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 Tesla CW 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 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)

ESR Resonance equation h n = g mB B B » 0.3T (3,000 Gauss) n » 9.5 GigaHz mB = 9.274 ´ 10-24 J / T

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 = hn / mB B (6.626´10-34 J s)(9.235´109 s-1)/(9.274´10-24 J T-1)(0.32948 T) Þ {(6.626´ 9.235) / (9.274´ 0.32948) } 10-34+9+24 Þ 2.00259 Typical: organics 2.00, inorganics 1.97-2.02, TMs 0-4

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= 2.0057 for [(O3S)2NO]2– g=1.981

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

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

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)

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

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‡

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

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

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 (100-13.5=86.5) % on C For 2s e on 13C computed field is 111.5 mT In 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 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)

Allyl radical Unpaired electron in non-bonding MO y2 = 0.707f1 - 0.707f3 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

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

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

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

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, 15.26 & 15.86 MHz n(ENDOR) = |nH ± (aH/2)| Average to nH = 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 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

ENDOR Example ENDOR of 1,4,5,8-tetrakis-(trimethylsiyl)-D-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 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) Þ14.580 hence a1=14.73-14.43=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