MS, Mass spectroscopy, HRMS

1. Mass spectrometry (MS)
For
M.Sc. I Chemistry
SPPU

2. Introduction
• Mass spectrometry is an analytical technique that measures the
mass-to-charge ratio of ions.
• Mass spectrometry is a powerful analytical technique used to
quantify known materials, to identify unknown compounds within a
sample, and to elucidate the structure and chemical properties of
different molecules.
• The complete process involves the conversion of the sample into
gaseous ions, with or without fragmentation, which are then
characterized by their mass to charge ratios (m/z) and relative
abundance.
• A mass spectrometer generates multiple ions from the sample
under investigation, it then separates them according to their
specific mass-to-charge ratio (m/z), and then records the relative
abundance of each ion type.
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3. The formed ions are separated by Deflection in magnetic
field according to their Mass to charge (m/z)
Ions formed Further break up into smaller ion
(Fragment ions or Daughter ions formed)
Mass Spectroscopy
Basic Principles
3
Molecules converted into highly energetic positively charged
ions (Molecular ions or Parent ions)
Organic Molecules are bombarded with electron
Mass Spectrum
m/z ratio Vs intensity (relative abundance of peak)
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4. Basic Principle
• The first step in the mass spectrometric analysis of compounds is the production of
gas phase ions of the compound, basically by electron ionization with bombarded
of electron to form molecular ions or parent ion.
• This molecular ion undergoes fragmentation to from positively charged ions and
neutral ions or radicals.
• Each primary product ion derived from the molecular ion, in turn, undergoes
fragmentation, and so on.
• The ions are separated in the mass spectrometer according to their mass-to-
charge ratio, and are detected in proportion to their abundance.
• A mass spectrum of the molecule is thus produced.
• It displays the result in the form of a plot of ion abundance versus mass-to-charge
ratio.
• Ions provide information concerning the nature and the structure of their
precursor molecule.
• In the spectrum of a pure compound, the molecular ion, if present, appears at the
highest value of m/z (followed by ions containing heavier isotopes) and gives the
molecular mass of the compound.
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5. • In MS technique, molecules are bombarded with a beam of energetic
electrons.
• The molecules are ionized into ions and broken up into many fragments,
some of which are positive ions.
• Each kind of ion has a particular ratio of mass to charge, i.e. m/e
ratio(value).
• For most ions, the charge is one and thus, m/e ratio is simply the
molecular mass of the ion.
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6. Instrumentation and Working
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7. Working of MS
• There are four key stages in the process for
Mass Spectrometry.
– Ionization
– Acceleration
– Deflection
– Detection
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8. Stage 1: Ionisation
8
• The initial sample may be a solid, liquid, or gas.
• The sample is vaporized into a gas and then ionized by the ion source, usually by
losing an electron to become a cation
• Electro Ionisation is the most common type of ionisation.
• The sample is bombarded by electrons which come from a heated filament.
• When the sample passes through the electron stream, the high energy electrons in
the stream knock electrons out of the sample to form ions.
• The ionization chamber is kept in a vacuum so the ions that are produced can
progress through the instrument without running into molecules from air.
Stage 2: Acceleration
• Acceleration is a simple step where the ions are placed between a set of charges
parallel plates.
• The ions will then be repelled by one plate and attracted to the other.
• There is a slit cut in the plate which the ions are attracted to. the force of
attraction and repulsion forces the ions through the slit at an accelerated rate.
• The speed of acceleration can be adjusted by changing the charge on the plates.
• The purpose of acceleration is to give all species the same kinetic energy, like
starting a race with all runners on the same line.
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9. Stage 3: Deflection
• Ions are deflected by the magnetic field surrounding the instrument.
• The amount of deflection depends on the mass and charge of the ions.
• Lighter components or components with more ionic charge will deflect in the field
more than heavier or less charged components.
