How to Check Whether a Fragment Fits a Molecular Formula

A compound has the molecular formula \(\ce{C4H9Cl}\). Its electron-impact mass spectrum contains molecular-ion peaks at \(m/z = 92\) and \(94\), with an intensity ratio of approximately 3:1. Further along the spectrum, a smaller pair appears at \(m/z = 49\) and \(51\), again with a ratio close to 3:1.

Does the fragment still contain chlorine?

Yes, probably. The two chlorine isotopes, chlorine-35 and chlorine-37, differ in mass by 2 u and occur naturally in approximately a 3:1 ratio. The matching pattern suggests that chlorine is still present in the fragment.

However, a single peak at \(m/z = 49\) would not establish this. For example, both \(\ce{CH2Cl+}\) and \(\ce{C4H+}\) have a nominal mass of 49. Only the first contains chlorine and would produce the characteristic chlorine isotope pair.

To check whether a proposed fragment fits a molecular formula, use four tests:

1. Check that the fragment contains no more atoms of each element than the original molecule.
2. Calculate its expected mass-to-charge ratio, \(m/z\).
3. Check that the difference between the parent and fragment masses corresponds to a possible neutral loss.
4. Compare the fragment’s isotope pattern with the elements it supposedly contains.

If all four checks agree, the proposed fragment is consistent with the evidence. That does not automatically prove it is the only possible structure.

01What a mass spectrometer actually measures

Before interpreting fragments, it helps to understand how the instrument produces them.

A mass spectrometer does not photograph an entire molecule and identify its shape. It creates ions, separates them according to their mass-to-charge ratio, and records how many ions reach a detector at each value.

The process occurs in three main stages.

1. Ionisation

The sample is introduced into the instrument as a gas. In electron-impact (EI) mass spectrometry, a beam of high-energy electrons strikes the molecules.

An incoming electron can remove an electron from a molecule, producing a positively charged radical ion called the molecular ion.

\[
\ce{M + e^- -> M^{+\bullet} + 2e^-}
\]

Here:

• \(\ce{M}\) represents the original neutral molecule.
• \(\ce{M^{+\bullet}}\) represents the molecular ion.
• The \(+\) indicates a positive charge.
• The dot indicates an unpaired electron.

The molecular ion still contains all the atoms of the original molecule.

However, it may have enough internal energy to break chemical bonds. When this happens, smaller charged particles called fragment ions form.

For example:

\[
\ce{M^{+\bullet} -> A+ + B^{\bullet}}
\]

One product retains the positive charge and can be detected. The other is neutral and is not directly detected by the mass analyser.

This is important: a mass spectrum records charged particles, not every piece produced during fragmentation.

2. Separation according to mass-to-charge ratio

The charged particles travel into a mass analyser.

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

• \(m\) is the mass of the ion, measured in atomic mass units (u).
• \(z\) is the positive charge number of the ion.

For the examples in this article, assume positive, singly charged ions, so \(z=1\).

This means the numerical \(m/z\) value is approximately equal to the ion’s nominal mass in u.

For example, a singly charged ion with a nominal mass of 45 u produces a peak at \(m/z=45\).

If an ion has a charge of \(+2\), its \(m/z\) value is approximately half its mass. This is why you must check the charge state before treating a peak position as the mass of an ion.

3. Detection and the mass spectrum

The detector records the abundance of ions reaching it at each \(m/z\) value.

The resulting mass spectrum has:

• A horizontal axis showing \(m/z\).
• A vertical axis showing relative abundance.

The most intense peak is called the base peak and is assigned a relative abundance of 100%.

Importantly, the base peak is not necessarily the molecular ion. It is simply the most abundant ion detected.

The molecular ion may be relatively weak or even absent because some molecules fragment extensively during ionisation.

What should the original measurement show?

If some molecules remain intact, we expect to detect molecular ions at the mass of the original molecule. If other molecules fragment, we expect additional peaks at lower \(m/z\) values.

Observing both types of peaks supports the idea that the molecule can survive ionisation intact while also undergoing fragmentation.

However, the instrument measures \(m/z\), not the original arrangement of chemical bonds. Different ions can have the same nominal mass.

