1. The intensity of the M+. ion is larger for the linear chain than the branched chain
Spectra 1
Spectra 2

2. Rule 1: Intensity of the M+. Is larger
for a linear chain than a branched one.
Spectra 1
M+.
Spectra 2

3. Rule 2: The intensity of the M+
Rule 2: The intensity of the M+. decreases with increasing molecular weight. (The carboxylic acid is the exception)
Spectra 1
Spectra 3

4. Rule 3: Cleavage is favoured at the branching point, this reflects the greater stability of the ion. The loss of the largest substituent is favoured also.
Loss of
MW=57
Spectra 4

5. Rule 4: Aromatic rings, double bonds and cyclic structures stabilise the M+. ions

6. Rule 4: Aromatic rings and cyclic structures stabilise the M+. ion
Spectra 6
Spectra 7

7. Tropylium ion
Spectra 8
A. is

8. Rule 5: Double bonds favour the allylic cleavage because resonance stabilises the
cation.

9. Spectra 9
Retro Diels-Alder

11. Oxonium ion
Spectra 10

13. Immonium ion
Spectra 11

14. Spectra

15. Result of McLafferty
arrangement
Spectra 12

18. McLafferty arrangement
MW=60

19. McLafferty arrangement
MW=60
Spectra dominated by the hydrocarbon

20. Halides Cleavage of C-X bonds Elimination of HX
α. β-fission with the formation of halonium ion
Remote cleavage with the formation of a cyclic halonium ion

23. The relative ion region is very complex for molecules containing more than one atom which has a significant isotope,
e.g. Cl, Br, C & S. An expression can be used to calculate the intensities.
(a+b)m
Where a= relative abundance of the lighter element
b= relative abundance of the heavier element
m= number of atoms of the element present
So if we have 2 atoms of the element we get:-
(a+b)2
= a2 + 2ab + b2
The first term is the relative intensity of the element containing only isotope a
The second term is the relative intensity of the 2 isotopes a and b
The third term is the relative intensity of the element containing only isotope b
So if we consider a molecule with 2 Chlorine atoms what are the relative intensities of the contributions to
the Cl atoms. Assume the isotopic ratios of Cl35:Cl37 = 3:1
The M values will be
Cl35Cl35Ξ M, Cl35Cl37Ξ M+2, Cl37Cl37Ξ M+4
(a+b)m = (a+b)2 = a2 + 2ab +b2 = 32 +2x3x =
So: M=9, M+2=6, M+4=1