Mass spectrometry

mass spectrometry

1. MASS SPECTROMETRY (MS)
Dr. Mishu Singh
Department of Chemistry
MaharanaParatapGovt. P.G College
Hardoi.

2. Mass Spectrometry
Atechniqueformeasuringandanalyzingmolecules,
thatinvolvesbombarmentofneutralmoleculesinvaporised
statewithhighenergyelectronbeamcausingionization.The
resultingprimaryionsandtheirfragmentsarethenanalyzed,
basedontheirmass/chargeratios,toproducea"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?
•Molecularweight
•Molecularformula(HRMS)
•Structure(fromfragmentationfingerprint)
•Isotopicincorporation/distribution
•Proteinsequence

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 10evforming a positively charged, odd-electron species,
called the molecular ion
e
–
beam
e
–
+
•

7. Molecularionpassesbetweenpolesofa
magnetandisdeflectedbymagneticfield
amount of
deflection depends
on m/z
highest m/z
deflected least
Lowest m/z
deflected most
+
•
 Iftheonlyionthatispresentisthemolecularion,mass
spectrometryprovidesawaytomeasurethemolecularweightof
acompoundandisoftenusedforthispurpose.
 However,themolecularionoftenfragmentstoamixtureof
speciesoflowerm/z.

8. Themoleculariondissociatestoacationandaradical.
+
•
+
•
Usuallyseveralfragmentationpathwaysareavailableanda
mixtureofionsisproduced.

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
HHC
H
H
C
H
H
HC
H
H
C
H
H
HC
H
H
+ e
-
HC
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.CH
3
+
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 cationformed by removal of a single
electron from a parent molecule in a mass spectrometer = MW.Peak that
corresponds to the unfragmentedradical cationis parent peakor 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.
Wewritethemolecularformulaoftheparentmoleculeinbracketswith:
-Aplussigntoshowthatitisacation.
-Adottoshowthatithasanoddnumberofelectrons.
TYPES OF IONS

15. base peak
M
+
The mass spectrum of ethanol
Massspectrum:Aplotoftherelativeabundanceofions
versustheirmass-to-chargeratio(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 unfragmentedradical cationis parent peak
or molecular ion (M
+
)
The Mass Spectrum
Propane
MW = M
+
= 44C
H
H
H
C
H
H
C
H
H
H

17. ANINSTRUMENT THATGENERATES IONSFROMMOLECULES
ANDMEASURESTHEIRMASSES
THE ESSENTIAL COMPONENTS OF A MASS SPECTROMETER :
SAMPLE
INLET
ION
SOURCE
ION
ACCELERATOR
ION
ANALYSER
ION
DETECTOR
signal
COMPUTERMASS SPECTRUM
DATABASE 0
50
100
01020304050607080
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 electronstypically 70 eV
:
M + e 2e + M
.+
fragments
Advantages
•inexpensive, versatile and reproducible
•fragmentation gives structural information
•large databases if EIspectra exist and are searchable
Disadvantages
•fragmentation at expense of molecular ion
•sample must be relatively volatile

21. Chemical Ionization(CI)
Vaporizedsamplereactswithpre-ionizedreagentgasviaprotontransfer.
–CommonCIreagents:
methane,ammonia,isobutane,hydrogen,methanol
•sample Mcollides with reagent ions present in excess e.g. 45 CHMHCHM 

The following may occur if analyteis a saturated HC4252 HCMHHCM 
 245 HCHH)-(MCHM 
 6252 HCH)-(MHCM 
 29)(Mm/zwith)HC(MHCM
5252 


22. •Resolution:Ameasureofhowwellamassspectrometerseparates
ionsofdifferentmass.
–lowresolution:Referstoinstrumentscapableofseparatingonly
ionsthatdifferinnominalmass;thatisionsthatdifferbyatleast1
ormoreatomicmassunits.
–highresolution:Referstoinstrumentscapableofseparatingions
thatdifferinmassbyaslittleas0.0001atomicmassunit.
Resolution
C
3H
8 C
2H
4O CO
2 CN
2H
4
44.06260 44.02720 43.98983 44.03740
•Amoleculewithmassof44couldbeC
3H
8,C
2H
4O,CO
2,orCN
2H
4.
•Ifamoreexactmassis44.029,pickthecorrectstructurefromthe
table:

23. –C
3H
6Oand C
3H
8Ohave nominal masses of 58and 60,and can be
distinguished by low-resolution MS.
–C
3H
8Oand C
2H
4O
2both have nominal masses of 60.
–Distinguish between them by high-resolution MS.C
2
H
4
O
2
C3H8O
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
12
C 100
13
C 1.11
Hydrogen
1
H 100
2
H 0.016
Nitrogen
14
N 100
15
N 0.38
Oxygen
16
O 100
17
O 0.04
18
O 0.20
Sulfur
32
S 100
33
S 0.78
34
S 4.40
Chlorine
35
Cl 100
37
Cl 32.5
Bromine
79
Br 100
81
Br 98.0

25. •Atomic mass of Carbon
–12.000 amufor
12
C but 13.3355 for
13
C
•Atomic mass of Chlorine
–34.9688 amufor
35
Cland 36.9659 for
37
Cl
•Atomic mass of Hydrogen
–1.00794 amufor Hand 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 thanM
+
•M+2= a peak that is two mass units higher thanM
+
•M+4 =a peak that is four mass units higher than M
+
The elements like, Cl; Br, S,I and Ncan be easily deducted from MS.

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

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

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

30. •Sulfuristheonlyotherelementcommontoorganiccompounds
thatgivesasignificantM+2peak.
–
32
S=95.02%and
34
S=4.21%ratio;M:M+2=4:1
M+2 and M+1 Peaks
M
+

31. Large gap
I
+
M
+ICH
2CN
•Iodine
–I
+
at 127
–Large gap

32. •Nitrogen:
–Odd number of N = odd MW
–CH
3CNM
+
= 41

33. The presence of heavier isotopes tell us:
•Hydrocarbons contain 13C,1.1% will be a small M+1peak.
•If Br is present, M+2is equal to M
+
.
•If Clis present, M+2is 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. MolecularFormulaasaCluetoStructure
Itisdifficulttoassignanentirestructurebasedonlyonthemass
spectra.However,themassspectragivesthemassand
formulaofthesamplewhichisveryimportantinformation
Nitrogenrule:Ingeneral,“small”organicmoleculeswithan
oddmassmusthaveanoddnumberofnitrogens.
Organicmoleculeswithanevenmasshavezerooran
evennumberofnitrogens
Ifthemasscanbedeterminedaccuratelyenough,thenthe
molecularformulacanbedetermined(high-resolution
massspectrometry)
Informationcanbeobtainedfromthemolecularformula:
Degreesofunsaturation:thenumberofringsand/or
-bondsinamolecule(IndexofHydrogen
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-3if, 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-6deduce 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
% Intn59 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 2.06and
0.08respectivley, 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 C
6H
12

37. 3.Mass spectral data
m/z 7879 80
% Intn23.60.797.55
Find out molecular composition
(RI=3.3)
4.Anorganiccompoundshowing[M
+.
]peakatm/z=70withrelative
intensitiessof[M+1]and[M+2]peaksat13.2and1.20rspectively,findout
molecularcomposition.
M.W.=170(even) Nmaybe0,2,4etc.
No.ofCatom=13.2/1.1=12,contributionof12carbon=144
Remainingpart=170-144=26
No.ofnitrogenatom=0(even)
NoofOxygenatom=1(16)
No.ofHydrogenatom=26-16=10
molecularcomposition=(C
12H
10O)

38. Relative intensities for more than one Halogen atoms
p-dichlorobenzene (146,148,150)
The intensity can be calculated as
HalogenM M+2 M+4 M+6
Cl 100 33 - -
2 Cl 100 65 11 -
3 Cl 100 97.8 31.9 3.47
SinceM:M+2foronechlorineatomis3:1,thereforethe
Expresion(3a+b)
n
onexpansionwillgivetheintensityratio
ForoneClatom(3a+b)
1
=3a+b
=3:1

39. FortwoClatoms,(3a+b)
2
=3a
2
+6ab+b
2
=9:6:1
ForthreeClatoms,(3a+b)
3
=27a
3
+27a
2
b+9ab
2
+b
3
=27:27:9:1
ForbromineatomHalogenM 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 oneBr atom(a+b)
1
= a + b
= 1 : 1
For two Br atom(a+b)
2
= a
2
+2ab+b
2
= 1 : 2 : 1
For threeBr atom(a+b)
3
= a
3
+3a
2
b+3ab
2
+b
3
= 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 alkanestend to fragment forming the most stable
carbocations.