• The heavier ions, are deflected the least (Ion stream C)
• The lightest ions are deflected the most (Ion Stream A)
• The ions at the correct mass and charge travel to the detector. (Ion Stream B)
• The mass to charge ratio (m/z) is determined from the ion that hits the detector.
9
Step 4: Detection
• When the ion stream reached the detector the hit a wire. On hitting the
wire they become neutralised by an electron jumping from the metal wire
to the ion.
• The amplifier picks up on this current being created between the wire and
the ion and amplifies the signal being detected.
• The computer picks up on this and converts it to mass/charge ratio and a
spectrum is produced.
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10. •Ionizing technique and their source in MS
There are many types of ionization methods are used in mass spectrometry
methods.
Gas Phase Sources. (Sources)
• Electron Impact Ionization (EI). (By Energetic electrons)
• Chemical Ionization (CI). (By reagent gaseous ions)
• Field Ionizations (FI). (By high potential electrodes)
Desorption Sources.
• Field Desorption (FD). (By high potential electrodes)
• Electrospray Ionization (ESI). ( by high electric field)
• Matrix assisted desorption/Ionisation (MALDI). (By laser beam)
• Plasma desorption (PD). (by fragments from 253Cf)
• Fast Atom Bombardment (FAB). (by energetic atomic beam)
• Thermospray Ionization (TS). (By high temperature)
• Secondary Ion Mass Spectrometry (SIMS). (energetic atomic beam)
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11. Mass Spectrum
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12. Types of Peaks in MS
• Molecular ion Peak
• Base Peak
• Metastable peak
• Fragment ion peak
• Rearrangement ion peak
• Multicharged ion
• Negative ion peak
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13. Molecular ion Peak
• When a any sample is bombarded with electrons of 9
to 15 eV energy, the molecular ion is produced by loss
of single electron. It is denoted by M (or M)
• When the vaporized organic sample passes into the
ionization chamber of a mass spectrometer, it is
bombarded by a stream of electrons. These electrons
have a high enough energy to knock an electron off an
organic molecule to form a positive ion. This ion is
called the molecular ion. (sometimes the parent ion.)
• Molecular ion peak gives molecular weight of
compounds
M M 2e+
e
molecular ion
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14. Molecular Ion peak in Ethanol
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15. Mass spectrum of Pentane
(Molecular weight = 72)
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16. High Resolution Molecular ion peak
• A unique molecular formula can often be derived accurate
mass measurement alone in High Resolution Mass
spectrometry (HRMS). So molecular ion in HRMS are called
high resolution molecular ion peak. In HRMS masses
obtained up to four or five decimal places.
• The unique molecular formula possible because the nuclide
masses are not integer.
• Exact masses of some nuclides
H 1.00783
C 12.0000
N 14.0031
O 15.9949
•
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17. Precise mass determination
• if we have molecular formula C6H12O then its HRMS molecular ion
mass is unique at 100.0888
• Mass = (number of Carbon X Mass of Carbon) + (number of hydrogen X
Mass of Hydrogen )+ (number of Oxygen X mass of Oxygen)
• Mass = 6 X Mass of Carbon + 12 X Mass of Hydrogen + 1 X mass of Oxygen
= 6 x 12.0000 + 12 x 1.00783 + 1 x 15. 9949
= 72.0000 + 12.09396 + 15.9949
=100.08886
• For molecular formula C5H8O2
• Mass = 5 x Mass of Carbon + 8 x Mass of Hydrogen + 2 x mass of Oxygen
= 5 x 12.0000 + 8 x 1.00783 + 2 x 15. 9949
= 60.0000 + 8.06264 + 31.9898
= 100.05244
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18. Significance / Use of the Molecular Ion (M)
• By using rule thirteen for molecular ion peak, one can determine the
molecular formula.
• From Molecular ion peak, it gives idea of isotopic element if present.
• It gives idea of odd nitrogen present in compound (nitrogen rule).
• The molecular ion along with other information from IR and NMR spectra
can allow the identity of an unknown to be determined.
• Now a days, Once the identity of the molecular ion has been determined
much can be learned about the compound. One extremely valuable piece
of information that can be determined from a high resolution mass
spectrometer is the molecular formula of an unknown analyte.
• If a molecular ion was identified to be at m/z 80 on an instrument with
unit resolution little could be determined about the molecular formula.
For example, some of the many possible molecular formulas include
C4H4N2 (80.0375), C5H4O(80.0262), and C6H8 (80.0626).
