, so they do not reach the ion detector as measurable ions.
The detected fragment therefore has a peak at 43, but the neutral chlorine fragments do not produce peaks at 35 or 37.
Key conclusion
The molecular-ion isotope pattern indicates chlorine in the original molecule, while the fragment at 43 is consistent with the charge remaining on the carbon-containing portion. The chlorine-containing complementary fragment is neutral and is not directly detected.
Notice that the assignment is consistent with the evidence, rather than uniquely proven by a single peak.
04Worked example 2: What if the detected fragment retains the halogen?
Now consider an illustrative spectrum of 2-bromobutane, \(\ce{C4H9Br}\).
The spectrum contains:
| Peak | \(m/z\) | Relative intensity pattern |
|---|---|---|
| Molecular ion | 136 | Approximately equal to 138 |
| Molecular ion | 138 | Approximately equal to 136 |
| Fragment ion | 121 | Approximately equal to 123 |
| Fragment ion | 123 | Approximately equal to 121 |
Assume singly charged ions and nominal masses of 79 and 81 for the two main bromine isotopes. A proposed assignment for the fragment is \(\ce{C3H6Br+}\).
Explain why the fragment retains a pair of isotope peaks, even though the molecule has broken apart.
Step 1
For the molecule containing bromine-79:
\[ \begin{aligned} m&=(4\times12)+9+79\\ &=136\ \mathrm{u} \end{aligned} \]
For the molecule containing bromine-81:
\[ m=48+9+81=138\ \mathrm{u} \]
The approximately 1:1 intensity ratio is consistent with one bromine atom.
Step 2
For the proposed fragment containing bromine-79:
\[ m=(3\times12)+6+79=121\ \mathrm{u} \]
For the corresponding fragment containing bromine-81:
\[ m=36+6+81=123\ \mathrm{u} \]
Both fragment ions are positively charged, so their \(m/z\) values are 121 and 123.
Step 3
Compare the matching molecular-ion and fragment-ion isotope peaks:
\[ 136-121=15\ \mathrm{u} \]
\[ 138-123=15\ \mathrm{u} \]
A neutral fragment with nominal mass 15 is lost.
The proposed neutral fragment is a methyl radical, \(\ce{CH3^{\bullet}}\).
The fragmentation pathway for the bromine-79 isotopologue is:
\[ \mathrm{C_4H_9{}^{79}Br}^{+\bullet} \rightarrow \mathrm{C_3H_6{}^{79}Br}^{+} +\mathrm{CH_3}^{\bullet} \]
Step 4
The detected fragment still contains bromine.
Therefore, molecules containing bromine-79 produce a fragment at 121, while molecules containing bromine-81 produce a fragment at 123.
The approximately 1:1 isotope pattern remains visible because the bromine atom is retained in the charged fragment.
The methyl radical is neutral, so it does not produce its own peak.
This example demonstrates an important point: fragmentation does not necessarily remove the element responsible for an isotope pattern. The isotope pattern depends on which atoms remain in the detected ion.
05How to interpret isotope patterns when fragments are neutral
Isotope patterns provide another way to work out which part of a molecule retains the charge.
Two elements commonly used in HSC mass-spectrometry questions are chlorine and bromine.
| Element | Main isotopes | Approximate isotope ratio |
|---|---|---|
| Chlorine | 35 and 37 | 3:1 |
| Bromine | 79 and 81 | 1:1 |
A molecule containing one chlorine atom typically produces molecular-ion peaks separated by two \(m/z\) units, with an approximate 3:1 ratio.
A molecule containing one bromine atom typically produces a pair separated by two \(m/z\) units, with approximately equal intensities.
A molecule containing two chlorine atoms can produce three molecular-ion peaks with an approximate 9:6:1 ratio.
However, there is a crucial detail:
An isotope pattern belongs to the detected ion, not automatically to every fragment of the original molecule.
Consider these two situations.
Situation 1: Chlorine leaves as a neutral fragment
Suppose the original molecular ion contains chlorine, but the detected fragment does not.
The parent molecular-ion peaks may show a 3:1 isotope pattern, while the fragment ion produces only one main peak.
This does not mean the original molecule lacked chlorine. It means the chlorine may have left with the neutral fragment.
Situation 2: Chlorine remains in the charged fragment
If the detected fragment retains chlorine, its isotope pattern should also show the characteristic pair.
