Chapter 3 Answers to Problem Questions

3.1 Basic NMR technique and resonance frequency problems

(a)  Calculate the resonance frequency of a ¹⁰³Rh nucleus in (i) a 11.744 T magnetic field and (ii) a 11.740 T magnetic field using the data in Table 3.1. Using Eqn 3.1 and E = hυ calculate the difference in resonance frequency (in Hz) of the ¹⁰³Rh nucleus in the two fields.

(b) Using the data in Table 3.2 discuss why the preferred nucleus for studying tin compounds by NMR is ¹¹⁹Sn.

(a) Using   and  gives  if the correct units of rad s⁻¹T⁻¹ are used for γ. From Table 3.2, for ¹⁰³Rh the magnitude of γ is 0.847 ×10⁷ rad s⁻¹T⁻¹ which gives for B₀ = 11.744 T, ν = 1.5831×10⁷ s^–1, which is 15.831 MHz; while for B₀ = 11.740 T, ν = 15.826 MHz. Therefore the difference is 5000 Hz

(b) Comparing the fundamental physical data for tin isotopes with I = ½

Isotopea	Natural abundanceb /%	Gyromagnetic ratio, γ/107 rad s–1 T–1	Frequency Ratio referenced to  1H = 100 MHz	Relative receptivity compared to 1Hc
(115Sn)	0.34	–8.801	32.718	1.21 ×10–4
(117Sn)	7.68	–9.589	35.632	3.54 ×10–3
119Sn	8.59	–10.031	37.290	4.53 ×10–3

 

Of the three isotopes with I=½ that could be studied using NMR the very low natural abundance of ¹¹⁵Sn  leads to a low relative receptivity compared to 1H of 1.2×10⁻⁴ and this isotope can be discounted. The other two isotopes have similar gyromagnetic ratios and natural abundances leading to similar relative receptivities but the slightly higher values for ¹¹⁹Sn mean this is the isotope of choice 

3.2 NMR and the number of environments

(a) Use VSEPR theory to predict shapes for the following species and hence determine the number of distinct fluorine resonances expected in their ¹⁹F NMR spectra. 

(i)  ClF₅     (ii) XeF₆ (assume a stereochemically inactive lone pair) (iii) XeOF₄ (iv) [ClOF₃]²⁻

(i) VSEPR theory applied to ClF₅ gives one lone pair and 5 bonding pairs around the central chlorine atom. The parent molecular shape with 6 electron pairs with be octahedral so the arrangement of fluorine atoms around chlorine will be a square based pyramid (with the other octahedron vertex occupied by the lone pair). This geometry has 2 different fluorine environments and (ignoring any spin spin coupling) would give two resonances in the ¹⁹F NMR spectrum with intensities 1:4 (apical :equatorial).

(ii) With a stereochemically inactive lone pair the shape of XeF₆ will be octahedral and all fluorine sites will be equivalent and one resonance will be observed in the  ¹⁹F NMR spectrum

(iii) XeOF₄ contains Xe(VI) so one lone pair, one double bond to oxygen, and four single bonds to fluorine need to be distributed around the xenon atom centre. This implies a parent octahedral distribution of these species and to minimise the most unfavourable electron density interaction, the Xe=O double bond with the lone pair, these will adopt positions trans to each other across the Xe centre.  For VSEPR theory the molecule with therefore be predicted to have a square-based pyramidal geometry with the oxygen atom in the apical position. In this geometry all the fluorine atoms are equivalent so a single resonance would be expected in the ¹⁹F NMR spectrum.

(iv) [ClOF₃]²⁻ _.   The oxidation state of chlorine is 3+ so there are two lone pairs, one oxygen and three fluorines to place around chlorine and the parent geometry with six areas of electron density will be octahedral. The lone pairs, with the highest electron density, will be directed trans to each other so the [ClOF₃] unit will display square planar geometry. One fluorine atom will be trans to oxygen and the other two trans to each other. The ¹⁹F NMR spectrum (ignoring spin-spin coupling) will, therefore, consist of two resonances with an intensity ratio of 1:2.     

(b) The ³¹P NMR spectrum of P₄O₈ shows two resonances of equal intensity; describe a possible structure for this oxide based on the structure of P₄O₉ shown in Figure 3.7(b).

