1. Electron Spin Resonance Spectroscopy ESR / EPR
Absolute requirement: 1 or more unpaired electrons
Stable species: O2, NO, [Fe(CN)6]3-
Transients: CH3, C6H6-, Ge[N{Si(CH3)3}2]3
Lifetime > 10 ns
High sensitivity: detect 1011 spins easily
MCWE region of spectrum: ca. 9.5 GHz (X-band)
Magnetic fields: ca Tesla
CW or FT

2. Physics of electron spin
The energy of an electron (mass m, charge -e) in a magnetic field B is: E = ± (1/2) g mB B
where mB is the Bohr magneton = 9.274´10-24 J T-1
mB is defined as = (e h ) / (4p m)
g is a dimensionless constant; the g-value (chemical shift)
For electrons the spin quantum number S=1/2 with components mS= ± (1/2)
Thus an unpaired electron whose spin orientation is opposite to that of the applied magnetic field has a lower energy (for a H-nucleus the converse is true)

3. ESR Resonance equation h n = g mB B B » 0.3T (3,000 Gauss)
n » 9.5 GigaHz
mB = ´ J / T

4. Example
The species, AlH3-, gives rise to a complex spectrum centered at mT with microwave radiation of frequency GHz.
Compute the g-value for AlH3- .
g = hn / mB B
(6.626´10-34 J s)(9.235´109 s-1)/(9.274´10-24 J T-1)( T)
Þ {(6.626´ 9.235) / (9.274´ ) }
Þ Typical: organics 2.00, inorganics , TMs 0-4

5. Calibration Use well known standard eg Mn2+ for which I=5/2
Doped into MgO at high dilution (I=0 for 24Mg )
Separation of two Mn2+ lines is 8.69 mT
Therefore aN = 1.30 mT and g= for [(O3S)2NO]2–
g=1.981

6. Hyperfine coupling Interaction between e & magnetic nucleus
Splitting of single line spectrum into a number of lines centred on the original
Key formula: If n nuclei of spin I interact equally with an unpaired electron then:
No. of lines = 2nI +1
If more than one set of identical nuclei then
No. of lines = P(2nI +1)
Examples: AlH3-
Al n=1 I=5/2
H n=3 I=1/2
Þ [2´1´ (5/2)+1][2´3´ (1/2)+1]
Þ [6] ´[4] = 24 lines
H· n=1 I=1/2
Þ [2´1´ (1/2)+1] = 2
CH3· n=3 I=1/2
Þ [2´3´ (1/2)+1] = 4

7. Intensities of lines?
For spin 1/2 nuclei only the binomial distribution predicts the intensities
Eg H· the 2 lines are 1:1
CH3· the 4 lines 1:3:3:1
·C6H6- radical anion the 7 lines 1:6:15:20:15:6:1
Other cases:
·VO2+ the 8 lines 1:1:1:1:1:1:1:1

8. Typical ESR spectrum H atoms; 2 lines Separation? Hyperfine coupling
50.7 mT
Independent of magnetic field
g-value measured in centre of multiplet
Note signal presented as derivative (dS/dB)

9. Zeeman effect n1 = ne + a/2 n2 = ne - a/2 d = a
Degenerate energy levels
Electron Zeeman effect
ne = 10 GHz
Nuclear Zeeman
nN = 14 MHz
Electron nuclear interaction, aH
n1 = ne + a/2
n2 = ne - a/2
d = a

10. Origin of hyperfine coupling
Dipole-dipole interaction (distant electron & magnetic nucleus)
If electron in s-orbital or molecule rotates rapidly (as in gas phase or solution) then dipole field averages to zero. Non-zero solid state; anisotropic hyperfine coupling‡.
Fermi contact interaction
Depends upon electron density at nucleus
a(nucleus) µ { ge gN |y (r=0)|2
Electrons in s-orbital only
Calculated values:
1H 1s 50 mT 19F 2s 1.7T
14N 2s 55 mT 14N 2p 3.4 mT‡

