C
3
H
5
BrO
2
MW = 152,97
g/mol
13
C NMR
(CDCl
3
, 125
MHz)
MS EI+
(70 eV)
Problem
1
C
3
H
5
BrO
2
MW = 152,97
g/mol
13
C NMR
(CDCl
3
, 125
MHz)
MS EI+
(70 eV)
Problem
1
C
3
H
5
BrO
2
MW = 152,97
g/mol
13
C NMR
(CDCl
3
, 125
MHz)
MS EI+
(70 eV)
Br-CH
2
-CH
2
-COOH
Problem 1
rozwiązanie
C
10
H
14
O
IR, KBr
EI+, 75 eV
EI, 75 eV
Hz ppm Int.
650.44 7.263 48
649.19 7.249
256 647.50 7.230
213 645.75
7.211 63 245.94
2.747 238
140.63 1.571 115
108.00
1.206 720
1
H NMR 89.56
MHz
0.04 ml : 0.5 ml
CDCl
3
IR, KBr
C
10
H
14
O cd
EI+, 75 eV
EI, 75 eV
Hz ppm Int.
650.44 7.263 48
649.19 7.249
256 647.50 7.230
213 645.75
7.211 63 245.94
2.747 238
140.63 1.571 115
108.00
1.206 720
1
H NMR 89.56
MHz
0.04 ml : 0.5 ml
CDCl
3
IR, KBr
C
10
H
14
O
rozwiązanie
OH
H
3
C
CH
3
EI+, 75 eV
6 atomów C
MW = 147,0
g/mol
MS EI+
(70 eV)
Problem 3
MS EI+
(70 eV)
6 atomów C
4 atomy H
MW = 147,0
g/mol
MS EI+
(70 eV)
MS EI+
(70 eV)
Problem 3 cd
6 atomów C
4 atomy H
2 atomy Cl
MW = 147,0
g/mol
MS EI+
(70 eV)
MS EI+
(70 eV)
Problem 3 cd
6 atomów C
4 atomy H
2 atomy Cl
MW = 147,0
g/mol
MS EI+
(70 eV)
MS EI+
(70 eV)
Problem 3 rozwiązanie
Cl
Cl
MS EI+
(70 eV)
1
H NMR
(CDCl
3,
500
MHz)
MS EI+
(70 eV)
13
C NMR
(CDCl
MHz)
Problem 4
MS EI+
(70 eV)
1
H NMR
(CDCl
3,
MHz)
MS EI+
(70 eV)
13
(CDCl
MHz)
8 atomów C
MW = 134.1 g/mol
MS EI+
(70 eV)
1
H NMR
(CDCl
3,
500
MHz)
MS EI+
(70 eV)
13
C NMR
(CDCl
MHz)
1
H NMR
(CDCl
3,
MHz)
13
(CDCl
MHz)
C
8
H
6
O
2
MW = 134.1 g/mol
Problem 4 cd
MS EI+
(70 eV)
1
H NMR
(CDCl
3,
500
MHz)
MS EI+
(70 eV)
13
C NMR
(CDCl
MHz)
1
H NMR
(CDCl
3,
MHz)
13
(CDCl
MHz)
C
8
H
6
O
2
MW = 134.1 g/mol
CHO
CHO
Problem 4 rozwiązanie
