Mass Spectrometry for Organic Compounds

Spec Mapping — OCR H432 Module 4.2.4(c) — Mass spectrometry of organic compounds, covering the principle of electron-impact mass spectrometry (ionisation, fragmentation, mass-to-charge separation, detection); identification of the molecular ion peak M⁺ as the highest-m/z peak corresponding to the relative molecular mass; the interpretation of fragmentation patterns through the masses of neutral fragments lost (M-15 = •CH₃, M-17 = •OH, M-29 = •CHO or •C₂H₅, M-43 = •C₃H₇ or •CH₃CO, M-45 = •COOH); the recognition of stable fragment cations (CH₃⁺ at m/z 15, CHO⁺/C₂H₅⁺ at 29, CH₃CO⁺ at 43, COOH⁺ at 45, tert-butyl C₄H₉⁺ at 57, phenyl at 77, tropylium at 91); the M+1 isotope peak from ¹³C natural abundance (~1.1 % per C); the M+2 isotope patterns diagnostic for chlorine (M:M+2 ≈ 3:1) and bromine (M:M+2 ≈ 1:1); the combined use of IR and MS to identify unknown structures (refer to the official OCR H432 specification document for exact wording).

Mass spectrometry (MS) is the second of OCR A-Level's two routinely-examined analytical techniques. Where infrared spectroscopy identifies the functional groups in an unknown (Lesson 9), mass spectrometry gives you the molecular mass directly (from the molecular ion peak M⁺ at the highest m/z) and structural clues from how the molecule fragments under high-energy electron-impact ionisation inside the instrument. The combination of IR + MS on a single sample, supplemented by the molecular formula and the question of degree of unsaturation, is the standard A-Level structure-determination workflow and is the basis of almost every spectroscopy-interpretation exam question. This lesson recaps the mass-spectrometer pipeline you met at AS (ionisation, fragmentation, acceleration, deflection, detection), develops the interpretation of fragmentation patterns through the lens of "what stable cation can the molecule produce when its weakest bond is broken?", explains the isotope-peak patterns at M+1 (¹³C carbon-counting), M+2 (chlorine and bromine diagnostic ratios), and M+4 (two halogen atoms), and concludes with two full IR + MS worked examples that take an unknown molecular formula plus a spectrum and produce a confident structural assignment. The general principle — fragmentation favours the most stable cation — is the same one that controls SN1/E1 selectivity in alcohol substitution (Lessons 3-6), so this lesson is also a synoptic re-encounter with carbocation stability.

Key Definition: the molecular ion M⁺ is the cation formed when the parent molecule loses one electron during electron-impact ionisation. Its mass is essentially equal to the relative molecular mass of the parent molecule (the electron mass is negligible). M⁺ appears at the highest m/z in the spectrum (apart from isotope peaks at M+1, M+2) and its value is the first piece of information you read off an unknown's mass spectrum.

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1. Recap: How the Mass Spectrometer Works

We met this at AS but it is worth a quick reminder:

1. Ionisation — the sample is vaporised and bombarded with high-energy electrons (in electron impact MS) or ionised more softly in electrospray MS. One electron is knocked off, giving a positive molecular ion, M⁺.
2. Fragmentation — the molecular ion has too much internal energy and breaks apart into smaller charged fragments and neutral pieces.
3. Acceleration and deflection — the charged fragments are accelerated by an electric field and deflected by a magnetic field by an amount that depends on their mass-to-charge ratio (m/z).
4. Detection — a detector records the abundance of ions at each m/z value.
5. Spectrum — a vertical bar-chart of abundance against m/z is produced.

In an A-Level spectrum, the y-axis is abundance as a percentage of the tallest peak (the "base peak"), and the x-axis is m/z (almost always equal to mass, because z = 1 for most ions).

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2. The Molecular Ion (M⁺) Peak

The molecular ion, M⁺, is formed by the loss of a single electron from the parent molecule:

$M + e^- \rightarrow M^+ + 2\,e^-$M+e−→M++2e−

It has the same mass as the parent molecule (to within one electron mass, which is negligible). Its m/z value therefore tells you the molecular mass of your unknown — a massively useful piece of data.