• A high resolution instrument measurement of this peak at 80.0372 ±
0.0005 would indicate that the empirical formula is C4H4N2.
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19. Base peak
• The most intense ion peak or tallest peak in mass
spectrum is assigned an abundance of 100, and it is
referred to as the base peak.
• The relative abundance of the tallest peak or
base peak ion is assigned a value of 100, and the
abundances of all other ions plotted in that mass spectr
um are normalised to that value.
• The base peak is not necessarily the same as the parent
molecular ion peak.
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20. Base peak in pentane
20
In the mass spectrum for pentane, the positive ion with m/z = 43 produces
the tallest spectrum. It is base of given spectrum.
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21. Mass Spectrum of ethyl Benzene
(Base peak = 91)
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22. Metastable peaks
22
• Metastable Ions: Fragment of a parent ion will give rise to a new ion (daughter)
plus either a neutral molecule or a radical.
• Fragment ion produced in the analyzer follows abnormal flight path on its way to
detector. This ions appear at an m/e ratio depend on its masses as well as the
masses of original ions from which it formed. Such ions are called Metastable ions.
• For example
• M1
+ M2
+ + non charged fragment
• An intermediate situation is possible; M1
+ may decompose to M2
+ while being
accelerated.
• The resultant daughter ion M2
+ will not be recorded at either M1 or M2, but at a
position M* as a rather broad, poorly focused peak. Such an ion is called a
metastable ion.
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23. • Metastable ions have lower kinetic energy than normal ions and
metastable peaks are smaller than the M1 and M2 peaks and also broader.
These metastable ions arise from fragmentation that takes place during
the flight down through ion rather than in the ionization chamber.
Molecular ions formed in the ionization
• Significance of Metastable ions:
• Metastable ions are useful in helping to establish fragments routes.
• Metastable ion peak can also be used to distinguish between
fragmentation Processes, which occur in few microseconds.
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24. Fragment ion peak
• Molecular ion peak in MS further undergo
fragmentation gives new daughter ion peaks
such peak produced in MS called as fragment
ion peaks.
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25. Negative ion peak
25
• Negative ions are formed from electron bombardment of
sample. These results due to the capture of electron by a
molecule during collision of molecules .
• Molecular ions observed in negative ion chemical ionization
mass spectra are usually M or [M-H]-.
• In the negative ion mode operation peaks corresponding to
deprotonated analyte molecules are observed.
• The formation of negative ions is very rare .
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26. *Nitrogen rule in Mass spectroscopy*
• When m/z for M has an odd mass (odd number
of amu), the corresponding molecular formula
has an odd number of nitrogen atoms (1, 3, 5,
etc.).
• When m/z for M has an even mass (even
number of amu), the corresponding molecular
formula has no / zero nitrogen or an even
number of nitrogen atoms (0, 2, 4, etc.).
• This is called nitrogen rule.
• Nitrogen rule is very useful for determining the
nitrogen content of an unknown compound.
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27. Example of Nitrogen rule
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28. Calculate Mass and tally with Nitrogen rule
28
NH2 NO2 NH2
mass
nitrogen
odd / even
odd / even odd / even odd / even
odd / even odd / even
H3C H3C
H2N
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29. Calculate Mass and nitrogen
29
NH2 NO2 NH2
NO2
mass
nitrogen
odd / even
odd / even odd / even odd / even
odd / even odd / even
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30. *Rule of 13*
• The Rule of 13 is a simple procedure for tabulating possible
chemical formula for a given molecular mass.
• The first step in applying the rule is to assume that only carbon and
hydrogen are present in the molecule and that the molecule
comprises some number of CH "units" each of which has a nominal
mass of 13.
• If the molecular weight of the molecule in question is M, the
number of possible CH units is n and
• where r is the remainder.
• The base formula for the molecule is CnHn+r
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31. 31AISC/SBS/MSc-I
Q2 ) for a m+ = 106,
n=8 (106/13) with r of 2.
A possible molecular formula for this ion C8H8+2 = C8H10

