Mass spectrometry II

What you'll learn

How to read an organic mass spectrum using the molecular ion peak and base peak.
How fragmentation produces useful structural clues.
How high-resolution mass spectrometry can identify a molecular formula.
How isotope patterns help spot carbon count, chlorine and bromine.

Starting point: what a mass spectrum shows

A mass spectrum is a graph showing the abundance of positive ions produced from a sample, separated according to their mass-to-charge ratio, written as $m / z$ .

For most A-Level organic mass spectra, the ions have a single positive charge, so $m / z$ is usually numerically equal to the relative mass of the ion.

Definition

Mass-to-charge ratio

The mass-to-charge ratio, $m / z$ , compares the mass of an ion with its charge. If the ion has charge +1, then its $m / z$ value is the same as its relative mass.

The vertical axis shows relative abundance. The tallest peak is always assigned 100% abundance, even if it is not the molecular ion.

Definition

Base peak

The base peak is the tallest peak in a mass spectrum. It is the most abundant ion detected and is assigned a relative abundance of 100%.

Ionisation: making positive ions

In many organic mass spectra, molecules are ionised by electron impact ionisation. A high-energy electron knocks an electron out of a gaseous molecule, forming a positive ion.

M(g)+e−→M+⋅(g)+2e−\mathrm{M(g) + e^- \to M^{+\cdot}(g) + 2e^-}

The ion $M+⋅\mathrm{M^{+\cdot}}$ is called the molecular ion. The dot shows that it is also a radical, but at A-Level you are usually focused on its mass.

Definition

Molecular ion

The molecular ion is the positive ion formed when a molecule loses one electron without breaking apart. Its peak gives the relative molecular mass, $M_r$ , of the compound.

Common Mistake

Base peak is not always the molecular ion

The tallest peak is the base peak, not necessarily the molecular ion peak. The molecular ion is usually the peak with the highest relevant $m / z$ value, ignoring small isotope peaks such as M+1.

Reading the molecular ion peak

The molecular ion peak, often labelled $M+\mathrm{M^+}$ , gives the relative molecular mass of the molecule.

For example, if the molecular ion peak is at $m / z = 58$ , then the molecule has $M_r = 58$ .

Mass spectrum of propanone showing molecular ion and fragment peaks

Example

Using a molecular ion peak with an empirical formula

A compound has empirical formula $C2H5O\mathrm{C_2H_5O}$ and a molecular ion peak at $m / z = 90$ . Find its molecular formula.

Calculate the empirical formula mass:
$\times 12.0) + (5 \times 1.0) + 16.0 = 45.0$
Compare the molecular mass from the molecular ion peak with the empirical formula mass:
$9045.0=2\frac{90}{45.0} = 2$
Multiply every atom in the empirical formula by 2:
$C2H5O×2=C4H10O2\mathrm{C_2H_5O} \times 2 = \mathrm{C_4H_{10}O_2}$

Fragmentation: why there are many peaks

The molecular ion is often unstable. It can break into smaller pieces. This process is called fragmentation.

A typical fragmentation forms:

one positive fragment ion, which is detected
one neutral radical or molecule, which is not detected

For example, propanone can fragment to form an acylium ion:

CH3COCH3+⋅→CH3CO++CH3⋅\mathrm{CH_3COCH_3^{+\cdot} \to CH_3CO^+ + CH_3^\cdot}

The peak at $m / z = 43$ is due to $CH3CO+\mathrm{CH_3CO^+}$ .

Key Idea

Only charged particles are detected

Mass spectrometers detect ions, not neutral fragments. When a molecule breaks apart, the peak you see comes from the fragment that keeps the positive charge.

Common fragment ions

You are not expected to memorise every possible fragment, but some common ones are very useful:

$m / z = 15$ : $CH3+\mathrm{CH_3^+}$
$m / z = 29$ : often $C2H5+\mathrm{C_2H_5^+}$ or $CHO+\mathrm{CHO^+}$
$m / z = 31$ : often $CH2OH+\mathrm{CH_2OH^+}$ , useful for alcohols
$m / z = 43$ : often $CH3CO+\mathrm{CH_3CO^+}$ or $C3H7+\mathrm{C_3H_7^+}$
$m / z = 57$ : often $C4H9+\mathrm{C_4H_9^+}$

Because different ions can have the same nominal mass, fragmentation gives clues rather than a complete structure by itself.

Example

Using fragment peaks to suggest a structure

A compound has a molecular ion peak at $m / z = 58$ and strong fragment peaks at $m / z = 15$ and $m / z = 43$ . Explain why propanone is a reasonable suggestion.

