1. Mass spectrometry
L.O.:
Understand how fragmentation can be useful to find the molecular structure

2. In pairs, discuss what you remember about mass spectrometry:
How does a mass spectrometer work?
What does a mass spectrum look like?
What is the molecular ion?

3. Ionisation, acceleration, deflection, detection.

4. butanone
CH3CH2COCH3
Mr = 72

5. THE MASS SPECTRUM - FRAGMENTATION
The rest of the spectrum provides additional information of the molecule’s structure.
Peaks appear due to characteristic fragments (e.g. 29 due to C2H5+) and differences between two peaks also indicates the loss of certain units (18 for H2O, 28 for CO). 43
Abundance % 29 57 71 85
114 .
m/z

6. THE MASS SPECTRUM - THE MOLECULAR ION
The small peak (M+1) at 115 due to the natural abundance (about 1%) of carbon-13. The height of this peak relative to that for the molecular ion depends on the number of carbon atoms in the molecule. The more carbons present, the larger the M+1 peak.
Abundance %
114 .
m/z

7. MOLECULAR MASS DETERMINATION USING MASS SPECTROMETRY
Nowadays, mass spectrometry is used to identify unknown or new compounds.
When a molecule is ionised it forms a MOLECULAR ION which can also undergo FRAGMENTATION or RE-ARRANGEMENT to produce particles of smaller mass.
Only particles with a positive charge will be deflected and detected.
The resulting spectrum has many peaks.
The final peak (M+) shows the molecular ion (highest m/z value) and indicates the molecular mass. The rest of the spectrum provides information about the structure.
IONISATION
MOLECULAR ION
FRAGMENTION
RE-ARRANGEMENT
FRAGMENTION

8. THE MASS SPECTRUM
Spectra obtained for organic molecules have many peaks. Each peak is due to a particular fragment with a certain m/z value.
highest m/z value usually corresponds to the molecular ion
its position provides information about the molecular mass of a substance
the tallest peaks come from the most stable species

9. differences between major peaks
THE MASS SPECTRUM
Spectra obtained for organic molecules have many peaks. Each peak is due to a particular fragment with a certain m/z value.
highest m/z value usually corresponds to the molecular ion
its position provides information about the molecular mass of a substance
the tallest peaks come from the most stable species
Interpretation of thousands of spectra has shown that many classes of organic compound show characteristic fragmentation patterns due to their functional groups.
It is possible to identify the type of compound from its spectrum by looking at the ...
position of peaks
differences between major peaks

10. THE MASS SPECTRUM - THE MOLECULAR ION
In the spectrum of octane, a signal occurs at 114 due to the species C8H18+
The species due to the final signal is known as the molecular ion and is usually corresponds to the molecular mass of the compound.
Abundance %
molecular ion
114 .
m/z

11. THE MASS SPECTRUM - THE MOLECULAR ION
The small peak (M+1) at 115 due to the natural abundance (about 1%) of carbon-13. The height of this peak relative to that for the molecular ion depends on the number of carbon atoms in the molecule. The more carbons present, the larger the M+1 peak.
Abundance %
114 .
m/z

12. FRAGMENTATION PATTERNS
ALKANES
The mass spectra of simple hydrocarbons have peaks at m/z values corresponding to the ions produced by breaking C-C bonds. Peaks can occur at ...
m/z etc CH3+ C2H5+ C3H C4H C5H C6H13+
• the stability of the carbocation formed affects its abundance
• the more stable the cation the higher the peak
• the more alkyl groups attached to the carbocation the more stable it is
most stable tertiary 3° > secondary 2° > primary 1° least stable
alkyl groups are electron releasing and stabilise the cation

13. This is the mass spectrum of EtOH. Identify:
a) the molecular ion peak.
b) the fragment responsible for causing the peak at m/z= 31.

14. Here is the mass spectrum of CH3CH2Cl.
Identify the molecular ion peaks for this molecule
Give the ratio of relative abundance you would expect these peaks to be found in.
Identify the fragments responsible for causing the pair of peaks at 49 and 51.

15. bromomethane
CH3Br
79Br:81Br = 1:1
[CH379Br]+.
[CH381Br]+.

16. butane or methylpropane.