32. Other way Rule of 13
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33. The nominal mass of a substance is 140.
What is its molecular formula?
• By using The rule of 13:
• Divide the nominal mass by 13:
• A hydrocarbon with this molecular weight
would have 10 C atoms
• Multiply the remainder by 13: 0.769*13 = 9.997 or 10;
• A hydrocarbon with this molecular weight
would have (n+r) (10+10) or 20 hydrogens
• ‘CH’Molecular formula will be C10H20
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34. By using rule of 13, molecular formula
determination with other element.
• From parent hydrocarbon which is derived
from rule of 13, other elements (like O, N) can
be added to that hydrocarbon.
• For addition of oxygen / nitrogen we have to
remove same mass of oxygen or nitrogen from
hydrocarbon formula.
• For addition of ‘O’ remove CH4
• For addition of ‘N’ remove CH2
• For addition of ‘C’ remove 12 H
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35. Q) A carboxylic acid of molecular ion peak (m/z) at 122. find its
molecular formula with the help of rule of 13.
• M = 122
• Therefore become
:. CH Molecular formula CnHn+r = C9H14
Carboxylic acid contain at least 2oxygen to form –COOH group
C9H14
+O -CH4
C8H10O
+O -CH4
C7H6O2
35
:. Molecular formula for give mass 122 is C7H6O2
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36. Q) A hydrocarbon compound with mass 128 and 8 hydrogen.
find its molecular formula with the help of rule of 13
36
• M = 128
• Therefore become
:. CH Molecular formula CnHn+r = C9H20
C9H20
remove12H add C
C10H8
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37. Q) A hydrocarbon compound with mass 109 and PMR shows
7 hydrogen. find its molecular formula with the help of rule
of 13
37
• M = 109
• Therefore become
:. CH Molecular formula CnHn+r = C8H13
mass is odd there it contain odd no. of Nitrogen (Nitrogen Rule)
C8H13
+N -CH2
C8H11N
+O -CH4
C7H7ON
:. Molecular formula for give mass 109 is C7H7ON
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38. Genesis of m/z fragments:
Alkanes (cyclic and acyclic), alcohols, amines
38AISC/SBS/MSc-I
• Genesis means mode of formation of something.
Here in MS, it is used for formation of fragments
from molecular ions. Fragments which having
positive charge (cations) which detected in MS
and shows peak in mass spectrum.
• In this we have to show how fragment for a given
value of m/z is formed from parent ion or by
daughter ions.
• Some compounds follows fragmentation rule for
particular function groups to gives new
fragments.