The molecular ion peak at $m / z = 58$ means the compound has $M_r = 58$ .
Propanone, $CH3COCH3\mathrm{CH_3COCH_3}$ , has relative molecular mass:
$\times 12.0) + (6 \times 1.0) + 16.0 = 58.0$
The peak at $m / z = 43$ can be explained by the fragment ion $CH3CO+\mathrm{CH_3CO^+}$ , formed by breaking a C-C bond next to the carbonyl group.
The peak at $m / z = 15$ can be explained by $CH3+\mathrm{CH_3^+}$ , showing that methyl fragments are present.

High-resolution mass spectrometry

A low-resolution mass spectrum gives whole-number $m / z$ values. This can identify $M_r$ , but different molecular formulae can have the same nominal mass.

High-resolution mass spectrometry measures masses much more precisely. This allows you to distinguish between possible molecular formulae.

Definition

High-resolution mass spectrometry

High-resolution mass spectrometry measures the accurate mass of an ion to several decimal places, allowing molecular formulae with the same nominal mass to be distinguished.

For accurate mass calculations, use precise isotopic masses, usually based on the most abundant isotopes:

$12C=12.0000\mathrm{^{12}C} = 12.0000$
$1H=1.0078\mathrm{^1H} = 1.0078$
$16O=15.9949\mathrm{^{16}O} = 15.9949$
$14N=14.0031\mathrm{^{14}N} = 14.0031$

Example

Choosing a molecular formula from accurate mass

A molecular ion has accurate mass 88.0524. Three possible formulae are $C4H8O2\mathrm{C_4H_8O_2}$ , $C3H4O3\mathrm{C_3H_4O_3}$ and $C5H12O\mathrm{C_5H_{12}O}$ . Identify the formula.

Calculate the accurate mass of $C4H8O2\mathrm{C_4H_8O_2}$ :
$Mr=(4×12.0000)+(8×1.0078)+(2×15.9949)=48.0000+8.0624+31.9898=88.0522\begin{aligned} M_r &= (4 \times 12.0000) + (8 \times 1.0078) + (2 \times 15.9949) \\ &= 48.0000 + 8.0624 + 31.9898 \\ &= 88.0522 \end{aligned}$
Calculate the accurate mass of $C3H4O3\mathrm{C_3H_4O_3}$ :
$Mr=(3×12.0000)+(4×1.0078)+(3×15.9949)=36.0000+4.0312+47.9847=88.0159\begin{aligned} M_r &= (3 \times 12.0000) + (4 \times 1.0078) + (3 \times 15.9949) \\ &= 36.0000 + 4.0312 + 47.9847 \\ &= 88.0159 \end{aligned}$
Calculate the accurate mass of $C5H12O\mathrm{C_5H_{12}O}$ :
$Mr=(5×12.0000)+(12×1.0078)+15.9949=60.0000+12.0936+15.9949=88.0885\begin{aligned} M_r &= (5 \times 12.0000) + (12 \times 1.0078) + 15.9949 \\ &= 60.0000 + 12.0936 + 15.9949 \\ &= 88.0885 \end{aligned}$
Compare each value with 88.0524. The closest match is $C4H8O2\mathrm{C_4H_8O_2}$ , so this is the molecular formula.

Tip

Accurate mass strategy

If several formulae have the same whole-number $M_r$ , calculate each accurate mass using the precise isotope masses given in the question, then choose the closest match to the measured molecular ion.

Isotope peaks: M+1 and M+2

Real samples contain naturally occurring isotopes, so a molecule can give small extra peaks near the molecular ion peak.

The M peak is the molecular ion made from the most common isotopes. The M+1 peak is one mass unit higher, often mainly due to one $13C\mathrm{^{13}C}$ atom instead of one $12C\mathrm{^{12}C}$ atom.

Since carbon-13 has a natural abundance of about 1.1%, the M+1 peak can estimate the number of carbon atoms:

M1.1\text{number of carbon atoms} \approx \frac{\text{M+1 abundance relative to M}}{1.1}

Example

Estimating the number of carbon atoms from M+1

A molecular ion peak has relative abundance 100. The M+1 peak has relative abundance 6.6. Estimate the number of carbon atoms in the molecule.

Use the carbon-13 approximation:
$atoms≈6.61.1\text{number of C atoms} \approx \frac{6.6}{1.1}$
Calculate the value:
$6.61.1=6\frac{6.6}{1.1} = 6$
The molecule is likely to contain 6 carbon atoms.

Chlorine and bromine patterns

Chlorine and bromine give very distinctive M and M+2 patterns because their common isotopes differ by two mass units.

