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3.1 Determine a spin category for a 139La nucleus.
139La has Z = 57 and n =82 which puts it in the odd ( protons) – even (neutrons) category so we would expect I = n/2. In fact I = 7/2
3.2 Calculate the resonance frequency of a 19F nucleus in an 11.744 T magnetic field using the data in Table 3.2.
From Table 3.2, γ for 19F is γ = 2.5181 x108 rad s−1T−1 with B0 = 11.744 T, ν = 4.707x108 s–1, which is 470.7 MHz
3.3 Explain why 125Te is the preferred over 123Te for NMR spectroscopic studies of tellurium compounds.
The key parameters that control the receptivity are the natural abundance and gyromagnetic ratio. From Table 3.1 the natural abundance of 125Te is 8 times that of 123Te while their gyromagnetic ratios are similar. The receptivity of 125Te is thus around 8 times that of 123Te, so is the preferred nucleus as NMR data collection times will be much shorter.
3.4 Calculate the resonance frequencies of 203Tl and 205Tl nuclei in an 11.744 T magnetic field using the data in Table 3.12 Comment on your answer.
From Table 3.2, γ for 203Tl is γ = 1.5539 x108 rad s−1T−1 with B0 = 11.744 T, ν = 4.707x108 s–1, which is 290.4MHz. γ for 205Tl is γ = 1.5692x108 rad s−1T−1 with B0 = 11.744 T, ν = 4.707x108 s–1, which is 293.3 MHz. There is only a small difference (~1%) in resonance frequencies and this means that signals from two different nuclei of the same element, 203Tl and 205Tl, can (unusually) be seen in the same NMR spectrum/ frequency range.
3.5 Predict the number of different fluorine environments and, therefore, the number of 19F resonances in (a) a trigonal bipyramidal molecule, such as PF5 and (b) an octahedral molecule, such as SF6.
(a) Trigonal bipyramidal PF5 has two axial and three equatorial fluorine environments so two resonances should be seen in the 19F NMR spectrum with relative intensities of 2:3. ( note that this prediction ignores the effect of spin-spin coupling)
(b) SF6. All the apical sites are equivalent in a perfect octahedron (there are no axial or equatorial sites in a regular octahedron). Therefore a single resonance will be seen in the 19F NMR spectrum.
3.6 Predict the 19F NMR spectra of the three possible XeO3F2 isomers shown in Figure 3.7(c). Is it possible to distinguish all these isomers using 19F NMR?
In (i) the two fluorine atoms both occupy equatorial positions and one resonance will be observed in the 19F NMR spectrum
In (ii) similarly the two fluorine atoms both occupy equivalent equatorial positions and one resonance will be observed in the 19F NMR spectrum.
In (iii) there are two distinct fluorine environments (one axial and one equatorial) so these will resonate at two distinct frequencies in the 19F NMR spectrum.
Therefore structure (iii) can be distinguished using 19F NMR but in terms of the number of resonances (i) and (ii) cannot be distinguished, though characteristic frequencies for axial and equatorial fluorine atoms could be used to suggest a likely structure.
3.7 Explain the chemical shift variation in the series Si-H, P-H and S-H shown in Table 3.4.
The values are Si-H (3 to 4 ppm), P-H (1.8 to 4 ppm) and S-H (1.5 to 2.5 ppm) so the direction of shift is to more negative ppm in the series Si→P→S. That is, as the element to which hydrogen is bonded becomes more electronegative the hydrogen becomes more shielded. This initially seems opposed to the view that a neighbouring electronegative atom would withdraw electrons and so become deshielded. However many other factors contribute to shielding and chemical shifts. For example the bond lengths decrease in the order Si-H (1.48 Å), P-H (1.44 Å) and S-H (1.34 Å) and this may increase the electronic shielding around H providing the observed trend. This analysis shows the difficulties in equating chemical shift values with chemical environment as many factors can contribute.
3.8 Predict the 19F NMR spectrum of the Sb2F11 – anion. Figure 3.9.
(a) There are three different environments for fluorine, a bridging fluorine atom ( ×1), two terminal fluorine atoms opposite ( through antimony) the bridging fluorine (×2) and 8 other equivalent terminal fluorine atoms. Therefore the 19F NMR spectrum will consist of three resonances with relative intensities 1:2:8 ( note that this prediction ignores the effect of spin-spin coupling between fluorine and antimony and fluorine and non-equivalent fluorine)
3.9 Calculate the separation in ppm of the 1J(19F - 31P) 680 Hz doublet in the 19F NMR spectrum of Me4PF on a 500 MHz 1H NMR spectrometer.