The two equal intensity resonances indicate that there are two different phosphorus environments present in the P₄O₈ molecule. This can be achieved if a further terminal oxygen atom is removed from the P₄O₉ molecule as in the Figure. The two different phosphorus sites are labelled

 

(c) Reaction of SF₄ with KF leads to a compound A whose solution ¹⁹F NMR spectrum contains two resonances with intensity ratios 4:1. Describe a structure for the anion in A consistent with these data.

F⁻ is a Lewis base and SF₄ a Lewis acid so a likely product is KSF₅ containing the adduct anion [SF₅]⁻  . The observation of the resonance intensity of 4:1 is also consistent with a product containing 5 fluorine atoms which exists as, 4 in equivalent environments and one is a different distinct environment. We can use VSEPR theory to predict a shape for the [SF₅]⁻  anion which has 5 fluorine atoms and a lone pair around sulfur producing a square-based pyramidal shape for the SF₅ unit ( with a lone pair in the other site of an octahedral distribution of electron pairs around sulfur). The apical fluorine atom (opposite a lone pair) is in a distinct environment from the other four equivalent fluorine atoms in the square pyramid basal plane.

3.3  First-order coupling spectra

(a) Draw out the possible orientations of magnetic moments on four equivalent nuclei and hence show that the multiplet formed from coupling to these nuclei consists of five resonances in the ratio 1:4:6:4:1

 

(b) Extend the prediction of the ¹⁹F NMR spectra in Problem F3.2(a) to include first order coupling and, hence, predict the high resolution ¹⁹F NMR spectra of

(i)  ClF₅     (ii) XeF₆   (iii) XeOF₄ (iv) [ClOF₃]²⁻

(Ignore coupling to nuclei other than ¹⁹F)

(i) ClF₅ adopts a square based pyramidal structure with an apical F, F(axial)), and 4 different fluorine environments in the square base (F(equatorial)). Coupling of the apical site to those in the square base will produce a quintet (1:4:6:4:1; total relative intensity of 1); coupling of the F(equatorial) to the single F(axial) with produce a doublet (1:1, total relative intensity 4)

(ii) As all the fluorine environments are equivalent in octahedral XeF₆ there will be no observed first order coupling between fluorine and a singlet resonance will be observed in the ¹⁹F NMR spectrum.

(iii) With all fluorine sites equivalent – no spin-spin coupling will occur and the ¹⁹F NMR spectrum shows a single resonance 

(iv) In square planar [ClOF₃]²⁻  the two equivalent fluorine atoms (trans to each other) will couple with the nonequivalent fluorine ( trans to oxygen) to give a doublet ( relative intensity 2) and the unique fluorine with couple with the trans fluorine pair to produce a triplet of relative intensity 1.

(c)  Interpret the following NMR data

(i)  [IF₄]^{–   19}F  δ  −120 (s) ;      [IF₄]^{+   19}F   δ −170 (t)  δ  −185 (t)           

A single resonance in the ¹⁹F NMR spectrum informs that the fluorine sites are all equivalent in the [IF₄]^–  anion. Likely geometries for the anion are, therefore, square planar and tetrahedral but NMR cannot distinguish directly between these (or indeed other shapes that have four equivalent sites, such as a flattened tetrahedron). VSEPR theory predicts square planar geometry for [IF₄]^–  and the NMR spectrum is consistent with this – but does not prove it. The two resonances in the ¹⁹F NMR spectrum of [IF₄]⁺  tells us that there are two distinct fluorine sites and as both triplets there are two fluorine atom in each of sites so that coupling to the nonequivalent sites produces the triplets. So any structure consistent with [I(F¹)₂(F²)₂ where  F¹ and F² are inequivalent environments would be feasible. VSEPR theory predicts a “saw-horse” shape based on trigonal bipyramidal distribution of electron pairs around iodine and an equatorial site occupied by a lone pair. F¹ is therefore an axial fluorine site and F² an equatorial position.            