One chlorine atom gives M : M+2 of about 3 : 1.
One bromine atom gives M : M+2 of about 1 : 1.

Isotope patterns for no halogen, one chlorine atom and one bromine atom

Common Mistake

More than one halogen

The simple 3 : 1 chlorine and 1 : 1 bromine patterns apply to molecules with one chlorine or one bromine atom. With two halogen atoms, you get a cluster of M, M+2 and M+4 peaks.

Pulling the evidence together

Mass spectrometry is most powerful when combined with other techniques.

A good structure-deduction sequence is often:

Use the molecular ion peak to find $M_r$ .
Use high-resolution mass spectrometry to find the molecular formula.
Use M+1 and M+2 peaks to check for carbon count and halogens.
Use fragmentation peaks to suggest parts of the structure.
Combine with IR spectroscopy and NMR spectroscopy to confirm functional groups and the carbon-hydrogen framework.

Key Idea

What mass spectrometry can and cannot do

Mass spectrometry is excellent for molecular mass, molecular formula and fragment clues, but it rarely proves a full organic structure on its own. Use it alongside IR and NMR evidence.

Exam technique

In the exam

Start with the molecular ion peak: it gives $M_r$ , but check whether nearby M+1 or M+2 isotope peaks are present.
For accurate mass questions, use the isotope masses given in the question and compare calculated values to the measured mass.
Treat fragment ions as evidence for possible groups, not absolute proof of a whole structure.
Watch for chlorine and bromine: M : M+2 ratios of about 3 : 1 and 1 : 1 are very high-value clues.

Self review

Check yourself

Why does the molecular ion peak give the relative molecular mass of the compound?
A compound has M and M+2 peaks in a 1 : 1 ratio. What element is strongly suggested?
How could high-resolution mass spectrometry distinguish between two formulae with the same nominal mass?

What you'll learn

How to read an organic mass spectrum using the molecular ion peak and base peak.
How fragmentation produces useful structural clues.
How high-resolution mass spectrometry can identify a molecular formula.
How isotope patterns help spot carbon count, chlorine and bromine.

Starting point: what a mass spectrum shows

A mass spectrum is a graph showing the abundance of positive ions produced from a sample, separated according to their mass-to-charge ratio, written as $m / z$ .

For most A-Level organic mass spectra, the ions have a single positive charge, so $m / z$ is usually numerically equal to the relative mass of the ion.

Definition

Mass-to-charge ratio

The mass-to-charge ratio, $m / z$ , compares the mass of an ion with its charge. If the ion has charge +1, then its $m / z$ value is the same as its relative mass.

The vertical axis shows relative abundance. The tallest peak is always assigned 100% abundance, even if it is not the molecular ion.

Definition

Base peak

The base peak is the tallest peak in a mass spectrum. It is the most abundant ion detected and is assigned a relative abundance of 100%.

Ionisation: making positive ions

In many organic mass spectra, molecules are ionised by electron impact ionisation. A high-energy electron knocks an electron out of a gaseous molecule, forming a positive ion.

M(g)+e−→M+⋅(g)+2e−\mathrm{M(g) + e^- \to M^{+\cdot}(g) + 2e^-}

The ion $M+⋅\mathrm{M^{+\cdot}}$ is called the molecular ion. The dot shows that it is also a radical, but at A-Level you are usually focused on its mass.

Definition

Molecular ion

The molecular ion is the positive ion formed when a molecule loses one electron without breaking apart. Its peak gives the relative molecular mass, $M_r$ , of the compound.

Common Mistake

Base peak is not always the molecular ion

The tallest peak is the base peak, not necessarily the molecular ion peak. The molecular ion is usually the peak with the highest relevant $m / z$ value, ignoring small isotope peaks such as M+1.

Reading the molecular ion peak

The molecular ion peak, often labelled $M+\mathrm{M^+}$ , gives the relative molecular mass of the molecule.

For example, if the molecular ion peak is at $m / z = 58$ , then the molecule has $M_r = 58$ .

Mass spectrum of propanone showing molecular ion and fragment peaks

Example

Using a molecular ion peak with an empirical formula

A compound has empirical formula $C2H5O\mathrm{C_2H_5O}$ and a molecular ion peak at $m / z = 90$ . Find its molecular formula.

Calculate the empirical formula mass:
$\times 12.0) + (5 \times 1.0) + 16.0 = 45.0$
Compare the molecular mass from the molecular ion peak with the empirical formula mass:
$9045.0=2\frac{90}{45.0} = 2$
Multiply every atom in the empirical formula by 2:
$C2H5O×2=C4H10O2\mathrm{C_2H_5O} \times 2 = \mathrm{C_4H_{10}O_2}$

Fragmentation: why there are many peaks

The molecular ion is often unstable. It can break into smaller pieces. This process is called fragmentation.

