P 100 %

P+1 10 %

Compound 1

148/13 = 11.38; .38*13 =5 C11H16

Compound 1

Compound 1

C11H16 C10H12O

Area : 5:2:2:3

m/e 105

+.

m/e 120

m.e = 148 m/e 120 P- 28 m/e 105 P - 43 m/e 77

Compound 2

P 100 %

P+1: 7.1%

P+2 : 33 %

126-35 = 91 91/13 = 7.0

91/13 = 7.0 C7H7Cl

Compound 2

Area: 5:2

Compound 3

m/e 114 114/13 = 8.769; 0.769*13 = 10 C8H18

Compound 3

C8H18 C7H14O C6H10O2

Compound 3

C8H18 C7H14O C6H10O2

Left to right: 1:1:2:3:3

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

m/e 86

m/e 69 m/e 41

OH

CH

CH3

H2C

CH2CH3C

H

CH3 1

2

34

5 6

7

1 1 2 332 3

3 6 4 215 7

OH

CH

CH3

H2C

CH2CH3C

H

CH3 1

2

34

5 6

7

3 coupled to 4. (1.2) 6 “ 5, 7 4 “ 3, 1,2, 5 1,2 “ 3, 4 5 “ 4, 6

HN C

O

OH

HC

HO Hb

He'

Hd

Hd'

He

Jcis 7-11 Jtrans 2 to 8

Compound 5

m/e 184 184-79 = 105 105/13 = 8.077; 0.077*13 = 1 C8H9Br

Compound 5

Compound 5

Area 5:2:2

6.

Compound 7

Compound 7

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

m/e 136 136/13 = 10.46 0.46*13 = 6 C10H16

Compound 7

C10H16

Compound 7 C10H16

heptet doublet

Empirical Rules for Dienes (π → π*) Homoannular(cisoid) Heteroannular (transoid) Parent λ = 253 nm λ = 214 nm Increments for: double bond extension 30 30 alkyl subst or ring 5 5 exocyclic double bond 5 5 Alkenes ethylene λ = 175 nm

λ = 253 +10 + 10 = 273 nm

λ = 253 +10 + 10 = 273 nm, λmax = 267 nm

Compound 8

m/e 108 108/13 = 8.3077 0.3077*13 = 4 C8H12

Compound 8

Compound 8

d

d t 13 C NMR

Compound 8

area = 1:1:2:2 left to right

P : m/e 108 m/e 80 = P-28

1H NMR: no signal

Problem 9

13 C NMR

282/13 = 21.692; 0.692*13= 9 C21H30 C18H31Cl C15H32Cl2 C12H33Cl3 C9H34Cl4 C6H35Cl5 C6Cl6 35Cl = 0.76 37Cl = 0.24 1 1 1 n = 1 1 2 1 n = 2 1 3 3 1 n = 3 1 4 6 4 1 n = 4 1 5 10 10 5 1 n = 5 1 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%

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

Compound 10

P-28

C11H14 C10H10O

Compound 10

2 1

C10H10O C9H6O2

Compound 10

160.63 S 153.99 S 143.48 D 131.79 D 127.95 D 124.43 D 118.81 S 116.70 D 116.56 D

C9H6O2

Compound 10

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

Compound 11

Compound 11

Neat liquid

Compound 11

C10H16 C9H12O

CCl4 solution

Compound 11

Area: 1:2:2:3 C9H12O

C8H8O2

190.70 D 164.63 S 131.93 D 129.97 S 114.33 D 55.53 Q

C8H8O2

Compound 11

Compound 12

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

Exact mass: 112.0347

Compound 12

C8H16

Degrees of unsaturation: 1

Area: 3:2:3 C8H16

Compound 12

147.28 S 126.63 D 123.24 D 122.62 D 23.25 T 16.03 Q

C8H16

C8H16 C7H12O C6H8O2

Exact mass: 112.0347 Exact mass of C8H16: 112.1253 Exact mass of C7H12O: 112.0888 Exact mass of C6H8O2: 112.0524

C8H16 Exact mass: 112.0347 Exact mass of C6H8O2: 112.0524 Exact mass of C6H8S: 112.0347

Problem 13 MW 172 C13H16 C12H12O C11H8O2

O

OH

H

6 Prob 13

Problem 13

O OCH2

CH3

C7H12O2

Problem 14

m/e 75 C2H3O3

+

C3H7O2+

m/e 73 C2HO3

+

C3H5O2+

C4H9O+

m/e 147 = P - 15 m/e 130 = P - 32 CH4O m/e 115 = P – 47 CH3O2; C2H7O

15.

6 16. The following compound has a nominal mass of 248 and contains a halogen.

Problem 17. The following spectra are of a chiral molecule, It has a molecular formula of C9H14O6. Its infrared, and 1H and 13 C NMR spectra are given below. What is the molecular formula and suggest a possible structure?

13 C NMR

H NMR C9H14O6.

1 3 3

S S D Q Q