Molecular Ion Peak & Fragmentation (Cambridge International AS Chemistry)

Mass Spectrometry: Deducing Molecular Formula

• Each peak in the mass spectrum corresponds to a certain fragment with a particular m/e **value
• The peak with the highest m/e value (ignoring the small [M+1] and [M+2] peaks) is the molecular ion (M+**) peak which gives information about the molecular mass of the compound
• The molecular ion is the entire molecule that has lost one electron when bombarded with a beam of electrons

MOLECULE$\overset{\mathrm{Electron} \mathrm{bombardment}}{\rightarrow}$MOLECULE⁺• + e^–

• 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 a 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

Worked Example Determine whether the following mass spectrum corresponds to propanal or butanal.

Answer:

• The mass spectrum corresponds to propanal as the molecular ion peak is at m/e = 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

Identifying Molecules using Fragmentation

• 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 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 peaks at M – 18 (loss of H₂O), M – 28 (loss of CO) and M – 44 (loss of CO₂)

Alkanes

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

m/e values of fragments table

Mass spectrum showing fragmentation of alkanes

*Straight chain alkanes show characteristic peaks *

Halogenoalkanes

• Halogenoalkanes often have multiple peaks around the molecular ion peak
• This is caused by the fact that there are different isotopes of the halogens

Mass spectrum for a bromine containing species

Mass spectra show different halogen isotopes in the molecular ion

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 of propan-1-ol

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

Worked Example Which of the following statements about the mass spectrum of CH₃Br is correct?

1. There is one peak for the molecular ion with an m/e value of 44
2. There is one peak for the molecular ion with an m/e value of 95
3. The last two peaks have abundances in the ratio 3:1 and occur at m/e values of 94 and 96
4. The last two peaks are of equal size and occur at m/e values of 94 and 96

Answer:

• The correct answer is 4 as bromomethane (CH₃Br) will fragment into 3 peaks

  • CH₃⁸¹Br → [CH₃⁸¹Br]⁺ + e⁻ at m/e 96
  • CH₃⁷⁹Br → [CH₃⁷⁹Br]⁺ + e⁻ at m/e 94
  • CH₃Br → [CH₃]⁺ + ^•Br at m/e 15

• The last two peaks (which correspond to the molecular ion peak) therefore are equal in size and occur at m/e values of 94 and 96

Worked Example Which alcohol is not likely to have a fragment ion at m/e at 43 in its mass spectrum?

1. (CH₃)₂CHCH₂OH
2. CH₃CH(OH)CH₂CH₂CH₃
3. CH₃CH₂CH₂CH₂OH
4. CH₃CH₂CH(OH)CH₃

Answer:

• The correct answer is 4 because a line at m/e = 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/e = 43 as it does not have either of these fragments in its structure