02The four checks for a proposed fragment

1. Check the atom inventory

Start with the molecular formula of the original compound.

Treat it as an inventory of available atoms. A fragment can contain fewer atoms than the parent molecule, but it cannot contain an element that was absent from the parent or more atoms of an element than the parent possessed.

For example, suppose the parent molecule has the formula:

\[
\ce{C4H9Cl}
\]

A proposed fragment is:

\[
\ce{CH2Cl+}
\]

Compare their atom counts:

The proposed fragment contains one carbon, two hydrogens, and one chlorine atom.

Every count is within the parent molecule’s available inventory.

Therefore, the proposed fragment passes the atom-count test.

Now consider a proposed fragment, \(\ce{C5H7+}\).

It contains five carbon atoms, but the parent only contains four.

Reject This fragment cannot have formed from the proposed parent molecule under ordinary fragmentation.

Think of the parent molecule as a collection of building blocks. You can rearrange the blocks or separate them into smaller groups, but you cannot create an additional carbon atom.

There is one limitation to this analogy: chemical fragments do not always remain continuous sections of the original structure. Rearrangements can occur before or during fragmentation.

This is why matching the atom inventory is necessary, but not sufficient.

If you need to refresh how to count hidden carbon and hydrogen atoms before checking a molecular formula, see How to Read Skeletal Formulas and Write Condensed Formulas.

2. Calculate the fragment’s expected mass

Once the atom counts are possible, calculate the nominal mass of the proposed fragment.

For common organic elements, use these approximate isotope masses:

For a singly charged ion, add the masses of all its atoms.

For example, the proposed fragment \(\ce{CH2Cl+}\), containing chlorine-35, has a nominal mass of:

\[
\begin{aligned}
m &= 12 + 2(1) + 35\\
&= 49\ \mathrm{u}
\end{aligned}
\]

Therefore:

\[
m/z = 49
\]

This agrees with a peak at \(m/z=49\).

But notice something interesting.

The alternative formula \(\ce{C4H+}\) also gives:

\[
4(12)+1=49\ \mathrm{u}
\]

Both formulas have the same nominal mass.

This is why matching the peak position alone is not enough to identify a fragment.

You need another piece of evidence.

3. Check the mass lost during fragmentation

A fragment must also be consistent with the amount of material removed from the original molecule.

Calculate the difference between the parent molecular-ion mass and the proposed fragment-ion mass.

For our example:

Parent:

\[
\ce{C4H9Cl}
\]

Nominal mass:

\[
4(12)+9(1)+35=92\ \mathrm{u}
\]

Proposed fragment:

\[
\ce{CH2Cl+}
\]

Nominal mass:

\[
12+2+35=49\ \mathrm{u}
\]

The mass difference is:

\[
92-49=43\ \mathrm{u}
\]

Now subtract the fragment’s atom counts from the parent formula.

\[
\ce{C4H9Cl – CH2Cl = C3H7}
\]

The neutral species removed has a nominal mass of:

\[
3(12)+7(1)=43\ \mathrm{u}
\]

The mass difference and atom difference agree.

One possible fragmentation pathway is:

\[
\ce{C4H9Cl^{+\bullet} -> CH2Cl+ + C3H7^{\bullet}}
\]

This equation conserves the atoms and the overall positive charge.

The neutral product is a radical, not necessarily a stable molecule. This matters because fragmentation can involve the loss of neutral molecules or neutral radicals.

If subtracting the proposed fragment formula from the parent produces a negative atom count, the assignment is impossible.

If the subtraction works but requires an implausible neutral species or an unsupported rearrangement, the assignment needs further evidence.

4. Check the isotope pattern

Isotopes are atoms of the same element with different numbers of neutrons.

They have the same chemical properties in most ordinary reactions but different masses.

Mass spectrometry can detect these mass differences, making isotope patterns particularly useful when checking proposed fragments.

Three isotope patterns are especially useful when analysing organic compounds.

The ratios are approximate and describe the natural isotope abundances for one atom of the relevant element.

Carbon-13: the M+1 peak

Carbon-13 is approximately 1.1% as abundant as carbon-12.

