mass spectrometry

1. MASS SPECTROMETRY (MS)
Dr. MishuSingh
Department of Chemistry
Maharana Paratap Govt. P.G College
Hardoi.

2. Mass Spectrometry
A technique for measuring and analyzing molecules,
that involves bombarment of neutral molecules in vaporised
state with high energy electron beam causing ionization. The
resulting primary ions and their fragments are then analyzed,
based on their mass/ charge ratios, to produce a "molecular
fingerprint‘‘

3. Mass Spectrometer
A mass spectrometer is designed to do three things
– Convert neutral atoms or molecules into a beam of
positive (or rarely negative) ions.
– Separate the ions on the basis of their mass-to-charge
(m/z) ratio.
– Measure the relative abundance of each ion.

4. What information can be determined?
• Molecular weight
• Molecular formula (HRMS)
• Structure (from fragmentation fingerprint)
• Isotopic incorporation / distribution
• Protein sequence

5. Pharmaceutical analysis
• Bioavailability studies
• Drug metabolism studies, pharmacokinetics
• Characterization of potential drugs
• Drug degradation product analysis
• Screening of drug candidates
• Identifying drug targets
Bio-molecular characterization
• Proteins and peptides
• Oligonucleotides
Environmental analysis
• Pesticides on foods
• Soil and groundwater contamination
Forensic analysis/clinical
Applications of Mass Spectrometry

6. THEORY AND PRINCIPLE
Atom or molecule is hit by high-energy electron from an electron
beam at 10ev forming a positively charged, odd-electron species,
called the molecular ion
e–beam
e–
+
•

7. Molecular ion passes between poles of a
magnet and is deflected by magnetic field
amount of
deflection depends
on m/z
highest m/z
deflected least
Lowest m/z
deflected most
+
•
 If the only ion that is present is the molecular ion, mass
spectrometry provides a way to measure the molecular weight of
a compound and is often used for this purpose.
 However, the molecular ion often fragments to a mixture of
species of lower m/z.

8. The molecular ion dissociates to a cation and a radical.
+
•
+
•
Usually several fragmentation pathways are available and a
mixture of ions is produced.

9. mixture of ions of
different mass
gives separate peak
for each m/z
+
+
+
+
+
+
intensity of peak
proportional to
percentage of each
ion of different
mass in mixture
separation of peaks
depends on relative
mass

10. mixture of ions of
different mass
gives separate peak
for each m/z
+ + + +
+ +
intensity of peak
proportional to
percentage of each
ion of different
mass in mixture
separation of peaks
depends on relative
mass

11. mixture of ions of
different mass
gives separate peak
for each m/z
+ + + +
+ +
intensity of peak
proportional to
percentage of each
ion of different
mass in mixture
separation of peaks
depends on relative
mass

12. Fragmentation of M
• Fragmentation of a molecular ion, M, produces a radical and a cation.
– Only the cation is detected by MS.
A-B
A
A
B
B
•
+
Molecular ion
(a radical cation)
+•
•+
+
+
Cation Radical
Radical Cation
(not detected by MS)
m/z = 29
molecular ion (M
+
) m/z = 30
+ C
H
H
H
+ H
HH C
H
H
C
H
H
H C
H
H
C
H
H
H C
H
H
+ e
-
H C
H
H
C
H
H
H

13. Fragmentation Pattern
• A great deal of the chemistry of ion fragmentation can be understood
in terms of the formation and relative stabilities of carbocations in
solution.
– Where fragmentation occurs to form new cations, the mode that
gives the most stable cation is favored.
– The probability of fragmentation to form new carbocations
increases in the order.
CH3
+
1°
2°
1° allylic
1° benzylic
< < <
3°
2° allylic
2° benzylic
< 3° allylic
3° benzylic

14. • Base ion: The most abundant peak. Assigned an arbitrary intensity of 100. The
relative abundance of all other ions is reported as a % of abundance of the base
peak. Tallest peak is base peak (100%)
• Molecular ion (M): A radical cation formed by removal of a single
electron from a parent molecule in a mass spectrometer = MW. Peak that
corresponds to the unfragmented radical cation is parent peak or molecular
ion (M+)
The molecular ion is usually the highest mass in the spectrum
– Some exceptions w/specific isotopes
– Some molecular ion peaks are absent.
We write the molecular formula of the parent molecule in brackets with:
-A plus sign to show that it is a cation.
-A dot to show that it has an odd number of electrons.
TYPES OF IONS

15. base peak
M+
The mass spectrum of ethanol
Mass spectrum: A plot of the relative abundance of ions
versus their mass-to-charge ratio (m/z).