2.1 Finding M⁺ in a Spectrum

• The molecular ion is (almost always) the peak with the highest m/z in the spectrum — because it has not lost any pieces, it is the heaviest ion present.
• It may be small compared to fragment peaks, because many molecular ions break apart before reaching the detector.
• Small isotope peaks at M+1 and M+2 appear to the right of M⁺ due to ¹³C, ³⁷Cl, ⁸¹Br etc. — these are discussed in section 5.

2.2 Worked Example — Reading M⁺

The mass spectrum of an unknown compound shows its highest-m/z peak at 60. Deduce the molecular mass.

Answer: Mᵣ = 60. Possible formulas with mass 60 include C₃H₈O (propan-1-ol or propan-2-ol), CH₃COOH (ethanoic acid), C₂H₄O₂ (methanoate), C₂H₈N₂, and so on. The next step is to use IR and the fragment pattern to narrow this down.

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3. Fragmentation — How and Why

When the molecular ion breaks apart, it does so by breaking a weak bond to give two pieces. One piece carries the positive charge and is detected; the other is neutral and is invisible to the detector.

General equation:

$M^+ \rightarrow F^+ + R\bullet$M+→F++R∙

Where F⁺ is the detected fragment and R• is a neutral radical.

3.1 Typical Mass Losses (Neutral Fragments)

When you go from M⁺ to a smaller peak, the difference in m/z tells you the mass of the neutral fragment that was lost. Common losses to recognise:

Mass lost	Neutral fragment	Common source
1	H•	C–H cleavage
15	•CH₃ (methyl)	Alkane, methyl branch
17	•OH	Alcohol, acid
18	H₂O	Alcohol dehydration in MS
28	CO or C₂H₄ or N₂	Aldehyde, ketone, alkene
29	•CHO or •C₂H₅	Aldehyde, ethyl branch
31	•OCH₃	Ester (methyl ester)
43	•C₃H₇ (propyl) or •CH₃CO (acetyl)	Propyl branch, methyl ketone
45	•COOH	Carboxylic acid

So if you see a peak 15 units below M⁺, you almost certainly have a methyl group being lost.

3.2 Common Fragment Ions

Conversely, some fragment cations are so stable they appear in many spectra. Learn to recognise them:

m/z	Fragment cation	Meaning
15	CH₃⁺	Methyl cation — methyl group present
17	OH⁺	Alcohol or acid
29	CHO⁺ or C₂H₅⁺	Aldehyde or ethyl branch
43	CH₃CO⁺ (acylium) or C₃H₇⁺	Methyl ketone or propyl
45	COOH⁺ or C₂H₅O⁺	Carboxylic acid or ether/alcohol
57	C₄H₉⁺ (tert-butyl) or CH₃CH₂CO⁺	Tert-butyl or ethyl ketone
77	C₆H₅⁺ (phenyl)	Benzene ring
91	C₇H₇⁺ (benzyl / tropylium)	Methylbenzene and derivatives
105	C₆H₅CO⁺ (benzoyl)	Benzoic acid and derivatives

3.3 Why Do These Fragments Dominate?

Cations are more stable when the positive charge can be stabilised:

• Methyl, ethyl, propyl cations — stabilised by alkyl hyperconjugation.
• Tert-butyl cation — very stable, 3° carbocation.
• Acylium (RCO⁺) — stabilised by resonance between C and O.
• Tropylium (C₇H₇⁺) — aromatic, very stable.
• Phenyl and benzoyl — aromatic ring stabilisation.