This means that a molecule containing carbon atoms has a small chance of containing one carbon-13 atom instead of carbon-12.

Replacing one carbon-12 atom with carbon-13 increases the mass by 1 u.

For small organic molecules, a useful approximation is:

\[
\frac{M+1}{M}\times100
\approx 1.1n_C\%
\]

Here, \(n_C\) is the number of carbon atoms.

For example, a fragment containing four carbon atoms should have an M+1 peak approximately 4.4% as intense as its main peak.

This provides an estimate of the number of carbon atoms in an ion.

Chlorine: the M+2 peak

Chlorine has two important naturally occurring isotopes:

• Chlorine-35, approximately 75.8%.
• Chlorine-37, approximately 24.2%.

The difference in mass is 2 u.

A molecule containing one chlorine atom therefore produces two molecular-ion peaks separated by two mass units, with an approximate intensity ratio of 3:1.

For example:

• The chlorine-35 molecular ion appears at \(m/z=92\).
• The chlorine-37 molecular ion appears at \(m/z=94\).

The same pattern applies to a chlorine-containing fragment.

Bromine: another M+2 pattern

Bromine has two important isotopes:

• Bromine-79, approximately 50.7%.
• Bromine-81, approximately 49.3%.

These isotopes produce two peaks separated by 2 u, with an approximate 1:1 intensity ratio.

A fragment containing one bromine atom should therefore produce a distinctive pair of peaks of approximately equal height.

The important point is that isotope patterns tell you which elements remain in a particular fragment.

A parent molecule may contain chlorine, but a fragment that has lost chlorine will not show the characteristic chlorine M+2 pair.

Also, an M+2 peak alone does not prove that chlorine or bromine is present. Other isotopes, including sulfur-34, can contribute to peaks two mass units higher. Always consider the complete pattern and the proposed molecular formula.

03Worked examples

The following examples use illustrative mass-spectrum data, not measurements from particular experimental samples.

04Common mistakes when checking fragment assignments

05Questions and solutions

The following questions move from straightforward formula checks to more demanding problems involving isotope patterns, alternative fragmentation pathways, and conservation of atoms and charge.

Question 1

A compound has the molecular formula \(\ce{C2H6O}\) and produces a molecular ion at \(m/z=46\).

A fragment peak appears at \(m/z=31\).

A proposed assignment is \(\ce{CH3O+}\).

Determine whether this fragment is consistent with the parent formula. Show the mass calculation and identify the neutral species lost.

Solution 1

Yes. The proposed fragment \(\ce{CH3O+}\) is consistent with the parent formula because its atom counts and calculated mass both agree with the available evidence.

Step 1: Calculate the proposed fragment mass.

\[
\begin{aligned}
m &=12+3(1)+16\\
&=31\ \mathrm{u}
\end{aligned}
\]

This matches the observed peak at \(m/z=31\).

Step 2: Check the atom inventory.

The parent contains two carbon atoms, six hydrogen atoms, and one oxygen atom.

The fragment contains one carbon atom, three hydrogen atoms, and one oxygen atom.

No element exceeds its available count.

Step 3: Calculate the neutral loss.

\[
46-31=15\ \mathrm{u}
\]

Subtracting the fragment formula from the parent formula gives:

\[
\ce{C2H6O – CH3O = CH3}
\]

The neutral loss is \(\ce{CH3^{\bullet}}\), with a nominal mass of 15 u.

A possible fragmentation equation is:

\[
\ce{C2H6O^{+\bullet} -> CH3O+ + CH3^{\bullet}}
\]

Conclusion: The proposed fragment passes the atom-count and mass checks. The neutral loss also balances correctly, making \(\ce{CH3O+}\) a consistent assignment.

Remember that a matching nominal mass is not unique proof of an ion’s structure. Other evidence is needed to distinguish possible arrangements.

Question 2

A compound has the molecular formula \(\ce{C3H7Cl}\).

Its illustrative EI mass spectrum contains the following peaks:

There is no significant chlorine-type isotope partner at \(m/z=45\).

A student proposes the following assignments:

• The peaks at 63 and 65 represent \(\ce{C2H4Cl+}\) containing chlorine-35 and chlorine-37.
• The peak at 43 represents \(\ce{C3H7+}\).

