Mass spectrometry problems-1.

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Dr. Jyotsna S. Meshram
Professor and Head,
Department of Chemistry,
Rashtrasant Tukadoji Maharaj,
Nagpur University, Nagpur, India
Mass Spectrometry problems- 1
7th June 2020

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t
Outline
IHD
Rule of 13
Problems
Solutions

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Index of Hydrogen Deficiency (IHD)
For the analysis of the molecular formula of unknown organic molecule, IHD
plays a important role. It is a calculation that determines the presence of the
multiple bonds and the rings present in the molecule
It is also called as
 Degree of Unsaturation ( DU) or
Double Bond Equivalence (DBE) or
Unsaturated Index (UI) or
Double bond ring( DBR) calculations

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IHD= Rings + multiple bond
= 2x+2-y-z-h
2
Where
x– No. of carbon atom in the molecule
y- No. of hydrogen atom in the molecule
z- No. of nitrogen atom in the molecule
h- No. of halogen atom in the molecule

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Oxygen and other divalent atoms do not contribute
to IHD
Halogens, F, Cl , Br and I are treated as H atoms
Thus CH2Cl2 and CH3Cl has the same IHD as CH4
For each Nitrogen, add 1 to Carbon and 1 to the
Hydrogen
Thus CH2NH2 should be treated as C2H6
 For CxHy : IHD= 2x+2-y
 2
 = x+1-y/2
C6H6 : IHD= 6+1-6/2 = 4, One ring and 3 pi bonds

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S.No. Molecule IHD
1. Benzene, C6H6 4
2. Chlorobenzene, C6H5Cl 4
3. Cyclohexane, C6H12 1
4. Cyclohexene, C6H10 2
5. Acetophenone, C6H5COCH3 5
6. Chloroacetophenone, ClC6H4COCH3 5
7. Aniline, C6H5NH2 , Nitroaniline, C6H5NO2 4
Table -1,Indexof Htdrogendeficiency

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Rule of thirteen: BASE FORMULA
 Mass spectral interpretation using rule of thirteen
 Rule of thirteen gives the quick supplementary information for the base
formula if the molecular ion M+ is known
 Rule of thirteen, Carbon- 12 and Hydrogen- 1
 Base formula = CxHy
 Where y= x+r

 M = x + r
 13 13

 Where M = Molecular weight of the molecule
 r- Remainder

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Calculate the base formula using rule of
thirteen for the hydrocarbon, if the
molecular ion peak is-
i) 78 m/z
ii) 82 m/z
iii) 84 m/z

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m/z x = M/13 r y Base Formula
78 6 0 6 C6H6
82 6 4 10 C6H10
84 6 6 12 C6H12
Table -2 Base Formula

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Solution m/z Base Formula/ Structure IHD
i) 78 C6H6
Benzene
4
One ring
3 pi bonds
ii) 82 C6H10
Cyclohexene
2
One ring
1 pi bond
iii) 84 C6H12
Cyclohexane
1
One ring
Table -3 Base Formula,IHD

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 If Oxygen is present in the molecule, remove CH4
 If Nitrogen is present in the molecule, remove CH2
 M = x + r
 13 13
If the Molecular ion peak = 58,
x= 4, r= 4, y=10, Base formula = C4H10
If one oxygen is present remove CH4 Actual MF = C3H6O
If one nitrogen is present remove CH2 Actual MF = C3H9N

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M+ x r y Base MF Actual MF IHD
58 4 4 10 C4H10 C3H6O 1
78 6 0 6 C6H6 C6H6 4
82 6 4 10 C6H10 C4H2O2 1
93 7 2 9 C7H9 C6H7N 4
120 9 3 12 C9H12 C8H8O 5
122 9 5 14 C9H14 C7H6O2 5
123 9 6 15 C9H15 C6H5NO2 4
Table-4, BaseFormulaandIndexofHydrogendeficiency

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Problem-1
If the molecule has MF C10H12O and the mass
spectral data is-
m/z 15, 43,57,91,105,148
1.Index of Hydrogen Deficiency
2.Identify the base peak
3.Fragmentation pattern
4.Identify the molecule

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Solution-1
 For CxHy : IHD= 2x+2-y
 2
 = x+1-y/2
 C10H12O, IHD = 5
 15 m/z confirms presence of CH3
+ methyl group
 91 m/z indicates presence of benzyl cation, C6H5CH2
+ =
C7H7
+ a stable aromatic carbocation, Tropylium
carbocation
 Stevenson’s rule- Most stable carbocation- Base peak

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43 m/z and 57m/z , a difference of 14 shows the
presence of – CH2 group
105 m/z and 148 m/z , a difference of 43, which can be
(15 +28), suggests presence of carbonyl group next to
methyl group, Acetyl group 43 m/z, CH3CO-
 Loss of acetyl group from the molecular ion 148 m/z
suggests a peak at 105 m/z, C6H5CH2CH2
+
A peak at 57 suggests 14 + acetyl group ( 14 + 43) = 57,
- CH2 group next to acetyl group
 Combing all the above information , the molecule is-

