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

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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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 142.17226 Exact 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 2 138.14094 Exact Mass b 3-butylcyclohex-1-ene C 10 H 18

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    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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 1 154.17226 Exact 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 5 190.13584 Exact Mass O c 2-methyl-1-phenylhexan-1-one C 13 H 18 O 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

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    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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 4 162.14094 Exact Mass e 1-(hexan-2-yl)benzene C 12 H 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 116.12018 Exact Mass OH f heptan-2-ol C 7 H 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

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    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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 1 186.16206 Exact Mass COOH g 3,5,7-trimethyloctanoic acid C 11 H 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 102.10452 Exact Mass OH h hexan-1-ol C 6 H 14 O 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

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    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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 130.13584 Exact 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 132.09738 Exact 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 4 192.15150 Exact 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 10 227.05827 Exact Mass O NO 2 l (4-nitrophenyl)(phenyl)methanone C 13 H 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 0 158.16716 Exact Mass O m 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 x N y O z IHD Bromine 1 78.9183 78.91830 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 1 234.06197 Exact Mass H O Br n 2-bromodecanal C 10 H 19 BrO 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 x N y O z IHD Bromine 0 78.9183 0.00000 Index = (n) - (m/2) - (x/2) + (y/2) + 1 = 2 142.09942 Exact Mass o C 8 H 14 O 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 C 7 H 14 O Exact Mass: 114.1045 IHD = 1, match with (k) A OH 3-Heptanol C 7 H 16 O Exact Mass: 116.1201 IHD = 0, match with (m) B Br 4-Bromotoluene C 7 H 7 Br Exact Mass: 169.9731 IHD = 4, match with (v) C Br 2-Bromopentane C 5 H 11 Br 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 C 6 H 12 O Exact Mass: 100.0888 IHD = 1, match with (h) E O OH Propionic Acid C 3 H 6 O 2 Exact Mass: 74.0368 IHD = 1, match with(c) F G H 2 N C 4 H 11 N Exact Mass: 73.0896 IHD = 0, match with (b) butan-1-amine OH prop-2-yn-1-ol C 3 H 4 O 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 3 H 7 NO 2 Exact Mass: 89.0477 IHD = 1, match with (d) I O OH 2-Phenoxylethanol J C 8 H 10 O 2 Exact Mass: 138.0681 IHD = 4, match with (q) Phenetole K C 8 H 10 O Exact Mass: 122.0732 IHD = 4, match with (o) O O O Methyl Butyrate L C 5 H 10 O 2 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 C 6 H 11 NO Exact Mass: 113.0841 IHD = 2, match with (j) N N 2-methylpyrazine N C 5 H 6 N 2 Exact Mass: 94.0531 IHD =4, match with (e) HO O O Methyl salicylate O C 8 H 8 O 3 Exact Mass: 152.0473 IHD = 5, match with (s) N + O - O Cl 1-Chloro-4-Nitrobenzene P C 6 H 4 ClNO 2 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 C 7 H 13 BrO 2 Exact Mass: 208.0099 IHD = 1, match with (w) HO 5-Hexyn-1-ol R C 6 H 10 O Exact Mass: 98.0732 IHD = 2, mathc with (g) O 6-Methyl-5-hepten-2-one S C 8 H 14 O Exact Mass: 126.1045 IHD = 2, match with (p) O HO Hexanoic acid T C 6 H 12 O 2 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 Cl HO 2,6-Dichlorophenol U C 6 H 4 Cl 2 O Exact Mass: 161.9639 IHD = 4, match with (u) HO 2,6-Dimethylphenol V C 8 H 10 O Exact Mass: 122.0732 IHD = 4, match with (n) O 2-Cyclohexen-1-one W C 6 H 8 O 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  2 c 2 v 2 m eff m eff = ( 𝑚1𝑚2 𝑚1+𝑚2 ) m eff = ( 1 H 19 F) = (1.0078 x 18.9984) (1.0078 + 18.9984) = 0.9570 μ m eff = ( 1 H 35 Cl) =0.9796 μ m eff = ( 1 H 81 Br) =0.9954 μ m eff = ( 1 H 127 I)=0.9999 μ ( k) ( 1 H 19 F) = 4 x 3.14 2 x (2.997 x 10 10 cm s -1 ) 2 x (4148.2 cm -1 ) 2 x (0.9570 x 1.66 x 10 -27 kg) =968.3 kg s -2 =968.3 N/m ( k) ( 1 H 35 Cl) = 514.6 N/m ( k) ( 1 H 81 Br) = 410.9 N/m ( k) ( 1 H 127 I) = 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) CH 3 CH 3 , (b) CH 4 , (c) CH 3 Cl. 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) C 6 H 6 has 3(12) – 6 = 30 normal modes (b) C 6 H 6 CH 3 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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