Evaluate both assignments. Explain why the second fragment does not need to show a chlorine isotope pair.

Solution 2

Both proposed assignments are consistent with the data. The first fragment retains chlorine, while the second has lost chlorine during fragmentation.

Step 1: Identify the molecular-ion isotope pattern.

For \(\ce{C3H7Cl}\), using chlorine-35:

\[
3(12)+7(1)+35=78
\]

Using chlorine-37:

\[
78+2=80
\]

The peaks at 78 and 80 have an intensity ratio of:

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

This is consistent with the approximately 3:1 chlorine isotope pattern.

Step 2: Check the proposed fragment at \(m/z=63\).

For \(\ce{C2H4Cl+}\):

\[
\begin{aligned}
m &=2(12)+4(1)+35\\
&=63\ \mathrm{u}
\end{aligned}
\]

Its chlorine-37 isotopologue appears at:

\[
63+2=65
\]

The observed peaks at 63 and 65 have a ratio of:

\[
\frac{48}{15}=3.2
\]

This is consistent with chlorine.

The atom inventory also works: the fragment contains two carbons, four hydrogens, and one chlorine atom, all available in the parent.

The mass difference is:

\[
78-63=15\ \mathrm{u}
\]

The neutral loss is \(\ce{CH3^{\bullet}}\).

Therefore, \(\ce{C2H4Cl+}\) is a reasonable assignment.

Step 3: Check the proposed fragment at \(m/z=43\).

For \(\ce{C3H7+}\):

\[
\begin{aligned}
m &=3(12)+7(1)\\
&=43\ \mathrm{u}
\end{aligned}
\]

The fragment contains all three carbon atoms and all seven hydrogen atoms, but no chlorine.

The mass difference from the parent is:

\[
78-43=35\ \mathrm{u}
\]

This corresponds to the loss of chlorine-35 as a neutral chlorine radical.

\[
\ce{C3H7Cl^{+\bullet} -> C3H7+ + Cl^{\bullet}}
\]

The fragment does not contain chlorine, so it should not produce the characteristic chlorine-35/chlorine-37 isotope pair.

Conclusion: Both assignments are supported. The peaks at 63 and 65 indicate a chlorine-containing fragment, whereas the peak at 43 is consistent with a fragment that has lost chlorine.

The important point is that different fragmentation pathways can produce ions containing different subsets of the original molecule’s atoms.

Question 3

A compound has the molecular formula \(\ce{C4H8O2}\) and a molecular-ion peak at \(m/z=88\).

A fragment appears at \(m/z=58\), accompanied by an M+1 peak at \(m/z=59\). After background correction, the M+1 peak has an intensity equal to 3.2% of the main fragment peak.

Two possible fragment formulas are proposed:

• \(\ce{C3H6O+}\)
• \(\ce{C2H2O2+}\)

Both have a nominal mass of 58 u.

Determine which formula is better supported by the isotope evidence. Show the neutral loss implied by each assignment and explain why the mass difference alone cannot distinguish them.

Assume the M+1 peak represents a carbon-13 isotope contribution and that the ions are singly charged.

Solution 3

\(\ce{C3H6O+}\) is better supported by the isotope evidence. Its three carbon atoms predict an M+1 contribution of approximately 3.3%, which is close to the observed 3.2%.

Step 1: Check the nominal masses.

For \(\ce{C3H6O+}\):

\[
3(12)+6(1)+16=58\ \mathrm{u}
\]

For \(\ce{C2H2O2+}\):

\[
2(12)+2(1)+2(16)=58\ \mathrm{u}
\]

Both formulas match the observed \(m/z=58\) peak.

Step 2: Check the parent atom inventory.

The parent contains four carbon atoms, eight hydrogen atoms, and two oxygen atoms.

Both proposed fragments contain fewer atoms of each element than the parent.

Therefore, both pass the atom-count test.

Step 3: Compare the predicted carbon-13 contributions.

Each carbon atom has approximately a 1.1% chance of being carbon-13.

