AQA A Level Chemistry复习笔记3.6.2 Mass spectrometry

• Mass spectroscopy is an analytical technique used to identify unknown compounds
• The molecules in the small sample are bombarded with high energy electrons which can cause the molecule to lose an electron
• This results in the formation of a positively charged molecular ion with one unpaired electron
  • One of the electrons in the pair has been removed by the beam of electrons

   

• The molecular ion can further fragment to form new ions, molecules, and radicals

Fragmentation of a molecule in mass spectroscopy

• These fragmented ions are accelerated by an electric field
• Based on their mass (m) to charge (z) ratio, the ion fragments are then separated by deflecting them into the detector
  • Most ions will only gain a charge of 1+ and therefore a ion with mass 12 and charge 1+ will have an m/z value of 12
  • It is, however, possible for a greater charge to occur. For example, an ion with mass 16 and charge 2+ will have a m/z value of 8

   

• The smaller and more positively charged fragment ions will be detected first as they will get deflected the most and are more attracted to the negative pole of the magnet
• Each fragment corresponds to a specific peak with a particular m/z value in the mass spectrum
• The base peak is the peak corresponding to the most abundant ion
• The m/z is sometimes referred to as the m/e ratio and it is almost always 1:1

Isotopes

• Isotopes are different atoms of the same element that contain the same number of protons and electrons but a different number of neutrons.
  • These are atoms of the same elements but with different mass number
  • For example, Cl-35 and Cl-37 are isotopes as they are both atoms of the same element (chlorine, Cl) but have a different mass number (35 and 37 respectively)

   

• Mass spectroscopy can be used to find the relative abundance of the isotopes experimentally
• The relative abundance of an isotope is the proportion of one particular isotope in a mixture of isotopes found in nature
  • For example, the relative abundance of Cl-35 and Cl-37 is 75% and 25% respectively
  • This means that in nature, 75% of the chlorine atoms is the Cl-35 isotope and 25% is the Cl-37 isotope

   

• The heights of the peaks in mass spectroscopy show the proportion of each isotope present

The peak heights show the relative abundance of the boron isotopes: boron-10 has a relative abundance of 19.9% and boron-11 has a relative abundance of 80.1%

Answer

• • 56Fe3+ has a smaller m/z ratio and will therefore be deflected more.
  • It also has the largest positive charge and will be more attracted to the negative pole of the magnet within the mass spectrometer.

   

• Each peak in the mass spectrum corresponds to a certain fragment with a particular m/z value
• The peak with the highest m/z value 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

• The [M+1] peak is a smaller peak which is due to the natural abundance of the isotope carbon-13
• The amount of naturally occurring C-13 is a little over 1%, so the [M+1] peak is very small
• 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 pent-1-ene as the molecular ion peak is at m/z = 70
    • The small peak at m/z = 71 is a C-13 peak, which does not count as the molecular ion peak
  • But-1-ene arises from the C4H8+ ion which has a molecular mass of 56
  • Pent-1-ene arises from the C5H10+ ion which has a molecular mass of 70

• 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 give rise to peaks at 18 (H2O), 28 (CO), and 44 (CO2)

   

Alkanes

• Simple alkanes are fragmented in mass spectroscopy 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 the fragmentation of C10H22

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 showing different isotopes of bromine 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 showing the fragmentation patterns in propan-1-ol (alcohol)

Answer

The correct answer is option D

• • Bromomethane (CH3Br) can produce 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

   

Answer

The correct answer is option D

• • 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.