p 100 % p+1 10 %

42
P 100 % P+1 10 % Compound 1 148/13 = 11.38; .38*13 =5 C 11 H 18

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Compound 1. P 100 % P+1 10 %. 148/13 = 11.38; .38*13 =5 C 11 H 18. Compound 1. Compound 1. C 11 H 18 C 10 H 14 O. Area : 5:2:2:3. m/e 120 m/e 105 m/e 77. +. m/e 120. m/e 105. Compound 2. P 100 % P+1: 7.1% P+2 : 33 %. 126-35 = 91 91/13 = 7.0 C 7 H 7 Cl. Compound 2. - PowerPoint PPT Presentation

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Page 1: P 100 % P+1 10 %

P 100 %

P+1 10 %

Compound 1

148/13 = 11.38; .38*13 =5C11H18

Page 2: P 100 % P+1 10 %

Compound 1

Page 3: P 100 % P+1 10 %

Compound 1

C11H18

C10H14O

Area : 5:2:2:3

Page 4: P 100 % P+1 10 %

m/e 105

+.

m/e 120

m/e 120m/e 105m/e 77

Page 5: P 100 % P+1 10 %

Compound 2

P 100 %

P+1: 7.1%

P+2 : 33 %

126-35 = 9191/13 = 7.0C7H7Cl

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Compound 2

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Area: 5:2

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Compound 3

m/e 114114/13 = 8.769; 0.769*13 = 10C8H18

Page 9: P 100 % P+1 10 %

Compound 3

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Compound 3

C8H18

C7H14OC6H10O2

Page 11: P 100 % P+1 10 %

m/e 99 P-15m/e 86 P-28m/e 69 P-45m/e 41

m/e 86

m/e 69m/e 41

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Compound 4

m/e 184184-79 = 105105/13 = 8.077; 0.077*13 = 1C8H9Br

Page 13: P 100 % P+1 10 %

Compound 4

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Compound 4

Area 5:2:2

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Compound 5

C8H18O3 mw 162

m/e 75C2H3O3

+

C3H7O2+

m/e 73C2HO3

+

C3H5O2+

C4H9O+

m/e 147 = P - 15m/e 130 = P - 32 CH4Om/e 115 = P – 47 CH3O2; C2H7O

Page 16: P 100 % P+1 10 %

Compound 6

Page 17: P 100 % P+1 10 %

Compound 7

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Compound 7

A comparison of two EIMS spectra obtained from different instruments and different samples

m/e 136136/13 = 10.460.46*13 = 6C10H16

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Compound 7

C10H16

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Compound 7

C10H16

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Empirical Rules for Dienes (π → π*)Homoannular(cisoid) Heteroannular (transoid)

Parent λ = 253 nm λ = 214 nmIncrements for:

double bond extension 30 30alkyl subst or ring 5 5exocyclic double bond 5 5

Alkenesethylene λ = 175 nm

λ = 253 +10 + 10 = 273 nm

Page 22: P 100 % P+1 10 %

Compound 8

m/e 108108/13 = 8.30770.3077*13 = 4C8H12

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Compound 8

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Compound 8

d

d t

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Compound 8

area = 2:2:4:4 left to right

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P : m/e 108

m/e 80 = P-28

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1H NMR: no signal13C NMR: 4 vinyl carbons

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282/13 = 21.692; 0.692*13= 9C21H30

C18H31ClC15H32Cl2

C12H33Cl3

C9H34Cl4

C6H35Cl5

C6Cl6

35Cl = 0.7637Cl = 0.24

1 1 1 n = 11 2 1 n = 2

1 3 3 1 n = 31 4 6 4 1 n = 4

1 5 10 10 5 1 n = 51 6 15 20 15 6 1 n = 6

n = 5 (0.76)5 = 0.254; 5(0.76)4(0.24)= 0.40 100% 157%n = 6 (0.76)6 = 0.193; 6(0.76)5(0.24) = 0.365

100% 189%

Page 30: P 100 % P+1 10 %

146/13 = 11.23; 0.23 *13 =3C11H14

Compound 9

P-28

Page 31: P 100 % P+1 10 %

C11H14 C10H10O

Compound 9

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2 1

C10H10O C9H6O2

Compound 9

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160.63 S 153.99 S143.48 D131.79 D127.95 D124.43 D118.81 S116.70 D116.56 D

C9H6O2

Compound 9

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136/13 = 10.462; 0.462*13 = 6C10H16

Compound 10

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Compound 10

Neat liquid

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Compound 10

C10H16

C9H12O

CCl4 solution

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Compound 10

Area: 1:2:2:3C9H12OC8H8O2

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190.70 D164.63 S131.93 D 129.97 S114.33 D 55.53 Q

C8H8O2

Page 39: P 100 % P+1 10 %

Compound 11

112/13 = 8.615; 0.615*13 = 8C8H16

Page 40: P 100 % P+1 10 %

Compound 11

C8H16

Degrees of unsaturation: 1

Page 41: P 100 % P+1 10 %

Area: 3:2:3C8H16

Page 42: P 100 % P+1 10 %

147.28 S126.63 D123.24 D 122.62 D23.25 T 16.03 Q

C8H16

C7H12OC6H8O2

Exact mass: 112.0347Exact mass of C6H8O2: 112.0524