For the three-carbon fragment:

\[
3\times1.1\%=3.3\%
\]

For the two-carbon fragment:

\[
2\times1.1\%=2.2\%
\]

The observed M+1 contribution is 3.2%.

This is much closer to the prediction for a three-carbon ion.

Therefore, the isotope evidence favours \(\ce{C3H6O+}\).

Step 4: Calculate the neutral losses.

For the first assignment:

\[
\ce{C4H8O2 – C3H6O = CH2O}
\]

The nominal mass of \(\ce{CH2O}\) is:

\[
12+2(1)+16=30\ \mathrm{u}
\]

For the second assignment:

\[
\ce{C4H8O2 – C2H2O2 = C2H6}
\]

The nominal mass of \(\ce{C2H6}\) is:

\[
2(12)+6(1)=30\ \mathrm{u}
\]

Both proposed fragments imply the same mass loss:

\[
88-58=30\ \mathrm{u}
\]

This is why calculating the mass difference alone cannot distinguish the two formulas.

Conclusion: The formula \(\ce{C3H6O+}\) is better supported because its predicted carbon-13 isotope contribution agrees with the observed M+1 intensity.

However, isotope abundance is an estimate rather than an exact structural fingerprint. Overlapping peaks, other isotopes, and experimental uncertainty can affect the measured ratio. The assignment should be checked against the rest of the spectrum.

Question 4

A compound has the molecular formula \(\ce{C5H10O2}\) and a molecular-ion peak at \(m/z=102\).

Two fragment peaks are observed:

• \(m/z=59\), proposed to be \(\ce{C3H7O+}\).
• \(m/z=43\), proposed to be \(\ce{C2H3O+}\).

A student notices that the two proposed fragment formulas add together to give the parent molecular formula:

\[
\ce{C3H7O + C2H3O = C5H10O2}
\]

The student concludes that both positive ions must be produced simultaneously when one molecular ion breaks apart.

Evaluate this conclusion using atom conservation and charge conservation. Show how each fragment could instead form through a separate fragmentation pathway.

Solution 4

The student’s conclusion is incorrect. Although the two fragment formulas add to the parent formula, a singly charged molecular ion cannot produce two separate singly charged positive ions in one ordinary fragmentation event.

Step 1: Check the individual fragment masses.

For \(\ce{C3H7O+}\):

\[
3(12)+7(1)+16=59\ \mathrm{u}
\]

For \(\ce{C2H3O+}\):

\[
2(12)+3(1)+16=43\ \mathrm{u}
\]

Both calculated masses agree with the observed peaks.

Step 2: Check the atom inventory.

The parent contains five carbon atoms, ten hydrogen atoms, and two oxygen atoms.

Each proposed fragment contains fewer atoms of each element than the parent.

Both are individually formula-compatible.

Step 3: Examine the first possible fragmentation pathway.

If \(\ce{C3H7O+}\) is the charged fragment, the remaining atoms form a neutral species:

\[
\ce{C5H10O2^{+\bullet} -> C3H7O+ + C2H3O^{\bullet}}
\]

The neutral complement has a nominal mass of 43 u.

Step 4: Examine the alternative pathway.

If \(\ce{C2H3O+}\) is the charged fragment, the remaining atoms form a different neutral species:

\[
\ce{C5H10O2^{+\bullet} -> C2H3O+ + C3H7O^{\bullet}}
\]

The neutral complement has a nominal mass of 59 u.

Both pathways conserve the atoms and the overall positive charge.

Step 5: Apply charge conservation.

The original molecular ion has a net charge of \(+1\).

If both products were positive ions, their combined charge would be:

\[
(+1)+(+1)=+2
\]

This does not conserve the original charge.

One ordinary fragmentation event involving a singly charged molecular ion therefore produces one charged fragment and one neutral product.

Conclusion: Both proposed fragment formulas are individually consistent with the parent molecular formula. They can appear in the same mass spectrum because different parent ions can undergo different fragmentation pathways.

However, they cannot both be produced as singly charged positive ions from the same singly charged parent ion in one ordinary fragmentation event.

This is an important distinction: two fragment formulas can add up perfectly to the parent formula without being simultaneous charged products.

Question 5

A compound has the molecular formula \(\ce{C4H9Br}\).

