Mass Spectrometry for Organic Compounds

Mass spectrometry (MS) is one of the most powerful analytical techniques available to chemists. It allows us to determine the molecular mass of a compound and, through fragmentation patterns, gain insight into its structural features. This lesson covers the principles of mass spectrometry, how to interpret spectra, and how to identify organic compounds from their fragmentation patterns.

How Mass Spectrometry Works

In a mass spectrometer, a sample is vaporised and then ionised. The most common ionisation method for organic molecules is electron impact (EI) ionisation:

Vaporisation — the sample is heated to produce a gas.
Ionisation — the gaseous molecules are bombarded with high-energy electrons (typically 70 eV). An electron is knocked out of the molecule, forming a molecular ion (also called the parent ion), M⁺˳:

M(g) + e⁻ → M⁺˳(g) + 2e⁻
Acceleration — the positive ions are accelerated by an electric field.
Deflection — the ions are deflected by a magnetic field. The degree of deflection depends on the mass-to-charge ratio (m/z). Lighter ions and more highly charged ions are deflected more.
Detection — the ions reach a detector, which records their m/z values and relative abundances.

Exam Tip: The molecular ion M⁺˳ is a radical cation — it has both an unpaired electron and a positive charge. It is formed by loss of one electron from the molecule. Always write it as M⁺˳ (with both the + and the ˳ dot), not simply M⁺.

The Molecular Ion Peak (M⁺˳)

The molecular ion peak appears at the highest m/z value in the spectrum (ignoring small M+1 and M+2 isotope peaks). Its m/z value equals the relative molecular mass (Mr) of the compound.

Isotope Peaks

The M+1 peak arises primarily from ¹³C (natural abundance ~1.1%). For a compound with n carbon atoms, the M+1 peak is approximately 1.1n% of the M⁺˳ peak intensity.
The M+2 peak is significant for compounds containing Br or Cl:
- Chlorine: ³⁵Cl:³⁷Cl ≈ 3:1, so M⁺˳ and M+2 appear in a roughly 3:1 ratio.
- Bromine: ⁹⁹Br:⁸¹Br ≈ 1:1, so M⁺˳ and M+2 appear in a roughly 1:1 ratio.

Exam Tip: If you see two peaks of roughly equal height separated by 2 m/z units, think bromine. If the ratio is approximately 3:1, think chlorine.

Fragmentation Patterns

The molecular ion often has enough internal energy to fragment — bonds break to produce smaller ions and neutral species. The fragmentation pattern provides structural information.

How Fragmentation Occurs

When a bond breaks in the molecular ion, one fragment retains the positive charge (and is detected) while the other is a neutral radical (and is not detected):

M⁺˳ → A⁺ + B˳ (A⁺ is detected; B˳ is not)

M⁺˳ → A˳ + B⁺ (B⁺ is detected; A˳ is not)

The more stable the cation formed, the more intense the corresponding peak.

Common Fragment Losses

Mass Lost	Fragment Lost	Likely Functional Group
15	CH₃˳	Methyl group
17	OH˳	Hydroxyl group (alcohols, carboxylic acids)
18	H₂O	Alcohols (dehydration)
28	CO	Carbonyl compounds, phenols
29	CHO˳ or C₂H₅˳	Aldehyde or ethyl group
31	CH₃O˳ (OCH₃)	Methyl ester / methoxy group
43	CH₃CO⁺ or C₃H₇⁺	Acetyl group (methyl ketones) or propyl group
45	OC₂H₅˳	Ethyl ester / ethoxy group
77	C₆H₅⁺	Phenyl group

Common Fragment Ions (Cations Detected)

m/z	Ion	Origin
15	CH₃⁺	Methyl cation
29	CHO⁺ or C₂H₅⁺	Formyl cation or ethyl cation
43	CH₃CO⁺ or C₃H₇⁺	Acylium ion (acetyl) or propyl cation
57	C₄H₉⁺ or C₃H₅CO⁺	Butyl cation or propanoyl cation
77	C₆H₅⁺	Phenyl cation
91	C₆H₅CH₂⁺ (tropylium)	Very stable; from methylbenzene derivatives
105	C₆H₅CO⁺	Benzoyl cation

Key Definition: The base peak is the most intense peak in the mass spectrum. It is assigned a relative abundance of 100%. It is NOT necessarily the molecular ion peak.

Worked Examples

Worked Example 1: Identifying Butanone

A compound has a molecular ion peak at m/z = 72 and a base peak at m/z = 43. Other significant peaks appear at m/z = 57 and m/z = 29.

