Organic Chemistry · Structure Determination: Mass Spectrometry and Infrared Spectroscopy
Interpreting Mass Spectra
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A mass spectrum is a puzzle with three kinds of clues. The molecular ion M+• tells you the molecular mass; the isotope peaks (M+1, M+2, and beyond) tell you how many carbons and whether chlorine, bromine, or sulfur is present; and the fragment peaks tell you what pieces broke off. Reading a spectrum is a systematic process: confirm the molecular ion, count the elements with the Nitrogen rule Odd # of N → odd nominal mass; even # → even mass Full entry → and isotope ratios, compute the degrees of unsaturation, then match the fragment losses to structural features.
The rules of this game are surprisingly few. Even-electron species (cations with no unpaired electron) dominate fragmentation; neutral fragments are lost in small, recognizable units (15, 18, 28, 29, 31…); and the molecular ion must be an odd-electron Radical cation Odd-electron positive ion (all molecular ions) Full entry →. Once you can name what was lost, you can name what was there.
Why this matters
Interpreting a mass spectrum is how chemists confirm they made what they intended. If your reaction supposedly produced C6H12O, the molecular ion should sit at m/z = 100, not 86 — a mismatch means the reaction went somewhere unexpected. In forensic labs, matching a spectrum identifies drugs and poisons; in metabolomics, interpreting fragment patterns identifies thousands of compounds in a single blood sample. For coursework, spectrum interpretation is a guaranteed exam skill: given a spectrum and a molecular formula, you must defend a structure with the isotope ratios, the nitrogen rule, and the fragment losses.
The college version
Core Concepts
Step 1: Find the molecular ion
The molecular ion M+• is the peak at the highest m/z in a cluster — usually, though not always, the tallest peak of the highest cluster. Three checks confirm it:
- It must be an odd-electron ion (a radical cation). If a fragment were heavier than the molecule, that would be impossible.
- The nitrogen rule: a molecule with an even number of nitrogen atoms (including zero) has an even nominal molecular mass; a molecule with an odd number of nitrogens has an odd Nominal mass Molecular mass rounded to whole numbers Full entry →. If the supposed molecular ion violates this, you have misidentified it — a common trap when the M⁺• peak is weak.
- The M+1 peak Peak one unit above M, mostly from 13C Full entry → should be small (roughly 1.1% of M per carbon atom), and M+2 should be small unless Cl, Br, or S is present.
Step 2: Count atoms with isotope ratios
The natural abundances of heavy isotopes create predictable satellite peaks:
- 13C (1.1%): each carbon contributes ~1.1% to the M+1 peak relative to M. So M+1/M × 100 ≈ 1.1 × (number of carbons). Counting carbons this way works best for molecules with up to ~10 carbons.
- 37Cl (24.2%): one chlorine gives an M+2 peak Peak two units above M from 37Cl, 81Br, 34S Full entry → about one-third the height of M (3:1 pattern).
- 81Br (49.3%): one bromine gives M and M+2 peaks of nearly equal height (1:1 pattern).
- 34S (4.4%): one sulfur adds ~4.4% to M+2.
For a compound containing n chlorines or bromines, the pattern is a binomial expansion: two chlorines give M : M+2 : M+4 in roughly 9 : 6 : 1.
Step 3: Compute degrees of unsaturation
Given a formula CaHbNcOdXe (X = halogen), the number of rings plus π bonds is:
DU = a - b + e2 + c2 + 1
Oxygen is ignored because it is divalent. A DU of 1 means one ring or one double bond; a DU of 4 with a formula that smells aromatic (e.g., C6H6) usually means a benzene ring. The DU must be a non-negative integer — if it is fractional or negative, the formula itself is wrong.
