The intensity of the M + . ion is larger for the linear chain than the branched chain
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1. The intensity of the M +. ion is larger for the linear chain than the branched chain H 3 C HC CH 3 CH 3 +Ÿ CH 3 Ÿ HC CH 3 CH 3 + + A lkanes: C leavage of the bond C H 3 CH 2 CH 2 CH 2 +Ÿ CH 3 Ÿ + C H 2 + CH 3 Spectra 1 Spectra 2
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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. - PowerPoint PPT Presentation
Transcript of The intensity of the M + . ion is larger for the linear chain than the branched chain
1. The intensity of the M+. ion is larger for the linear chain than the branched chain
H3C HC
CH3
CH3
+Ÿ
CH3Ÿ HC
CH3
CH3
+ +
Alkanes: Cleavage of the bond
CH3
CH2
CH2
CH2
+Ÿ
CH3Ÿ + CH2
+CH3
Spectra 1
Spectra 2
Rule 1: Intensity of the M+. Is larger for a linear chain than a branched one.
M+.
Spectra 1
Spectra 2
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.
CH3 C+
CH3
CH3
Loss of MW=57
Spectra 4
CH+
AŸCH
+
CH2
+ AŸ
Benzylic cleavage
+
C+ Z
Ÿ+ ZŸ
Vinylic cleavage
Aromatic Hydrocarbons
C
CH+
XY
ZH
Ÿ
C+
X
H
H
+Ÿ+ Z Y
McLafferty rearrangement: H atoms present
X,Y and Z can be any combination of C,N,O or S
CX
YZ
AH
+
Ÿ
X
AŸ +
+Y
Z
HX,Y and Z can be any combination of C,N,O or S
Elimination of neutral fragments from 2- substituted compounds
+Ÿ
CH2
CH2+Ÿ
+ C2H4
Retro Diels- Alder
Z=alkyl,aryl of heteroatom
A=alkyl,aryl of heteroatom
Rule 4: Aromatic rings, double bonds and cyclic structures stabilise the M+. ions
Rule 4: Aromatic rings and cyclicstructures stabilise the M+. ion
Spectra 7
Spectra 6
Tropylium ion
CH+
AŸCH
+
CH2
+ AŸ
Benzylic cleavage
+
A. is CH2 CH3
propyl radical
Spectra 8
H2C CH2
CH2 CH2
R
+ŸCH2
+
CH2
+
Allylic cleavage: not very frequent
C+ H
H
HŸ
CH3
CH2+Ÿ
+ C2H4C
+ HH
HŸ
CH3
CH2
+ CH2+
CH2Ÿ
McLafferty rearrangement: atom present
CH2 CH3
CH+Ÿ
CH2
CH2+
Ÿ + C2H4
Retro Diels- Alder
Alkenes
(R)
H2C RŸ
Rule 5: Double bonds favour the allylic cleavage because resonance stabilises the cation.
Retro Diels-Alder
C+
C
CH2
CH2+
+ C2H4
Retro Diels- Alder
Spectra 9
CH3OH
HCH3 CH2
- H 2O
T hermal
CH3 C CH2 CH2
H
H
OH+
CH3 CH CH2 CH2+
this may be a 1,3 or 1,4 elimination
CHCH2+
CH3 C2H4 +CH2
+CH
CH3
the 1,4- dehydration product may undergo further cleavage
Alcohols
CH3 CH3
OH+
CH3
O+
H
+ CH3
,- cleavage to form oxonium ions
ethylideneoxonium
R
R2
R1
OH+ L oss of largest group
- R2
CH3
CH3
O+
H
CH3C
+
CH3
OH
HOH
+R
R
R
R
CH OH+R
R
R
R
H- H 2O CH
CH+
R
R
R
RCH
+
R
R
R
R
HOH
+R
R
R
R
- H 2O
- CH R=CH R
R
R
OH+
H
O+ H
HCH2 CH
O+ H
HCH3
CH2
+
O+
CH2
H
Oxonium ion
Spectra 10
CH3
CH3
NH2+
CH3
NH2+
+ CH3
Amines
,- Cleavage to form an immonium ion
With primary amines the M+ ion is odd
CH3
NH2+
cleaves here
CH2 NH2+
propan- 1- amine
+ CH3 CH2
CH3 NH+
CH2 H
N- ethylethanamine
NH2+
CH3+ CH2 CH2
There can be further cleavage fission with H transfer
Secondary amines
Immonium ion
Spectra 11
Aldehyde and ketones
CH3
R
O+
CH3 + RO+ C
-O
+
R+
+
- cleavage
HO
+
R
CH2 CH2 + CH2
R
OH+
McLafferty arrangement
O+ O
+
CH2H CH
O+
CH3
CH3
CH2
+
O+
CH2
Complex fissions for cyclic ketones
C+
O
CC
+
O
C CH+
+ CO
Bridged aromatic ketones emit CO
Spectra
Result of McLafferty arrangement
Spectra 12
CH2+
O
OH
McLafferty arrangementMW=60
CH3
O
OH
29 43 57 71 85 99
115 101 87 73 59 45
Charge on alkyl
Charge on oxygen
Spectra dominated by the hydrocarbon
McLafferty arrangementMW=60
Cl+
CH3
Cl+
chlorolanium
+CH3
CH2
MW=91.5
Cl+
CH3H
- H C l
CH 3CH =CH 2
CH3
CH2
+
MW=56
CH3 X+.CH3 + X
+
CH3 X+.
CH3+
+ X.
Elimination of HX
Halides
H2CCH2
CH2CH2
CH2
H X+.
HX + CH2CH2
+
Cleavage of C-X bonds
CH3
CH3
X+.
CH3
X+
+ CH3
α. β-fission with the formation of halonium ion
Remote cleavage with the formation of a cyclic halonium ion
Cl+
CH3
Cl+
chlorolanium
+CH3
CH2
MW=91.5
Cl+
CH3H
- H C l
CH 3CH =CH 2
CH3
CH2
+
MW=56
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 aThe second term is the relative intensity of the 2 isotopes a and bThe 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
