Spin Systems , 1D Spectra principles quantum mechanical theory

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

Multiplet Splitting H H H C C
B0 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.

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

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

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

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

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

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

Convolution 1 1 1 1 1 2 3 2 1 1 3 6 7 6 3 1

(Splitting Pattern , Convolution)
FT * FT * FT

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

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

Beispiel H H Me OH H H H H H H H H H

H H H CH2 CH Ring Olefin.

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

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

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

QM Description of Coupled Spin Systems
= ¾ h2 I 2 I z = m h I = ½ : Two states m = ½ m = - ½ I 2 = ¾ 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

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

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

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

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

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

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

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

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

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

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 )

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

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

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 coupling partners (network) e.g. Methyl groups ( rotating freely ) Magnetically equivalent protons cause triplet and quartet splitting Equivalente 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

Protons which aren't Equivalent
Freely rotating methylene groups CH2 next to a stereo centre Ra Rc Rb H R H Rc Rb R Ra H and 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. 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

Non Equivalent Protons
SR P1_1E5

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

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

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

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

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

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

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

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

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

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

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

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.

AA´BB´ System (Examples)
pyrrole

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 )

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

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

pyrrole

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

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

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

Protons which aren't Equivalent
Protonen die nicht äquivalent sind Freely rotating methylene groups CH2 next to a stereo centre Ra Rc Rb H R H Rc Rb R Ra H and Never see the same surrounding. H and Will therefore constitute an AB system (e.g. ascorbic acid). hold as well if Ra equals CH2 – R , diastereotopic protons H R Rc Rb e.g. glycerine H and often appearing as AB System with double intensity or as AA‘BB‘ System chemically equiv. but magnetically non equivalent holds as well for methylene and methyl groups, ( CH3 )2

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

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

1,4- Butandiol A2A2´B2B2´ System

Proposed Structure

A2A2´B2B2´ System

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

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

TIPS 500MHz

TIPS 700MHz (sim.)

Spin Systems , 1D Spectra principles quantum mechanical theory

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