N
X
C
O
N
S
Hal
Solvents
Natural
Products
P Si
C
C
C C
C
C
200
5 1 H NMR
H 2
H 4
H 5
H 6
H 2
H 3
Substituent
in position 3 or 5
in position 4
Z 32
Z 56
Z 34
Z 54
Z 35
Z 53
Z 36
Z 52
Z 42
Z 46
Z 43
Z 45
–CH 3
-0.15 -0.16 -0.07 -0.17
-0.13 -0.13
–CH 2 CH 3
-0.13 -0.14 -0.06 -0.17
-0.12 -0.14
–CH 2 –phenyl
-0.08 -0.18 -0.04 -0.14
0.00 -0.15
–phenyl
0.25 0.20 0.08 -0.03
–CH=CH 2
-0.12 -0.08
–F
-0.05 -0.21 0.04 -0.13
-0.07 -0.03
–Cl
0.09 0.00 0.05 -0.05
0.00 0.05
–Br
0.09 0.18 -0.04 -0.07
0.09 0.35
–OH
-0.31 -0.29 0.06 -0.50
–OCH 3
-0.27 -0.37 -0.04 -0.40
-0.16 -0.42
–OCOCH 3
-0.15 -0.15 0.08 -0.13
–NH 2
-0.51 -0.65 -0.20 -0.60
-0.15 -0.74
–N(CH 3 ) 2
-0.38 -0.77
–NHCOCH 3
0.37 0.50 0.06 -0.16
-0.19 0.16
–C– – – N
0.32 0.38 0.25 0.26
0.24 0.32
–S–phenyl
0.05 -0.16
–S(O) 2 OH
0.70 1.14 0.81 0.70
–CHO
0.52 0.58 0.30 0.28
0.31 0.49
–COCH 3
0.58 0.61 0.20 0.20
0.21 0.50
–CO–phenyl
0.23 0.35
–COOH
0.54 0.57 0.20 0.24
0.20 0.45
–COOCH 3
0.64 0.67 0.16 0.19
0.19 0.61
–COO–phenyl
0.24 0.75
–CONH 2
0.49 0.50 0.15 0.15
–CSNH 2
0.68 0.67 0.24 0.26
0.35 0.68
–CH=NOH
0.39 0.43 0.19 0.15
0.06 0.32
–Si(CH 3 )
0.08 0.00 -0.21 -0.11
-0.08 0.01
C
X
O
N
S
O
||
C
Si
δ H-2 = 8.59 + Z i2
δ H-3 = 7.25 + Z i3
δ H-4 = 7.62 + Z i4
δ H-5 = 7.25 + Z i5
δ H-6 = 8.59 + Z i 6
2
3
4
6
5
N