For example, a fragment containing one chlorine atom may produce peaks separated by two mass units with an approximate 3:1 ratio.
This provides evidence that chlorine remains in the detected fragment.
The same reasoning applies to bromine, using its approximately 1:1 isotope pattern.
There are limits to this method. Weak peaks, overlapping signals, and other isotope contributions can make a pattern difficult to recognise. An isotope pattern is useful evidence, but it does not uniquely establish a complete molecular structure.
06A reliable method for interpreting neutral losses
When a mass spectrum shows a molecular ion and one or more fragment ions, use the following sequence.
| Step | What to examine | What it tells you |
|---|---|---|
| 1 | Molecular-ion peak | The approximate mass of the original molecule |
| 2 | Molecular-ion isotope pattern | Which isotopic elements may be present |
| 3 | Fragment-ion \(m/z\) | The mass-to-charge ratio of the detected fragment |
| 4 | Parent-minus-fragment mass difference | The approximate mass of the neutral loss |
| 5 | Fragment isotope pattern | Whether an isotope-bearing element remains in the detected ion |
For the usual HSC examples, assume singly charged ions unless stated otherwise.
Under this assumption:
\[ m_{\text{neutral}} = m_{\text{parent ion}} - m_{\text{fragment ion}} \]
This calculation is valid when comparing the appropriate singly charged parent and fragment ions and matching their isotope compositions.
For example, if a molecular ion has a mass of 78 u and produces a fragment ion of 43 u:
\[ m_{\text{neutral}}=78-43=35\ \mathrm{u} \]
The neutral partner has a nominal mass of 35 u.
However, this calculation does not automatically establish the neutral fragment’s chemical formula. Different combinations of atoms can have the same nominal mass.
To check whether a proposed fragment formula is chemically possible, use How to Check Whether a Fragment Fits a Molecular Formula.
For more practice distinguishing isotope peaks from separate fragment peaks, see How to Distinguish Isotopic Peaks from Fragment Peaks.
Common mistakes to avoid
| Incorrect interpretation | Correct interpretation |
|---|---|
| Every fragment produced must have a peak. | Only detected ions produce the usual mass-spectrum peaks. |
| A neutral fragment has no mass. | It has mass, but no measurable \(m/z\) in the standard ion analyser. |
| No chlorine isotope pair means the original molecule contains no chlorine. | Chlorine may have left as a neutral fragment. |
| Two positive fragment peaks must come from the same fragmentation event. | They may come from different fragmentation pathways involving different molecular ions. |
| The mass difference proves the exact neutral fragment formula. | It establishes a nominal mass difference; further evidence is needed to assign a formula. |
07Questions and solutions
Question 1
A positive-EI mass spectrum contains a molecular-ion peak at \(m/z=72\) and a fragment-ion peak at \(m/z=57\).
Assume both ions are singly charged and the fragment is produced by a simple fragmentation of the molecular ion.
Calculate the nominal mass of the neutral fragment. Explain why it does not produce a separate peak.
Solution 1
The neutral fragment has a nominal mass of 15 u, and it does not produce a peak because it has no electrical charge.
For singly charged ions, the difference between the parent-ion mass and fragment-ion mass gives the neutral loss:
\[ \begin{aligned} m_{\text{neutral}}&=72-57\\ &=15\ \mathrm{u} \end{aligned} \]
The molecular ion has broken into a charged fragment of mass 57 u and a neutral fragment of mass 15 u.
The charged fragment is accelerated and separated by the mass analyser, producing the peak at \(m/z=57\).
The neutral fragment is not steered into the ion detector, so it produces no separate signal.
A methyl radical would have a nominal mass of 15 u, but the mass difference alone does not prove that this is the neutral fragment’s formula.
Question 2
An illustrative mass spectrum of a compound with molecular formula \(\ce{C3H7Cl}\) contains:
- Molecular-ion peaks at \(m/z=78\) and \(m/z=80\), with an approximate 3:1 intensity ratio.
- Fragment-ion peaks at \(m/z=49\) and \(m/z=51\), also with an approximate 3:1 ratio.
A proposed fragment assignment is \(\ce{CH2Cl+}\).
Assume all ions are singly charged and use nominal isotope masses.
(a) Explain what the molecular-ion isotope pattern indicates.
(b) Calculate the mass of the neutral partner for each corresponding isotope combination.
(c) Explain why the fragment ion also produces an isotope pair.
