Lecture 4

Question 1

• What is the McConnell equation?

Answer 1

• Relates the magnitude of isotropic hyperfine coupling to the amount of spin density on an α-proton
• Used to describe spin polarized systems where an unpaired electron is localised in a p-orbital on a carbon that has no spin density at the nucleus, therefore hyperfine cannot be explained via fermi contact interaction
• Hunds rule says that electrons near this radical will preferentially have the same spin orientation
• This perturbs spin density in C-H sp³ hybrid orbitals, which have s-character localised on hydrogen atoms (hyperconjugation)
• Therefore, it is possible to measure hyperfine due to fermi contact interactions between a radical and a hydrogen atom

Question 2

• What is the largest hyperfine splitting in this molecule?

Answer 2

• All dihydrogen environments produce a 1:2:1 splitting pattern except for H_para (1:1)
• Largest splitting from radical as any unpaired electron leaking on to the ring giving further splitting (consequence of McConnell equation

Question 3

• How does bond polarizability change with bond angle of the hyperfine of an adjacent carbon?

Answer 3

• Isotropic β-hyperfine constants explained by (hyperconjugation) overlap of the p-orbital bearing the unpaired electron with the sp³-orbital of the C-H bond at the adjacent carbon
• |a| = Bcos²θ – implies bond is polarised better is parallel and poorer if perpendicular

Question 4

• Originally said … hyperfine originates from unpaired electron localised on … we are … …
• However now know it does not have to be localised there, can be somewhere else and create a net … of … …/… on atom looking at resulting in a …

Answer 4

• Originally said isotropic hyperfine originates from unpaired electron localised on atom we are looking at
• However now know it does not have to be localised there, can be somewhere else and create a net excess of spin up/down on atom looking at resulting in a hyperfine

Question 5

• How is the hyperfine interaction of a p orbital related to the g tensor in the Hamiltonian describing the anisotropic Zeeman interaction?

Answer 5

• G tensor is different along all axis, such that varying direction of B₀ will give a different resonance along any given position to the next (as seen before)
• In a powder spectrum with axial symmetry the hyperfine interaction (A) can also be anisotropic as electrons in a p orbital will have a stronger hyperfine interaction in the z than xy axis

Question 6

• Describe the resulting EPR spectra of the following conditions
• i) a­_iso­ << b
• ii) a_iso = b
• iii) a_iso >> b

Answer 6

• i) get regular isotropic spectrum as tensor is just a­_iso down diagonal (d)
• ii) 0 0 3b on diagonal …?
• iii) 2 lines separated
• intensities depend on how B₀ aligns with axis
• c is a pake pattern

Question 7

• To calculate hyperfine interactions, one must average over unpaired electron wavefunction to attain a distribution, what are the resulting values of the tensor in this interaction and why do they take the form they do?

Answer 7

• Magnitude of hyperfine along x/y = -b, z = 2b
• Must = 0 overall for anisotropic hyperfine as when tumbling fast, averages to 0 to give isotropic hyperfine.

Question 8

• What comprises real world measurements of the hyperfine interaction?

Answer 8

• A combination of isotropic and anisotropic components
• A_exp = experimentally observed hyperfine
• A_iso = calculated isotropic (Fermi contact) hyperfine
• b = calculated anisotropic (dipolar) hyperfine

Question 9

• Describe the result of increasing the microwave frequency on the powder spectrum of a system with anisotropic g and hyperfine values

Answer 9

• System not tumbling quick enough to average out anisotropy –> get anisotropic interaction
• Changing magnetic field for a given microwave frequency allows interrogation of anisotropy (hyperfine interaction stays constant as is field independent)
• As increase from 1.5 –> 9.75 –> 94 GHz g anisotropy resolved out
• At very low field (plot 1) cant resolve hyperfine as all in middle
• At very high field (plot 3) get three equally spaced lines where anisotropy has resolved. Smallest g value occurs at highest field –> g_zz pushed to right
• g_xx/yy and A_xx/yy somewhere mixed together in spectrum still, must move to higher field to separate similar g values wth identical hyperfine splitting

Question 10

• Determine the spin Hamiltonian parameters from the spectrum, assuming a microwave frequency of 9.25 GHz. Given that a₀ = 557G and b₀ = 33.35 G for ¹⁴N, and assuming b = a_iso – A_T, calculate the electron spin density on the ¹⁴N atom.

Answer 10

• I_14N = 1, 2I + 1 = 3 lines
• Must be an anisotrpic g tensor and hyperfine splitting from N as 2 sets of lines one of which is the resolved A_II (z-axis/principle)
• A_T = 55G (measured off plot)
• A_II = 20G (measured off plot - accidentally wrote A_T)
• a_iso = (A­_T + A_T + A_II)
• a_iso = (55 + 55 + 20)/3
• a_iso = 43.33 G
• c_s² = a_iso/a₀
• c_s² = 43.33/557G = 8%; s character:1
• b = a_iso - A_T
• c_p² = b/b₀
• c_p² = 11.67/33.5 = 35%; p character:4
• indicates large amount of localisation spread across molecule
• orbital hybridisation: λ²= c_s²/c_p² = 8/35 = 0.23

Question 11

• What is a major drawback of EPR and what is a solution?

Answer 11

• Small number of paramagnetic systems
• Introduce a paramagnetic tracer molecule (e.g. a nitroxide)

Question 12

• Why are nitroxides so stable?

Answer 12

• Spin density delocalized favours electronegative O(40 % N, 60% O)
• Full methyl substitution in the β-position also provides steric hindrance as well as removal of structural motif of a β-proton
• This inhibits typical radical reactions like dimerization’s and disproportionation’s

Question 13

• What is a spin label?

Answer 13

• Stable radical (often a nitroxide) that can be attached to a non-paramagnetic molecule to probe structure, dynamics and polarity
• E.g. attach a label to all sulphur groups on a specific amino acid in a protein, do ESR
• Looking at dynamics results, degree of rotation speed gives insight into whether the label is inside (hindered) or outside the molecule.

Question 14

• What effect does spin labelling with a nitroxide have on an EPR powder spectrum

Answer 14

• Motion of spin probe is strongly influenced by dynamics and local structure of its surroundings.
• EPR is sensitive to rotational diffusion rather than translational
• If the spin probe rotates fast, hyperfine couplings in different directions average to an isotropic value, a_iso giving a spectrum of three equal lines
• When rotational motion is prohibited, an anisotropic spectrum forms, indicating something about environment surrounding probe and therefore structure.

Question 15

• What links the tumbling time of a solution to the rotational motion of a spin label and its affect of an EPR powder spectrum?

Answer 15

• τ_c = 1/6D_R
• when correlation time low enough, rotational diffusion occurring fast enough so tumbling produces isotropic spectra
• as correlation time slows, anisotropy introduced as rotation of nitroxide hindered in several directions

Question 16

• What is a spin trap?

Answer 16

• Solution to trapping specific short -lived radicals and boost detection in EPR
• Produced via chemical reaction (e.g. DMPO – a nitroxide) with molecule trying to trap
• Radical adduct stable for EPR but still short lived
• O is an EWG so depending on what radical is bound, will change in magnitude of H_β hyperfine splitting.