(ii)   Square planar [RhF₂Cl(CO)]²⁻  

 Isomer B        ¹⁹F    δ  −200 (d, J = 420 Hz) ;  ¹⁰³Rh   δ  −9200 (t,   J = 420 Hz)

Isomer  C        ¹⁹F    δ  −190 (dd, J = 420 Hz, J´ =30 Hz) , δ  −196 (dd, J = 390 Hz, J´=30Hz);

 ¹⁰³Rh   δ  -9750 (dd,   J = 420 Hz, J´ = 390 Hz )

In isomer B a single Rh environment can be inferred coupled to two equivalent fluorine sites – this is also consistent with the ¹⁹F NMR where a single fluorine environment has the same coupling constant to the rhodium nucleus. This must be the trans isomer.

The NMR data from isomer C is consistent with the cis isomer which has two distinct fluorine environments – one fluorine is trans to Co and the other trans to Cl. Heteronuclear coupling between Rh and the two distinguishable fluorine sites will have different coupling constants (420 and Hz)  and homonuclear coupling (J´ =30 Hz) will give rise to the dd features seen in the ¹⁹F NMR     

(d) Predict the form of the ¹²⁹Xe NMR spectra of (i) [XeOF₃]⁺ and (ii) XeOF₂   

(i) VSEPR theory can be used to predict a shape for [XeOF₃]⁺ and hence its ¹²⁹Xe NMR spectrum. The oxidation state of Xe in [XeOF₃]⁺ is 6+ so one pair of electrons on xenon remains as a lone pair and the parent shape of [XeOF₃]⁺ with be a trigonal bipyramid. The lone pair and oxygen will occupy equatorial positions in the molecular ion to minimized 90° O to lone pair (most unfavourable)  and 90° O to F and lone pair to F interactions ( next most unfavourable interactions). The predict geometry of [XeOF₃]⁺ is shown in the figure.

The ¹²⁹Xe nuclei with couple with the ¹⁹F_eq to produce a doublet with ¹J(¹²⁹Xe-¹⁹F_eq) and with the two axial positions with ¹J´(¹²⁹Xe-¹⁹F_ax) to form triplets. Therefore the overall ¹²⁹Xe NMR spectrum will consist of a doublet of triplets.

(ii) Similarly application of VSEPR theory predicts the following structure for XeOF₂ avoiding 90° interactions between lone pairs and the double bond.

 

The ¹²⁹Xe NMR spectrum will consist of a triplet from coupling to the two equivalent fluorine nuclei.

3.4 Coupling to I  >  ½ nuclei. Use Table 3.3

(a) Predict the form of the ¹⁹F NMR spectrum of the [NbF₆]^– anion using the data in Table 3.3.

⁹³Nb          I = 9/2 100%.

There is one fluorine environment in the octahedral [NbF₆]^– anion and coupling to ⁹³Nb           I = 9/2 100 % will  produce a multiplet with 2nI+1 = 10 resonances, all of equal intensity.

(b) The ¹⁹F NMR spectrum of the [AsF₄]⁺ cation consists of four lines of equal intensity. Explain this observation.

The [AsF₄]⁺ cation with have tetrahedral symmetry and all fluorine positions will be equivalent. Heteronuclear coupling to ⁷⁵As I = 3/2, 100% abundant will produce a 2nI+1 = 4 multiplet, the quartet observed in the ¹⁹F NMR spectrum. These resonances will be of equal intensity representing the probabilities of occurrence of the 3/2, 1/2,−1/2 and −3/2 m_I  states on As.

(c) The ¹⁹F NMR spectrum of a sample containing the [BF₄]⁻ anion consisted of a quartet of equal intensity peaks (total relative intensity 80%) and a septet of equal intensity resonances (total relative intensity 20%). Explain this observation.

The two stable isotopes of boron and there abundances are ¹⁰B, I =3, 19.9% and ¹¹B I =3/2, 80.1%. All the fluorine atoms are equivalent so the multiplets will derive from coupling to the two different quadrupolar boron nuclei. Coupling to ¹¹B will produce a 2nI+1=4 multiplet, a quartet in 80% of molecular [BF₄]⁻ ions; coupling to ¹⁰B will produce a 2nI+1=7 multiplet, a septet in 20% of molecular [BF₄]⁻ ions. Note that as the gyromagnetic ratios of ¹⁰B and ¹¹B are different, 2.875  and 8.585 ×107 rad s–1 T–1  the coupling constants for these multiplets will be different and the resonances will not overlap.

(d) Predict the form of the ¹⁹F NMR spectrum of ⁸¹BrF₃ .

