1. Spin Systems , 1D Spectra principles quantum mechanical theory
selected spin systems

2. Multiplet Splitting H H H C CB0 H H H C C
Energetically favourable
Electrons in equal spin states side step each other. -> lower energy (exchange integrals) (Hund‘s rule) ->
electrons of equal spin are most likely to be found at the carbon (highest probability).
At the position of the second proton the external field will be weakened or strengthened
depending on the spin of the first one.
One spin seems to take effect directly at the position of it‘s coupling partner.
The spin-spin interaction will not be averaged out.

3. Multiplet Splitting n0 1 coupling partner H J1 J2 2 coupling partners
singlet H J1
V0 - J1 /2 J1 J2
3 coupling partners J3 H
duplet
V0 + J1 /2 J2 J1
doublet of duplets J2 J3 J3
duplet of duplets of duplets J1 J2 J3

4. Multiplet Splitting H J2 = J3 J1 J2 J2 1 : 1 J3 J3 J3 J3 DDD DT
1 : 1 J3 J3 J3 J3
DDD DT
J1 = J2 J1 J1 J1
1 : 2 : 1 J2 J2 J3 J3 J3 J3
DDD TD
J1 = J2 = J3
1 : 3 : J1 J3 J1 J2 J1 J1
DDD Q

5. Multiplet Splitting
Each coupling partner causes a doubling of splitting.
Equal couplings yield Triplets, quartets ….
General multiplicity rule = 2I + 1 1
Number of coupling partners Pascal‘s triangle 1 1 1 1 2 1 2 1 3 3 1 3 1 4 6 4 1 4 1 5 10 10 5 1 5 1 6 15 20 15 6 1 6
Isopropyl-
Similar coupling constants deliver „pseudo triplets“ , „pseudo quartets“ … 2 1 1 1

6. Multiplet Splitting I=1
in general M = 2I + 1

7. Multiplet Splitting DMSO 13C 13C2D3 I=3*1 multiplicity = 2*3+1
DMSO 1H
12C 1H1 2D2
I=2*1 multiplicty = 2*2+1
CD2Cl2 13C
I=2*1 multiplicity = 2*2+1

8. Convolution F ( t´ ) G( t - t´) dt´ + - (F # G ) i = S F j G ( i - j)
(F # G ) ( t) = 1 1 1 1 2 1 1 3 3 1

9. Convolution 1 1 1 1 1 2 3 2 1 1 3 6 7 6 3 1

10. (Splitting Pattern , Convolution)FT * FT * FT

11. (Splitting Pattern , Convolution)
DMSO 13C
DMSO 1H
CD2Cl2 13C

12. Multiplet Splitting e.g.H H2 H H2
Olefine
CH2 CH
Ring

13. Multiplet Splitting e.g.
Beispiel H H Me OH H H H H H H H H H

14. Multiplet Splitting e.g.H H H
CH2 CH
Ring
Olefin.

15. Weitreichende Kopplung, Dihydrobenzol
Long Range Coupling , Dihydrobenzene
Weitreichende Kopplung, Dihydrobenzol H H H H H

16. Spinsysteme höherer Ordnung (ABC)
Higher Order Spin Systems (AMX - ABC)
Spinsysteme höherer Ordnung (ABC)
ABC
15 lines
ABX
AMX

17. Spectroscopy, Quantum Mechanics
Hamiltonian H energy
H = h2 /2 m D V (r)
wave function Y (r) describes state
eigenfunction Y (r) belongs to eigenvalue E
H Y (r) = E Y (r)
expectation value E = < Y (r) | H | Y (r) >
Eigenfunctions Y i and Y j belonging to distinct eigenvalues are orthogonal to each other.
< Y i (r) | Y j (r) > = d i j
prediction of a spectrum (1D NMR crossed coils setup)
Expectation value for Hamiltonian
E = < Y (r) | H | Y (r) >
Transition probability , squared absolute value of transition operator
D E = E Y - E F W Y F = < Y (r) | D | F (r) > 2

18. QM Description of Coupled Spin Systems
= ¾ h2
I 2
I z
= m h
I = ½ : Two states m = ½ m = - ½ r r
wave functions a b
orthonormality : < a , a > = 1 , < b , b > = 1 , < a , b > = 0 , < b , a > = 0
Hamiltonian H = B0 * I = ( 0, 0 , B0) * ( I x =
I y
I z )
B0 I z
operator of z component of angular momentum I z
½ a = I z a , ½ b = I z b
a , b eigenfunctions with respect to I z
matrix representation:
a g 1
b g
eigenfunctions of I z constitute a basis of I2, (Iz), H
I z g
½ 0
0 - ½
I 2 g
¾ 0 ¾

