Solution Manual For Spectrometric Identification of Organic Compounds, 8th Edition

Name: Solution Manual For Spectrometric Identification of Organic Compounds, 8th Edition
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Take the stress out of textbook problems with Solution Manual For Spectrometric Identification of Organic Compounds, 8th Edition, a complete guide to solving every question.

Jackson Roberts

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Answers for Student Exercises 1.1 to 1.5 1
Atom Type # of Atoms Mass Total Mass
Hydrogen 22 1.00783 22.17226
Carbon 10 12.00000 120.00000
Nitrogen 0 14.0031 0.00000
Oxygen 0 15.9949 0.00000
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 142.17226Exact Mass
a
3-ethyloctane
C 10 H 22
- (C 2 H 5 )
Rule 3
- (C 5 H 11 )
Rule 3
- (CH 2 =CH 2 )
Rule 9
H
or
- (C 4 H 9 )
Rule 5 and 6
- (CH 2 =CH 2 )
Rule 6
- (C 3 H 7 )
Rule 3
Atom Type # of Atoms Mass Total Mass
Hydrogen 18 1.00783 18.14094
Carbon 10 12.00000 120.00000
Nitrogen 0 14.0031 0.00000
Oxygen 0 15.9949 0.00000
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 2 138.14094Exact Mass
b
3-butylcyclohex-1-ene
C 10 H 18

Answers for Student Exercises 1.1 to 1.5 2
Atom Type # of Atoms Mass Total Mass
Hydrogen 22 1.00783 22.17226
Carbon 11 12.00000 132.00000
Nitrogen 0 14.0031 0.00000
Oxygen 0 15.9949 0.00000
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 1 154.17226Exact Mass
d
(E)-undec-3-ene
C 11 H 22
or
- (CH 3 )
Rule 5
- (C 6 H 13 )
Rule 5
H - (C 7 H 14 )
Rule 9
Atom Type # of Atoms Mass Total Mass
Hydrogen 18 1.00783 18.14094
Carbon 13 12.00000 156.00000
Nitrogen 0 14.0031 0.00000
Oxygen 1 15.9949 15.99490
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 5 190.13584Exact Mass
O
c
2-methyl-1-phenylhexan-1-one
C 13H 18O
O
- (C 6 H 13 )
Rule 8
O O
O - (C 6 H 5 )
Rule 8
O
O H
- (C 4 H 8 )
Rule 9
O
H
OH
O
- (CO)
Rule 8+9

Answers for Student Exercises 1.1 to 1.5 3
Atom Type # of Atoms Mass Total Mass
Hydrogen 18 1.00783 18.14094
Carbon 12 12.00000 144.00000
Nitrogen 0 14.0031 0.00000
Oxygen 0 15.9949 0.00000
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 4 162.14094Exact Mass
e
1-(hexan-2-yl)benzene
C 12H 18
- (CH 3 )
Rule 7
- (C 4 H 9 )
Rule 7
- (C 4 H 8 )
Rule 9
H
H H
Atom Type # of Atoms Mass Total Mass
Hydrogen 16 1.00783 16.12528
Carbon 7 12.00000 84.00000
Nitrogen 0 14.0031 0.00000
Oxygen 1 15.9949 15.99490
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 116.12018Exact Mass
OH
f
heptan-2-ol
C 7H 16 O
OH
OH
OH
- (C 5 H 11 )
Rule 8
- (CH 3 )
Rule 8
- (H )
Rule 8
OH
OH
OH
H
OH
H - (H 2 O)
Rule 9