J(Hz) = [chemical shift separation (in ppm)] x [spectrometer frequency (when given as a number of MHz)]
From Table 3.2 the 19F resonance frequency equivalent to 500 MHz (1H) 94.094 x 5 MHz = 470.47 MHz
680 = δ (ppm) × 470.47 give δ = 1.45 ppm
3.10 Interpret the 19F NMR spectra in Figure 3.20 from (a) SF5+ (S is NMR inactive) and (b) Me2PF3 (with both methyl groups equatorial in a trigonal bipyramid) in terms of the structure of the species (31P, I = ½, 100%).
(a) The 19F NMR spectrum of SF5+ is a triplet intensity 3 and a quartet intensity 2. The intensities and the two resonance positions can be explained by the presence of two different fluorine environments in the ratio 3:2. The spin-spin coupling schemes observed would also fit this description with the 3 equivalent fluorine sites coupling to the 2 nonequivalent ones to produce a triplet and the intensity 2 resonance coupling with the three non equivalent one to produce a quartet. So any structure with two types of fluorine position in the ratio 3:2 would fit the NMR data. The most common molecular geometry for an AX5 species with this arrangement is a trigonal bipyramid with 3 equatorial X and 2 axial X. Note that SF5+ is isoelectronic with PF5 and VSEPR theory would predict a trigonal bipyramidal shape.
(b) We need to consider the structure of the [PF3] unit, with an equatorially placed fluorine and 2 axial fluorine sites (as we can ignore the methyl groups). Therefore there are two fluorine environments and the 19F NMR spectrum shows two resonances in the intensity ratio 2(axial):1(equatorial). Spin-spin coupling to the directly bonded phosphorus atom (31P, I = ½, 100%) will produce doublets with a coupling constant 1J (19F-31P)in the spectrum. Coupling to the non-equivalent fluorine atom through two bonds, 2J(F(axial)-F(equatorial)), will cause each component of the 1J (F-P) doublet from F(axial) to split into a further doublet producing a doublet of doublets (total intensity 2); and for F(equatorial), 2J(F(equatorial)-F(axial))coupling to the two non-equivalent fluorine atoms will produce triplets from each 1J (F-P) doublet giving a doublet of triplets.
3.11 Predict the 19F NMR spectrum of the [123Sb2F11]– anion; 123Sb I = 7/2.
We can now extend the answer to self test 3.8 to include spin-spin coupling but shall only consider those up to 2J.
Without spin-spin coupling the 19F NMR spectrum consists of three resonances in the intensity ratio 1:2:8 corresponding to the bridging (x1) and terminal (trans to bridging x2 and cis to bridging ×8) fluorine sites.
Considering first the trans to bridging site this will couple to antimony,1J(F-Sb), to produce a 2nI+1 = (2 × 7/2)+1 = 8 octet with all the signals being the same intensity. Coupling to the single trans fluorine will produce doublets and to the 4 cis fluorine sites a quintet form each of these signals. Therefore, is all resonances are resolved this fluorine would produce an octet of doublets of quintets!
Similar analysis for the cis terminal fluorine atoms gives an octet of doublets of doublets
For the bridging fluorine site coupling to equivalent two antimony sites would produce a 2×2×7/2+1 – 15 line multiplet with each of these coupling to cis and trans fluorine atoms give very complex multiple line patterns if all the resonances are resolved.
3.12 Predict the form of the 11B{1H} NMR spectrum of decaborane, 11B10H14, (Figure 3.25). Comment on how this spectrum would compare with that obtained without decoupling; each boron atom has a terminal B-H bond and bridging hydrogen bonds exist along the B-B edges shown in Figure 3.25 (consider only 1J (B-H) couplings).
There are four different boron sites in decaborane, 11B10H14 as shown in the figure so four resonances would be expected in the 11B{1H} NMR spectrum
For the 11B NMR spectrum with 1J(11B-1H) coupling . Considering B(1) first coupling to the terminal hydrogen atom would give a doublet and to the two adjacent bridging hydrogen atoms triplets – so a doublet of triplets for B(1). B(2) has one bridging and one terminal hydrogen site giving a doublet of doublets. B(3) has just a terminal hydrogen yielding a doublet and B(4) similarly a doublet. Overall the 11B NMR spectrum would consist of two doublets, a doublet of doublets and a doublet of triplets so 11B{1H} decoupling simplifies the spectrum considerable.
3.13 Predict the 19F NMR spectrum of the XeF3+ cation which has the same shape as ClF3 shown in Figure 3.17 with Xe replacing Cl.
