Mass Spectrum Example 5

Example 5: identifying an unknown halogenoalkaneA compound gives two molecular ion peaks of equal height at m/z 108 and 110, and a large peak at m/z 29. Identify it.Step 1: look at the molecular ion peaks. There are two peaks 2 units apart with heights in the ratio 1:1. Bromine has two isotopes, bromine-79 and bromine-81, which occur in a ratio of about 1:1. So the compound contains one bromine atom. (A chlorine compound would give a 3:1 ratio.)Step 2: find the rest of the molecule. Subtract the mass of the lighter isotope from the lower molecular ion value:108 – 79 = 29A mass of 29 corresponds to C₂H₅.Step 3: use the fragment peaks. The peak at m/z 29 is C₂H₅⁺, formed when the molecular ion loses a bromine atom. It is a single peak because it contains no bromine.C₂H₅Br⁺ → C₂H₅⁺ + Br•Small peaks at m/z 79 and 81 (equal height) come from Br⁺ ions.Answer: the compound is bromoethane, CH₃CH₂Br.Check: CH₃CH₂⁷⁹Br = 24 + 5 + 79 = 108 and CH₃CH₂⁸¹Br = 24 + 5 + 81 = 110, which matches the spectrum.Summary of halogen patterns: one Cl atom gives M and M+2 peaks in a 3:1 ratio; one Br atom gives M and M+2 peaks in a 1:1 ratio.