Electron Spin Resonance Spectroscopy ESR / EPR Absolute requirement: 1 or more unpaired electrons –Stable species: O 2, NO, [Fe(CN) 6 ] 3 –Transients:

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. 0.35 TeslaCW or FT

Physics of electron spin The energy of an electron (mass m, charge -e) in a

magnetic field B is: E = (1/2) g B B

where B is the Bohr magneton = 9.27410-24 J T-1

B is defined as = (e h ) / (4 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)

ESR

Resonance equation h g BB

B Gauss)9.5 GigaHz

B = 9.274 10-24 J / T

S

K lys tro n D etec to r

N

E ne rg y M ag n e t i c fi e l d

B Z

Example

The species, AlH3 gives rise to a complex spectrum

centered at 329.48 mT with microwave radiation of frequency 9.235 GHz.

Compute the g-value for AlH3 .

g = h / B B

(6.62610-34 J s)(9.235109 s-1)/(9.27410-24 J T-1)(0.32948 T)

(6.6269.235) / (9.2740.32948) } 10-34+9+24

2.00259 Typical: organics 2.00, inorganics 1.97-2.02, TMs 0-4

Calibration

Use well known standardeg 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= 2.0057 for [(O3S)2NO]2–

3300.00 3320.00 3340.00 3360.00 3380.00 3400.00

g=1.981

Hyperfine couplingInteraction 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 = (2nI +1)

Examples: AlH3

– Al n=1 I=5/2– H n=3 I=1/2 [21 (5/2)+1][23

(1/2)+1] [6] [4] = 24 lines

H n=1 I=1/2

[21 (1/2)+1] = 2 CH3

n=3 I=1/2

[23 (1/2)+1] = 4

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

-500

-400

-300

-200

-100

0

100

200

300

400

500

3340 3342 3344 3346 3348 3350 3352 3354 3356 3358 3360

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)3300 3350 3400

-400

-200

0

200

400 50.7 mT

Zeeman effect Degenerate

energy levels Electron

Zeeman effect e = 10 GHz

Nuclear Zeeman N = 14 MHz

Electron nuclear interaction, aH

1 = e + a/2

2 = e - a/2 = a

Z E R O F IE L D

F I E L D B Z

m S = + 1 /2

m S = - 1 /2

m I

- ( 1 /2 )

+ ( 1 /2 )

- ( 1 /2 )

+ ( 1 /2 )

e

N

N

2

1

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) { e N | (r=0)|2

– Electrons in s-orbital onlyCalculated values:– 1H 1s 50 mT 19F 2s 1.7T– 14N 2s 55 mT 14N 2p 3.4 mT‡

ESR Spectral analysis (WinSim)

13CH3

Two quartets

Two coupling

constants

aH = 2.3 mT

aC = 4.1 mT-500

-400

-300

-200

-100

0

100

200

300

400

500

3340 3342 3344 3346 3348 3350 3352 3354 3356 3358 3360

Simulated ESR spectraWinSim, 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

Mapping unpaired e densityFor 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 (100-13.5=86.5) % on CFor 2s e on 13C computed field is 111.5 mTIn 13CH3

aH = (+)4.1 mT so (4.1/111.5) 100=3.7%

So (100-3.7)=96.3% on 2p C AOs Hence methyl radical is planar

Spin polarisation

How come? Electron in 2p AO polarizes electrons so that neighbouring

e has same spin, the other e has opposite spin Near H atom bond 1s AO so e can interact with

magnetic nucleus Note sense of interaction is different (-ve coupling)

C H C H

Allyl radical Unpaired electron in non-

bonding MO = 0.707 - 0.707

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

C C C

C C C

1 2 3

Line shape analysis

Electron transfer reactions(TCNE) + TCNE TCNE + (TCNE)

(NC)2C=C(CN)2

4 equivalent Ns with I=1 Slow:

– (241 + 1) = 9 nonet Fast:

– collapse to single line Intermediate:

– Line fit kinetics of reaction

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

N

NH2

O

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 + CH2OH

CH3O not observed

N

O

C(CH3)3

N-tert-Butyl Phenyl-nitrone

N

O

5,5-Dimethyl-1-pyrroline N-oxide

Electron-Nuclear Double ResonanceUse ESR spectrometer Pick a specific B & 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

Anthraquinone radical anion has 4 -Hs & 4 -Hs and thus a 25 line ESR

ENDOR spectrum consists of just 4 lines at 13.12, 13.72, 15.26 & 15.86 MHz(ENDOR) = |H aH/2)|

Average to H = 14.49 MHz– a1 = 15.86 - 13.12 = 2.74 MHz

– a2 = 15.26 - 13.72 = 1.54 MHz

– Or a1 = 9.8 mT & a2 = 5.5 mT

ENDOR Example ENDOR of 1,4,5,8-tetrakis-(trimethylsiyl)--octalin

radical anion has 10 lines (all in MHz):4.07 13.27 14.31 14.43 14.73 14.86 15.73 15.90 21.53 33.23

In this spectrometer H14.6 MHz & Si2.9 MHzHow many sets of equivalent nuclei in this species?Pick pairs of lines which average to 14.6 MHz (2.9??) (14.73, 14.43) 14.580 hence a1=14.73-14.43=0.30 MHz (14.86, 14.31) 14.585, (15.73, 13.27) 14.50 no! (15.90, 13.27) 14.585, (33.23, 4.07) difference is 2H sum=a a4=37.30 MHz