Fragmentation always prefers pathways that lead to the most stable cation.

graph TD
  A[Molecular ion M+] --> B{Weakest bond?}
  B --> C["C-H next to branch -<br/>loss of H, minor"]
  B --> D["C-C alpha to C=O -<br/>loss of R as radical,<br/>gives acylium"]
  B --> E["C-C at branch -<br/>gives stable branched cation"]
  B --> F["Alcohol -<br/>loss of OH or H2O"]

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4. Worked Example — Full Interpretation

The mass spectrum of an organic compound shows the following major peaks:

• m/z = 72: medium abundance (M⁺)
• m/z = 57: very strong
• m/z = 43: strong
• m/z = 29: medium
• m/z = 15: weak

The IR spectrum shows a strong sharp peak at 1715 cm⁻¹, a C–H peak around 2900, and no O–H or N–H peaks. Identify the compound.

Step 1: Molecular mass. M⁺ = 72 ⇒ Mᵣ = 72.

Step 2: IR.

• 1715 cm⁻¹ → C=O (aldehyde, ketone, acid, ester, amide).
• No O–H → rules out acid, alcohol.
• No N–H → rules out primary amine, amide.
• So this is a ketone or aldehyde or ester.

Step 3: Molecular formula.

• Mᵣ = 72 and contains C=O. Possible formulas: C₄H₈O (Mᵣ = 72, butanone or butanal) or C₃H₄O₂ (Mᵣ = 72, methyl methanoate) or C₃H₈N₂.
• Given no N–H and a simple spectrum, C₄H₈O is the strong candidate.

Step 4: Fragment analysis.

• Loss from 72 to 57 = loss of 15 (CH₃). So the compound has a methyl group attached to the carbonyl or adjacent position.
• Peak at 57 could be CH₃CH₂CO⁺ (mass = 57), i.e. the acylium from butanone after loss of methyl. Consistent.
• Peak at 43 is the acylium CH₃CO⁺, i.e. loss of C₂H₅ (mass 29) from M⁺. Consistent with butanone fragmenting either side of the carbonyl.
• Peak at 29 is CHO⁺ or C₂H₅⁺.
• Peak at 15 is CH₃⁺.

Fragmentation diagram for butanone:

$CH_3-CO-CH_2-CH_3 \rightarrow \underbrace{CH_3CO^+}_{m/z=43} + \underbrace{\bullet CH_2CH_3}_{\text{lost}}$CH3​−CO−CH2​−CH3​→m/z=43CH3​CO+​​+lost∙CH2​CH3​​​

$CH_3-CO-CH_2-CH_3 \rightarrow \underbrace{CH_3CH_2CO^+}_{m/z=57} + \underbrace{\bullet CH_3}_{\text{lost}}$CH3​−CO−CH2​−CH3​→m/z=57CH3​CH2​CO+​​+lost∙CH3​​​

The compound is butanone (CH₃COCH₂CH₃). (If it were butanal, the characteristic CHO• loss would give a stronger m/z = 43 and the molecular ion pattern would differ slightly, plus the IR would show the aldehyde C–H doublet at 2700–2900 cm⁻¹, which was not mentioned.)

graph TD
  A[Butanone M+ 72] --> B[- CH3 15]
  A --> C[- C2H5 29]
  B --> D[CH3CH2CO+ 57]
  C --> E[CH3CO+ 43]

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5. Isotope Peaks: The Fingerprint of Cl and Br

The M+1 peak is almost always present, about 1.1% of the M⁺ peak per carbon atom, because of the natural abundance of ¹³C (1.1%). This lets you estimate the number of carbons in the molecule: if M+1 is about 5.5% of M⁺, there are about 5 carbons.

Much more diagnostic are the M+2 peaks caused by chlorine and bromine:

• Chlorine: ³⁵Cl is 75.8% abundant, ³⁷Cl is 24.2%. An organic compound with one Cl shows an M+2 peak about one-third the height of the M⁺ peak.
• Bromine: ⁷⁹Br is 50.7%, ⁸¹Br is 49.3%. An organic compound with one Br shows an M+2 peak about the same height as M⁺.

Halogen present	M : M+2 ratio
None	100 : 0 (just ¹³C)
One Cl	100 : 33
One Br	100 : 98
Two Cl	100 : 65 : 11 (M+4 also)
Two Br	100 : 195 : 95 (M+4 dominates)

5.1 Worked Example — Detecting Halogen

An unknown compound shows M⁺ = 78 (abundance 100) and M+2 = 80 (abundance 32). What halogen, if any, does it contain?

