Mass Spectrum Example 1

Example 1: ethanol, CH₃CH₂OH (M_r = 46)To interpret a mass spectrum, first find the molecular ion peak, then work out what each fragment peak has lost from the molecular ion.m/z 46: the molecular ion, CH₃CH₂OH⁺. This confirms M_r = 46. (A tiny peak at 47 is caused by carbon-13.)m/z 45: 46 – 45 = 1, so the molecular ion has lost one hydrogen atom, giving C₂H₅O⁺.m/z 31: the base peak. 46 – 31 = 15, the mass of CH₃. The molecular ion has lost a methyl radical, leaving CH₂OH⁺.CH₃CH₂OH⁺ → CH₂OH⁺ + CH₃•m/z 29: 46 – 29 = 17, the mass of OH. The molecular ion has lost a hydroxyl radical, leaving C₂H₅⁺.CH₃CH₂OH⁺ → C₂H₅⁺ + OH•m/z 15: CH₃⁺.The peak at m/z 31 is typical of primary alcohols that contain the CH₂OH group. Its appearance as the base peak, together with the molecular ion at 46, is strong evidence that the compound is ethanol and not its isomer methoxymethane, CH₃OCH₃, which has no CH₂OH group.