Mass Spectrometry (OCR A Level Chemistry)

• When a compound is analysed in a mass spectrometer, vaporised molecules are bombarded with a beam of high-speed electrons
• These knock off an electron from some of the molecules, creating molecular ions:

• The relative abundances of the detected ions form a mass spectrum: a kind of molecular fingerprint that can be identified by computer using a spectral database
• The peak with the highest m/z value is the molecular ion (M⁺) peak which gives information about the molecular mass of the compound
• This value of m/z is equal to the relative molecular mass of the compound

The M+1 peak

• The [M+1] peak is a smaller peak which is due to the natural abundance of the isotope carbon-13
• The height of the [M+1] peak for a particular ion depends on how many carbon atoms are present in that molecule; The more carbon atoms, the larger the [M+1] peak is
  • For example, the height of the [M+1] peak for an hexane (containing six carbon atoms) ion will be greater than the height of the [M+1] peak of an ethane (containing two carbon atoms) ion

Answer:

• • The mass spectrum corresponds to propanal as the molecular ion peak is at m/z = 58
  • Propanal arises from the CH₃CH₂CHO⁺ ion which has a molecular mass of 58
  • Butanal arises from the CH₃CH₂CH₂CHO⁺ ion which has a molecular mass of 72

• The molecular ion peak can be used to identify the molecular mass of a compound
• However, different compounds may have the same molecular mass
• To further determine the structure of the unknown compound, fragmentation analysis is used
• Fragments may appear due to the formation of characteristic fragments or the loss of small molecules
  • For example, a peak at 29 is due to the characteristic fragment C₂H₅⁺­­
  • Loss of small molecules gives rise to differences between peaks of, for example, 18 (H₂O), 28 (CO), and 44 (CO₂)
  • An alcohol can typically dehydrate in a MS, so one peak to look for is M-18

Alkanes

• Simple alkanes are fragmented in mass spectroscopy by breaking the C-C bonds
• M/z values of some of the common alkane fragments are given in the table below

m/z values of Fragments Table

Mass spectrum showing fragmentation of alkanes

Alcohols

• Alcohols often tend to lose a water molecule giving rise to a peak at 18 below the molecular ion
• Another common peak is found at m/e value 31 which corresponds to the CH₂OH⁺­­ fragment
• For example, the mass spectrum of propan-1-ol shows that the compound has fragmented in four different ways:
  • Loss of H^• to form a C₃H₇O⁺ fragment with m/e = 59
  • Loss of a water molecule to form a C₃H₆⁺ fragment with m/e = 42
  • Loss of a ^•C₂H₅ to form a CH₂OH⁺ fragment with m/e = 31
  • And the loss of ^•CH₂OH to form a C₂H₅⁺ fragment with m/e = 29

Mass spectrum showing the fragmentation patterns in propan-1-ol (alcohol)

Answer

The correct answer is option D

• • Because a line at m/z = 43 corresponds to an ion with a mass of 43 for example:
    • [CH₃CH₂CH₂]⁺
    • [(CH₃)₂CH]⁺
  • 2-butanol is not likely to have a fragment at m/z = 43 as it does not have either of these fragments in its structure.