Nuclear Magnetic Resonance Transition Moment Integral Probability - PowerPoint PPT Presentation

Nuclear Magnetic Resonance

Transition Moment Integral Probability of an excitation  P a -> b |∫ b * Ĥ 1 a dr| 2 d (E b -E a -h n )  Hamiltonian Ĥ 1 defined as  Ĥ 1 = - m B 1  Ĥ 1 = -g N b N /(hbar) I * B 1  Ĥ 1 = -g N b N /(hbar) I * (cos(2 pn t)) * B 1 as  Molecular Spectroscopy CEM 484 2

Selection Rules Selection rules developed using perturbation theory as  before (rotations and vibrations) Ĥ 1 = - m B 1 : B 1 is magnetic field direction  Assume field lies along each direction and determine  excitation probability B||z, Ĥ 1 = - m B 1 = - g *B 1 * Ī z  P z = ∫ b *Ĥ 1 a d t = - g B 1 ∫ b * Ī z a d t = - g B 1 (hbar /2) ∫ b * a d t = 0  Molecular Spectroscopy CEM 484 3

Selection Rules Selection rules developed using perturbation theory as  before (rotations and vibrations) Ĥ 1 = - m B 1 : B 1 is magnetic field direction  Assume field lies along each direction and determine  excitation probability B||x, Ĥ 1 = - m B 1 = -g n b n /(hbar)*B 1 * Ī x  P x = ∫ b *Ĥ 1 a d t = - g B 1 ∫ b * Ī x a d t = - g B 1 (hbar /2) ∫ b * b d t = -  g B 1 (hbar/2) B||x, Ĥ 1 = - m B 1 = -g n b n /(hbar)*B 1 * Ī x  P x = ∫ b *Ĥ 1 a d t = - g B 1 ∫ b * Ī x a d t = - g B 1 (hbar /2) ∫ b * b d t = -  g B 1 (hbar/2) Molecular Spectroscopy CEM 484 4

Ladder Operators a and b are eigenstates of I 2 and I z but not I x and I y .  Rewrite in terms of raising and lowering operators.  I + = I x + iI y I + (I,m z )=hbarsqrt[(I(I+1)-m(m+1)] (I,m z +1)  I - = I x – iI y I - (I,m z )=hbarsqrt[(I(I+1)-m(m+1)] (I,m z -1)  Redefine I x .  I x = ½(I + + I - )  Molecular Spectroscopy CEM 484 5

Transition Moment Integral Use ladder operator to evaluate integral.  P a -> b  ∫ b * Ĥ 1 a dr = ∫ b *{-g N b N B 1 I x /(hbar)} a dr  P a -> b  (-g N b N B 1 /hbar )∫ b *I x a dr  P a -> b  (-g N b N B 1 /hbar )∫ b *(1/2 (I + + I - )) a dr  Ladder operator results.  I + a = 0 – can’t raise already at max  I + b = hbar *sqrt[(1/2*(1/2+1)-(-1/2)*(-1/2+1)] (1/2,1/2) =  hbar a I - b = 0 – can’t lower already at min  I + a = hbar *sqrt[(1/2*(1/2+1)-(-1/2)*(-1/2+1)] (1/2,1/2) =  hbar b Molecular Spectroscopy CEM 484 6

Transition Moment Integral Evaluate integral.  P a -> b  (-g N b N B 1 /hbar )∫ b *(1/2 (I + + I - )) a dr  P a -> b  (-g N b N B 1 /2hbar) (∫ b *I + a dr + ∫ b *I - a dr)  P a -> b  (-g N b N B 1 /2hbar) (∫ b *I - a dr)  P a -> b  (-g N b N B 1 /2hbar) (∫ b *I - a dr) = (-g N b N B 1 /2)  Selection Rules  m z +- 1  B 1 must be perpendicular to B 0  Molecular Spectroscopy CEM 484 7

Shielding NMR spectroscopy is useful based on sensitivity to “local”  chemical environment. Consider benzene molecule  Static B 0 field generates a current in pi electron system  Current generates a magnetic field B elec  B elec opposes applied magnetic field  Introduce the concept of a chemical shift  Modify the magnetic field by (1- s )  s is shielding constant and depends on chemical environment  D E = h n = g N b N (1- s )B 0  n = g N b N (1- s )B 0 /h  Molecular Spectroscopy CEM 484 8

Chemical Shift Desire to compare data from machines with different  magnetic fields. Calibrate spectra in ppm relative to standard reference  TMS is reference  Chemical shift scale  d H = ( n H – n TMS )/v spec * 10 6 ppm  Current generates a magnetic field B elec  B elec opposes applied magnetic field  Chemical shift example  TMS at 90 MHz and n H at -100 MHz then  d H = -100/90*10 6 Hz * 10 6 = -1.11 ppm  Molecular Spectroscopy CEM 484 9

Chemical Shift Difference between chemical shifts is independent of field  strength n 1 = g N b N /hbar (1- s 1 )B o  n 2 = g N b N /hbar (1- s 2 )B o  d 1 - d 2 = ( n 1 - n 2 / n spec )*10 6  d 1 - d 2 = (1- s 1 – 1 + s 2 ) *10 6 = ( s 2 - s 1 ) *10 6  NMR difference example  Peak at 8.6 ppm and 2.5 ppm. What is difference if data were  collected at 300 MHz instrument. d 1 - d 2 = ( n 1 - n 2 / n spec )*10 6 = (8.6ppm-2.5ppm) = 6.1 ppm  6.1ppm*300MHz/10 6 ppm = n 1 - n 2 = 1.8310 -3 MHz = 1830 Hz  If 900 MHz machine difference is 5490 Hz  Molecular Spectroscopy CEM 484 10

Representative Chemical Shifts Different types of chemical environments show different  chemical shifts (Table 14.3 from book). Compound Proton Example d 1.2 – 1.4 Alkane R 2 CH 2 (CH 3 ) 2 CH 2 6.0 – 8.5 Aromatic ArH Benzene 3.4 – 3.8 Chloroalkane RCH 2 Cl CH 3 CH 2 Cl 3.3 – 3.9 Ether ROCH 2 R CH 3 OCH 2 CH 3 Molecular Spectroscopy CEM 484 11