Structure Determination
Mass Spectrometry
Aims
• How can a mass spectrometer be used
to find molecular mass?
• What is fragmentation?
• How can fragmentation be used to
help find molecular structure?
Mass spectrometry is a highly sensitive technique that can
be used to help identify compounds. Samples of only a few
milligrams are required.
Mass spectrometry is useful in the identification of
unknown materials, because the mass spectrum of an
unknown sample can be compared with a database of
known spectra.
When a compound is analysed in a mass spectrometer,
gaseous molecules are bombarded with high-speed
electrons from an electron gun.
These knock out an
electron from some
of the molecules,
creating molecular
ions, which travel to
the detector plates:
M(g) + e-  M+(g) + 2eThe peak with the highest mass-to-charge ratio (m/z) is formed by the
molecular ion, and the value of m/z is equal to the relative molecular mass of
the compound.
The peak at the highest m/z on the mass spectrum is
formed by the heaviest ion that passes through the
spectrometer. Unless all molecules of the original
substance break up, this corresponds to the molecular ion
of the sample substance.
abundance (%)
100
mass spectrum
of paracetamol
80
molecular ion peak
60
40
20
0
0
40
80
120
160
m/z
Mass Spectrometry
•
A mass spectrum is essentially a graph of abundance (vertical axis) against mass/charge
ratio (m/z), (horizontal axis), but since the charge on the ions is normally +1, the
horizontal axis is effectively relative mass.
•
When a sample of an organic compound (M) is introduced into a mass spectrometer,
provided that the molecule is not completely fragmented (see below), the peak at the
maximum value of m/z (furthest to right on horizontal axis) corresponds to the
molecular ion, M+ ● .
•
Equation:
•
The value of m/z for this ion is equal to the relative molecular mass , Mr of the
compound.
•
Thus the molecular ion peak can be used to determine the Mr of an unknown
•
Note 1: The molecular ion peak is not necessarily the tallest peak (often called the
base peak), as it may not be the most abundant positive ion present.
•
Note 2: There may be small peaks of greater mass (further to right) than the
molecular ion in the mass spectrum. These are caused by the presence of isotopes such
as 13C in the molecular ion. They are of small abundance as only 1% of carbon is 13C.
M
molecule
M+ ●
molecular ion
+
e-
Fragmentation of the molecular ion
•
The high energy of the ionising beam in the mass spectrometer causes
some of the molecular ions to break apart or fragment. The molecular ion,
also referred to as the parent ion, splits up to give smaller, positively
charged ions that are detected, and uncharged radicals that are not
detected.
•
Equations:
Or
M+ ●
M+ ●
X+ +
Y●
(Y● is not detected)
Y+
X●
(X● is not detected)
+
•
Note: Both X+ and Y+ species may be detected, but they cannot be
formed from the same molecular ion. They are rarely formed in equal
amounts.
•
A variety of fragmentation pathways is usually possible and, for each
route, one of the fragments retains the positive charge and is detected.
Further fragmentation of X+ and Y+ may also occur. The result is a
characteristic relative abundance mass spectral fragmentation pattern (a
mass spectrum).
•
Although only the molecular ion peak needs to be identified to give the
relative molecular mass of the compound, the other peaks can give useful
information about the structure of the compound.
Butane C4H10
Give the m/z of the molecular ion peak and three
fragments.
m/z
Ion causing peak
How formed
58
CH3CH2CH2CH3+ ●
43
CH3CH2CH2+
CH3CH2CH2CH3+ ●
CH3CH2CH2+ + CH3●
29
CH3CH2+
CH3CH2CH2CH3+ ●
CH3CH2+ + CH3CH2●
15
CH3+
CH3CH2CH2CH3+ ●
CH3+ + CH3CH2CH2●
CH3CH2CH2CH3
CH3CH2CH2CH3+ ● + e-
Methyl Propane C4H10
Give the m/z of the molecular ion peak and two
fragments.
m/z
Ion causing peak
How formed
58
CH3CH(CH3)CH3+ ●
CH3CH(CH3)CH3
CH3CH(CH3)CH3+ ● + e-
43
CH3CHCH3+
CH3CH(CH3)CH3+ ●
CH3CHCH3+ + CH3●
15
CH3+
CH3CH(CH3)CH3+ ●
CH3+ + CH3CHCH3●
Note 1: Fragmentation is more likely to take place at weaker bonds.
Note 2: The more stable the fragment formed, the greater its abundance in
the spectrum.
Note 3: The stability of the positive ions formed from hydrocarbons
(carbocations) is in the order 3ᴼ › 2ᴼ › 1ᴼ.
Note 4: The mass spectra of hydrocarbons are fairly complex, due to the
presence of small peaks associated with 13C and with loss of hydrogen atoms.
Fragmentation of Carbonyl Compounds
The fragmentation patterns of carbonyl compounds,
especially ketones, are often useful in structural
identifications. The predominant decomposition
pathway is at the carbonyl group to give an alkyl radical
and a stable acylium cation :–
+
Acylium cation
RCOR + ●
RCO+ + R●
Propanal C3H6O
Give the m/z of the molecular ion peak
and two fragments.
m/z
Ion causing peak
How formed
58
CH3CH2CHO+ ●
CH3CH2CHO
CH3CH2CHO+ ● + e-
29
CH3CH2+ and
CHO+
CH3CH2CHO+ ●
CH3CH2+ + CHO●
CH3CH2CHO+ ●
CHO+ + CH3CH2 ●
Propanone C3H6O
Give the m/z of the molecular ion peak and
two fragments.
m/z
Ion causing peak
58
CH3COCH3+ ●
How formed
CH3COCH3
CH3COCH3+ ● + e-
43
CH3CO+
CH3COCH3+ ●
CH3CO+ + CH3●
15
CH3+
CH3COCH3+ ●
CH3+ + CH3CO●
Isotope peaks
• As 1% of all carbon atoms are carbon-13 atoms, there may be a
small peak at one mass unit to the right of the molecular ion peak.
The height of this peak, often called the M+1 peak, relative to the
molecular ion peak, gives the number of carbon atoms in the
molecule. For example, if the height of M+1 is 5% of the height of
M, there are 5 carbon atoms per molecule.
• Chlorine has two isotopes, 35Cl and 37Cl, with natural abundances
75% and 25% respectively. Peaks with intensity ratio 3 to 1 two
mass units apart in a spectrum therefore suggest the presence of
chlorine in the compound.
• Bromine has two isotopes, 79Br and 81Br, with natural abundances
50% and 50%. Peaks with intensity ratio 1 to 1 two mass units
apart in a spectrum therefore suggest the presence of bromine in
the compound.
• Every organic compound has a typical mass spectrum which means
that an unknown compound can be identified by comparing it with
the mass spectra of known compounds.
Isotope peaks: heights are in the same ratio
of abundance for particular elements.
Pairs of peaks correspond to
isotopes of 35Cl and 37Cl in
the ratio of 75%:25% ie. 3:1.
Highlight these.
mass
To do
• Summary questions page 136
• Homework- Worksheet 13.11 Exercise 1Mass Spectra