Step 4: Read the fragment losses
Fragmentation of the radical cation produces an even-electron cation plus a neutral fragment (either a radical or a small stable molecule). Subtract the neutral fragment's mass from M to get the fragment ion's m/z. Classic losses:
| Loss | Mass (Da) | What left |
|---|---|---|
| 15 | CH3• | Methyl group |
| 18 | H2O | Water (alcohols) |
| 28 | CO or C2H4 | Carbon monoxide or ethylene |
| 29 | CHO• or C2H5• | Aldehyde H–C=O or ethyl group |
| 31 | CH3O• or CH2OH• | Methoxy or hydroxymethyl (alcohols/ethers) |
| 35/37 | Cl• | Chlorine atom |
| 79/81 | Br• | Bromine atom |
| 44 | CO2 | Carbon dioxide (carboxylic acids) |
A peak at M−18 immediately suggests an alcohol; M−35 with a 3:1 M:M+2 pattern says the molecule contains chlorine; a strong M−29 says ethyl or aldehyde. The most stable carbocation tends to win, so fragments cluster where a stable cation can form (e.g., m/z 43, 57, 71 for alkyl chains).
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| M+1 from 13C | M+2 from 37Cl/81Br | M+1 ≈ 1.1% per carbon (always present); M+2 large only with Cl, Br, or S |
| Nitrogen rule on nominal mass | Nitrogen rule on exact mass | The rule is about nominal (integer) masses; exact masses can be even or odd regardless |
| Odd number of nitrogens → even mass | Correct pairing | Odd N count → odd mass; even N count (incl. zero) → even mass |
| Base peak | Molecular ion | Base peak = most abundant (often a fragment); M⁺• = intact molecule at highest m/z |
| M−18 (water loss, alcohol) | M−17 (NH₃ loss, amine; or OH loss) | A difference of one mass unit — check for N by the nitrogen rule first |
| Degree of unsaturation = number of double bonds | Rings + double bonds | A ring counts as one DU; benzene counts as 4 (3 π + 1 ring) |
| DU for CaHbNcOdXe ignoring O | Including O in the formula | Oxygen is divalent and drops out of the DU formula — do not add it in |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine a toy castle made of blocks. The mass spectrum weighs the whole castle (that's the molecular ion) and then shows the weight of the biggest pile of blocks after it was smashed (the base peak). The tiny "extra weight" peaks tell you how many carbon blocks are in the castle, and the missing-block weights tell you which towers broke off. If the whole castle weighs an odd number of pounds, it must contain an odd number of "nitrogen blocks" — that's the nitrogen rule.
Worked example
Example 1: From spectrum to formula
A compound gives a molecular ion at m/z = 100 (M), with M+1 = 6.6% of M and a tiny M+2. The mass is even, so the nitrogen rule says zero or an even number of nitrogens — try zero. Estimate carbons:
carbons ≈ M+1 (%)1.1 = 6.61.1 = 6
Formula first: 6 carbons account for 72 Da, leaving 100 - 72 = 28 Da for the rest. Hydrogen + oxygen can fill that: C6H12O = 72 + 12 + 16 = 100 ✓.
Verify with DU:
DU = 6 - 122 + 02 + 1 = 6 - 6 + 0 + 1 = 1
One degree of unsaturation — one ring, one C=C, or one C=O. A strong M−18 peak (82) would confirm an alcohol, but M−18 is absent here; a peak at M−28 (72) would suggest CO loss from a ketone/aldehyde. The data point to a monounsaturated C6H12O — e.g., a ketone such as 3-hexanone — pending IR confirmation of the carbonyl.
Example 2: The halogen fingerprint
A compound shows M at m/z = 122 and an M+2 peak at 124 with equal intensity (~1:1). The even mass suggests no nitrogen. What is the formula?
Interpret the isotope pattern: 1:1 M:M+2 = one bromine atom (79 or 81 Da). Subtract Br from the nominal mass: 122 - 79 = 43. The remainder, 43 Da, must be C3H7 (36 + 7 = 43). So the formula is C3H7Br, bromopropane.
Check the DU:
DU = 3 - 7 + 12 + 0 + 1 = 3 - 4 + 1 = 0
Zero degrees of unsaturation — a saturated compound, consistent with bromopropane. The two isomers (1-bromopropane vs 2-bromopropane) would be distinguished by fragments: loss of CH3• (M−15 = 107) is more favorable from the 2-isomer, giving a stronger m/z 107 peak.