Answers for Student Exercises 1.1 to 1.5 4
Atom Type # of Atoms Mass Total Mass
Hydrogen 22 1.00783 22.17226
Carbon 11 12.00000 132.00000
Nitrogen 0 14.0031 0.00000
Oxygen 2 15.9949 31.98980
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 1 186.16206Exact Mass
COOH
g
3,5,7-trimethyloctanoic acid
C 11H 22 O 2
OH
O - (OH )
Rule 8
- (C 10 H 21 )
Rule 8
- (C 9 H 18 )
Rule 9
C
O
C
O
OH
O
C OH
O
C OH
O
HO
O
H
HO
O
HO
OH
H
Atom Type # of Atoms Mass Total Mass
Hydrogen 14 1.00783 14.10962
Carbon 6 12.00000 72.00000
Nitrogen 0 14.0031 0.00000
Oxygen 1 15.9949 15.99490
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 102.10452Exact Mass
OH
h
hexan-1-ol
C 6 H 14O
OH
OH
- (CH 2 =CH 2 ) - (H 2 O)
Rule 9
- (C 5 H 11 )
Rule 8
- (H )
Rule 8
H 2 C OH
O
H
H
OH

Answers for Student Exercises 1.1 to 1.5 5
Atom Type # of Atoms Mass Total Mass
Hydrogen 18 1.00783 18.14094
Carbon 8 12.00000 96.00000
Nitrogen 0 14.0031 0.00000
Oxygen 1 15.9949 15.99490
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 130.13584Exact Mass
OH
i
2-methylheptan-2-ol
C 8 H 18 O
OH
- (C 5 H 11 )
Rule 8
- (CH 3 )
Rule 8
Rule 8
OH
OH OH
H O
H
- (H 2 O)
Rule 8
- (CH 2 =C(CH 3 ) 2 )
Atom Type # of Atoms Mass Total Mass
Hydrogen 16 1.00783 16.12528
Carbon 7 12.00000 84.00000
Nitrogen 0 14.0031 0.00000
Oxygen 0 15.9949 0.00000
Sulfur 1 31.9721 31.97210
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 132.09738Exact Mass
S
j
butyl(propyl)sulfane
C 7 H 16 S
S
- (C 3 H 7 )
Rule 8
- (C 2 H 5 )
Rule 9
- (CH 2 =CH 2 )
Rule 8
S S
SH
S
S
H

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Answers for Student Exercises 1.1 to 1.5 6
Atom Type # of Atoms Mass Total Mass
Hydrogen 20 1.00783 20.15660
Carbon 13 12.00000 156.00000
Nitrogen 0 14.0031 0.00000
Oxygen 1 15.9949 15.99490
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 4 192.15150Exact Mass
OH
k
(2-hexylphenyl)methanol
C 13 H 20 O
Atom Type # of Atoms Mass Total Mass
Hydrogen 9 1.00783 9.07047
Carbon 13 12.00000 156.00000
Nitrogen 1 14.0031 14.00310
Oxygen 3 15.9949 47.98470
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 10 227.05827Exact Mass
O
NO 2
l
(4-nitrophenyl)(phenyl)methanone
C 13H 9 NO 3
O
NO 2
- (C 6 H 4 NO 2 )
Rule 7&8
- (C 6 H 5 )
Rule 8&9
- (CO)
Rule 7&8
O
O
NO 2
O
O
NO 2
OH
OH
OH
H
CH 2
OH
CH 2
- (C 5 H 11 )
Rule 7
- (H 2 O)
Rule 7
- (OH )
Rule 9
OH OH
or

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Answers for Student Exercises 1.1 to 1.5 7
Atom Type # of Atoms Mass Total Mass
Hydrogen 22 1.00783 22.17226
Carbon 10 12.00000 120.00000
Nitrogen 0 14.0031 0.00000
Oxygen 1 15.9949 15.99490
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 158.16716Exact Mass
Om
3-(ethoxymethyl)heptane
C 10 H 22 O
O - (CH 3 )
Rule 8
- (C 4 H 9 )
Rule 8
- (C 7 H 15 )
Rule 3
O
O
O
CH 2
O
O
Atom Type # of Atoms Mass Total Mass
Hydrogen 19 1.00783 19.14877
Carbon 10 12.00000 120.00000
Nitrogen 0 14.0031 0.00000
Oxygen 1 15.9949 15.99490
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 1 78.9183 78.91830
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 1 234.06197Exact Mass
H
O
Br
n
2-bromodecanal
C 10H 19BrO
H
O
Br
- (H )
Rule 8
- (C 8 H 16 )
Rule 9
- (CO )
Rule 9
O
Br
H
O
Br
O
Br
Br
H H
O
Br
H H