16. • Plot mass of ions (m/z) (x-axis) versus the intensity of the signal (roughly
corresponding to the number of ions) (y-axis)
• Tallest peak is base peak (100%)
– Other peaks listed as the % of that peak
• Peak that corresponds to the unfragmented radical cation is parent peak
or molecular ion (M+)
The Mass Spectrum
Propane
MW = M+ = 44 C
H
H
H
C
H
H
C
H
H
H

17. AN INSTRUMENT THAT GENERATES IONS FROM MOLECULES
AND MEASURES THEIR MASSES
THE ESSENTIAL COMPONENTS OF A MASS SPECTROMETER:
SAMPLE
INLET
ION
SOURCE
ION
ACCELERATOR
ION
ANALYSER
ION
DETECTOR
signal
COMPUTERMASS SPECTRUM
DATABASE
0
50
100
0 10 20 30 40 50 60 70 80
2
15
27
41
53
69
84
1-Butene, 3,3-dimethy l-
MASS SPECTROMETER

18. Fig. 13.39

19. 1. Electron Ionization (EI)
most common ionization technique, limited to relatively low MW compounds (<600
amu)
2. Chemical Ionization (CI)
ionization with very little fragmentation, still for low MW compounds (<800 amu)
3. Desorption Ionization (DI)
for higher MW or very labile compounds
4. Spray ionization (SI)
for biomolecules, etc.
Ionization Methods: Neutral species  Charged species

20. Electron impact (EI)
- vapor of sample is bombarded with electrons typically 70 eV
:
M + e 2e + M.+ fragments
Advantages
• inexpensive, versatile and reproducible
• fragmentation gives structural information
• large databases if EI spectra exist and are searchable
Disadvantages
• fragmentation at expense of molecular ion
• sample must be relatively volatile

21.  Chemical Ionization (CI)
Vaporized sample reacts with pre-ionized reagent gas via proton transfer.
– Common CI reagents:
methane, ammonia, isobutane, hydrogen, methanol
• sample M collides with reagent ions present in excess e.g.
45 CHMHCHM  
The following may occur if analyte is a saturated HC
4252 HCMHHCM  
245 HCHH)-(MCHM  
6252 HCH)-(MHCM  
29)(Mm/zwith)HC(MHCM 5252  

22. • Resolution: A measure of how well a mass spectrometer separates
ions of different mass.
– low resolution: Refers to instruments capable of separating only
ions that differ in nominal mass; that is ions that differ by at least 1
or more atomic mass units.
– high resolution: Refers to instruments capable of separating ions
that differ in mass by as little as 0.0001 atomic mass unit.
Resolution
C3H8 C2H4O CO2 CN2H4
44.06260 44.02720 43.98983 44.03740
• A molecule with mass of 44 could be C3H8, C2H4O, CO2, or CN2H4.
• If a more exact mass is 44.029, pick the correct structure from the
table:

23. – C3H6O and C3H8O have nominal masses of 58 and 60, and can be
distinguished by low-resolution MS.
– C3H8O and C2H4O2 both have nominal masses of 60.
– Distinguish between them by high-resolution MS.
C2 H4 O2
C3 H8 O
60.02112
60.05754
60
60
Molecular
Formula
Nominal
Mass
Precise
Mass
Resolution
– High resolution MS can replace elemental analysis for chemical
formula confirmation

24. Relative Isotope Abundance of Common Elements:
Element Isotope
Relative
Abundanc
e
Isotope
Relative
Abundanc
e
Isotope
Relative
Abundanc
e
Carbon 12C 100 13C 1.11
Hydrogen 1H 100 2H 0.016
Nitrogen 14N 100 15N 0.38
Oxygen 16O 100 17O 0.04 18O 0.20
Sulfur 32S 100 33S 0.78 34S 4.40
Chlorine 35Cl 100 37Cl 32.5
Bromine 79Br 100 81Br 98.0

25. • Atomic mass of Carbon
– 12.000 amu for 12C but 13.3355 for 13C
• Atomic mass of Chlorine
– 34.9688 amu for 35Cl and 36.9659 for 37Cl
• Atomic mass of Hydrogen
– 1.00794 amu for H and 2.0141 for D!