11. ESR Spectral analysis (WinSim)
13CH3·
Two quartets
Two coupling constants
aH = 2.3 mT
aC = 4.1 mT

12. Simulated ESR spectra Fire up, click on Simulations menu, parameters
WinSim, by Dave Duling (1996) National Institute of Environmental Health Sciences
Fire up, click on Simulations menu, parameters
Toggle “Calculate this species?” on
Enter appropriate values for:
Set Hyperfine Spin Number
Change Simple LW from 0.50 as required
Now Simulate; green ESR display onscreen
Pick Display menu, toggle FT Imaginary off, Update & Close
Use Edit menu & Copy data to paste into Excel & plot

13. Mapping unpaired e density
For H· aH = (+)50.68 mT 100% in 1s AO
But in CH3· aH = (-)2.3 mT so e only spends (2.3/50.68) ´100% on 1s AO on a H
4.5% in a specific 1s H AO or 13.5% on all 3 Hs
Therefore ( =86.5) % on C
For 2s e on 13C computed field is mT
In 13CH3· aH = (+)4.1 mT so (4.1/111.5) ´100=3.7%
So ( )=96.3% on 2p C AOs
Hence methyl radical is planar

14. Spin polarisation How come?
Electron in 2p AO polarizes s electrons so that neighbouring e has same spin, the other e has opposite spin
Near H atom s bond » 1s AO so e can interact with magnetic nucleus
Note sense of interaction is different (-ve coupling)

15. Allyl radical Unpaired electron in non-bonding MO
y2 = 0.707f f3
So e interacts only with 4 terminal Hs not central H
But from ESR spectrum
a(terminal H) = (-) 1.39 mT
a(central H) = (+) 0.40 mT
Enhanced unpaired spin at terminals and at central H

16. Line shape analysis Electron transfer reactions (NC)2C=C(CN)2
(TCNE·)- + TCNE ® TCNE + (TCNE·)-
(NC)2C=C(CN)2
4 equivalent Ns with I=1
Slow:
(2´4´1 + 1) = 9 nonet
Fast:
collapse to single line
Intermediate:
Line fit Þ kinetics of reaction

17. Spin labels
If compound of interest does not have an unpaired electron it has no ESR.
Attach (by reaction) a compound (spin label) that does; its ESR spectrum will hopefully change reflecting the new environment that the spin label is in.
Piperidinyloxy free radical

18. Spin traps
The compound of interest has a very small lifetime & is therefore invisible in the ESR spectrometer.
Trap it with a diamagnetic reactant to form a new more stable compound but still containing an unpaired electron.
CH3OH + g ® ·CH2OH
CH3O· not observed

19. Electron-Nuclear Double Resonance
Anthraquinone radical anion has 4 a-Hs & 4 b-Hs and thus a 25 line ESR
ENDOR spectrum consists of just 4 lines at 13.12, 13.72, & MHz
n(ENDOR) = |nH ± (aH/2)|
Average to nH = MHz
a1 = = 2.74 MHz
a2 = = 1.54 MHz
Or a1 = 9.8 mT & a2 = 5.5 mT
Use ESR spectrometer
Pick a specific B & n
Increase MCWE power until that ESR signal disappears
Irradiate with a second transmitter from 0 to 50 MHz
ESR signal reappears momentarily at two frequencies
Each set of equivalent spin 1/2 nuclei gives rise to just 2 ENDOR lines

20. ENDOR Example
ENDOR of 1,4,5,8-tetrakis-(trimethylsiyl)-D-octalin radical anion has 10 lines (all in MHz):
In this spectrometer nH»14.6 MHz & nSi»2.9 MHz
How many sets of equivalent nuclei in this species?
Pick pairs of lines which average to 14.6 MHz (2.9??)
(14.73, 14.43) Þ hence a1= =0.30 MHz
(14.86, 14.31) Þ14.585, (15.73, 13.27) Þ14.50 no!
(15.90, 13.27) Þ14.585,
(33.23, 4.07) difference is 2´nH sum=a a4=37.30 MHz