structure determination mass spectrometry aims how can a mass spectrometer be used to find molecular...

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.

M(g) + e- M+(g) + 2e-

The 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.

These knock out an electron from some of the molecules, creating molecular ions, which travel to the detector plates:

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.

molecular ion peakmass

spectrum of paracetamol

40 80 120 160

20

40

60

80

100

00

ab

un

dan

ce (

%)

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: M M+ ● + e-

molecule molecular ion • 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.

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: M+ ● X+ + Y● (Y● is not detected)

Or M+ ● 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+ ● CH3CH2CH2CH3

CH3CH2CH2CH3+ ● + e-

43 CH3CH2CH2+ CH3CH2CH2CH3

+ ● CH3CH2CH2

+ + CH3●

29 CH3CH2+ CH3CH2CH2CH3

+ ● CH3CH2

+ + CH3CH2●

15 CH3+ CH3CH2CH2CH3

+ ● CH3

+ + CH3CH2CH2●

Methyl Propane C4H10

Give the m/z of the molecular ion peak and two fragments.


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 CompoundsThe 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 :–

+

RCOR + ● RCO+ + R●

Acylium cation

Propanal C3H6O



58 CH3CH2CHO+ ● CH3CH2CHO

CH3CH2CHO+ ● + e-

29 CH3CH2+ and

CHO+

CH3CH2CHO+ ● CH3CH2

+ + CHO●

CH3CH2CHO+ ● CHO+ + CH3CH2

●

Propanone C3H6O



58 CH3COCH3+ ● 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

1-Mass Spectra