An illustrative electron-impact mass spectrum contains the following selected peaks:

No significant bromine-type M+2 isotope partners are resolved for the fragment peaks at 57 and 43.

Assume:

• The molecular ions and fragments are positively charged and singly charged.
• The sample contains only the stated compound.
• The peak positions are correctly calibrated.
• The small M+1 peaks represent carbon-13 contributions.

Three proposed fragment formulas are:

A student draws the following fragmentation sequence:

\[
\text{Parent}\rightarrow121\rightarrow57\rightarrow43
\]

Their reasoning is that the fragment peaks become progressively smaller in mass, so each lower-mass ion must have formed from the previous higher-mass ion.

Evaluate this proposed sequence.

Your response should:

• Check all three fragment formulas against the parent.
• Use the isotope evidence to assess the proposed carbon and bromine counts.
• Identify which fragmentation relationships are impossible.
• Explain which pathways are possible but cannot be established from the spectrum alone.

Solution 5

The proposed sequence is impossible because the fragment at \(m/z=121\) cannot produce the fragment at \(m/z=57\) through ordinary fragmentation: the proposed product contains more carbon atoms than its supposed precursor.

The three proposed fragment formulas are individually consistent with the parent, but their formation pathways need to be examined separately.

Step 1: Identify the parent molecular-ion isotope pattern.

For bromine-79:

\[
\begin{aligned}
m &=4(12)+9(1)+79\\
&=136\ \mathrm{u}
\end{aligned}
\]

For bromine-81:

\[
136+2=138\ \mathrm{u}
\]

The peaks at 136 and 138 have relative intensities of 100 and 98.

This is consistent with the approximately 1:1 isotope pattern of bromine.

The peak at 137 is also consistent with the carbon-13 contribution from a molecule containing four carbon atoms:

\[
4(1.1\%)=4.4\%
\]

The parent molecular-ion assignment is therefore consistent with the isotope data.

Step 2: Evaluate the fragment at \(m/z=121\).

The proposed formula is \(\ce{C3H6Br+}\).

Using bromine-79:

\[
\begin{aligned}
m &=3(12)+6(1)+79\\
&=36+6+79\\
&=121\ \mathrm{u}
\end{aligned}
\]

Replacing bromine-79 with bromine-81 produces an ion at:

\[
121+2=123
\]

The observed peaks at 121 and 123 have intensities of 31 and 30.

Their ratio is approximately 1:1, consistent with bromine.

The M+1 contribution can also be checked:

\[
\frac{1.0}{31}\times100
\approx3.2\%
\]

This is close to the predicted 3.3% for three carbon atoms.

The proposed formula is therefore supported by both its mass and isotope pattern.

Its atom inventory also fits the parent.

The neutral loss from the parent is:

\[
\ce{C4H9Br – C3H6Br = CH3}
\]

This corresponds to a methyl radical with a nominal mass of 15 u.

A possible fragmentation pathway is:

\[
\ce{C4H9Br^{+\bullet} -> C3H6Br+ + CH3^{\bullet}}
\]

Therefore, formation of the \(m/z=121\) fragment directly from the parent is consistent with the available evidence.

Step 3: Evaluate the fragment at \(m/z=57\).

The proposed formula is \(\ce{C4H9+}\).

Its nominal mass is:

\[
\begin{aligned}
m &=4(12)+9(1)\\
&=57\ \mathrm{u}
\end{aligned}
\]

The observed M+1 contribution is:

\[
\frac{4.4}{100}\times100=4.4\%
\]

This is consistent with a fragment containing four carbon atoms.

The proposed fragment contains no bromine, so it should not produce a bromine-type M+2 isotope pair.

The mass difference between the parent and fragment is:

\[
136-57=79\ \mathrm{u}
\]

This corresponds to the loss of bromine-79 as a neutral radical.

A possible pathway is:

\[
\ce{C4H9Br^{+\bullet} -> C4H9+ + Br^{\bullet}}
\]

The assignment is therefore consistent with a direct fragmentation pathway from the parent.

Step 4: Evaluate the fragment at \(m/z=43\).

The proposed formula is \(\ce{C3H7+}\).