A typical fragmentation forms:

one positive fragment ion, which is detected
one neutral radical or molecule, which is not detected

For example, propanone can fragment to form an acylium ion:

CH3COCH3+⋅→CH3CO++CH3⋅\mathrm{CH_3COCH_3^{+\cdot} \to CH_3CO^+ + CH_3^\cdot}

The peak at $m / z = 43$ is due to $CH3CO+\mathrm{CH_3CO^+}$ .

Key Idea

Only charged particles are detected

Mass spectrometers detect ions, not neutral fragments. When a molecule breaks apart, the peak you see comes from the fragment that keeps the positive charge.

Common fragment ions

You are not expected to memorise every possible fragment, but some common ones are very useful:

$m / z = 15$ : $CH3+\mathrm{CH_3^+}$
$m / z = 29$ : often $C2H5+\mathrm{C_2H_5^+}$ or $CHO+\mathrm{CHO^+}$
$m / z = 31$ : often $CH2OH+\mathrm{CH_2OH^+}$ , useful for alcohols
$m / z = 43$ : often $CH3CO+\mathrm{CH_3CO^+}$ or $C3H7+\mathrm{C_3H_7^+}$
$m / z = 57$ : often $C4H9+\mathrm{C_4H_9^+}$

Because different ions can have the same nominal mass, fragmentation gives clues rather than a complete structure by itself.

Example

Using fragment peaks to suggest a structure

A compound has a molecular ion peak at $m / z = 58$ and strong fragment peaks at $m / z = 15$ and $m / z = 43$ . Explain why propanone is a reasonable suggestion.

The molecular ion peak at $m / z = 58$ means the compound has $M_r = 58$ .
Propanone, $CH3COCH3\mathrm{CH_3COCH_3}$ , has relative molecular mass:
$\times 12.0) + (6 \times 1.0) + 16.0 = 58.0$
The peak at $m / z = 43$ can be explained by the fragment ion $CH3CO+\mathrm{CH_3CO^+}$ , formed by breaking a C-C bond next to the carbonyl group.
The peak at $m / z = 15$ can be explained by $CH3+\mathrm{CH_3^+}$ , showing that methyl fragments are present.

High-resolution mass spectrometry

A low-resolution mass spectrum gives whole-number $m / z$ values. This can identify $M_r$ , but different molecular formulae can have the same nominal mass.

High-resolution mass spectrometry measures masses much more precisely. This allows you to distinguish between possible molecular formulae.

Definition

High-resolution mass spectrometry

High-resolution mass spectrometry measures the accurate mass of an ion to several decimal places, allowing molecular formulae with the same nominal mass to be distinguished.

For accurate mass calculations, use precise isotopic masses, usually based on the most abundant isotopes:

$12C=12.0000\mathrm{^{12}C} = 12.0000$
$1H=1.0078\mathrm{^1H} = 1.0078$
$16O=15.9949\mathrm{^{16}O} = 15.9949$
$14N=14.0031\mathrm{^{14}N} = 14.0031$

Example

Choosing a molecular formula from accurate mass

A molecular ion has accurate mass 88.0524. Three possible formulae are $C4H8O2\mathrm{C_4H_8O_2}$ , $C3H4O3\mathrm{C_3H_4O_3}$ and $C5H12O\mathrm{C_5H_{12}O}$ . Identify the formula.

Calculate the accurate mass of $C4H8O2\mathrm{C_4H_8O_2}$ :
$Mr=(4×12.0000)+(8×1.0078)+(2×15.9949)=48.0000+8.0624+31.9898=88.0522\begin{aligned} M_r &= (4 \times 12.0000) + (8 \times 1.0078) + (2 \times 15.9949) \\ &= 48.0000 + 8.0624 + 31.9898 \\ &= 88.0522 \end{aligned}$
Calculate the accurate mass of $C3H4O3\mathrm{C_3H_4O_3}$ :
$Mr=(3×12.0000)+(4×1.0078)+(3×15.9949)=36.0000+4.0312+47.9847=88.0159\begin{aligned} M_r &= (3 \times 12.0000) + (4 \times 1.0078) + (3 \times 15.9949) \\ &= 36.0000 + 4.0312 + 47.9847 \\ &= 88.0159 \end{aligned}$
Calculate the accurate mass of $C5H12O\mathrm{C_5H_{12}O}$ :
$Mr=(5×12.0000)+(12×1.0078)+15.9949=60.0000+12.0936+15.9949=88.0885\begin{aligned} M_r &= (5 \times 12.0000) + (12 \times 1.0078) + 15.9949 \\ &= 60.0000 + 12.0936 + 15.9949 \\ &= 88.0885 \end{aligned}$
Compare each value with 88.0524. The closest match is $C4H8O2\mathrm{C_4H_8O_2}$ , so this is the molecular formula.