We now have to account for the 1J(19F-129Xe) coupling with 129Xe (26.5% abundant, I = ½). As before there are two fluorine environments which can labelled F(axial) and F(equatorial). The single F(equatorial) couples to 129Xe in 26.5% of molecules to produce a doublet ( each with intensity ~13%) either side of the main uncoupled resonance (intensity ~74%); each of these signals becomes a 1:2:1 triple by coupling to the two non-equivalent F(axial)s. Similar reasoning for F(axial) produces the three signals with intensity 13:74:13 from 1J(19F-129Xe) 26% of the time and each of the resonances observed as a doublet due to 2J(19F(axial)-19F(equatorial) coupling ; note that the total intensity in these signals will be twice that of those derived from F(equatorial)
3.14 The 1H NMR spectrum, as a function of temperature between 218 and 346 K, of Ge(η1-C5H5)(CH3)3, Figure 3.31(a), is shown in Figure 3.31(b). Interpret the changes observed as a function of temperature.
The low temperature spectrum shows distinct resonances from the methyl and cyclopentadienyl hydrogen atoms at their characteristic chemical shifts. At the very lowest temperature the different hydrogen environments on the η1- Cp can be distinguished; there are three different hydrogen environments. On heating these Cp 1H signal coalesce simultaneously with the disappearance of the methyl hydrogen signal to produce a single average resonance at the weighted midpoint chemical shift at room temperature – demonstrating that all the hydrogen atoms have become equivalent. On the NMR time scale at room temperature the hydrogen atoms are exchanging rapidly between the methyl groups and the Cp ring and also the Cp ring is cycling rapidly between configurations where each of the carbon atoms is bonded to germanium.
3.15 Explain the following 1J(13C-29Si) values R2Si=C(OR)(R), 84 Hz, and R3Si-COCMe3, 34 Hz.
In hybridization terms the orbitals involved in forming the Si-C and Si=C bonds in each of the bonding in the compounds is sp3 and sp2 respectively So the s-character in the bond is 25%, and 33% which follows the expected trend of increased coupling constant with increasing s-character in the bond.
3.16 Determine the expected form of the 31P NMR spectrum of [F2P2O5]2–.
The phosphorus atoms are chemically equivalent, as are the fluorine sites. However with respect to either P1 or P2 the fluorine atoms are magnetically inequivalent. Therefore the 31P NMR spectrum with consist of a doublet of doublets from the 1J(31P - 19F) and 3J(31P - 19F) coupling , with 1J(31P - 19F) >> 3J(31P - 19F).
3.18 Predict the 11B-11B COSY NMR spectra of closo-2-CB6H8 shown in Figure 3.44.
There are three different boron sites, labelled 1, 2 and 3 in the figure. B1 is bonded to 2 x B3 and 2 x B2, B2 to 1 x B1 and 1 x B3 and B3 to 1 x B1, 1x B2 and 1x B3. As each boron site is bonded directly to at least one of each non equivalent boron there will be in the 11B-11B COSY NMR spectra cross terms for all resonances, see Figure.
3.20 Predict the 29Si MASNMR spectrum of a linear trisilicate ion of the formula [Si3O10]8−, [O3Si-O-SiO2-O-SiO3] 8−
This linear trisilicate has two distinct silicon sites, two equivalent terminal Q1 sites and a central Q2 site. Therefore the 29Si MASNMR spectrum should show two resonances one at around −80 ppm (Q1) and the other at around −80 ppm (Q2). Intensities should be in the ratio 2:1.
3.21 Calculate the form of the 29Si MASNMR spectrum expected for a zeolite with a Si:Al ratio of 2:1.
The probability of each neighbouring tetrahedral site around a central Si site being Al is 1/3. The probability of each silicon environment Si(OSi)4 (Q4) to Si(OAl)4 (Q0), can be calculated as follows
Si(OAl)4 Q0 |
Si(OAl)3(OSi) Q1 |
Si(OAl)2(OSi)2 Q2 |
Si(OAl)(OSi)3 Q3 |
Si(OSi)4 Q4 |
(⅓)4 |
4×(⅓)3 (⅔) |
6×(⅓)2 (⅔)2 |
4×(⅓) (⅔)3 |
(⅔)4 |
0.0123 |
0.0987 |
0.296 |
0.395 |
0.197 |
–78 to –86 |
–85 to –91 |
– 92 to –98 |
–97 to –102 |
–104 to –110 |
The 29Si MASNMR spectrum will therefore be expected to have the form shown in the Figure