Most elements have more than one stable isotope.
– For example, most carbon atoms have a mass of 12amu, but in nature,
1.1% of C atoms have an extra neutron, making their mass 13 amu.

26. Easily Recognized Elements in MS
• Most elements occur naturally as a mixture of isotopes.
– The presence of significant amounts of heavier isotopes leads to
small peaks that have masses that are higher than the parent ion
peak.
• M+1 = a peak that is one mass unit higher than M+
• M+2 = a peak that is two mass units higher than M+
• M+4 =a peak that is four mass units higher than M+
The elements like, Cl; Br, S,I and N can be easily deducted from MS.

27. Molecular Ions and Interpreting a mass spectrum
• The only elements to give significant M + 2 peaks are Cl and Br.
• If no large M + 2 peak is present, these elements are absent.
• Is the mass of the molecular ion odd or even?
• Nitrogen Rule: If a compound has
• zero or an even number of nitrogen atoms, its molecular ion will
have an even m/z value.
• an odd number of nitrogen atoms, its molecular ion will have an
odd m/z value.

28. • The most common elements giving rise to significant M + 2 peaks are chlorine
and bromine.
– Chlorine in nature is 75.77% 35Cl and 24.23% 37Cl.
– A ratio of M to M + 2 of approximately 3:1 indicates the presence of a single
chlorine in a compound.
M+2 and M+1 Peaks
Cl
M+
M+2

29. – Bromine in nature is 50.7% 79Br and 49.3% 81Br.
– A ratio of M to M + 2 of approximately 1:1 indicates the presence of a
single bromine in a compound.
M And M+2 Peaks

30. • Sulfur is the only other element common to organic compounds
that gives a significant M + 2 peak.
– 32S = 95.02% and 34S = 4.21% ratio; M: M+2= 4:1
M+2 and M+1 Peaks
M+

31. Large gap
I+
M+
ICH2CN
• Iodine
– I+ at 127
– Large gap

32. • Nitrogen:
– Odd number of N = odd MW
– CH3CN
M
+
= 41

33. The presence of heavier isotopes tell us:
• Hydrocarbons contain 13C,1.1% 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+.

34. Molecular Formula as a Clue to Structure
It is difficult to assign an entire structure based only on the mass
spectra. However, the mass spectra gives the mass and
formula of the sample which is very important information
Nitrogen rule: In general, “small” organic molecules with an
odd mass must have an odd number of nitrogens.
Organic molecules with an even mass have zero or an
even number of nitrogens
If the mass can be determined accurately enough, then the
molecular formula can be determined (high-resolution
mass spectrometry)
Information can be obtained from the molecular formula:
Degrees of unsaturation: the number of rings and/or
-bonds in a molecule (Index of Hydrogen
Deficiency)

35. Determination of molecular formula from M S
It requires the following steps
Step-1 find out the, Base peak; Molecular ion peak ( mol.wt.) from the MS
Step-2 find out the intensities of M+, M+1, M+2 peaks.
Step-3 if, molecular ion peak is not base peak, then recalculate the intensities of
M+1 and M+2 peaks relative to base peak
taken as 100%.
Relative intens. of [M+1] = [M+1] / [M] x 100
Relative intens .of [M+2] = [M+2] / [M] x 100
Step-4 find out the number of C-atoms
No of C-atoms = [M+1] / 1.1
No of Cl-atoms = [M+2 ] , [M+4 ] [M+6 ] etc.
No of S-aton = [M+2] / 4.4
Step-5 deduce the contribution of C, Cl or S atoms from the mole.Wt.
Step-6 deduce the contribution of N-atom ( apply N– rule ).
Step-7 find out then number of Hatoms.