42. Alkanes
–Mass spectrum of octane.

43. 57=Loss of CH
3CH
2(29)
71=Loss of CH
3(15)
Hexane(m/z = 86 for parent) has peaks at
m/z = 71, 57, 43, 29

44. More stable carbocations will be more abundant.CH
3CH
2CH
2CHCH
3
CH
3 m/z86
+
•CH
3CH
2CH
2CHCH
3
CH
3
m/z86
+
•
Facile
Facile
More difficultCH
3CH
2CH
2
CHCH
3
CH
3
+
• +
m/z43
CH
3 CH
3CH
2CH
2CH
CH
3
CH
3CH
2CH
2CH
CH
3
m/z71
+•
+
+
• +
m/z57
CH
3
CH
2 CH
2
CHCH
3
CH
3

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 allylcationif possible to form resonance-stabilized allylic
cationsR - R [CH
2=CHCH
2CH
2CH
3] CH
2=CHCH
2
+
+ •CH
2CH
3
+
•

49.  AlkyneFragmentation
–Molecular ion readily visible
–Terminal alkynes readily lose hydrogen atom
–Terminal alkynes lose propargylcationif possibleHCC=CH
2
HCC-CH
2
+ +
3-Propynyl cation
(Propargyl cation)

50.  ReteroDiel’s–Alder Fragmentation
 InCycloalkenes
Retro Diels-Alder cleavage ( unsaturated six member ring under go retro Diels-
Alder fragmentation to produce radical cationof adieneand a neutral alkene.
Cyclohexenesgive a 1,3-dieneand an alkene, a process that is the reverse of a
Diels-Alder reaction.
+
+
•
+
•
+
+
•
Observed!
Observed!

51. A radical cation
(m/z 68)
+
•
•
+
+
CH
3
C
H
3C CH
2
C
CH
2H
3C
CH
3
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/z108
m/z107
m/z79
[C6H7]
+ m/z77
[C
6H5]
+
CH2OH
OH
HH
H
H
H
H
H

+
7 memberd ring
Benzoniumion
Phenyl ion
m/z91
+ 
+
+ 
H
H-CH
3

CH
3
CH
3
CH
3
Benzene Compounds
Tropylium ion

56. Tropylium ion
Mass Spectrum of n-Octylbenzene

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

58. NO
2
77 77
M
+
= 123
Aromatics may also have a peak at m/z = 77for the benzene ring.

59. •
Molecular ion
(a radical cation)
A radical
••
•
+
+OHC R
••
••
+
R'-COH
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
–Commonlyloseanalkylgroupattachedtothecarbinolcarbonforminganoxoniumion.
(largestRgrouplostasradical).
–1
o
alcoholusuallyhasprominentpeakatm/z=31correspondingtoH
2C=OH
+
Molecularionstrengthdependsonsubstitution
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 (C
3H
7)
MS of 1-butanolis 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 iminiumion1°, 2°, and
3°give intense peaks at.m/z=30;. m/z= 58and.m/z =86 ,respectively
Iminium ion
Amines

63. CH
3CH
2CH
2N
H
CH
2CH
2CH
2CH
3 CH
3CH
2CH
2NCH
2
H
m/z =72 iminium ion 86
CH
3CH
2CH
2N
H
CH
2CH
2CH
2CH
3
72

64. Aliphatic 3°Amines

65. Aromatic Amines

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

67. MS of diethylether(CH
3CH
2OCH
2CH
3)HOCH
2 HOCHCH
3 CH
3CH
2OCH
2

68. Fragmentation of Aryl EthersO
CH
3
O
C
5H
5
+
+ . +
-
.
CH
3 - CO + .
O
CH
3
+ H
2CO
H-rearrangement and loss of neutral molecule
C-O bond cleavage

69. Examples of Aromatic Ether Cleavages

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

71. m/z 128
+
•
-cleavage
m/z 43
+
•
+
+
m/z 113
CH
3
O
O
•
O
+  Aldehydesand Ketones
Characteristic fragmentation patterns are:
–Cleavage of a bond to the carbonyl group.CO
R
R'
COR'
COR'
+ R
Acylium ion
Note that an a cleavage of an aldehydecould produce a peak at M –1 by eliminating H
atom.
This is useful in distinguishing between aldehydesand ketones.