Solution 2
The detected fragment retains chlorine, while the neutral partner has a nominal mass of 29 u.
(a) The molecular-ion peaks at 78 and 80 are separated by two mass units and have an approximate 3:1 ratio.
This indicates that the original molecule contains one chlorine atom, with the difference arising from chlorine-35 and chlorine-37.
(b) The proposed fragment containing chlorine-35 has a nominal mass of:
\[ 12+2+35=49\ \mathrm{u} \]
The corresponding fragment containing chlorine-37 has a nominal mass of:
\[ 12+2+37=51\ \mathrm{u} \]
Now calculate the neutral losses:
\[ 78-49=29\ \mathrm{u} \]
\[ 80-51=29\ \mathrm{u} \]
The neutral partner therefore has a nominal mass of 29 u.
Subtracting the atoms in \(\ce{CH2Cl}\) from the original molecular formula gives \(\ce{C2H5}\).
The proposed neutral fragment is therefore \(\ce{C2H5^{\bullet}}\).
(c) The fragment retains the chlorine atom.
Consequently, the chlorine-35-containing fragment produces the peak at 49, while the chlorine-37-containing fragment produces the peak at 51.
Their approximately 3:1 ratio reflects the natural abundance of the two chlorine isotopes.
The neutral \(\ce{C2H5^{\bullet}}\) fragment does not produce a peak because it is not charged.
The isotope pattern therefore provides evidence that chlorine remains in the detected fragment, rather than leaving as a neutral species.
Question 3
Ethyl bromide, \(\ce{C2H5Br}\), produces the following illustrative positive-EI mass-spectrum data:
| Peak | \(m/z\) | Approximate intensity relationship |
|---|---|---|
| Molecular ion | 108 | 1 |
| Molecular ion | 110 | 1 |
| Fragment ion | 29 | Strong |
| Fragment ion | 79 | 1 |
| Fragment ion | 81 | 1 |
Assume singly charged ions, nominal isotope masses, and simple cleavage of the carbon-bromine bond.
(a) Calculate the molecular-ion masses for the two bromine isotopes.
(b) Propose two different fragmentation pathways that explain the fragment peaks.
(c) Identify the neutral partner in each pathway.
(d) Explain how both fragment families can appear in the same mass spectrum without violating conservation of charge.
Solution 3
Both fragment families can appear in the same spectrum because different molecular ions can retain their positive charge on different sides of the carbon-bromine bond.
(a) For bromine-79:
\[ \begin{aligned} m&=(2\times12)+5+79\\ &=108\ \mathrm{u} \end{aligned} \]
For bromine-81:
\[ m=24+5+81=110\ \mathrm{u} \]
The approximately 1:1 ratio is consistent with one bromine atom.
(b) There are two possible charge-retention pathways.
In the first pathway, the ethyl group retains the positive charge:
\[ \mathrm{C_2H_5Br}^{+\bullet} \rightarrow \mathrm{C_2H_5}^{+} +\mathrm{Br}^{\bullet} \]
The ethyl ion has a nominal mass of:
\[ (2\times12)+5=29\ \mathrm{u} \]
This explains the fragment peak at \(m/z=29\).
In the second pathway, the positive charge remains on bromine:
\[ \mathrm{C_2H_5Br}^{+\bullet} \rightarrow \mathrm{C_2H_5}^{\bullet} +\mathrm{Br}^{+} \]
Bromine-79 and bromine-81 produce ions at \(m/z=79\) and \(m/z=81\), respectively.
(c) The neutral partners are different in the two pathways.
| Charged fragment | Neutral partner |
|---|---|
| \(\ce{C2H5+}\) at 29 | \(\ce{Br^{\bullet}}\), mass 79 or 81 u |
| \(\mathrm{Br}^{+}\) at 79 or 81 | \(\ce{C2H5^{\bullet}}\), mass 29 u |
Neither neutral partner produces its own mass-spectrum peak.
(d) A single singly charged molecular ion cannot produce two positive singly charged fragments through one simple two-body cleavage.
However, the spectrum represents many separate fragmentation events.
Some molecular ions retain their charge on the ethyl group, producing \(\ce{C2H5+}\). Others retain their charge on bromine, producing \(\mathrm{Br}^{+}\).
Both fragment families can therefore appear in the same spectrum without violating conservation of charge.