BrF₃ is a T-shaped molecule with two equivalent (axial) fluorine sites and a distinct equatorial fluorine site that has a shorter Br-F distance. Ignoring spin-spin coupling the ¹⁹F NMR spectrum would consist of two resonances with an intensity ratio 2:1. ⁸¹Br has I =3/2. ¹J(¹⁹F - ⁸¹Br) coupling will produce equal intensity (within the multiplet) quartets from each of the resonances. Each of the signals in each of the quartets will be further split by ²J(¹⁹F- ¹⁹F ) to the non-equivalent fluorine atom(s) to product a quartet of doublets ( from the axial sites, overall relative intensity 2) and a quartet of triplets (from the equatorial site, overall relative intensity 1)

3.5 Decoupled spectra

(a) Predict the form of the ¹³C NMR and ¹³C{¹H} spectra of the cis and trans forms of W(CH₃)₂(CO)₄  (take W as NMR inactive).

In the trans form the methyl groups are symmetry equivalent as are the four carbonyl groups (all trans to CO). Therefore the ¹³C{¹H} spectrum of trans W(CH₃)₂(CO)₄ will just show two resonances, one from methyl carbon atoms ( typically −25 < δ < +25 ppm for metal coordinated –CH₃ groups) and one from the carbonyl group carbon (typically +190 < δ < +210 ppm in tungsten carbonyls). In the ¹³C NMR spectrum of trans W(CH₃)₂(CO)₄  spin-spin coupling, ¹J(¹³C-¹H) will occur producing a quartet signal from the methyl ¹³C resonance; it is possible that ³J(¹³C-¹H) would also be observed for the carbonyl carbon resonances though this would be very small in frequency terms. 

In the cis form of W(CH₃)₂(CO)₄   the methyl groups remain equivalent but the carbonyl groups divide into two sets – an equivalent pair trans to each other and a second pair which are trans to -CH₃. The ¹³C{¹H}NMR spectrum will, therefore, consist of three resonances – one in the methyl chemical shift region with −25 < δ < +25 ppm and two in the carbonyl region +190 < δ < +210 ppm. In the non-decoupled spectrum the signals in the methyl region would become quartets. 

3.6  Isotopomers

(a) Extend the prediction of the ¹⁹F NMR spectra in Problems F3.2(a) and F3.3(b) to include coupling to ¹²⁹Xe (I = ½, 26% abundant) and, hence, predict the high resolution ¹⁹F NMR spectra of

(i) XeF₆   (ii) XeOF₄ 

What would the ¹²⁹Xe NMR spectra of these compounds show?

(i) Ignoring ¹J(¹⁹F - ¹²⁹Xe) coupling one resonance will be observed in the  ¹⁹F NMR spectrum. The coupling to ¹²⁹Xe in ~26% of molecules will give rise to satellites with the characteristic 13:74:13 intensity pattern 

(ii) For the square-based pyramidal geometry with the oxygen atom in the apical position all the fluorine atoms are equivalent so a single resonance would be expected in the ¹⁹F NMR spectrum. Coupling to ¹²⁹Xe in ~26% of molecules will give rise to satellites with the characteristic 13:74:13 intensity pattern 

(b) Explain why the ³¹P NMR spectrum of trans-Pt[PMe₃]₂Cl₂ consists of three resonances with intensity ratios 1:4:1. Predict the expected features in the ³¹P NMR spectrum of cis-Pt[PMe₃]₂Cl₂. (¹⁹⁵Pt, I = ½, 33.8% abundant).    

In trans-Pt[PMe₃]₂Cl₂ the two phosphorus sites are equivalent and give rise to a single resonance. In 33.8% of molecules coupling to ¹⁹⁵Pt will give rise to satellites (separated by ¹J(³¹P-¹⁹⁵Pt) each at 16.9% of the intensity of the main resonance of total intensity 66.2%. The ratio 16.9:66.2:16.9 is close to the observed 1:4:1.

The NMR spectrum of cis-Pt[PMe₃]₂Cl₂ will look very similar to that of  trans-Pt[PMe₃]₂Cl₂  as the phosphorus atoms are again in equivalent positions, trans to Cl, and the same satellites from coupling to ¹⁹⁵Pt will be observed.