Physical methods of structural determination

Question 1

What does nuclear spin depend on and give examples of elements with nuclear spin I = 0, integar, half integar?

Answer 1

Nuclear spin depends on Z and the isotoic mass

^even_evenE I = 0 e.g. ¹²₆C, ¹⁶₈O

^even_oddE I = integar e.g. ¹⁴₇N, ²₁H

^oddE I = half integar e.g. ¹³₆C I = 1/2, ¹⁹₉F I = 1/2

When I ≥ 1 the nucleus is quadropolar

Question 2

How many spin states does any given element have?

Answer 2

2I + 1

Question 3

How does isotope abundance affect the NMR?

Answer 3

When the isotopic abundance is below 100% only a fraction of the nuclei show up in the specra - less receptive

Question 4

What is receptivity?

Answer 4

How sensitive the nucleus is

Proportional to γ³NI(I+1)

γ is the gyromagnetic ratio, N is the natural abundance, I is the nuclear spin

Question 5

What is the Larmor frequency?

Answer 5

E = m_IγBħ

Refers to ¹H in a specific field (400 MHz)

γ is the gyromagnetic ratio which is unique to each isotope

Question 6

How sensitive is NMR?

Answer 6

Very insensitive

Question 7

What are the mechanisms of relaxation?

Answer 7

spin-lattice relaxation - T₁ - energy of the excited state is dissipated to the surroundings

Spin-spin relaxation - T₂ - energy of the excited state dissipated to other nuclear spins

In high resolution, solution state NMR in most cases T₁ = T₂

Question 8

How does the Heisenberg uncertainty principle affect NMR

Answer 8

ΔEΔt = ħ/2 (constant)

radiatve decay gives a spectrum - when there is very fast decay there is no radiatve process and no specrum

ΔE = uncertainty in energy - line width

Δt =fast ΔE large which implies broad spectrum

For ¹H Δt = long so ΔE is small = high resolution spectrum

Question 9

How does the Heisenberg uncertainty principle affect Paramagnetic species?

Answer 9

Δt - short when there is an unpaired electron

γe (of the electron) ≈ 10³γ_¹H

Relaxation α γ₁² x γ₂²

Very fast relaxation = broad lines and change in chemical shift

Question 10

What are quadropolar nuclei?

Answer 10

Nuclei with spin ≥ 1

In a nucleus where I = 1/2 the charge distribution is spherical

When I ≥ 1 there is a prolate distribution (non spherical) - there is an electric field gradient at the nucleus

Quadropole initiates relaxation - charge couples with the non spherical charge distribution in the molecule

Question 11

What is the rate of quadropolar relaxation?

Answer 11

When q_z does not equal 0 R_Q does not equal 0 and realxation is fast

When q_z = 0 R_QR = 0 and relaxation is slow and a spectrum is observed

Question 12

How many lines are seen for NMR actuve nuclei in the spectra?

Answer 12

2nI + 1 lines

For aluminium flouride (AlF₄)

Al: I = 5/2

(2 x 1 x 5/2) = 6 lines

F: I = 1/2

(2 x 4 x 1/2) = 5 lines

Question 13

How does couplingto quadropolar nuclei show up

Answer 13

Non binomial distibution

For CH₂D₂ the ¹H and ¹³C{¹H} spectra show 1:2:3:2:1 quintet

Question 14

Why do we often decouple a spectra?

Answer 14

Decoupling can suppress magnetic inequivelence

Question 15

What are the spins, isotopes and abundances of Pd, Rh and Pt?

Answer 15

Pd (22% ¹⁰⁵Pd I = 5/2 (quadropolar) the rest is I = 0) - no coupling - quadropolar nucleus means relaxation is too fast

Rh (100% ¹⁰³Rh, I = 1/2) - every Rh centre will couple

Pt (33% ¹⁹⁵Pt, I = 1/2) The other are all I = 0

Question 16

What are satellites?

Answer 16

When a nucleus such as phosphorus couples to Pt 66% of the nuclei have I = 0 and therefore do not couple and therefore the phosphorus environments show up as singlets

33% of the Pt nuclei couple and show up as a doublet either side of the singlet - The J value between them is J_Pt-P - These are satellites

Satellites show up in spin dilute systems

Question 17

What is partial hydrogen decoupling?