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Answers for Student Exercises 1.1 to 1.5 8
O - (C 6 H 11 O )
Rule 8
- (CH 2 =C=O)
Rule 8
- (CH 3 )
Rule 9
O
O O
O O
O
O H
O O
H
Atom Type # of Atoms Mass Total Mass
Hydrogen 14 1.00783 14.10962
Carbon 8 12.00000 96.00000
Nitrogen 0 14.0031 0.00000
Oxygen 2 15.9949 31.98980
Sulfur 0 31.9721 0.00000
C n H m X xN yO z IHD Bromine 0 78.9183 0.00000
Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 2 142.09942Exact Mass
o
C 8H 14O 2
O
O
cyclohexyl acetate

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Answers for Student Exercises 1.6 to 1.11 9
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
71 114 99 / C6H11O 15 / CH3
86 / C5H10 28 / C2H4 R
71 / C4H7O 43 / C3H7
58 / C3H5O 56 / C4H8 R
43 / C3H7 71 / C4H7O
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
59 116 by CI 115 / C7H15O 1 / H
98 / C7H14 18 / H2O R
87 / C5H11O 29 / C2H5
69 / C5H9 18 / H2O, 29 / C2H5 R
59 / C3H7O 57 / C4H9
45 / C2H5O 71 / C5H11
43 / C3H7 73 / C4H9O
41 / C4H5 18 / H2O, 57 / C4H9 R
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
91 170 91 / C7H7 79 / Br
Tropylium
Ion
75 / C6H3 80 / HBr, 16 / CH4 R
65 / C5H5 79 / Br, 26 / C2H2 R
50 / C4H2 120 / C7H5Br
39 / C3H2 131 / C4H5Br
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
71 150 121 / C3H6Br 29 / C2H5
107 / C2H4Br 43 / C3H7
71 / C5H11 79 / Br
55 / C4H7 80 / HBr, 16 / CH4 R
43 / C3H7 C2H4Br
4-Heptanone
O
C7H 14O
Exact Mass: 114.1045
IHD = 1, match with (k)
A
OH
3-Heptanol
C 7H16 O
Exact Mass: 116.1201
IHD = 0, match with (m)
B
Br
4-Bromotoluene
C7 H7Br
Exact Mass: 169.9731
IHD = 4, match with (v)
C
Br
2-Bromopentane
C 5H11Br
Exact Mass: 150.0044
IHD = 0, match with (r)
D

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Answers for Student Exercises 1.6 to 1.11 10
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
43 100 85 / C5H9O 15 / CH3
71 / C4H7O 29 / C2H5
58 / C4H10 42 / C2H2O R
57 / C3H5O 43 / C3H7
43 / C3H7 57 / C3H5O
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
74 74 73 / C3H5O2 1 / H
57 / C3H5O 17 / OH
56 / C3H4O 18 / H2O R
45 / CHO2 29 / C2H5
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
73 73 56 / C4H8 17 / NH3 R
55 / C4H7 18 / NH4 R
44 / C2H6N 29 / C2H5
43 / C3H7 30 / CH4N
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
55 56 by CI 55 / C3H3O 1 / H
39 / C3H3 17 / OH
38 / C3H2 18 / H2O
O
2-Hexanone
C6H12O
Exact Mass: 100.0888
IHD = 1, match with (h)
E
O
OH
Propionic Acid
C 3H6O2
Exact Mass: 74.0368
IHD = 1, match with(c)
F
G
H 2N
C4H11N
Exact Mass: 73.0896
IHD = 0, match with (b)
butan-1-amine
OH
prop-2-yn-1-ol
C3H4O
Exact Mass: 56.0264
IHD = 2, match with (a)
H

