The intensity of the M+. ion is larger for the linear chain than the branched chain Spectra 1 Spectra 2
Rule 1: Intensity of the M+. Is larger for a linear chain than a branched one. Spectra 1 M+. Spectra 2
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
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
Rule 4: Aromatic rings, double bonds and cyclic structures stabilise the M+. ions
Rule 4: Aromatic rings and cyclic structures stabilise the M+. ion Spectra 6 Spectra 7
Tropylium ion Spectra 8 A. is
Rule 5: Double bonds favour the allylic cleavage because resonance stabilises the cation.
Spectra 9 Retro Diels-Alder
Oxonium ion Spectra 10
Immonium ion Spectra 11
Spectra
Result of McLafferty arrangement Spectra 12
McLafferty arrangement MW=60
McLafferty arrangement MW=60 Spectra dominated by the hydrocarbon
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
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 +2x3x1 + 12 = 9 + 6 + 1 So: M=9, M+2=6, M+4=1