36. Examples
1.Find out the molecular composition of an organic compound from the
following mass spectral data.(4.5 &0.27)
m/z 27 28 29 41 43 44 72 73 74
% Intn 59 15 54 60 79 100 73 3.3 0.2
2.An organic compound give a molecular ion peak at m/z = 84 with
intensity 31.2 % . The intensities of [M+1] and [M+2] peaks are 2.06 and
0.08 respectivley, find out molecular composition.
Solution :M.W. = 84 (even); N=0,2,4 etc.
R.I. M+1 = (2.06/31.2) x 100 = 6.6
M+2 = (0.08/31.3) x 100 = 0.25
Molecular Formula C6H12

37. 3.Mass spectral data
m/z 78 79 80
% Intn 23.6 0.79 7.55
Find out molecular composition
(RI=3.3)
4. An organic compound showing [M+.] peak at m/z = 70 with relative
intensitiess of [M+1] and [M+2] peaks at 13.2 and 1.20 rspectively , find out
molecular composition.
M.W. = 170(even) N may be 0,2,4 etc.
No. of C atom =13.2/1.1=12, contribution of 12 carbon =144
Remaining part =170-144=26
No. of nitrogen atom =0(even)
No of Oxygen atom=1 (16)
No. of Hydrogen atom =26-16=10
molecular composition = (C12H10O)

38. Relative intensities for more than one Halogen atoms
p-dichlorobenzene (146,148,150)
The intensity can be calculated as
Halogen M M+2 M+4 M+6
Cl 100 33 - -
2 Cl 100 65 11 -
3 Cl 100 97.8 31.9 3.47
Since M:M+2 for one chlorine atom is 3:1, therefore the
Expresion (3a+b) n on expansion will give the intensity ratio
For one Cl atom (3a+b) 1 = 3a+b
= 3 : 1

39. For two Cl atoms, (3a+b) 2 = 3a2+6ab+b2
= 9: 6 : 1
For three Cl atoms, (3a+b) 3 = 27a3+ 27a2b+9ab2 +b3
= 27 : 27 : 9 : 1
For bromine atom Halogen M M+2 M+4 M+6
Br 100 98 - -
2 Br 100 195 95.4 -
3Br 100 293 286 94.4

40. Calculation: Br 79 and Br 81 have equal natural abundance (1:1)
Coefficient of the expression (a+b) n
For one Br atom (a+b) 1 = a + b
= 1 : 1
For two Br atom (a+b) 2 = a2+2ab+b2
= 1 : 2 : 1
For three Br atom (a+b) 3 = a3+3a2b+3ab2 +b3
= 1 : 3 : 3 : 1

41. .
Fragmentation Patterns
Alkanes
• - Strong molecular ion
– Fragmentation often splits off simple alkyl groups:
• Loss of methyl M+ - 15
• Loss of ethyl M+ - 29
• Loss of propyl M+ - 43
• Loss of butyl M+ - 57
– Branched alkanes tend to fragment forming the most stable
carbocations.

42. Alkanes
– Mass spectrum of octane.

43. 57=Loss of CH3CH2 (29)
71=Loss of CH3 (15)
Hexane (m/z = 86 for parent) has peaks at
m/z = 71, 57, 43, 29

44. More stable carbocations will be more abundant.
CH3CH2CH2CHCH3
CH3
m/z 86
+
•
CH3CH2CH2CHCH3
CH3
m/z 86
+
•
Facile
Facile
More difficult
CH3CH2CH2
CHCH3
CH3
+• +
m/z 43
CH3 CH3CH2CH2CH
CH3
CH3CH2CH2CH
CH3
m/z 71
+•
+
+• +
m/z 57
CH3CH2 CH2CHCH3
CH3

45.  Cycloalkanes
– Loss of side chain
–Loss of ethylene fragments
.
Ionization followed by fragmentation
Splitting out of ethylene.
Ejection of
electron
Radical/cation
fragmentation

46. – Mass spectrum of methylcyclopentane.
+ CH3
amu = 41

47.  Alkenes Fragmentation
– Fairly prominent M+
– Fragment ions of CnH2n+ and CnH2n-1+
– Terminal alkenes lose allyl cation if possible to form resonance-stabilized allylic
cations
R - R
[CH2=CHCH2 CH2 CH3 ] CH2 =CHCH2
+
+ • CH2 CH3
+
•

49.  Alkyne Fragmentation
– Molecular ion readily visible
– Terminal alkynes readily lose hydrogen atom
– Terminal alkynes lose propargyl cation if possible
HC C=CH2HC C-CH2
+ +3-Propynyl cation
(Propargyl cation)

50.  Retero Diel’s –Alder Fragmentation
 In Cycloalkenes
Retro Diels-Alder cleavage ( unsaturated six member ring under go retro Diels-
Alder fragmentation to produce radical cation of adiene and a neutral alkene.
Cyclohexenes give a 1,3-diene and an alkene, a process that is the reverse of a
Diels-Alder reaction.
+
+•
+•
+
+•
Observed!
Observed!