Step 1: Mr = 72. Possible molecular formula: C₄H₈O (Mr = 72).

Step 2: Base peak at m/z = 43 corresponds to CH₃CO⁺ (the acylium ion, mass 43). This strongly suggests a methyl ketone.

Step 3: Loss of 72 − 43 = 29 (loss of C₂H₅˳). This is consistent with loss of an ethyl group from the other side of the carbonyl.

Step 4: Peak at m/z = 57: loss of 72 − 57 = 15 (loss of CH₃˳). Loss of a methyl group.

Step 5: Peak at m/z = 29: C₂H₅⁺ (ethyl cation) or CHO⁺.

Conclusion: The compound is butanone (CH₃COCH₂CH₃). The dominant fragmentation is cleavage either side of the C=O group, producing CH₃CO⁺ (m/z = 43) and C₂H₅⁺ (m/z = 29).

Worked Example 2: A Compound with Mr = 60

A mass spectrum shows M⁺˳ at m/z = 60, with significant peaks at m/z = 43 (loss of 17), m/z = 45 (loss of 15), and m/z = 15.

Loss of 17 from 60 = OH˳ lost → suggests an –OH group.
Loss of 15 from 60 = CH₃˳ lost → suggests a methyl group.
Peak at m/z = 45 = loss of CH₃ from M⁺˳. The fragment CHO₂⁺ (m/z = 45) or C₂H₅O⁺.
With Mr = 60 and formula C₂H₄O₂ (acetic acid) or C₃H₈O (propan-1-ol).

If the compound is ethanoic acid (CH₃COOH, Mr = 60): loss of OH gives m/z = 43 (CH₃CO⁺), loss of CH₃ gives m/z = 45 (COOH⁺). This fits perfectly.

Worked Example 3: Detecting Halogens

A compound shows peaks at m/z = 78 and m/z = 80 in approximately equal intensity.

The 1:1 ratio of M and M+2 peaks indicates the presence of bromine. Mr = 78 could not contain Br (Br alone is 79). Looking more carefully, if the molecular ion is actually at m/z = 108 and 110 (1:1 ratio), then Mr = 108. Subtracting Br (79) leaves 29, which could be C₂H₅. The compound is bromoethane (CH₃CH₂Br).

If instead M⁺˳ appears at m/z = 92 and 94 in a 3:1 ratio, this indicates chlorine. Subtracting Cl (35) leaves 57, possibly C₄H₉. The compound could be 1-chlorobutane.

High-Resolution Mass Spectrometry

High-resolution mass spectrometry measures m/z values to several decimal places, allowing determination of the exact molecular formula.

This works because exact atomic masses are not whole numbers:

¹H = 1.00794
¹²C = 12.00000 (by definition)
¹⁴N = 14.00307
¹⁶O = 15.99491

For example, both CO and C₂H₄ have a nominal mass of 28, but:

CO: 12.000 + 15.995 = 27.995
C₂H₄: 2(12.000) + 4(1.008) = 28.032

High-resolution MS can distinguish between these.

Exam Tip: In AQA exams, you may be given a high-resolution molecular mass and asked to determine the molecular formula. Use the exact masses of ¹²C, ¹H, ¹⁶O, and ¹⁴N to work out possible combinations. The degree of unsaturation (IHD) can then help confirm the structure.

The Nitrogen Rule

If a compound has an odd Mr, it contains an odd number of nitrogen atoms. If Mr is even, it contains zero or an even number of nitrogen atoms.

This is because nitrogen has an even atomic mass (14) but is trivalent (odd valency), which affects the overall molecular formula.

Common Misconceptions

"The tallest peak is always the molecular ion." Incorrect — the tallest peak is the base peak, which may or may not be the molecular ion. The molecular ion is the peak at the highest m/z (excluding isotope peaks).
"Every fragment detected is a positive ion." Correct — only positive ions are detected. Neutral fragments and radicals are not detected.
"You can always see a molecular ion peak." Not always — some molecules fragment so readily that the molecular ion peak is very small or absent (e.g., some highly branched alkanes).

Summary

Mass spectrometry determines molecular mass from the molecular ion peak M⁺˳.
Fragmentation patterns reveal structural features.
Learn the common fragment losses (15, 17, 18, 28, 29, 31, 43, 45) and what they indicate.
Isotope patterns reveal Cl (3:1) and Br (1:1).
High-resolution MS gives exact molecular formulae.
The base peak is the most intense peak and is not necessarily the molecular ion.