39. Formation of molecular ion M+
and
Fragmentation of the molecular ion
AISC/SBS/MSc-I 39
• When a any molecule / sample is bombarded with electrons, the
molecular ion is produced by loss of single electron with same mass of its
molecular weight. Molecular ion is also denoted by a radical cation as
below
A B
e
A B
M o le c u la r io n
M
2 e
• These Molecular ion or a radical cation undergo fragmentation to break a
bond between A&B to form a radical and a cation as given below.
• Only cations or fragment with positive charge is detected in MS.
Molecular ion
a radical cation
A B
A + B
A + B
radical
radical cation
cation

40. Fragmentation of Alkanes
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• Fragmentation of straight chain alkane is observed by
breaking of any C-C bond from molecular ion.
• Alkane gives following m/z fragments due loose of
group
M-15 (due to loose of -CH3 methyl group)
M-29 (due to loose of –CH2CH3 ethyl group)
M-43 (due to loose of -CH2CH2CH3 propyl group)
• Lose of largest substituent is more fevered due to
stability .
• Stability of cation :
• methyl < primary < secondary < tertiary

41. Genesis of butane
Mass (m/z): 15, 29, 43(100%), 58.
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CH3CH2CH2 CH3
m/z = 58
CH3CH2CH2 CH3+
m/z = 43
CH3CH2CH2 CH3
m/z = 58
CH3CH3CH2 CH3+
m/z = 43
CH3CH2 CH2CH3
CH3CH2 + CH2CH3
m/z = 58 m/z = 29
CH3CH2CH2CH3
molecular ion
m/z = 58
CH3CH2CH2CH3
Molecular Weight = 58
i)
ii)
iii)
iv)
C4H10

42. Q.2) Give the genesis for pentane
Mass (m/z): 15, 29, 43(100%), 57, 72
42AISC/SBS/MSc-I
Molecular ion
m/z = 72
CH3CH2CH2CH2CH3
e
CH3CH2CH2CH2CH3
MW =72
Answer

43. Another way it is also written like this
AISC/SBS/MSc-I 43

44. Genesis for Hexane
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45. Fragmentation of Branched alkane
• Genesis of 2-methyl pentane:
• m/z = 15, 29, 43 (100%),71, 86
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46. Q) Show the genesis for following branched alkanes.
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47. Fragmentation of Cycloalkanes
• Loss of side chains if present, by  cleavage
• Loss of ethylene or double bond fragment
from ring
• Cleavage takes place by homolytic fasion
(show single headed arrow mechanism)
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48. Fragmentation in cyclopentane
AISC/SBS/MSc-I 48
e- +
MW = M/Z =
molecular ion
70 70
+
M/Z = 70
+
+
CH2
CH2
m/z =42
28C3H6
C5H10
C5H10

49. Fragmentation in cyclohexane
AISC/SBS/MSc-I 49
e- +
MW =
M/Z =
molecular ion
84
84
+
M/Z = 84
+ +
CH2
CH2
m/z =56
28C4H8
C6H12
+ +
m/z =42
C3H6
i)
ii)
iii)
C6H12

50. Genesis of 1-metyl cyclopentane
AISC/SBS/MSc-I 50
e-
+
MW = M/Z =
molecular ion
84 84
+
+ + CH2
CH2
m/z =56
28C4H8
C6H12
CH3 CH3
+
M/Z = 84
C6H12
+
CH3
m/z = 69
M/Z = 84
C6H12
i)
ii)
iii)

51. Genesis for
1-methyl
cyclohexane
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e- +
MW =
M/Z =
molecular ion98
98
+
M/Z = 98
+ +
CH2
CH2
m/z =70
28C5H10
C6H12
+
+m/z =56
i)
ii)
iii)
+
M/Z = 98
+
CH3
m/z = 83
iv)

52. Fragmentation of Alcohols
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• The molecular ion of alcohols is usually small and
sometimes undetectable especially in tertiary alcohols.
• The identification of the molecular ion is complicated
by the prevalence of a M-1 peak caused by the loss of a
single hydrogen from the α carbon in primary and
secondary alcohols.
• Fragmentation pattern in alcohol
– -Bond Cleavage to oxygen
– loss of water molecule Dehydration
– Rearrangement and loss of water molecule
– -Bond Cleavage to oxygen giving m/z= 31

53. -cleavage in alcohol
• The most important fragmentation reaction in
alcohols is the loss of an alkyl group which are
.
• Example
or
AISC/SBS/MSc-I 53
e-
molecular ion
i)
ii)
OH
+
+
OH
+
OH OHa-cleavage