18. FRAGMENTATION PATTERNS
HALOGENOALKANES
Multiple peaks occur in the molecular ion region due to different halogen isotopes.
There are two peaks for the molecular ion of C2H5Br, one for the molecule containing the isotope 79Br and the other for the one with the 81Br isotope. Because the two isotopes are of similar abundance, the peaks are of similar height.
m/z
Abundance %
molecular ion contains...79Br Br

19. FRAGMENTATION PATTERNS
ALDEHYDES AND KETONES
Cleavage of bonds next to the carbonyl group (C=O) is a characteristic fragmentation of aldehydes and ketones. A common fragment is carbon monoxide (CO) but as it is a molecule and thus uncharged it will not produce a peak of its own. However, it will produce an m/z drop of 28 somewhere in the spectrum.
The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group
the more stable the acylium ion RCO+, the more abundant it will be and
the more abundant the species the taller its peak in the mass spectrum

20. FRAGMENTATION PATTERNS
Aldehydes and ketones
The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. O
CH3 C C4H9
MOLECULAR ION
has m/z = 100
• +

21. FRAGMENTATION PATTERNS
Aldehydes and ketones
The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. O
CH3 C C4H9
MOLECULAR ION
has m/z = 100
• + O
C4H9 C+
CH3•
Breaking the bond between the methyl group and the carbonyl group produces two possible ions, depending on how the bond breaks.
Two peaks at m/z values 15 and 85 will appear in the mass spectrum.
m/z = 85 O
C4H C•
CH3+
m/z = 15

22. FRAGMENTATION PATTERNS
Aldehydes and ketones
The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. O
CH3 C C4H9
MOLECULAR ION
has m/z = 100
• + O
CH3 C+
Breaking the bond between the butyl group and the carbonyl group produces two further ions, depending on how the bond breaks.
Two peaks at m/z values 43 and 57 will appear in the mass spectrum.
C4H9•
m/z = 43 O
CH3 C•
C4H9+
m/z = 57

23. FRAGMENTATION PATTERNS
Aldehydes and ketones
The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. O
CH3 C C4H9
Example;
MOLECULAR ION
has m/z = 100
• + O
C4H9 C+ O
CH3 C+
CH3•
C4H9•
m/z = 85
m/z = 43 O
C4H C• O
CH3 C•
CH3+
C4H9+
m/z = 15
m/z = 57
A further peak occurs at m/z = 72 (100-28) due to loss of CO

24. Identify as many fragment ions as you can.
Extension: write the corresponding equations showing the formation of the ions.
CH3CH2COCH3.

25. CH3CH2COCH3. m/z ratio Ion responsible
Equation to show formation of ion 72
[CH3CH2COCH3]+· 15
[CH3]+
[CH3CH2COCH3]+·  [CH3]+ + [CH2COCH3]· 29
[CH3CH2]+
[CH3CH2COCH3]+·  [CH3CH2]+ + [COCH3]· 57
[CH3CH2CO]+
[CH3CH2COCH3]+·  [CH3CH2CO]+ + [CH3]· 43
[COCH3]+
[CH3CH2COCH3]+·  [COCH3]+ + [CH3CH2]·
CH3CH2COCH3.

26. Suppose you have to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra.

27. 4.11 Exercise 1

28. IDENTIFY THE COMPOUNDS

29. Mass spectrometry
L.O.:
Understand how fragmentation can be useful to find the molecular structure

30. IDENTIFY THE COMPOUND Abundance % m/z 43 29 122 124 79 81 29 43
Abundance %
m/z

31. C3H7Br IDENTIFY THE COMPOUND Abundance % m/z 43 29 122 124 79 81 29 43
Abundance %
m/z
C3H7Br

32. IDENTIFY THE COMPOUND Abundance % m/z 105 77 51 120 43 28
Abundance %
m/z

33. C6H5COCH3 IDENTIFY THE COMPOUND Abundance % m/z 105 77 51 120 43 28
Abundance %
m/z
C6H5COCH3

34. IDENTIFY THE COMPOUND Abundance % m/z 28 105 106 77 57 43 51
m/z
Abundance % 28 77 57 43 51

35. C6H5CHO IDENTIFY THE COMPOUND Abundance % m/z 28 105 106 77 57 43 51
m/z
Abundance % 28 77 57 43 51
C6H5CHO

36. IDENTIFY THE COMPOUND Abundance % m/z 56 57 113 43 71 142
m/z
Abundance %
142
113 71 43

37. C10H22 IDENTIFY THE COMPOUND Abundance % m/z 56 57 113 43 71 142
m/z
Abundance %
142
113 71 43
C10H22