51. A radical cation
(m/z 68)
+
•
•
+
+
CH3
C
H3 C CH2
C
CH2H3C
CH3
Limonene
(m/z 136)
A neutral diene
(m/z 68)
EXAMPLES

52. 4-terpineol
(MW 154)
OH
OH
O
mz 86
+
+
mz 68
EI Mass Spectrum
+
Mass spectrum of 4-terpineol as a good example for Retro Diels Alder
fragmentation

53. Aromatic Hydrocarbon Fragmentation
– Molecular ion usually strong
- Alkylbenzenes will often form intense ions at m/z 91
– Tropylium ion
– 7-membered ring favored
- Tropylium ion can fragment by successive losses of acetylene
– 91  65  39
– Phenyl ions (C6H5)+• decompose the same way e.g.(77  51)

55. +
+
+

-H
-CO
-H2
m/z 108
m/z 107
m/z 79
[C6H7]+ m/z 77
[C 6 H5 ]+
CH2OH
OH
HH
H
H
H
H
H

+
7 memberd ring
Benzonium ion
Phenyl ion
m/z 91
+ 
+
+ 
H
H - CH3

CH3
CH3
CH3
Benzene Compounds
Tropylium ion

56. Tropylium ion
Mass Spectrum of n-Octylbenzene

57. Br
Bromine pattern
Tropylium ion
Mass Spectrum of benzyl bromide
CH
H
CH Br
H
Fragment at the benzylic
carbon, forming a resonance
stabilized benzylic carbocation
(which rearranges to the
tropylium ion)

58. NO2
77
77 M+ = 123
Aromatics may also have a peak at m/z = 77 for the benzene ring.

59. •
Molecular ion
(a radical cation)
A radical
• •
•
+
+O HC R
• •
• •
+
R'-C O H
A resonance-stabilized
oxonium ion
R
R"
R'
R"
R'-C=O-H
R"
+
– Fragment easily resulting in very small or missing parent ion peak
– May lose hydroxyl radical or water
• M+ - 17 or M+ - 18
–Commonly lose an alkyl group attached to the carbinol carbon forming an oxonium ion.
(largest R group lost as radical).
–1o alcohol usually has prominent peak at m/z = 31 corresponding to H2C=OH+
Molecular ion strength depends on substitution
primary alcohol weak M+
secondary alcohol VERY weak M+
tertiary alcohol M+ usually absent
 Alcohol Fragmentation

60. Elimination of water.
74 – 56 = 18 (water).
Elimination of propyl radical.
74 – 31 = 43 (C3H7)
MS of 1-butanol is an example demonstrating both processes.

61. Mass Spectrum of 3-methyl-1-butanol

62.  Nitrogen Rule: Amines with odd # of N’s have Odd M+
 Amines undergo -cleavage, dominates forming an iminium ion 1°, 2°, and
3°give intense peaks at.m/z=30;. m/z= 58 and.m/z =86 ,respectively
Iminium ion
Amines

63. CH3CH2 CH2 N
H
CH2 CH2CH2CH3 CH3CH2CH2N CH2
H
m/z =72
iminium ion
86
CH3CH2 CH2 N
H
CH2 CH2CH2CH3
72

64. Aliphatic 3°Amines

65. Aromatic Amines

66. Ethers
– a-cleavage forming oxonium ion
– Loss of alkyl group forming oxonium ion
– Loss of alkyl group forming a carbocation

67. MS of diethylether (CH3CH2OCH2CH3)
H O CH2
H O CHCH3
CH3CH2O CH2

68. Fragmentation of Aryl Ethers
O
CH3
O
C5H5
+
+ . +
-
.
CH3 - CO
+ .
O
CH3
+ H2CO
H-rearrangement and loss of neutral molecule
C-O bond cleavage

69. Examples of Aromatic Ether Cleavages

70. Carbonyl Compounds Fragmentation
Dominant fragmentation pathways are :
• α-cleavage
• β-cleavage
• McLafferty rearrangement

71. m/z 128
+•
-cleavage
m/z 43
+
•
+
+
m/z 113
CH3
O
O
•
O
+
 Aldehydes and Ketones
Characteristic fragmentation patterns are:
– Cleavage of a bond to the carbonyl group.
C O
R
R'
C OR'
C OR'
+ R
Acylium ion
Note that an a cleavage of an aldehyde could produce a peak at M – 1 by eliminating H
atom.
This is useful in distinguishing between aldehydes and ketones.

