492
M. Rashid et al.
Calculation of 19 F NMR peaks.
ClF 3
Cl
F
F
F
a
a
b
doublet, triplet
BrF 5
Fa
2nI + 1= 2x1x1/2 + 1= 2
Fb
2nI + 1= 2x2x1/2 + 1= 3
F a
F b
Br
F e
F e
F e
F e
Fa
Br
F e
F e
Fa
F e
Fa
Fe
2nI + 1= 2x1x1/2 + 1= 2
Fa
2nI + 1= 2x4x1/2 + 1= 5
F a
F e
Fa
2nI + 1= 2x3x1/2 + 1= 4
Fe
2nI + 1= 2x2x1/2 + 1= 3
F a
F e
SF 4
S
S
Fa
Fa
F e
F e
Fa
2nI + 1= 2x2x1/2 + 1= 3
Fe
2nI + 1= 2x2x1/2 + 1= 3
F a
F e
T d singlet (wrong)
sp
3 d
6 31 P-Nuclear Magnetic Resonance
31 P nucleus has natural abundance of 100% and have NMR-active spin quantum
number I = 1/2. There are no rules for the prediction of
31 P shifts and the range
is fairly large but the multiplicities of
31 P can be predicted using the same rule as
applied for
1 H and
19 F NMR-spectrum [20].
Considering the spectra of the compound such as (Ph 3 )P, H 3 PO 4 , (C 2 H 5 O) 2 POCl.
All P-atoms differ widely respect to valence state and groups attached to it yet every
M. Rashid et al.
Calculation of 19 F NMR peaks.
ClF 3
Cl
F
F
F
a
a
b
doublet, triplet
BrF 5
Fa
2nI + 1= 2x1x1/2 + 1= 2
Fb
2nI + 1= 2x2x1/2 + 1= 3
F a
F b
Br
F e
F e
F e
F e
Fa
Br
F e
F e
Fa
F e
Fa
Fe
2nI + 1= 2x1x1/2 + 1= 2
Fa
2nI + 1= 2x4x1/2 + 1= 5
F a
F e
Fa
2nI + 1= 2x3x1/2 + 1= 4
Fe
2nI + 1= 2x2x1/2 + 1= 3
F a
F e
SF 4
S
S
Fa
Fa
F e
F e
Fa
2nI + 1= 2x2x1/2 + 1= 3
Fe
2nI + 1= 2x2x1/2 + 1= 3
F a
F e
T d singlet (wrong)
sp
3 d
6 31 P-Nuclear Magnetic Resonance
31 P nucleus has natural abundance of 100% and have NMR-active spin quantum
number I = 1/2. There are no rules for the prediction of
31 P shifts and the range
is fairly large but the multiplicities of
31 P can be predicted using the same rule as
applied for
1 H and
19 F NMR-spectrum [20].
Considering the spectra of the compound such as (Ph 3 )P, H 3 PO 4 , (C 2 H 5 O) 2 POCl.
All P-atoms differ widely respect to valence state and groups attached to it yet every
