Mass Spectrum
X -axis: m/z
mass -based on
12
C 12.0000
Z –charge on ion
Y -axis: relative abundance
usually normalized with respect to the
highest peak
(base peak)
0 -100 %
Relative Abundance is Expressed as percentage of the base peak.% relative abundance =
Peak height
Peak height of base peak
X 100
Types of Mass Spectrometry
ESI (Electro spray Ionization)
FAB (Fast Atom Bombardment)
MALDI (Matrix -Assisted Laser Desorption
Ionization)
GC-MS ( Gas Chromatography Mass
Spectrometry)
Fast Atom Bombardment
Worksbestfororganiccompounds with
polarityandeitheracidicandorbasic
functionalgroups.
Basicgroupsrunwellinpositiveionization
mode.
Acidiccentersrunwellinnegativeionization
mode.
ClassesthatuseFABare:peptides,proteins,
fattyacids,organometallics,surfactants,
carbohydrates,antibiotics,andgangliosides.
Major Steps in Analysis of Mass
Spectral Data
Identification of molecular ion
Base peak
Analysis of fragmentation pattern
Propose possible structures
Compare postulated species to available reference
spectra
Type of ions
Molecular ion
Fragment ion
Rearrangement ion
Metastable ion
Multiple charged ion
Isotope ion
Negative ion
Molecular Ion
Ion whose mass equals that calculated from
the molecular formula using the masses for
each element which have the highest natural
abundance; often tallest peak in highest m/z
group
Molecular ion represents the molecular
weight of the sample.
Base peak-most intense peak in spectrum;
not necessarily the molecular ion peak!M + e M
+.+ 2e
Fragment ion
Fragmentionsareformedbysimplecleavage
andrearrangementprocesses.
e.gCH3CH2
CH3CH2Cl
+
+
.
+.CH3CH2Cl
Cl
Fragmentation ion
Themolecularionsareenergetically
unstable,andsomeofthemwillbreakup
intosmallerpieces.Thesimplestcaseisthat
amolecularionbreaksintotwoparts-one
ofwhichisanotherpositiveion,andthe
otherisanunchargedfreeradical.CH2
H
CH2
Cl
CH2CH2
+.
+HCl
O.E E.E
The Nitrogen Rule
Molecule of even number MW contain no nitrogen or even number
of nitrogen
An odd number MW contain odd number of nitrogen atom
Problem
Which of the following ions will
appear at an even mass number?
NH
3 MW = 17
C
4H
9N MW = 71
C
2H
5NH
2MW = 45
CH
2N
2 MW = 42
ANS: CH
2N
2
Satellite peaks : (M+1) and (M+2) peaks
(M+1) peak can arise by the abstraction of a proton from the neutral
molecule by molecular ion (ion molecule reaction)
Examples: ether, amines and nitriles
(M-1) peak is obtained by the loss of hydrogen
Example: aldehyde
Aromaticcompoundsgiveintensive(presenceof-electronsystem)
Conjugatedolefins-intense peakthanthecorrespondingnon-conjugated
olefinswiththesamenumberofunsaturation.
Aliphatic compounds -branching results decrease in the abundance of
peak
Introduction of groups which lowers the I.P (methoxy, hydroxy, amino etc)
increases the molecular ion abundance
Introduction of groups which increases the I.P (cyano, nitro etc) lowers the
molecurlar ion abundance. M
+. M
+. M
+.
Rearrangement ion
Intramolecular reorganization processes involving the migration of hydrogen or
other atoms or groups
Example: Mclafferty rearrangement
Electron impact induced hydrogen transfer rearrangement
“-hydrogen is transferred through a six membered transtion state to an electron
deficient centre followed by the cleavage of the bond”
Reactionresultsintheeliminationofneutralmolecule
Charge will be carried by the oxygen containing fragment
Characteristics of ketones, aldehydes, acids, esters, olfins etc.
McLafferty Rearrangement
+
+
+
+
O
CCH
3O
CH
2
CH
2
CR
2
H
R
2
CCH
2
O
H
CCH
3O
CH
2
O
H
CCH
3O
CH
2
O
H
CCH
3O
CH
2
Molecularionpeakofastraightchainmonocarboxylicacidis
weakbutusuallydiscernible.Themostcharacteristicpeak
(sometimesthebasepeak)isatm/z=60duetoMcLafferty
rearrangement.