54. • In tertiary alcohol, there are chances of
braking each all three -bond to alcohol
• Alcohols also undergo dehydration to loss
water molecule either 1,2 elimination oe 1,4
elimination resulting (M-18) peak.
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1,2 elimination
1,4 elimination

55. • Alcohols also frequently undergo the rearrangement resulting in a
M-18 peak from the loss of water. This peak is most easily visible in
primary alcohols but can be found in secondary and tertiary
alcohols as well. Primary alcohols also can lose both water and an
alkene.
• Alcohol containing four or more carbon may undergo such
rearrangement simultaneous loss water and ethylene as
AISC/SBS/MSc-I 55
Or

56. • Alcohols also frequently cleave to give resonance
stabilized cations due to the breaking of the β
bond. As a result of this cleavage, alcohols show a
prominent peak at m/z 31
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57. Genesis of 1-butanol
mass (m/z) = 31, 43, 73, 74.
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58. AISC/SBS/MSc-I 58
• Q1. Give Genesis of 1-pentanol
mass (m/z) = 31, 42 (100%), 55, 70, 87,88.
• Q2. Give Genesis of 2-pentanol
mass (m/z) = 42, 45 (100%), 55, 73, 87,88.
• Q3. Give Genesis of 2methyl-2-butanol
mass (m/z) = 31, 43, 55, 59 (100%),70, 73, 88.

59. Q4) Give the genesis for following base
peak fragments for their molecular ion.
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60. Cyclic alcohol may undergo fragmentation by at least
three different pathway
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61. Fragmentation of Amines
61AISC/SBS/MSc-I
The fragmentation reaction in amines are similar to
alcohol
• -cleavage in amines:

62. • Primary amine that are not branched at the
carbon next to nitrogen, the most intense
peak occurs at m/e = 30. it arises from
cleavage.
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63. Genesis of dipropyl amine
m/z = 30, 58, 72, 100 ,101
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64. Questions to solve
64AISC/SBS/MSc-I
• Q1. Give Genesis for Ethyl amine
mass (m/z) = 30 (100%) , 44, 45.
• Q2. Give Genesis for diethyl amine
mass (m/z) = 30, 58 (100%) , 72, 73.
• Q1. Give Genesis for triethyl amine
mass (m/z) = 30, 58, 86 (100%) , 100, 101.

65. Applications of Mass Spectrometry
• Mass Spectrometry as a technique can be coupled with
other techniques such as HPLC and GC.
• As it is used in the identification of compounds it is
used in all areas of science.
• Some of its uses are:
• Trace Gas Analysis
• Pharmaceutical Industry
• Space Exploration
• Forensic Toxicology
• Archaeological Dating.
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66. Question from question papers
66AISC/SBS/MSc-I
i)
ii)
iii)
iv)

67. AISC/SBS/MSc-I 67
i)
iii)

68. AISC/SBS/MSc-I 68

69. Calculation of intensity of isotopic (M+1) peak
• M+1 peak possible due to isotopes of C, H, and N.
M+1 peak intensity can be calculated by number of
atoms present with their isotopic abundance
• The formula for calculating the intensity of the
M+1 peak is as follows
• % (M+1) = 1.1 X number of C atoms +
0.016 X number of H atoms +
0.38 X number of N atoms +…
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70. What will be the M+1 intensity for
propene C3H6
• Propene C3H6
• % (M+1) = 1.1 X number of C atoms + 0.016 X number of H atoms +
0.38 X number of N atoms +…
% (M+1) = 1.1 x 3 + 0.016 x 6
= 3.3 + 0.096
= 3.396 %
Calculate intensity of M+1 peak for Aniline
Aniline C6H7N
% (M+1) = 1.1 X number of C atoms + 0.016 X number of H atoms +
0.38 X number of N atoms +…
% (M+1) = 1.1 x 6 + 0.016 x 7 + 0.38 x 1
= 6.6 + 0.112 + 0.38
= 7.1 %
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