Its nominal mass is:

\[
\begin{aligned}
m &=3(12)+7(1)\\
&=43\ \mathrm{u}
\end{aligned}
\]

The M+1 contribution is:

\[
\frac{0.66}{20}\times100=3.3\%
\]

This agrees with the approximate 3.3% contribution expected for three carbon atoms.

The proposed fragment contains no bromine, consistent with the absence of a bromine-type isotope pair.

The assignment therefore fits the mass and isotope evidence.

Step 5: Test whether the fragment at 121 can produce the fragment at 57.

This is where the student’s proposed sequence fails.

The proposed precursor is:

\[
\ce{C3H6Br+}
\]

The proposed product is:

\[
\ce{C4H9+}
\]

The supposed product contains four carbon atoms, but the proposed precursor contains only three.

Fragmentation cannot create an additional carbon atom.

The difference is not simply a neutral loss. The proposed product requires atoms that are absent from the supposed precursor.

Therefore:

\[
\ce{C3H6Br+ -> C4H9+}
\]

is impossible as an ordinary fragmentation pathway.

The fact that 57 is smaller than 121 does not make it a daughter ion of 121.

Step 6: Determine whether the fragment at 57 could produce the fragment at 43.

This pathway is more interesting.

The proposed precursor contains:

\[
\ce{C4H9+}
\]

The proposed product contains:

\[
\ce{C3H7+}
\]

Subtracting the atom counts gives:

\[
\ce{C4H9 – C3H7 = CH2}
\]

The mass difference is:

\[
57-43=14\ \mathrm{u}
\]

The atom and mass balances therefore permit the loss of neutral \(\ce{CH2}\).

A possible secondary fragmentation equation is:

\[
\ce{C4H9+ -> C3H7+ + CH2}
\]

However, this does not prove that the \(m/z=43\) peak actually forms through this pathway. The chemical plausibility of the neutral loss and the molecular structure must also be considered.

There is another possibility: the parent could produce \(\ce{C3H7+}\) directly.

Subtracting its formula from the parent gives:

\[
\ce{C4H9Br – C3H7 = CH2Br}
\]

The mass of the neutral complement is:

\[
12+2(1)+79=93\ \mathrm{u}
\]

This gives another atom- and mass-balanced pathway:

\[
\ce{C4H9Br^{+\bullet} -> C3H7+ + CH2Br^{\bullet}}
\]

The original mass spectrum alone cannot establish whether the \(m/z=43\) fragment forms directly from the parent, through secondary fragmentation of the \(m/z=57\) ion, or through a combination of pathways.

Conclusion: All three proposed fragment formulas are individually consistent with the parent molecular formula and their respective mass and isotope data.

However:

• The \(m/z=121\) fragment cannot produce the \(m/z=57\) fragment because that would require gaining a carbon atom.
• The \(m/z=57\) fragment could potentially produce the \(m/z=43\) fragment by losing \(\ce{CH2}\).
• The \(m/z=43\) fragment could also form directly from the parent by losing \(\ce{CH2Br^{\bullet}}\).

The original spectrum does not establish which of the possible pathways actually occurs.

This final example illustrates why fragment assignments require more than matching peak positions. Atom conservation, charge conservation, isotope patterns, and possible fragmentation pathways must all agree.

06What to do after a fragment passes the checks

Once a proposed fragment is consistent with the molecular formula, the next task is to decide whether its formation makes chemical sense for the proposed molecular structure.

Consider:

• Does the structure contain a bond that could break to produce the fragment?
• Is the neutral species lost consistent with the remaining atoms?
• Could a rearrangement explain the fragment?
• Do other fragment peaks support the same structural interpretation?
• Do infrared and NMR data support the proposed functional groups and molecular arrangement?

A mass spectrum provides evidence about molecular mass, fragment composition, and possible structural features. It does not normally identify every bond or uniquely establish a complete structure from one peak.

For the next step, connect fragment assignments to the bonding and functional groups in Covalent Molecular Structures and Intermolecular Forces.

The goal is to build one consistent structural explanation from several independent pieces of evidence, rather than forcing every peak to fit a structure you’ve already guessed.