Undergraduate · 15 min

Reading isotope patterns in mass spectra

Compare the M/M+2 signatures of chlorinated and brominated compounds and deduce which halogen — and how many — a molecule contains.

Goal

Identify the halogen in each spectrum from the M:M+2 ratio and predict the full isotope cluster of Cl₂ and Br₂.

Apparatus and reagents

Electron-impact mass spectrum datasets for CH₃Cl, CH₃Br, Cl₂ and Br₂; isotope-abundance table (³⁵Cl/³⁷Cl ≈ 3:1, ⁷⁹Br/⁸¹Br ≈ 1:1).

Procedure

Open the chloromethane spectrum and measure the heights of the M and M+2 peaks.
Repeat with bromomethane and compare the M:M+2 ratio (~3:1 for Cl, ~1:1 for Br).
Predict the three-line cluster of Cl₂ (70/72/74 ≈ 9:6:1) before displaying it, then check.
Do the same for Br₂ (158/160/162 ≈ 1:2:1) and explain the difference with binomial coefficients.

What to observe

CH₃Cl shows M:M+2 ≈ 3:1 (m/z 50/52); CH₃Br shows two nearly equal peaks at 94/96.
For two halogens the cluster widens: Cl₂ gives 9:6:1, Br₂ gives 1:2:1 — the combinatorics of two isotope pairs.

Explanation

³⁵Cl and ³⁷Cl have abundances ≈ 75.8 % and 24.2 %, so one chlorine splits the molecular ion into M and M+2 in a ~3:1 ratio; ⁷⁹Br/⁸¹Br ≈ 51:49 gives ~1:1. For n halogens, peak intensities follow the binomial expansion of the two abundances — the basis of halogen counting in routine MS interpretation.

History of the experiment

Francis Aston built the first mass spectrograph in 1919 and used the separated neon isotopes to prove that atomic weights need not be integers — isotope patterns have been structure clues ever since.

Chemists behind it

Francis William Aston John B. Fenn