72. Aldehydes(RCHO)-Fragmentation may form acyliumion
Common fragments:
M
+
-1 for R (i.e. RCHO - CHO) RCO CCCH
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 acyliumion:
Loss of R forming
Loss of R’ formingR'CO RCO RCR'
O CH
3CH
2CH
2CO CH
3CH
2CH
2CO CH
3CO
M
+CH
3CCH
2CH
2CH
3
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 distonicradical cation(charge and radical separate)
–6-membered intermediate is stericallyfavorable
–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-octanonedisplays both alpha cleavage and McLafferty
CH
3CO
+
resulting from
cleavage.
CH
3CH
2CH
2CH
2CH
2CH
2CO
+
resulting from cleavage.

79.  Carboxylic Acids
Characteristic fragmentation patterns are:
–cleavage to give the ion [CO
2H]
+
of m/z45.
–McLaffertyrearrangement if gamaH 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 44is the loss of CO
2

80. Esters (RCO
2R’)
–Common fragmentation patterns include:
•Loss of OR’ Loss of R’
•peak at M
+
-OR’ peak at M
+
-R
’C
O
OCH
3
105
77
M
+
= 136
77
105

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

82. –Amoleculewithanoddnumber
ofnitrogenshasanodd
molecularweight.
–AmoleculethatcontainsonlyC,
H,andOorwhichhasaneven
numberofnitrogenshasaneven
molecularweight.
NH
2
93
138 NH
2
O
2
N
183
NH
2
O
2
N
NO
2
The Nitrogen Rule

83. –For a molecular formula, C
c H
hN
nO
oX
x, 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. –KnowingthatthemolecularformulaofasubstanceisC
7H
16tellsus
immediatelythatisanalkanebecauseitcorrespondstoC
nH
2n+2
–C
7H
14lackstwohydrogensofanalkane,thereforecontainseitheraringora
doublebond
 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)
C
7H
14
1
2
(C
7H
16–C
7H
14)
=
=
1
2
(2) = 1=
Therefore, one ring or one double bond.
Example 1

86. C
7H
12
1
2
(C
7H
16–C
7H
12)=
1
2
(4) = 2=
Therefore, two rings orone triple bond or
two double bonds, orone double bond + one ring.
Example 2

87. CH
3(CH
2)
5CH
2OH(1-heptanol,C
7H
16O)hassamenumberofHatomsas
heptane
Index of hydrogen deficiency =
1
2
(C
7H
16–C
7H
16O)
= 0
No rings or double bonds
Oxygen has no effect

88. Index of hydrogen deficiency =
1
2
(C
5H
12–C
5H
8O
2)= 2
One ring plus one double bond
CH
3CO
O
Cyclopropylacetate

89. Treat a halogenas if it were hydrogen.
CC
CH
3
ClH
H
C
3H
5Cl
same index of hydrogen
deficiency as for C
3H
6
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-quadrupoleion trap
1992-R.A. Marcus-RRKM theory of unimoleculardissociation
1996-Curl, Kroto, and Smalley-fullerenes (used MS)
2002-J. Fenn-electrosprayionization 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: C
8H
10
•If a heteroatom is present,
–Subtract the mass of each heteroatom from the MW
–Calculate the formula for the corresponding hydrocarbon
–Add the heteroatomsto the formula

93. •The “Rule of Thirteen” can be used to identify
possible molecular formulas for an unknown
hydrocarbon, C
nH
m.
–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 hydrocarbonC
nH
2n+2
cycloalkane (1 ring) C
nH
2n
alkene (1 -bond) C
nH
2n
alkyne (2 -bonds) C
nH
2n-2
For each ring or -bond, -2H from the formula of the saturated alkaneHH
H
H
H
H
HH
H
H
H
H
C
6H
14
-C
6H
12
H
2
2 = 1
1
2
C
6H
14
-C
6H
6
H
8
8 = 4
1
2H
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. 100CC
H
3C
H
3C
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
CDCl
3
61.2
41.7
22.6
Problems
1.

97. 101
Problem 2

98. 102
13
C NMR:
C
5H
10O

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)
C
10H
14
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.Inperformingthefollowingreactiontomake5-heptynoicacid.asideproduct-
H–isisolatedIts1H-NMRspectrumisgivesthefollowing:
1H-NMR Data of compound Hare:
Tripletat 1.0 (3 H)
Quintetat 1.5 (2 H)
Singletat 1.6 (3H)
Tripletat 2.2 (2H)