CHO
CH2
OH
HOC
O
H
CH2
+
+
O
COH
CH2
CH2
C
HR
H
H
2CCHR
+
m/z 60
H
+
CH2
CH2
+. CH
CH2
CH2
C
CH3O
H
CH2
C
CH3OH
+.
+
CH CH2
+.
Structural requirements
1.Side chain containing atleast 3 carbon atoms,
the last bearing a hydrogen atom
2.A double bond (carbonyl, olefin or aromatic
system)
3.Djerassi et al. in a detailed study of steriod
ketones showed that the interatomic distance
for McLafferty rearrangement should be less than
1.8 A°
Metastable ion
life time few microseconds
appear as a broad and diffuse peak
If ion of mass m
1decomposes to give a daughter ion of mass m
2, then
the metastable peak m* is given by
Meatstable peaks are not symmetrical
Flat top metastable peak shows K.E is released during the process
From the pattern of the metastable peak it is possible to find out
whether the polyisotopic element is present in the neutral or in the
resulting daughter ion.m*
2
m
m
2
1
Acetophenoneshowsamolecularionpeakat
m/z=120anditshowsadaughterionpeak
atm/z=105.Calculatethemetastableion?
m*=(105)
2
/120
=91.88
Acetophenoneshowsapeakatm/z=92.
Problem m*
2
m
m
2
1
General Fragmentation Modes
Simple cleavage
Retro-Diels-Alder reaction
Hydrogen transfer rearrangement
Redro-Diels-Alder reaction
Multicentred fragmentation mode
involving the cleavage of two bonds
of cyclic system resulting in the
formation of two stable unsaturated
fragments in which two new bonds
are formed.
In the mass spectrum of 3,5,6-triphenyl-4-nitro-
1
-
cyclohexene the base peak is due to the diene ion (m/z
206)Ph
Ph
Ph
NO2
+.
Ph
Ph
+. +.
+.
Ph
Ph
+
Ph
Ph
Ph
NO2
Ph
Ph
+
NH2
CH
CH
Ph
NH2
CH
CH
Ph
m/z 206 (100%)
In the spectrum of 3,5,6-triphenyl-4-amino-1-cyclohexene,
the diene and dienophile ions are in the ratio 1:4
100%
25%
O
+.
O
+.
+ CH2C
CH3
CH3
-ionone
- ionone
+
+.
+
O O
CH3
. -ionone and -ionone can be differentiated by mass
spectrometry
-ionone undergoes RDA fragmentation to give an intense peak
at m/z 122. This peak is insignificant in mass spectrum of -
ionone
In -ionone , loss of one geminyl methyl group takes place
Among natural products, Norborene, flavones, triterpenes, steroids and
alkaloids undergo RDA fragmentation which is a valuable tool for
structural elucidation
Simple cleavage
i)Heterolytic cleavage at the cationic site
Example :
Alkyl halidesCH3CH2 CH2CH
ClH
-HCl
+.
CH3CH2CHCH2
+.
m/z 56
m/z 92 C4H9I
+.
C4H9+I
+.
C4H9
+.
C4H9+
+.
Br Br
m/e 57
m/e 57
m/z 184
m/z 136
Homolytic cleavage induced by a radical site
Oddelectronionshaveanunpairedwhichis
capableofnewbondformationwhichis
exothermicandaccompaniedbybondcleavage.
Thesereactionsaremorecommon.R.
+RCR
+.
X
CR
X
+
RCH
OH
+.
R.
+CH
R
R
OH
+
Alcohols
Sec
PriRCH2
OH
+.
R
.
+
CH2
OH
+
Ter
R
R.
+RCR
+.
X
R
CR
X
+
R
CR
X
+
R
R.
+
+
Sec
Pri
R R
.
+
Ter
R.
+RCR
+.
+.
R
+.
R
CR
+
CH2
NH2
NH2
CH2
CHR
CHR
NH2
+
NH2
NH2
NH2
Amines
m/z 94
m/z 66
.
OH O
H
H
-CO
-CHO
H
-H
+.
----
+
m/z 65
+.
+
+
+
-H
-CO
-H2
m/z108
m/z107
m/z79
m/z77
5
CH
2OH
OH
HH
H
H
H
H
H
+
Benzyl alcohol
7 memberd ring
Benzonium ion
Phenyl ion+
C4H3
+
m/z 51
Ortho effect
Insuitablysubstitutedaromaticcompoundsorcis-olefins
thesubstituentandhydrogenatomcancomeinclose
proximitysoastohelpeliminationofaneutralmolecule.