72. Aldehydes (RCHO)- Fragmentation may form acylium ion
Common fragments:
M+ - 1 for
R (i.e. RCHO - CHO)
RC O
C C C H
H
H
H
H
O
133
105
91
M+ = 134
M+ - 29 For

73. Example of an aliphatic aldehyde

74. Ketones
– Fragmentation leads to formation of acylium ion:
Loss of R forming
Loss of R’ forming
R'C O
RC O
RCR'
O
CH3CH2CH2C OCH3CH2CH2C O
CH3C O
M+
CH3CCH2CH2CH3
O

75. Example of an aromatic ketone

76. • A hydrogen on a carbon 4 atoms away from the carbonyl
oxygen is transferred
– The “1,5 shift” in carbonyl-containing ions is called the McLafferty
rearrangement
– Creates a distonic radical cation (charge and radical separate)
– 6-membered intermediate is sterically favorable
– Such rearrangements are common
• Once the rearrangement is complete, molecule can fragment by
any previously described mechanism
Mc-Lafferty Rearrangement

77. R= alkyl / aryl -ald / ketones
= COOH ,COOR,CONH2,COX etc.
•
+
m/z 58
McLafferty
rearrangement
Molecular ion
m/z 114
•
+
O
H
O
H
+

78. Mass Spec of 2-octanone displays both alpha cleavage and McLafferty
CH3CO+
resulting from
 cleavage.
CH3CH2CH2CH2CH2CH2CO+
resulting from  cleavage.

79.  Carboxylic Acids
Characteristic fragmentation patterns are:
–cleavage to give the ion [CO2H]+ of m/z 45.
– McLafferty rearrangement if gama H present.
-cleavage
• +
m/z 45
O= C-O- H
OH
O
Molecular ion
m/z 88
+
•+•+
+
McLafferty
rearrangement
m/z 60
O
H
OH
O
H
OH
Molecular ion
m/z 88
– Loss of water,
– Loss of 44 is the loss of CO2

80. Esters (RCO2R’)
– Common fragmentation patterns include:
• Loss of OR’ Loss of R’
• peak at M+ - OR’ peak at M+ - R’
C
O
O CH3
105
77
M+ = 136
77
105

81.  Halide Fragmentation
– Loss of halogen atom
– Elimination of HX
– a-cleavage

82. – A molecule with an odd number
of nitrogens has an odd
molecular weight.
– A molecule that contains only C,
H, and O or which has an even
number of nitrogens has an even
molecular weight.
NH2
93
138NH2O2N
183NH2O2N
NO2
The Nitrogen Rule

83. – For a molecular formula, Cc HhNnOoXx, the degree of unsaturation
can be calculated by:
index of hydrogen deficiency = ½ (2c + 2 - h - x + n)
Index of Hydrogen Deficiency
Degree of Unsaturation
– Index of hydrogen deficiency tells us
the sum of rings plus multiple bonds.
- Using catalytic hydrogenation, the number of
multiple bonds can be determined.

84. – Knowing that the molecular formula of a substance is C7H16 tells us
immediately that is an alkane because it corresponds to CnH2n+2
– C7H14 lacks two hydrogens of an alkane, therefore contains either a ring or a
double bond
 Molecular Formulas
index of hydrogen deficiency =
1
2
(molecular formula of alkane – molecular formula of compound)

85. Index of hydrogen deficiency
1
2
(molecular formula of alkane – molecular formula of compound)
C7H14
1
2
(C7H16 – C7H14)
=
=
1
2
(2) = 1=
Therefore, one ring or one double bond.
Example 1

86. C7H12
1
2
(C7H16 – C7H12)=
1
2
(4) = 2=
Therefore, two rings or one triple bond or
two double bonds, or one double bond + one ring.
Example 2

