Topics

Isotope patterns and fragmentation

Why some spectra have two peaks at the top

Most elements have more than one isotope. If a heavier isotope is common enough, molecules containing it show up as a separate peak two units to the right of the molecular ion.
This gives an immediate, unmistakable test for chlorine and bromine.

Element	Isotopes	Natural ratio	Peak pattern
Chlorine	³⁵Cl and ³⁷Cl	about 3 : 1	M and M+2 in a 3 : 1 ratio
Bromine	⁷⁹Br and ⁸¹Br	about 1 : 1	M and M+2 in a 1 : 1 ratio
Carbon	¹²C and ¹³C	about 99 : 1	small M+1 peak, ~1.1% per carbon

A pair of peaks two units apart, roughly equal in height → bromine.
A pair of peaks two units apart, the right one about a third as tall → chlorine.
Note this identifies which halogen, not just that a halogen is present — which IR cannot do, since C–Cl and C–Br absorptions both sit in the crowded fingerprint region.

The nitrogen rule

Nitrogen has an even atomic mass (14) but an odd valency (3). This combination makes it detectable from the molar mass alone.
A molecule of C, H and O only always has an even molar mass.
An odd molar mass means an odd number of nitrogen atoms — usually one.

Molar mass	What it suggests
Even	no nitrogen, or an even number of nitrogens
Odd	one (or three) nitrogen atoms

Combined with an IR showing N–H, this pins down an amine or an amide immediately.

Fragmentation

Inside the spectrometer, the molecular ion breaks apart. The mass lost identifies the piece that broke off.
Work with the difference between the molecular ion and the fragment peak.

Mass lost	Fragment lost	Suggests
15	CH₃	a methyl group
17	OH	a carboxylic acid or alcohol
18	H₂O	an alcohol
28	CO or C₂H₄	a carbonyl or an ethyl chain
29	CHO or C₂H₅	an aldehyde, or an ethyl group
31	CH₂OH or OCH₃	a primary alcohol, or a methyl ester
43	C₃H₇ or CH₃CO	a propyl group, or an ethanoyl group
45	COOH or OC₂H₅	a carboxylic acid, or an ethyl ester

Fragments themselves also appear as peaks, so both readings are available:
- A peak at m/z 43 is often the CH₃CO⁺ ion, strongly suggesting a methyl ketone or an ethanoate.
- A peak at m/z 29 is often CHO⁺, suggesting an aldehyde.
Where a mass loss is ambiguous — 43 could be C₃H₇ or CH₃CO — the IR resolves it: CH₃CO requires a C=O and C₃H₇ does not.

Using fragments to locate a group

Fragmentation tells you not just what groups exist but where they are attached.
Two isomers with the same formula fragment differently, because different bonds break most easily.
- Butan-2-one, CH₃COCH₂CH₃, loses CH₃ (15) to give m/z 57, and loses C₂H₅ (29) to give m/z 43.
- Butanal, CH₃CH₂CH₂CHO, loses CHO (29) to give m/z 43, and loses C₃H₇ (43) to give m/z 29.
The pattern of losses is therefore evidence about the skeleton, not only about composition.

Worked ExampleIdentifying a halogen from the isotope pattern

A compound shows peaks at m/z 108 and m/z 110 of approximately equal height, and a strong fragment at m/z 29. Its IR shows C–H absorptions just below 3000 cm⁻¹ and nothing above 3000 cm⁻¹ or near 1700 cm⁻¹. Identify the halogen present and determine the structure.

Step 1 — Interpret the isotope pattern

Two peaks two mass units apart of approximately equal height is the signature of bromine, whose isotopes ⁷⁹Br and ⁸¹Br occur in a roughly 1 : 1 ratio.

(A 3 : 1 ratio would have indicated chlorine instead.)

So the molecular ion is at m/z 108, with the M+2 peak at 110 arising from molecules containing ⁸¹Br.

Step 2 — Use the IR to eliminate functional groups

Nothing above 3000 cm⁻¹ → no O–H and no N–H, so not an alcohol, carboxylic acid, amine or amide.
Nothing near 1700 cm⁻¹ → no C=O, so not an aldehyde, ketone, ester, amide or acid chloride.
C–H absorptions just below 3000 cm⁻¹ → a saturated hydrocarbon skeleton.

The only family left from the eleven is a haloalkane — consistent with the bromine already identified.

Step 3 — Work out the rest of the molecule

Subtracting one bromine atom, using the lighter isotope to match the m/z 108 peak:

$108 - 79 = 29$

Twenty-nine mass units of carbon and hydrogen is C₂H₅, since $2(12) + 5(1) = 29$ .

Step 4 — Assemble the formula

$\text{C}_2\text{H}_5\text{Br}$

Check: $2(12) + 5(1) + 79 = 24 + 5 + 79 = 108$ . ✓

Step 5 — Confirm with the fragment

The strong fragment at m/z 29 is the C₂H₅⁺ ion, produced by loss of the bromine atom ( $108 - 79 = 29$ ). This is the expected dominant fragmentation for a haloalkane, since the C–Br bond is the weakest bond in the molecule.

Step 6 — Check for isomers

C₂H₅Br has only one possible structure — with two carbons there is no choice about where the bromine sits. No ¹³C data is needed to resolve isomers here, though a ¹³C spectrum would confirm two signals, with the carbon bearing the bromine shifted downfield.

Answer: bromoethane, CH₃CH₂Br, molecular formula C₂H₅Br. Bromine is identified by the 1 : 1 M / M+2 pattern, and the absence of any absorption above 3000 cm⁻¹ or near 1700 cm⁻¹ rules out every family except a haloalkane.