19. QM Description of Coupled Spin Systems
matrix representation of angular momentum operators (needed for transition moment operators):
I 2 g
¾ 0 ¾
I z ->
½ 0
0 - ½
I x ->
½ ½ 0
I y ->
0 - i/2
i/2 0
commutator relations:
[ I x , I y ]
= i I z
[ I z , I x ]
= i I y
[ I y , I z ]
[ I x , I y ] -> ½ ½
0 -i/2 i/ - =
-> i I z
I x I y = i/2 I z , I y I x = -i/2 I z , I y I z = i/2 I x , I z I y = -i/2 I x , I z I x = i/2 I y , I x I z = -i/2 I y
I x I x = 1/ , I y I y = 1/4 1 , I z I z = 1/4 1
E, mz
Hamiltonian : H = w I z a
Transition probability, transition operator , transition amplitude
½ w
½ b = I x a , ½ b = I x a
D E = w b
D E = w , W a b = < a| I x | b > 2 = ¼
- ½ w
i ½ b = I y a , i ½ b = I y a

20. 2 Spins without Coupling
product basis
a( s1 ) a( s2 ) , a( s1 ) b( s2 ) , b( s1 ) a( s2 ) , b( s1 ) b( s2 )
a a , a b , b a , b b
Hamiltonian : H = wI I z wS S z
H a a = ( wI I z wS S z ) a a = wI I z a a wS S z a a
H a a = wI a I z a + wS a S z a = ½ wI a a ½ wS a a = ½ ( wI + wS ) a a
H a b = wI b I z a + wS a S z b = ½ wI a b ½ wS a b = ½ ( wI - wS ) a b
E, mz a a
½ wI + ½ wS a b
½ wI - ½ wS
½ wS - ½ wI b a
- ½ wI - ½ wS b b

21. 2 Spins Transition Matrix Elements
W a a , a b =
< a a | I x S x | a b > 2 = < a a | I x a b + S x a b > 2 = ( ½< a a | b b > + ½ < a a | a a >) 2 = 1/4
½< a a | b b > =
½< a | b > *
< a | b >
W a a , b b =
< a a | I x S x | b b > 2 = < a a | I x b b + S x b b > 2 = ( ½< a a | a b > + ½ < a a | b a >) 2 = 0
E, mz a a
½ wI + ½ wS wS wI a b
½ wI - ½ wS
wI - wS
½ wS - ½ wI b a
wS + wI wI wS wS wI
- ½ wI - ½ wS b b

22. Scalar Coupling (weak, AX Approximation)
Hamiltonian : H = wI I z wS S z J I S
H = wI I z wS S z J ( I x S x + I y S y + I z S z )
AX approximation : H = wI I z wS S z J I z S z
J I z S z a a = J I z S z a a = J I z a S z a = J ½ a ½ a = J/4 a a
J I z S z a b = J I z S z a b = J I z a S z b = J ½ a (- ½ ) b = - J/4 a a J J
E, mz a a
½ wI + ½ wS
+ J/4 wS
+ J/2 wI
+ J/2 a b
½ wI - ½ wS wS
- J/4 wI
½ wS - ½ wI
- J/4 b a wI
- J/2 wS
- J/2
- ½ wI - ½ wS
+ J/4 b b

23. Scalar Coupling (Strong , A2 , AB System)
Hamiltonian : H = wI I z wS S z J I S
H = wI I z wS S z J ( I x S x + I y S y + I z S z )
J I x S x a a = J I x S x a a = J I x a S x a = J ½ b ½ b = J/4 b b
J I y S y a a = J I y S y a a = J I y a S y a = J i ½ b i ½ b = - J/4 b b
J I x S x a b = J I x S x a b = J I x a S x b = J ½ b ½ a = J/4 b a
J I y S y a b = J I y S y a b = J I y a S y b = J i ½ b (- i ) ½ a = + J/4 b a
½ wI + ½ wS + J/4
½ wI - ½ wS - J/4
+ J/2
½ wS - ½ wI - J/4
+ J/2
- ½ wI - ½ wS + J/4