87. CH3(CH2)5CH2OH (1-heptanol, C7H16O) has same number of H atoms as
heptane
Index of hydrogen deficiency =
1
2
(C7H16 – C7H16O) = 0
No rings or double bonds
Oxygen has no effect

88. Index of hydrogen deficiency =
1
2
(C5H12 – C5H8O2) = 2
One ring plus one double bond
CH3CO
O
Cyclopropyl acetate

89. Treat a halogen as if it were hydrogen.
C C
CH3
ClH
H
C3H5Cl same index of hydrogen
deficiency as for C3H6
If halogen is present
– Index of hydrogen deficiency tells us the sum of rings plus multiple bonds.
-Using catalytic hydrogenation, the number of multiple bonds can be determined.
Rings versus Multiple Bonds

90. Nobel Prizes in Mass Spectrometry
1906- J.J. Thomson- m/z of electron
1911- W. Wien- anode rays have positive charge
1922- F. Aston- isotopes (first MS with velocity focusing)
1989- H. Dehmelt, W. Paul- quadrupole ion trap
1992- R.A. Marcus- RRKM theory of unimolecular dissociation
1996- Curl, Kroto, and Smalley- fullerenes (used MS)
2002- J. Fenn- electrospray ionization of biomolecules
K. Tanaka- laser desorption ionization of biomolecules

91. THANK YOU

92. • Example: The formula for a hydrocarbon with M+ =106 can
be found:
– Step 1: n = 106/13 = 8 (R = 2)
– Step 2: m = 8 + 2 = 10
– Formula: C8H10
• If a heteroatom is present,
– Subtract the mass of each heteroatom from the MW
– Calculate the formula for the corresponding hydrocarbon
– Add the heteroatoms to the formula

93. • The “Rule of Thirteen” can be used to identify
possible molecular formulas for an unknown
hydrocarbon, CnHm.
– Step 1: n = M+/13 (integer only, use remainder in
step 2)
– Step 2: m = n + remainder from step 1
Rule of Thirteen

94. 98
Degrees of unsaturation
saturated hydrocarbonCnH2n+2
cycloalkane (1 ring) CnH2n
alkene (1 -bond) CnH2n
alkyne (2 -bonds) CnH2n-2
For each ring or -bond, -2H from the formula of the saturated alkane
HH
H
H
H
H
H H
H
H
H
H
C6H14
- C6H12
H2
2 = 11
2
C6H14
- C6H6
H8
8 = 41
2
H
H
H
H
H
H
Hydrogen Deficiency
Degrees of Unsaturation

95. The “Nitrogen Rule”
•Molecules containing atoms limited to C,H,O,N,S,X,P
of even-numbered molecular weight contain either NO
nitrogen or an even number of N
•This is true as well for radicals as well.
• Not true for pre-charged, e.g. quats, (rule inverts) or
radical cations.
•In the case of Chemical Ionization, where [M+H]+ is
observed, need to subtract 1, then apply nitrogen rule.
•Example, if we know a compound is free of nitrogen and
gives an ion at m/z=201, then that peak cannot be the
molecular ion.

96. 100
C C
H3C
H3C
H
H
H
C
H
H
OH
= 3.65,
t, 2H
= 2.25,
br s, 1H
= 1.7,
m, 1H
= 1.5,
q, 2H
= 0.9,
d, 3H
CDCl3
61.2
41.7
22.6
Problems
1.

97. 101
Problem 2

98. 102
13C NMR:
C5H10O

99. 103
-CHOHCH3

100. 104
= 7.4-7.1
(m, 5H)
= 2.61
(sextet,
J=7, 1H)
= 1.60
(quintet,
J=7, 2H)
= 1.21
(d, 3H)
=0.91
(t, 3H)
C10H14
147.6
128.2
127.0
125.7
41.7
31.2
21.8
12.3

101. Problem

102. Problem
MASS

103. 13C
1H-NMR

104. PROBLEM
MASS

105. 13C
1HNMR

106. 110
Ques. In performing the following reaction to make 5-heptynoic acid. a side product -
H – is isolated Its 1H-NMR spectrum is gives the following :
1H-NMR Data of compound H are:
Triplet at 1.0 (3 H)
Quintet at 1.5 (2 H)
Singlet at 1.6 (3H)
Triplet at 2.2 (2H)