Thiseffectisknownasorthoeffect.
Example:
Elimination of methanol from cis-methyl crotonateCH
CH
CH2
H
C
O
OCH3
CH
CH
CH2
CO
+.
+ CH3OH
Orthoeffectisusefultodistinguishbetweencis-
andtranscompoundsandorthofrommetaand
paraisomers.O
C
OH
X
H
+.
C
O
X
+H2O
X = O, NH
Tropylium ion
Mass Spectrum of n-Octylbenzene
Tropylium ion
Tropylium ion
Naphthalene
Molecules with Heteroatoms
Isotopes: present in their usual abundance.
Hydrocarbons contain 1.1% C-13, so there will be
a small M+1 peak.
If Br is present, M+2 is equal to M
+
.
If Cl is present, M+2 is one-third of M
+
.
If iodine is present, peak at 127, large gap.
If N is present, M
+
will be an odd number.
If S is present, M+2 will be 4% of M
+
. =>
Isotopic Abundance
=>
81
Br
Mass Spectrum with Sulfur
=>
Mass Spectrum with Chlorine
=>
Mass Spectrum with Bromine
=>
Mass Spectra of Alkanes
More stable carbocations will be more abundant.
=>
Mass Spectra of Alkenes
Resonance-stabilized cations favored.
=>
Mass Spectra of Alcohols
Alcohols usually lose a water molecule.
M
+
may not be visible.
=>
Problem-1
A hydrocarbon with MF C
7H
12(Molecular ion 96 shows a large peak due to
M-15 and at m/z 54. What structure assigned to this compound?
Answer
Theindexofhydrogendeficiencyinthiscompoundistwo(analkane
withsevencarbonatomswouldbeC
7H
16,thedifferenceC
7H
16-C
7H
12
isfourhydrogenatoms,thustheindexofhydrogendeficiencyistwo).
Thelarge(M-15)pointstothepresenceofabranchedmethylgroup.
Thepeakatm/z54isduetoC
4H
6,i.e.butadiene.ARDAreactionwill
eliminatebutadienefragmentm/z54fromacyclicalkaneandthe
presenceofacycloalkeneisinkeepingwiththeindexofhydrogen
deficiency.Thecompoundis4-methylcyclohexene.CH3
+.
+.
+
CH3
m/z 96
4-methylcyclohexene m/z 54
Problem-2
How will you differentiate 3-methylcyclohexene from 4-methylcyclohexene
on the basis of mass spectroscopy?
Answer
IncyclicalkenesaRETRO-DIELS-ALDERreactiongives
achargeddienefragmentandanunchargedethylene
fragment.Them/zofthedienefragmentgivesuseful
informationofthepositionofdoublebond.Thusunlike4-
methylcyclohexene,3-methylcyclohexenegivesadiene
fragment,withm/z=68.CH3
+.
CH3
+.+
m/z 96
m/z683-methylcyclohexene
Problem 3
A compound C
9H
10O
2(MW 156.0681) shows a strong peak m/z 119, 118,
91 and 65. What is the structure of the compound?
Answer:
Thecompoundhasanalkylaromaticsystem,thusstrongpeaksat
m/z=91and65willappear.Largepeaksatm/z=118and119willbe
duetolossof32and31amu.Them/z=119isthelossofOCH
3andthe
lossofCH
3OHtogivem/z=118canoccuronlyifthecompoundis
orthoester.CH2
H
C
O
CH3
O C
O
+.
CH2
+CH3OH
m/z 150
m/z 118
Problem 3
Ethylbutanoate in its mass spectrum shows two charcteristic
peaks due to odd electron ions at m/z 88 and 60 and
abundant ion at m/z 71. Explain the fragmentation.
Theionsatm/z=88and60arisefromtwosuccessiveMcLafferty
rearrangementswhilethe-cleavagegivestheabundantpeakatm/z=71.m/z 71
CH2
CH2
CH2
C
O
CH2
O
CH3
H
+.
H2C
C
+
CH2
CH2
O
H
O
H
+.
H2C
C
CH2
CH2
O
H
m/z 88
C
OH
H2C OH
+.
+
m/z 60