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Answers for Student Exercises 1.6 to 1.11 11
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
43 89 by CI 90 / C3H8NO2 Addition of H
72 / C3H6NO 17 / OH R
43 / C3H7 46 / NO2
42 / C3H6 47 / HNO2 R
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
94 138 107 / C7H7O 31 / CH3O
94 / C6H5O 44 / C2H4O R
77 / C6H5 61 / C2H5O2
66 / C5H6 72 C3H5O2 R
51 / C4H3 87 / C4H7O2
39 / C3H2 99 / C5H8O2
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
94 122 94 / C6H5O 28 / C2H4 R
77 / C6H5 45 / C2H5O
66 / C5H6 56 / C3H4O
51 / C4H3 71 / C4H7O
39 / C3H2 83 / C5H8O
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
43 102 87 / C4H7O2 15 / CH3
74 / C3H5O2 28 / C2H4 R
71 / C4H7O 31 / CH3O
59 / C2H3O2 43 / C3H7
43 / C3H7 59 / C2H3O2
N +
O
O-
1-Nitropropane
C 3H7NO2
Exact Mass: 89.0477
IHD = 1, match with (d)
I
O
OH
2-Phenoxylethanol
J
C 8H10O2
Exact Mass: 138.0681
IHD = 4, match with (q)
Phenetole
K
C 8H10O
Exact Mass: 122.0732
IHD = 4, match with (o)
O
O
O
Methyl Butyrate
L
C 5H10O2
Exact Mass: 102.0681
IHD = 1, match with (i)

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Answers for Student Exercises 1.6 to 1.11 12
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
113 113 85 / C5H11N 28 / CO R
84 / C5H10N 29 / CHO
56 / C4H8 57 / C2H3NO
55 / C4H7 58 / C2H4NO
42 / C3H6 71 / C3H5NO R
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
94 94 67 / C4H5N 27 / CHN
53 / C3H3N 41 / C2H3N
42 / C2H4N 52 / C3H2N
41 / C2H3N 53 / C3H3N
40 / C2H2N 54 / C3H4N
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
152 152 135 / C8H7O2 17 / OH
122 / C7H6O2 30 / CH2O R
107 / C7H7O 45 / CHO2 R
105 / C7H5O 17 / OH, 30 / CH2O R
92 / C6H4O 60 / C2H4O2 R
77 / C6H5
17 / OH, 58 /
C2H2O2
63 / C5H3 89 / C3H5O3
39 / C3H2 113 / C5H6O3
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
75 or 157 157
141 /
C6H4ClNO 16 / O
127 / C6H4O 30 / NO R
111 / C6H4Cl 46 / NO2
99 / C5H4Cl 58 / CNO2
75 / C6H3 46 / NO2, 36 / HCl R
50 / C4H2 107 / C2H2ClNO2
O
HN
Caprolactam
M
C6H11NO
Exact Mass: 113.0841
IHD = 2, match with (j)
N
N
2-methylpyrazine
N
C5H6 N2
Exact Mass: 94.0531
IHD =4, match with (e)
HO
O
O
Methyl salicylate
O
C8H8O3
Exact Mass: 152.0473
IHD = 5, match with (s)
N +
O
- O
Cl
1-Chloro-4-Nitrobenzene
P
C 6H4ClNO2
Exact Mass: 156.9931
IHD = 5, match with (t)

