Nuclei with I=1/2 and <100% abundance and quadrupolar nuclei

Question 1

What does the ¹⁹F NMR spectrum look like for WF₆ (Oₕ symmetry)
¹³⁸W has I = 1/2, abundance = 14.4%
Coupling to all other W can be ignored

Answer 1

• 100-14.4 = 85.6% of all molecules of WF₆ will not contain ¹³⁸W
• Since all fluorines are equivalent and no couplings to W is observed
• A singlet in the ¹⁹F NMR will be observed (85.6% spin-inactive W)
• The other 14.4% have ¹³⁸W (with I = 1/2) where coupling will be seen as a doublet (satellite peaks)

Question 2

What does the ¹H NMR spectra of Sn[N(CH₃)₂]₄ (Td symmetry at Sn) look like?
¹¹⁷Sn (7.6% abundant) and ¹¹⁹Sn (8.6% abundant), I = 1/2 for both

Answer 2

• 100 - (7.6+8.6) = 83.8% are spin-inactive and will produce a singlet
• ¹¹⁷Sn: (2x1x0.5) + 1 = 2 = doublet (7.6/2 = 3.8%)
• ¹¹⁹Sn: (2x1x0.5) + 1 = 2 = doublet (8.6/2 = 4.3%)

Question 3

What does the ¹⁹F NMR spectra of W(CO)₅(PF₃) look like?
³¹P (100% abundant), I = 1/2, ¹³⁸W (14.4% abundant), I = 1/2

Answer 3

• 1J FP: (2x1x0.5) + 1 = 2 = doublet
• 2J FW: (2x1x0.5) + 1 = 2 = doublet

Question 4

What does the ³¹P NMR spectra of W(CO)₅(PF₃) look like?
¹⁹F (100% abundant), I = 1/2, ¹³⁸W (14.4% abundant), I = 1/2

Answer 4

• 1J PF: (2x3x0.5) +1 = 4 = quartet
• 1J PW: (2x1x0.5) + 1 = 2 = doublet

Question 5

Quadropolar nuclei are where I =

Answer 5

> 1/2

Question 6

How can we differentiate between planar (sp²) boron (D₃ₕ) and tetrahedral (sp₃) boron (Td) using NMR

Answer 6

• Planar (sp²) boron compounds are more deshielded (higher resonance ppm) compared to the tetrahedral ones
• This is due to boron in sp² compounds being electron deficient as it does not have enough valance electrons to fill all of its valance orbitals
• e.g. BMe₃ ¹¹B resonance is at +86.2ppm while BH₄⁻ ¹¹B is at -38.2ppm

Question 7

Explain the following trend

Answer 7

• Consistent with sequence of shielding (hence electron density of B atom):
• BI₃ > BF₃ > BBr₃ > BCl₃
• Oder parallels electronegativity of halide, except for BF₃
• Reason for this is that in the B-F bond there is a match in size between the (full) pz orbtial on the F and the (empty) pz orbital on the p
• Backbonding of electron density to B leads to greater than expected shielding

Question 8

I = 1 has how many spin state

Answer 8

3 possible spin states
all with equal intensities

Question 9

What does the ¹H NMR look like of ¹⁴NH₄⁺
¹⁴N I = 1

Answer 9

(2x1x1) + 1 = 3 = triplet
Intensity of 1:1:1

Question 10

What does the ¹H NMR look like of ¹¹BH₄⁻
¹¹B I = 3/2

Answer 10

(2x1x3/2) + 1 = 4 = quartet
with equal intensities of 1:1:1:1

Question 11

Coupling to one ¹¹B give the expected equal intensity quartet
Interaction with two ¹¹B nuclei with I = 3/2 gives?

Answer 11

• 7 possible different arrangements, with relative likelihoods in the order:
• 1:2:3:4:3:2:1

Question 12

Coupling to two I = 3/2 nuclei gives 7 likelihoods in the order: 1:2:3:4:3:2:1
What does coupling to 3 I = 3/2 nuclei give?

Answer 12

1:3:6:10:12:12:10:6:3:1