Further consideration for Quadrupolar nuclei and fluxionality

Question 1

What would the ¹H NMR look like for isotopically unenriched KBH₄
¹¹B (I = 3/2; 80.4%) ¹⁰B (I = 3; 19.6%)

Answer 1

The ¹H NMR spectrum will therefore contain sets of resonances from both isotopes
¹⁰B: (2x1x3) + 1 = 7 = septet
¹¹B: (2x1x3) + 1 = 4 = quartet
1:4:6:4:1 ratio

Question 2

What is a quadropole moment (eQ) ?

Answer 2

Measure of distortion from spherical charge distribution
I > 1/2

Question 3

All I = 1/2 nuclei have eQ =

Answer 3

0
all I > 1/2 nuclei have eQ ≠ 0 (quadropolar nuclei)

Question 4

What creates a defined line width in a NMR spectra

Answer 4

• Any nucleus with I > 0, absorption of radiation occurs, raising the nucleus to a higher energy level
• If the energies of the inital and final states are well defined the resulting line in the spectrum will be very sharp

Question 5

How does Heinsenburg uncertainity principle relate to NMR spectra lines?

Answer 5

• Heisenburg consideres ΔE = uncertainity in energy state and Δt = uncertainity in time specified
• If the nucleus occupies a state for a long time, then Δt is very large so ΔE is very small, and energy can be defined very precisely
• Ground state is always long lived - so well defined (the case for I = 1/2)
• But for I - >1/2 the excitied state is only occupied for a short time, therefore a much larger uncertainity with ΔE = uncertainity in energy state, which gives a much broader line (high res may not be possible)

Question 6

How can we get high resolution spectra for quadropolar nuclei?

Answer 6

For many quadrupolar nuclei, we only get high-resolution spectra (narrow lines) when they are in a very symmetrical situation (low or no field gradient at the nucleus because of lost dipole moment)

Question 7

Why is there no spin-spin splitting by quadrupolar nuclei for a spectra of a I = 1/2 nuclei

Answer 7

• looking at spectra of I = 1/2 nuclei - if any quadrupolar nuclei are present in the molecule, these nuclei will not give any spin-spin splitting if they are in an unsymmetrical environment, and they can be ignored
• This is due to the nuclei rapidly transitioning between their spin states
• e.g. the Cl in ClO₃⁻

Question 8

What do you expect the NMR to look like for the following molecule?

Answer 8

• SF₄ is an AB₄L shape (see-saw) predicted by VSEPR
• Results in 2 axial Fs, and 2 equatorial F
• Hence we would expect two ¹⁹F resonances of equal intensity, split into a 1:2:1 triplet

Question 9

What does the NMR to look like for the following molecule look like in reality

Answer 9

• A singlet due to fluxionality between the axial and equatorial positions (NMR is slow)
• However upon cooling the sample within the NMR, preventing the fluxionality, the expected outcome is seen instead

Question 10

What is the difference between the ³¹P NMRs for the following molecule where there is static vs rapid exchange?

Answer 10

rapid exchange: (2x5x0.5)+1 = 6 = sextet
static exchange: (2x2x0.5)+1 = 3 = triplet AND (2x3x0.5)+1 = 4 = quartet SO a quartet or triplets is produced

Question 11

What is the difference between the ¹⁹F NMR for the following molecule where there is static vs rapid exchange

Answer 11

rapid exchange: (2x1x0.5)+1 = 2 = doublet
static exchange: 1J FeqP: (2x1x0.5)+1 = 2 AND 2J FeqFax: (2x2x0.5)+1 = 3, SO doublet of triplets
1J FaxP: (2x1x0.5)+1 = 2 AND 2J FaxFeq: (2x3x0.5)+1 = 4, SO doublet of quartets