Answer: The ratio M : M+2 ≈ 100 : 32 ≈ 3 : 1 is the classic signature of a single chlorine atom. Mᵣ = 78 with one Cl (mass 35) leaves 43 for the rest of the molecule — consistent with C₃H₇Cl (propyl chloride, Mᵣ = 78.5 for ³⁵Cl isotope = 78).

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6. Combining IR and MS

This is the most common exam scenario: you are given an IR spectrum and a mass spectrum and asked to identify the compound. The workflow is:

graph TD
  A[Read IR - identify functional groups] --> B[Note C=O, O-H, N-H, C-H etc]
  B --> C[Read MS M+ value = Mr]
  C --> D["Consider possible molecular formulas<br/>consistent with Mr and functional groups"]
  D --> E["Look for fragment ions and losses:<br/>15 CH3, 17 OH, 29 CHO or C2H5,<br/>43 CH3CO or C3H7, etc"]
  E --> F[Check for Cl Br M+2 isotope pattern]
  F --> G[Propose a structure]
  G --> H["Cross-check: does it fit ALL<br/>the evidence?"]

6.1 Worked Example — Full IR + MS Question

An unknown C₄H₈O₂ compound has:

• IR: broad O–H stretching 2500–3300 cm⁻¹; strong sharp C=O at 1710 cm⁻¹.
• MS: M⁺ = 88; peaks at 73 (weak), 60 (strong), 45 (very strong, base peak), 43 (strong), 29 (medium).

Identify the compound.

Step 1: Functional groups (IR). Broad O–H 2500–3300 + C=O 1710 = carboxylic acid.

Step 2: Molecular formula. C₄H₈O₂, Mᵣ = 88. Confirmed by M⁺ = 88.

Step 3: Fragment losses.

• 88 → 73 = loss of 15 (•CH₃).
• 88 → 45 = loss of 43 (•C₃H₇?).
• m/z = 45 is COOH⁺ (very characteristic of carboxylic acids).
• m/z = 43 is C₃H₇⁺ (propyl) or CH₃CO⁺.

Step 4: Put it together. A C₄ carboxylic acid: candidates are butanoic acid (CH₃CH₂CH₂COOH) or 2-methylpropanoic acid ((CH₃)₂CHCOOH).

Both would give a strong 45 (COOH⁺). Butanoic acid would give a 73 peak (loss of CH₃) but more prominently would give 60 (McLafferty rearrangement — loss of C₂H₄ = 28 to give CH₂=C(OH)₂⁺ at m/z = 60, diagnostic of straight-chain carboxylic acids with ≥ γ-hydrogens).

2-Methylpropanoic acid would give stronger 73 (loss of methyl, since it has two equivalent methyls on the α-carbon) but weaker 60 (no γ-H for McLafferty).

The observation of a strong 60 peak points to butanoic acid.

Structure: CH₃CH₂CH₂COOH (butanoic acid).

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7. Worked Calculation — Using Isotope Ratios for Chlorine

A compound has molecular ion at 92 with abundance 100, and a peak at 94 with abundance 33. It also has a peak at 96 with abundance near zero. Calculate the number of chlorine atoms and suggest a formula.

Answer:

The ratio M : M+2 = 100 : 33 ≈ 3 : 1 → exactly one chlorine atom. The absence of a significant M+4 confirms there is only one Cl.

Mᵣ = 92 with one ³⁵Cl = 35 leaves 57 for the rest: C₄H₉ has mass 57 exactly. So the molecular formula is C₄H₉Cl — either 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane or 2-chloro-2-methylpropane.

Further fragment analysis would distinguish these — e.g. 2-chloro-2-methylpropane gives a very strong C₄H₉⁺ (57) peak due to the stable tertiary cation, while 1-chlorobutane gives a weaker 57 peak and more small alkyl fragments.

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