24. Block Structure E, mz Hamiltonian : H = wI I z + wS S z + J I S
H = wI I z wS S z J ( I x S x + I y S y + I z S z )
F z = S z I z
E, mz
[ H , F z ] = F z and H commute => same subsets of eigenvectors
=> Matrix representations posses the same block structure
4 spins
aaab
aaaa
aaba
abaa
f z = 2
f z = 1
baaa
aabb
abab
baab
bbaa
abba
baba
abbb
babb
bbab
bbba
bbbb
f z = 0
f z = -1
f z = -2 1 2 3 4 6
Pascal‘s triangle again Number of coupling spins
2 spins ab aa ba bb
3 spins
aab
aaa
aba
baa
bbb
abb
bba
bab
f z = 1
f z = 0
f z = -1
f z = 3/2
f z = 1/2
f z = -1/2
f z = -3/2

25. Scalar Coupling (Strong , A2 System)
Hamiltonían : H = wI I z wS S z J I S
A2 system : wI = wS
½ wI + ½ wS+ J/4
wI + J/4
½ wI - ½ wS - J/4
- J/4
+ J/2
½ wS - ½ wI - J/4
+ J/2
- J/4
½ wI - ½ wS + J/4
- wI + J/4
symmetric eigenfunctions : a a , b b , 1/Ö 2 ( a b + b a )
antisymmetric eigenfunction : 1/Ö 2 ( a b - b a )
wI + J/4
- wI + J/4
+ J/4
- 3/4 J
J ( I x S x + I y S y ) a b = J/2 b a
J ( I x S x + I y S y ) b a = J/2 a b
J I x S x b a = J I x S x b a = J I x b S x a = J ½ a ½ b = J/4 a b
J I y S y b a = J I y S y b a = J I y b S y a = J (- i ) ½ a i ½ b = + J/4 a b

26. Scalar Coupling (Strong , A2 System)
E, mz
wI + J/4
- J/4
+ J/2
a a
+ J/2
- J/4
- wI + J/4 wI
wI + J/2
1/Ö 2 ( a b + b a )
1/Ö 2 ( a b - b a )
wI + J/4
- wI + J/4
+ J/4
- 3/4 J wI
wI - J/2
b b
H 1/Ö 2 ( a b + b a ) = 1/Ö 2 ( - J/4 a b + J/2 b a - J/4 b a + J/2 a b ) = J/4 1/Ö 2 ( a b + b a )
H 1/Ö 2 ( a b - b a ) = 1/Ö 2 ( - J/4 a b + J/2 b a + J/4 b a - J/2 a b ) = - J 3/4 1/Ö 2 ( a b - b a )
W a a , ( a b - b a ) =
< a a | I x S x |( a b - b a ) > 2 = (< a a | I x a b + S x a b > - (< a a | I x b a + S x b a > ) 2
= (½< a a | b b > + ½ < a a | a a > - ½< a a | a a > + ½ < a a | b b > ) 2 = 0

27. Scalar Coupling (Strong , AB System )
AB system : wI = wS wS - wI << J
Hamiltonian : H = wI I z wS S z J I S
H = wI I z wS S z J ( I x S x + I y S y + I z S z )
½ wI - ½ wS - J/4
½ wS - ½ wI - J/4
½ wI + ½ wS + J/4
- ½ wI - ½ wS + J/4
+ J/2
a a
b b
a b
b a
½ ( wI + wS ) + J/4 = e 1 , - ½ ( wI + wS ) + J/4 = e 4
½ D w- J/4
- e
+ J/2
½ wI - ½ wS = ½ D w
Det
= 0
+ J/2
- ½ D w - J/4
- e
e 2 - ¼ D w2 + ( J/4) 2 + e J/ J 2 /4 = 0
e e J/ /16 J ¼ D w = 0
1/2 Ö (J 2 + D w2 ) = C
- J/4 +/- 1/2 Ö (J 2 + D w2 ) = e 2,3

28. Transition Probability (AB System)
E, mz
1/2 Ö (J 2 + D w2 ) = C a a
½ ( wI + wS ) + J/4
½ ( wI + wS ) + J/2 - C a b
- J/4 + C
½ ( wI + wS ) + J/2 + C b a
- J/4 - C
½ ( wI + wS ) - J/2 + C
- ½ ( wI + wS ) + J/4 b b
½ ( wI + wS ) - J/2 - C
½ D w- J/4
- e
+ J/2
cos q
cos q
= e 2,3
+ J/2
- ½ D w - J/4
- e
sin q
sin q
cos q a b + sin q b a = Y 2
cos q a b - sin q b a = Y 3