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Answers for Student Exercises 1.6 to 1.11 13
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
60 208 by CI 191 / C7H12OBr 17 / OH
149 / C5H10Br 59 / C2H3O2
129 / C7H13O2 79 /Br
111 / C7H11O 79 /Br, 18 / H2O R
83 / C6H11 125 / CH2BrO2 R
60 / C2H4O2 147 / C5H9Br R
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
70 98 97 / C6H9O 1 / H
83 / C5H7O 15 / CH3
79 / C6H7 19 / H3O
70 / C4H6O 28 / C2H4
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
43 126 111 / C7H11O 15 / CH3
108 / C8H12 18 / H2O R
93 / C7H9 33 / CH5O R
83 / C6H11 43 / C2H3O
69 / C5H9 57 / C3H5O
58 / C3H6O 68 / C5H8 R
55 / C4H7 71 C4H7O
43 / C2H3O 83 / C6H11
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
60 116 by CI 87 / C4H7O2 29 / C2H5
73 / C3H5O2 43 / C3H7
60 / C2H4O2 56 / C4H8 R
45 / CHO2 71 / C5H11
O
HO Br
7-Bromo Heptanoic Acid
Q
C7 H13BrO2
Exact Mass: 208.0099
IHD = 1, match with (w)
HO
5-Hexyn-1-ol
R
C6H10O
Exact Mass: 98.0732
IHD = 2, mathc with (g)
O
6-Methyl-5-hepten-2-one
S
C8 H14O
Exact Mass: 126.1045
IHD = 2, match with (p)
O
HO
Hexanoic acid
T
C6H12O2
Exact Mass: 116.0837
IHD = 1, match with (l)

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Answers for Student Exercises 1.6 to 1.11 14
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
162 162 126 / C6H3OCl 36 / HCl R
98 / C5H3Cl 64 / CHOCl R
63 / C5H3 99 / CHOCl2
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
122 122 121/C8H9O 1 / H
107/C7H7O 15 / CH3
91 / C7H7 31 / CH3O
77 / C6H5 45 / C2H5O
65 / C5H5 57 / C3H5O
Base Peak
Molecular
Ion Fragement Ion Fragment Loss Rearrangment
68 96 68 / C4H4O 28 / C2H4 R
55 / C3H3O 41 / C3H5
53 / C4H5 43 / C2H3O
42 / C2H2O 54 / C4H6 R
Cl
ClHO
2,6-Dichlorophenol
U
C 6H4Cl2O
Exact Mass: 161.9639
IHD = 4, match with (u)
HO
2,6-Dimethylphenol
V
C8H10 O
Exact Mass: 122.0732
IHD = 4, match with (n)
O
2-Cyclohexen-1-one
W
C6 H8O
Exact Mass: 96.0575
IHD = 3, match with (f)

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CHAPTER 2
2.1
𝑣 = 1
2𝜋c √𝑘 ( 𝑚1𝑚2
𝑚1+𝑚2)
(k) force constant = 42c2 v2 meff
meff = ( 𝑚1𝑚2
𝑚1+𝑚2)
meff = (1H19F) = (1.0078 x 18.9984)
(1.0078 + 18.9984) = 0.9570 μ
meff = (1H35Cl) =0.9796 μ
meff = (1H81Br) =0.9954 μ
meff = (1H127I)=0.9999 μ
(k) (1H19F) = 4 x 3.142 x (2.997 x 1010 cm s-1)2 x (4148.2 cm-1)2 x (0.9570 x 1.66 x 10-27kg)
=968.3 kg s-2 =968.3 N/m
(k) (1H35Cl) = 514.6 N/m
(k) (1H81Br) = 410.9 N/m
(k) (1H127I) = 313.6 N/m
Assume that the force constant for halides are the same when substituting the deuterium -- the calculated
bond stretch frequency for halides:
𝑣 = 1
2𝜋c √𝑘 ( 𝑚1𝑚2
𝑚1+𝑚2)
2.2. The gross selection rule for infrared activity is that motion corresponding to a normal mode should be
accompanied by a change of dipole moment. So those molecules in which a vibration gives rise to a
change in dipole moment are infrared active compounds:
(a) CH3CH3, (b) CH4, (c) CH3Cl.
It is helpful to write down the structural formulas of the compounds
2.3. A nonlinear molecule has 3N-6 normal modes of vibration, where N is the number of atoms in the
molecule; a linear molecule has 3N-5.
(a) C6H6 has 3(12) – 6 = 30 normal modes
(b) C6H6CH3 has 3(16) – 6 = 42 normal modes
(c) HC≡C-C≡CH is linear; it has 3(6) – 5 = 13 normal modes
2.4. From top to bottom the: o-xylene, m-xylene, then p-xylene

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Solution Manual For Spectrometric Identification of Organic Compounds, 8th Edition

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