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105 57
148 m/z
C6H5CH2-----CH2-----COCH3
91 14 43

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Problem-2
If IR spectrum gives a significant information about a strong peak at 1688
cm-1 and C- H stretching and bending at 2879 to 2979 cm-1
Identify-
1. Base formula
2. Index of Hydrogen Deficiency
3. Fragmentation pattern
4. Identify the molecule

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Solution -2
Solution
 Molecular ion peak- 134 m/z;
Rule of13, Molecular ion peak = [M] =134
 Molecular ion peak = x+ r
13 13
Base formula= CxHx+r
gives the base formula- C10H14

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 IR spectra gives the information regarding the presence of the
carbonyl group at 1688 cm-1 - presence of oxygen
 For the presence of each oxygen add O and subtract - CH4
 Now actual formula C10H14 - CH4 = C9H10O
 Index of the hydrogen deficiency IHD
 Presence of oxygen does not affect IHD
 For the molecule CxHy, IHD= 2x+2-y
2
For the unknown molecule C9H10O, IHD = 5

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 The base peak(100%) at 105 m/z and 77 m/z is attributed to benzoyl cation and phenyl cation
which further loses a molecule of acetylene to give C4H3
+ (51 m/z)
 IHD 5 suggests the presence of phenyl ring, 3 DB and 1 carbonyl group
 The MF C9H10O suggests-
 ethyl phenyl ketone, 1-Phenylpropan-1-one
Solution -2 contd-----
- C2H2
C4H3
+
m/z 51

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Problem-3
If IR spectrum gives a significant information about a strong peak at 1680 cm-1 and a very strong
broad peak extending from 2500 to 3500 cm-1 and a strong peak at 740 cm-1
1. Base formula
2. IHD
3. Fragmentation pattern
4. Assign the structure

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Rule of13, Molecular ion peak = [M] =136
 Molecular ion peak = x+ r
13 13
136/13 , Base formula= CxHx+r
gives the base formula- C10H16

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 IR spectra gives the information regarding the presence
of the carbonyl group at 1680 cm-1 - presence of oxygen
 IR spectra gives the information regarding the presence
of strong broad peak 2500-3500 cm-1 , this indicates
presence of -OH group in the molecule
 The molecule has carboxylic group, COOH group
 For the presence of each oxygen add O and subtract - CH4
 Now base formula C10H16 – 2(CH4 ) = C8H8O2
 Index of the hydrogen deficiency IHD
 Presence of oxygen does not affect IHD
 For the molecule CxHy, IHD= 2x+2-y
2
For the molecule C8H8O2 , IHD = 5

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 Stevenson’s rule- More stable carbocation (
resonance stabilized) is the most abundant
fragment
 The base peak 91(100%) confirms the presence
of-
 C6H5CH2
+ = C7H7
+
 Stable aromatic carbocation, Tropylium
carbocation

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 The MF C8H8O2 may be –Phenyl acetic acid 136 m/z
 Peak at 119 m/z indicates loss of –OH
 IHD = 5, one ring, 3 DB and one carbonyl ring
C3H3
+
m/z 91 m/z 91 m/z 65 m/z 39

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IHD = 5, one ring, 3 DB and one
carbonyl ring
Peak at 119 m/z indicates loss of –OH
The MF C8H8O2 suggests the isomers
136 m/z,
Phenyl acetic acid (136 m/z)
o-toluic acid
 p-toluic acid

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 IHD = 5, one ring, 3 DB and one carbonyl ring
 119 m/z is a strong peak, thus structure Phenyl acetic
acid is ruled out
 IR 740 cm-1 indicated ortho disubstituted benzene
 The structure p-toluic acid is also ruled out
 Thus the structure of the molecule is o- toluic acid
o- toluic acid p- toluic acid Phenyl acetic acid

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Problem-4
Identify the molecule

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From the above table from the MS ratio of two isotopes 3:1
m/z Ratio
111 and 113 3:1
139 and 141 3:1
154 and 156 3:1
Isotopes 35Cl 37Cl
Abundance % 100 32.5
Ratio 3 1
This indicates presence of Chlorine in unknown molecule

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IR spectrum shows a strong peak at 1710 cm-1 confirms the presence of
carbonyl group
Molecular ion peak is at 154 m/z
Now the possible structures can be –
Phenacylchloride 2-chloroacetophenone 4-chloroacetophenone

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M- 15 = 139 is the base peak, so the possibility of phenacyl chloride
is ruled out
Prominent MS peak at 111 m/z corresponds to ClC6H4
+
IR- strong peak at 850 cm-1 confirms p-chloroacetophenone

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References-
 Pavia, Lampman and Kriz,
Introduction to
Spectroscopy, Cengage
Learning
 William Kemp, Organic
Spectroscopy, Palgrave, 3rd
Edition
 Reg Davis, Martin Frearson,
Mass Spectrometry, John
Wiley and Sons
Thanks

Mass spectrometry problems-1.

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