29. Transition Probability (AB System)
1/2 Ö (J 2 + D w2 ) = C
½ D w- J/4
+ J/2
cos q
cos q =
- J/4 + C
- ½ D w - J/4
sin q
sin q
+ J/2
( ½ D w - C )´cos q + J/2 sin q = 0
cos q a b + sin q b a = Y 2
-( ½ D w + C )´sin q + J/2 cos q = 0
cos q a b - sin q b a = Y 3
cos 2 q = ½ D w / C - sin 2 q = - J/ 2 C
cos q / sin q = - J/ 2 ( ½ D w - C )
cos q sin q = J/ 4 C
W a a , Y =
< a a | I x S x |(cos q a b + sin q b a ) > 2 = < a a | I x b a sin q + S x a b cos q > 2
= (½ sin q + ½ cos q ) 2 = ¼ ( sin 2 q cos q cos q sin q ) = ¼ ( 1 + sin 2 q )

30. Higher Order Spin Systems
J =1 Hz
Dd varied
Dd 1 Hz
A2 system
Dd 2 Hz
higher order
AB system
Dd 5 Hz
Dd 10 Hz
Dd 20 Hz
roof effect
Dd 40 Hz
1st - Order
AX system
Dd 50 Hz

31. Sign of Coupling Constants
In the case of ABC systems signs of coupling constants matter
sign varied
n a = Hz
n b = Hz
n c = Hz
J ab = 10 Hz
J ac = 15 Hz
J bc = 20 Hz
J ab = Hz
J ac = Hz
J bc = Hz

32. Forbidden Transitions
ABC
15 lines E
W = I x M x + S x
aaa
mz = 3/2
AMX
W baa , b a b = < baa | S x |b a b > = 0
baa
aba
mz = 1/2
aab
W baa , b a b = < aba | I x |b a b > = 0
W baa , b a b = < aba |M x |b a b > = 0
abb
W baa , b a b = < aba | S x |b a b > = 0
bab
bba
mz = -1/2
ABC
W baa , b a b = < aba | I x +M x+ S x |b a b > 2 = 0
mz = -3/2
bbb

33. Äquivalente Protonen Equivalent Protons
Arise from equal chemical surroundings
- Chemically equivalent : equal shifts , equal couplings constants, to distinct coupling partners
- Magnetically equivalent : equal shifts , equal coupling constants and equal (same) coupling partners (network)
e.g. Methyl groups
( rotating freely )
Magnetically equivalent protons cause triplet and quartet splitting
Equivalent Protons
don‘t interact with each other
don‘t show a splitting due to their mutual interaction
higher order spin systems
D d = 0
AX-System : D d >> J
AB-System : D d < J A2-System : D d = 0
D d >> J
D d = J
D d < J

34. Protons which aren't Equivalent
Freely rotating methylene groups
CH2 next to a stereo centre H Rc Rb R Ra Rc Rb H R Ra Rc Ra H R Ra Rc Rb H R Rc Ra H R H
and H
Never see the same surrounding. H
and
Will therefore constitute an AB system (e.g. ascorbic acid).
A stereo centre in the vicinity (somewhere) suffices. H R Rc Rb
holds as well for methylene and methyl groups, ( CH3 )2 - CH - close to a stereo centre
hold as well if Ra equals CH2 – R , diastereotopic protons

35. Non Equivalent Protons
SR P1_1E5

36. Non Equivalent Protons
MNOVA Prediction

37. Non Equivalent Protons
SR P1_1E5

38. ABMX- System (Ascorbic Acid)H H H2 H2 * H H * H

39. ABM System - JBM Can be viewed as composed of two AB systems.
1st systems M point up
2nd systems M point down nA nB
A spin
B spin
M spin
JAM
JAB
JBM
JAB
- JBM

40. ABMX- System (Ascorbic Acid)F1 F2 F3 F4
JAB = F1 – F2 = F3 –F4 = F1 – F2 = F3 – F4
n0d = ( F1 – F4) * ( F2 –F3 )
Z = ( F1 + F2 + F3 + F4 ) / 4
nA = Z – ½ n0d
nB = Z + ½ n0d
nA = Hz nB = Hz JAB = 11.6
nA = Hz nB = Hz JAB = 11.6
nB = Hz nA = Hz JAB = 11.6
nA nA = nA +/- JAX
nB nB = nB +/- JBX
JAX = nA - nA JBX = nB - nB
JAX = nA - nB JBX = nB - nA
JAX = 7.71
JAX = -0.01
JBX = 5.49
JBX = 13.2

41. ABMX System (Two Solutions)
JAX = 7.71
JBX = 5.49
JAB = 11.6
JAX = -0.01
JBX = 13.2
sim.
sim.
exp.

42. AA´BB´ System (ODCB)
chemically equivalent protons (magnetically non equivalent )
example ODCB
AA´BB´ System
JAB JAB´
JA´A´ JB´B
JA´B´ JA´B

43. AA´BB´ System (Assignment of Resonances)
Assignment of lines is the most demanding task (also in automated analysis, DAVINS, LAOCOON).