Example 3: Using M+1 to count carbons precisely
A molecule shows M at m/z = 86 and M+1 at 5.7% of M. Estimate the carbon count.
Formula first: carbons ≈ (M+1%)/1.1 = 5.7/1.1 ≈ 5.2.
Interpret: round to 5 carbons (predicted M+1 ≈ 5.5%; the small excess could come from a sulfur atom or measurement noise). With m/z = 86: 86 - 5(12) = 26 Da remaining. The candidate formula C5H10O = 60 + 10 + 16 = 86 fits, with DU = 5 - 10/2 + 1 = 1 — one double bond or ring. The alternative C6H14 (hexane, DU = 0) predicts M+1 ≈ 6.6%, higher than observed. So M+1 points to C5H10O — but it is an estimate: a high-resolution exact mass (86.0732 Da for C₅H₁₀O vs 86.1096 Da for C₆H₁₄) settles the question definitively. This example shows the limit of isotope ratios: they narrow the options, and exact mass picks the winner.
Key takeaways
- Molecular ion = highest-m/z cluster; must be a radical cation; must obey the nitrogen rule.
- Nitrogen rule: odd number of N atoms → odd nominal mass; even number (incl. 0) → even nominal mass.
- M+1 ≈ 1.1% × (# carbons) relative to M; used to estimate the carbon count.
- Chlorine: M : M+2 ≈ 3 : 1. Bromine: M : M+2 ≈ 1 : 1. Two halogens → binomial pattern (9:6:1 for two Cl).
- Degrees of unsaturation: DU = a - (b+e)/2 + c/2 + 1 for CaHbNcOdXe; must be a non-negative integer.
- Fragment ions are even-electron cations; neutral losses are small molecules or radicals (15, 18, 28, 29, 31, 35, 44, 79…).
- The base peak is the most abundant fragment — usually the most stable cation, not the molecular ion.
- A weak or missing molecular ion is common for alcohols and branched alkanes; look for M−18 (alcohol) or M−15 (methyl loss).
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
A molecule has nominal mass 73. How many nitrogen atoms does it contain?
Show answer
An odd nominal mass means an odd number of nitrogens (likely 1). With zero or two nitrogens the mass would be even.
A compound's M+1 peak is 6.6% of M. About how many carbons does it have?
Show answer
6.6 / 1.1 = 6 carbons (each carbon contributes ~1.1% to M+1).
A spectrum shows M and M+2 in a 3:1 ratio. What element is present?
Show answer
One chlorine atom: 35Cl (75.8%) vs 37Cl (24.2%) gives M : M+2 ≈ 3 : 1.
Compute the degrees of unsaturation for C6H6.
Show answer
DU = 6 − 6/2 + 0 + 1 = 4 — consistent with a benzene ring (3 double bonds + 1 ring).
Why is a fragment ion never a radical cation?
Show answer
Fragmentation of the radical cation M⁺• splits it into an even-electron cation and a neutral fragment (radical or molecule); the charge stays on the even-electron species, which is more stable.
An alcohol often shows a strong M−18 peak. What was lost, and why?
Show answer
Water (H₂O, 18 Da). Under electron impact, alcohols readily lose water — typically with a neighboring hydrogen — so the molecular ion is often weak and M−18 is diagnostic.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Molecular ion (M+•)
- Intact molecule minus one electron; odd-electron radical cation
- Base peak
- Most abundant ion, set to 100%
- Nitrogen rule
- Odd # of N → odd nominal mass; even # → even mass
- M+1 peak
- Peak one unit above M, mostly from 13C
- M+2 peak
- Peak two units above M from 37Cl, 81Br, 34S
- Degrees of unsaturation (DU)
- Number of rings + π bonds implied by the formula
- Even-electron rule
- Stable fragment ions have no unpaired electrons
- Neutral loss
- Mass difference between a fragment ion and its parent
- Radical cation
- Odd-electron positive ion (all molecular ions)
- Nominal mass
- Molecular mass rounded to whole numbers
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