Question 4
1,2-dichloroethane has the structural formula \(\ce{ClCH2CH2Cl}\).
An illustrative positive-EI spectrum contains:
- Molecular-ion peaks at \(m/z=98\), 100, and 102, with an approximate 9:6:1 intensity ratio.
- Fragment-ion peaks at \(m/z=49\) and 51, with an approximate 3:1 intensity ratio.
Assume that:
- The molecular formula is \(\ce{C2H4Cl2}\).
- The fragment forms by cleavage of the central carbon-carbon bond.
- The proposed charged fragment is \(\ce{CH2Cl+}\).
- All detected ions are singly charged.
- No rearrangement occurs during fragmentation.
(a) Explain the 9:6:1 molecular-ion isotope pattern.
(b) Explain why the fragment produces a 3:1 isotope pair.
(c) Determine the formula and possible nominal masses of the neutral partner.
(d) Explain why the mixed-isotope molecular ion at \(m/z=100\) requires particular care when interpreting the fragment masses.
Solution 4
The molecular ion contains two chlorine atoms, while each fragment contains one chlorine atom. The complementary fragment is neutral and therefore is not directly detected.
(a) Chlorine has two main isotopes, with an approximate abundance ratio of 3:1.
For a molecule containing two chlorine atoms, there are three possible isotope combinations:
- Two chlorine-35 atoms.
- One chlorine-35 atom and one chlorine-37 atom.
- Two chlorine-37 atoms.
Their approximate relative abundances are:
\[ 9:6:1 \]
The corresponding molecular-ion masses are:
\[ 98,\quad100,\quad102 \]
The three peaks therefore indicate two chlorine atoms in the original molecule.
(b) The proposed fragment is \(\ce{CH2Cl+}\).
With chlorine-35:
\[ m=12+2+35=49\ \mathrm{u} \]
With chlorine-37:
\[ m=12+2+37=51\ \mathrm{u} \]
Each fragment contains one chlorine atom, so the two fragment peaks have an approximate 3:1 intensity ratio.
(c) Cleavage of the central carbon-carbon bond separates the original molecule into two \(\ce{CH2Cl}\) groups.
If one group retains the positive charge, the other becomes neutral:
\[ \mathrm{ClCH_2CH_2Cl}^{+\bullet} \rightarrow \mathrm{CH_2Cl}^{+} + \mathrm{CH_2Cl}^{\bullet} \]
The neutral partner can contain either chlorine isotope.
Its possible nominal masses are:
\[ 12+2+35=49\ \mathrm{u} \]
or
\[ 12+2+37=51\ \mathrm{u} \]
It is not directly detected because it has no electrical charge.
(d) This is where simply subtracting the nearest-looking peaks can lead to an incorrect interpretation.
A molecular ion at \(m/z=100\) contains one chlorine-35 atom and one chlorine-37 atom.
If the charged fragment retains chlorine-35, its peak appears at 49, and the neutral partner has mass 51.
Alternatively, if the charged fragment retains chlorine-37, its peak appears at 51, and the neutral partner has mass 49.
Both combinations conserve mass:
\[ 49+51=100 \]
The spectrum contains the combined signals from many fragmentation events. It does not tell you which individual neutral partner accompanied each detected ion.
This is why isotope patterns must be interpreted alongside the molecular formula and the proposed fragmentation pathway.
Question 5
An unknown compound has the molecular formula \(\ce{C4H9Cl}\).
Its illustrative positive-EI mass spectrum contains:
| Ion | \(m/z\) | Approximate isotope pattern |
|---|---|---|
| Molecular ion | 92 and 94 | 3:1 |
| Fragment A | 57 | No significant corresponding peak at 59 |
| Fragment B | 35 and 37 | 3:1 |
For this question, assume the proposed assignments are:
- Fragment A is \(\ce{C4H9+}\).
- Fragment B is \(\mathrm{Cl}^{+}\).
Assume all ions are singly charged and fragmentation involves simple cleavage into two complementary pieces, without rearrangement or secondary ionisation.
Two students offer different explanations.
Student 1: Since \(57+35=92\) and \(57+37=94\), each molecular ion must produce both positive fragments together.
Student 2: A single molecular ion cannot produce both positive fragments, so the two fragment families must have come from two different compounds.
Evaluate both explanations.
Your response should:
- Write the two possible fragmentation pathways for each chlorine isotope.