44. AA´BB´- System AA´BB´ System
After all lines have been assigned the theoretical transitions, a set of equations has to be solved.
AA´BB´- System
K = g-h = i-j = 7.83
g-j = 18.0
h-i = 2.35
l = (( h-i ) * ( g –h )) 1/2
c-f = 16.88
d-e = 2.56
l = ( (c-f) * (d-e) )1/2
M = c-d =e –f =7.15
N =a-b =k-l =9.6

45. AA´BB´ System (Simulation)
8.05
1.55
0.34
7.49
Which multiplet belongs to ortho or meta position ?
simulation

46. AA´BB´ System (Partial Spectra)
partial spectra are symmetric (mirror images)

47. AA´BB´- System (Satelliten)
AA´BB´ System (13C Satellites)
AA´BB´- System (Satelliten)

48. 13C Satellites, Splitting Pattern
7.08 – J13C (ppm)
13C
8.04 Hz
7.47 Hz
1.55 Hz

49. AA´BB´- System (Simulation)
exp.
satellites
satellites
ortho position 13C: AA‘BB‘ splitting in satellites.
meta position 13C: Additional coupling and isotope effect perturb AA‘BB‘ system.
sim.

50. AA´BB´ System (Examples)
pyrrole

51. Electric Quadrupol Moment 11B , 14N…
magnetic moment || nuclear angular momentum B0
electric quadruple moment for spin quantum number I > ½
electron
x,y- plain
T1 relaxation
T2 relaxation
angle enclosed between molecular axis and external field
( cos 2 q )

52. Relaxation and Scalar Coupling1H / 14N, 11B
14N (3 states I=1 11B (4 states I=3/2) ) 1H 1H
14N (11B) relaxation 1H
14N (11B ) decoupling

53. 15N- ,14N- 1H- Couplings Pyrrol in DMSO
electr. quadrupol.
14N I = %
15N I = ½ %
50 Hz
100 Hz Ho Hm
14N decoupled NH

54. AA´BB´ System (Examples)
pyrrole

55. 19F- 1H -Kopplungen 19F- 1H- Couplings 7 Hz 10 Hz 7 Hz
Simulation JAB = 10 JAA´ = 2 JAB´ = 0.5 J19F = 7 bzw. 10 Hz dshift =100 Hz dshift19F=3000 Hz

56. 19F- 1H- Couplings (19F Splitting)
10 Hz
10 Hz
7 Hz
19F

57. Weitere AA´BB´- Systeme
AA´BB´ System (Examples)
Weitere AA´BB´- Systeme
pyrrole

58. A2B2C- System (Glycerin)
AA´BB´C System glycerine)
A2B2C- System (Glycerin)
chemically equivalent protons (magnetically nonequivalent)
sim.

59. 1,4- Butandiol A2A2´B2 B2´- System
8 Hz
0.5 Hz
10 Hz
0.5 Hz

60. 1,4- Butandiol A2A2´B2B2´ System

61. Proposed Structure

62. A2A´2B2B´2 System

63. AA´BB´ System
Simulation/spin analysis
MNOVA prediction
Exp.

64. 5 Spin Systems (700MHz) Simulation
Even at 700 MHz higher order spectra may appear ( therefore spin simulation) 1 2 3 4 4a
4b, 3, 2 1

65. TIPS 500MHz

66. TIPS 700MHz (sim.)

67. AA´BB´ System Both conformers can give rise to AA’BB’ Systems 700 MHz
Splitting is due to Coupling (independent of B0)
400MHz

68. Two AA´BB´ Systems needed
Population 1 A A´ B B´ A B B´ A´
Population 2 A A´ B
Best match is achieved by the assumption of two conformers B´

69. (1E,3E)-1,4-diphenylbuta-1, 3-diene
Population 2 A A´ B
Best match is achieved by the assumption of two conformers

70. (1E,3E)-1,4-diphenylbuta-1, 3-diene
Population 2 A A´ B
Best match is achieved by the assumption of two conformers

71. (1E,3E)-1,4-diphenylbuta-1, 3-diene
Population 2 A
Starting from MNOVA
prediction -> modifying A´ B
Best match is achieved by the assumption of two conformers