Tip

Accurate mass strategy

Isotope peaks: M+1 and M+2

Real samples contain naturally occurring isotopes, so a molecule can give small extra peaks near the molecular ion peak.

Since carbon-13 has a natural abundance of about 1.1%, the M+1 peak can estimate the number of carbon atoms:

M1.1\text{number of carbon atoms} \approx \frac{\text{M+1 abundance relative to M}}{1.1}

Example

Estimating the number of carbon atoms from M+1

A molecular ion peak has relative abundance 100. The M+1 peak has relative abundance 6.6. Estimate the number of carbon atoms in the molecule.

Use the carbon-13 approximation:
$atoms≈6.61.1\text{number of C atoms} \approx \frac{6.6}{1.1}$
Calculate the value:
$6.61.1=6\frac{6.6}{1.1} = 6$
The molecule is likely to contain 6 carbon atoms.

Chlorine and bromine patterns

Chlorine and bromine give very distinctive M and M+2 patterns because their common isotopes differ by two mass units.

One chlorine atom gives M : M+2 of about 3 : 1.
One bromine atom gives M : M+2 of about 1 : 1.

Isotope patterns for no halogen, one chlorine atom and one bromine atom

Common Mistake

More than one halogen

The simple 3 : 1 chlorine and 1 : 1 bromine patterns apply to molecules with one chlorine or one bromine atom. With two halogen atoms, you get a cluster of M, M+2 and M+4 peaks.

Pulling the evidence together

Mass spectrometry is most powerful when combined with other techniques.

A good structure-deduction sequence is often:

Use the molecular ion peak to find $M_r$ .
Use high-resolution mass spectrometry to find the molecular formula.
Use M+1 and M+2 peaks to check for carbon count and halogens.
Use fragmentation peaks to suggest parts of the structure.
Combine with IR spectroscopy and NMR spectroscopy to confirm functional groups and the carbon-hydrogen framework.

Key Idea

What mass spectrometry can and cannot do

Mass spectrometry is excellent for molecular mass, molecular formula and fragment clues, but it rarely proves a full organic structure on its own. Use it alongside IR and NMR evidence.

Exam technique

In the exam

Start with the molecular ion peak: it gives $M_r$ , but check whether nearby M+1 or M+2 isotope peaks are present.
For accurate mass questions, use the isotope masses given in the question and compare calculated values to the measured mass.
Treat fragment ions as evidence for possible groups, not absolute proof of a whole structure.
Watch for chlorine and bromine: M : M+2 ratios of about 3 : 1 and 1 : 1 are very high-value clues.

Self review

Check yourself

Why does the molecular ion peak give the relative molecular mass of the compound?
A compound has M and M+2 peaks in a 1 : 1 ratio. What element is strongly suggested?
How could high-resolution mass spectrometry distinguish between two formulae with the same nominal mass?

What you'll learn

Starting point: what a mass spectrum shows

Ionisation: making positive ions

Reading the molecular ion peak

Fragmentation: why there are many peaks

Common fragment ions

High-resolution mass spectrometry

Isotope peaks: M+1 and M+2

Chlorine and bromine patterns

Pulling the evidence together

Review Mass spectrometry II by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mass spectrometry II

What you'll learn

Starting point: what a mass spectrum shows

Ionisation: making positive ions

Reading the molecular ion peak

Fragmentation: why there are many peaks

Common fragment ions

High-resolution mass spectrometry

Isotope peaks: M+1 and M+2

Chlorine and bromine patterns

Pulling the evidence together

Review Mass spectrometry II by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mass spectrometry II

What you'll learn

Starting point: what a mass spectrum shows

Ionisation: making positive ions

Reading the molecular ion peak

Fragmentation: why there are many peaks

Common fragment ions

High-resolution mass spectrometry

Isotope peaks: M+1 and M+2

Chlorine and bromine patterns

Pulling the evidence together

Review Mass spectrometry II by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mass spectrometry II Revision Guide

Mass spectrometry II

What you'll learn

Starting point: what a mass spectrum shows

Ionisation: making positive ions

Reading the molecular ion peak

Fragmentation: why there are many peaks

Common fragment ions

High-resolution mass spectrometry

Isotope peaks: M+1 and M+2

Chlorine and bromine patterns

Pulling the evidence together

Review Mass spectrometry II by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mass spectrometry II Revision Guide