Answer 17

When only some of the coupling to hydrogen is excluded

For example to hydrogens are decoupled leaving any hydrides

Question 18

How does the transeffect affect the J coupling constant?

Answer 18

The substituents trans to each other are competeing for the same orbital

If one substituent outcompetes another the bond to that substituent will be stringer and the J value will be higher

The weaker the trans influence of the substituent transto an NMR active substituent the higher the J value

Water, Cl have weak trans influences

CO has string trans influence

Question 19

How does oxidation state affect the J couplig constant?

Answer 19

Generally when oxidation state increases the J constant decreases

This is due to the hybidisation - coupling is larger with greater s character as the s penetrates the nucleus better than the p and d orbitals

Question 20

How do we calculate timescale?

Answer 20

Timescale = 1/frequency of experiment

NMR - MHz - 10⁶ s^-1 so timescale is 10^-6 s

IR - THz - 10¹² s ^-1 so timescale is 10^-12 s

Individual species involved in exchanges can be ‘seen’ if rate f exchange is less than the experimental timescale

Berry pseudo rotation happens too fast on the NMR timescale and cannot be seen - only 1 environment observed

IR happens on faster timescale so can show BPR

Question 21

Is it ever possible to see Berry Pseudo rotation in NMR?

Answer 21

If we cool the solution down so that rotation is on an NMR timescale we can observe multiple environments

Have to cool to about -132°C

Question 22

What is EPR?

Answer 22

Electron Paramagnetic Resonance

Akin to NMR spectroscopy but uses spin of unpaired electron (I = 1/2)

0.34

No liquid helium or nitrogen needed

Frequencies 9.5 —> 35 GHz

uses microwaves

Question 23

What is the g value in EPR?

Answer 23

‘The chemical shift’ of EPR

typical organic radicals:

1.99 ≤ g ≤ 2.01

Transition metals 0 < g < 6 - spin orbit coupling

• > 1 unpaired electron

Question 24

What is hyperfine coupling?

Answer 24

electron spin will interact wth the magnetic dipole of neighbouring spins

Energy of electron spin-state, E, is modified to

E = gμ_BB₀M_s + aM_sm_I

a - hyperfine coupling (mT) (G) - gaus

m_I - spin of the interacting nucleus

Δm_s - ±1

Δm_I = 0

Question 25

What is observed an electron couples to a single nucleus with spin 1/2?

Answer 25

Same rules as NMR apply - doublet shows Shape of peaks different

Question 26

Can unpaired electrons couple to quadropolar nuclei?

Answer 26

Yes The timescale of the EPR spectra is much faster (10⁹ s^-1) so rapid relaxation not a probelm

Question 27

What can EPR tell us about the unpaired electron?

Answer 27

Can give us informaton about which atom/s the electron is on and therefiore wich molecular orbital the electron is in The NO logand can exist as NO**^.** keeping the unpaired electron, NO⁺ where the electron is donated or NO^- where the electron is accepted. NO**^.** - is diamagnetic so should not show peaks in EPR NO⁺ - should show coupling to Fe but not N NO^- - should show coupling to N but not Fe

Question 28

WHat are EXAFs?

Answer 28

Etended X-ray Absorption Fine Structure A synchrotron-based technique Incoming X-rays eject a core electron and the electron is observed as a spherical wave which can be backscattered

Question 29

What happens in an EXAFS experiment?

Answer 29

An electron is ejected The electron sees the electron density of neighbouring atoms and is back scattered Back scattering can inpinge on teh original atom in phase or out of phase This affects the absorption characteristics of the atom being observed

Question 30

What is the difference between in phase and out of phase?

Answer 30

Question 31

What is scattering dependent on?

Answer 31

Atomic number of scatterng nuclei Their distance from the absorbing atom The extent of bond vibration Scattering generates an absorption spectrum with peaks showing constructive interference and trophs showing destructive interference The line eventually dies away

Question 32

What are the experimental considerations of EXAFS?

Answer 32

High beam intensities access elements with approx Z ≥ 15 (P) do not need single crystals use on liquids use on biological sample used widelu in the study of heterogenous catalysts

Question 33

How do you process EXAFS spcetra?

Answer 33

Correct for the slope Change X axis from energy to k using k α √E-E₀ where E is the observed energy and E₀ is the energy at absorption edge Remove background data amd normalise

Question 34

WHat is a limitation of EXAFS data?

Answer 34

DAta looks identical for cis and trans No radical information