- Explain why both fragment families can appear in the same spectrum.
- Use the isotope patterns to explain which fragment contains chlorine.
- Decide whether the evidence proves a particular structural isomer.
Solution 5
Neither student is correct. A single singly charged molecular ion cannot produce two positive singly charged fragments through one simple two-body cleavage, but the same compound can produce both fragment families through different fragmentation events.
Step 1: Establish the molecular-ion masses.
For chlorine-35:
\[ m= (4\times12)+9+35 =92\ \mathrm{u} \]
For chlorine-37:
\[ m=48+9+37 =94\ \mathrm{u} \]
The molecular-ion pair at 92 and 94, with an approximate 3:1 ratio, indicates one chlorine atom.
Step 2: Consider the first charge-retention pathway.
The positive charge can remain on the carbon-containing fragment:
\[ \mathrm{C_4H_9{}^{35}Cl}^{+\bullet} \rightarrow \mathrm{C_4H_9}^{+} + {}^{35}\mathrm{Cl}^{\bullet} \]
The detected ion is \(\ce{C4H9+}\).
Its mass is:
\[ m=(4\times12)+9=57\ \mathrm{u} \]
The complementary chlorine-35 fragment is neutral and has a mass of 35 u.
The corresponding chlorine-37 pathway produces the same detected ion at 57, but the neutral chlorine fragment has a mass of 37 u.
This explains why the detected fragment has no corresponding chlorine isotope peak at 59.
The chlorine isotope has left with the neutral fragment.
Step 3: Consider the alternative charge-retention pathway.
The positive charge can instead remain on chlorine:
\[ \mathrm{C_4H_9{}^{35}Cl}^{+\bullet} \rightarrow \mathrm{C_4H_9}^{\bullet} + {}^{35}\mathrm{Cl}^{+} \]
The detected chlorine ion has \(m/z=35\).
For the chlorine-37 isotopologue:
\[ \mathrm{C_4H_9{}^{37}Cl}^{+\bullet} \rightarrow \mathrm{C_4H_9}^{\bullet} + {}^{37}\mathrm{Cl}^{+} \]
The detected ion now has \(m/z=37\).
The two peaks at 35 and 37 have an approximate 3:1 ratio because the charged fragment contains chlorine.
The complementary \(\ce{C4H9^{\bullet}}\) fragment is neutral and has a nominal mass of 57 u.
Step 4: Resolve the charge-conservation problem.
Student 1’s argument uses correct mass arithmetic but incorrect charge reasoning.
For the chlorine-35 molecular ion:
\[ 57+35=92 \]
This shows that the two fragments have masses that add to the original molecular-ion mass.
However, mass conservation does not mean both fragments carry positive charges.
The two possible outcomes are:
| Pathway | Detected positive ion | Neutral partner |
|---|---|---|
| Charge retained by carbon fragment | \(\ce{C4H9+}\), \(m/z=57\) | \(\ce{Cl^{\bullet}}\), 35 or 37 u |
| Charge retained by chlorine | \(\mathrm{Cl}^{+}\), \(m/z=35\) or 37 | \(\ce{C4H9^{\bullet}}\), 57 u |
Each pathway conserves the original charge of \(+1\).
Producing both positive fragments simultaneously would give a total product charge of \(+2\), which is inconsistent with the assumed simple fragmentation of a singly charged molecular ion.
Student 2 also makes an incorrect assumption.
The two fragment families do not need to come from different compounds. They can come from different molecular ions of the same compound undergoing alternative fragmentation pathways.
The mass spectrum combines the signals from these separate events.
Step 5: Decide what the evidence establishes about the structure.
The spectrum provides evidence that:
- The original molecule contains one chlorine atom.
- The fragment at 57 is consistent with a carbon-containing ion that has lost chlorine.
- The pair at 35 and 37 is consistent with chlorine ions.
- The charge can remain on either side of the original carbon-chlorine bond in different fragmentation events.
However, these observations do not prove a unique structural isomer of \(\ce{C4H9Cl}\).
Different structural arrangements can produce ions with the same nominal masses. Additional fragment peaks, their relative abundances, and other structural evidence would be needed to distinguish the possibilities.
The central lesson is that a mass spectrum records charged fragments, not every product of molecular fragmentation. Once you understand where the charge goes, you can use fragment masses, neutral losses, and isotope patterns together to test possible structures without claiming more than the evidence supports.