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Structural Methods in Molecular Inorganic Chemistry
David W. H. Rankin, Norbert W. Mitzel and Carole A. Morrison
First Edition, 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
http://www.wiley.com/go/rankin/structural
Supplement to Chapter 4. NMR Spectroscopy
Worked examples
Example 1 Coupling patterns
The Figure below shows the195
Pt NMR spectrum of a compound prepared by bubbling H2 through a
mixture of [Pt(PPh3)3] and [AuPPh3][NO3]. The molecular composition of the product was shown by mass
spectrometry to be PtAu7H(PPh3)8(NO3)2. What can you deduce from the NMR spectrum?
a) b)
Figure 1
a)195
Pt NMR spectrum of [Pt(H)(PPh3)(AuPPh3)7]2+
, showing the coupling with H and the two different P sites. b)
Structure of the complex core. The spectrum and structure are taken, with permission, from [1]. Copyright 1989 The
American Chemical Society.
The spectrum has one large doublet coupling (ca. 2300 Hz), and the rest of the pattern can be explained as a
doublet (ca. 550 Hz) of multiplets (ca. 400 Hz). The multiplets clearly have even numbers of peaks, so the
coupling must be to an odd number of nuclei. With five nuclei, the intensity ratio for the six lines would be
1:5:10:10:5:1, with seven it would be 1:7:21:35:35:21:7:1, and with nine it would be
1:9:36:84:126:126:84:36:9:1. The observed intensities fit best for the eight-line pattern, so we deduce thatthe platinum atom is probably bound directly to one PPh 3ligand and to the unique hydrogen atom, and is
surrounded by seven AuPPh3groups. This is confirmed by the crystal structure depicted below. All the
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metal and phosphorus atoms are shown, but the position of the hydrogen atom has to be inferred from the
relatively long separations between Au(5), Au(6) and Au(9).
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Example 2 Isomers and 13C NMR
Account for the forms of the1H-decoup1ed
13C NMR spectra of the cis and trans isomers of
[RuCl(NO)(bpy)2](PF6) (bpy = 2,2'-bipyridine) illustrated below. Assume that each of the peaks marked with
arrows includes two overlapping lines. [2]
Figure 21H-decoup1ed
13C NMR spectra of the cis and trans isomers of [RuCl(NO)(bpy)2](PF6). Taken, with permission, from
[2]. Copyright 1989 The American Chemical Society.
In trans-[RuCl(NO)(bpy)2]+there are two planes of symmetry, so each bpy ligand is bisected by one mirror
plane, and the two ligands are related by the second plane [structure (a)below]. There are just five
chemically distinct carbon atoms, and so five resonances are seen in the 13C NMR spectrum. The resonance
at highest frequency is less intense than all the others. This is because all the other resonances arise from
carbon bound to hydrogen, and their intensities are increased by the Nuclear Overhauser Effect (see Section
4.11.4).
Figure 3
(a) cisand (b) transisomers of [RuCl(NO)(bpy)2]+.
In cis-[RuCl(NO)(bpy)2]+[Figure 3, structure (a)]there are no planes of symmetry, and so the carbon
nuclei are all chemically distinct. There are thus 20 resonances in all, of which four are due to quaternary
carbons, and are less intense than the others. Each line in the spectrum of the trans isomer has become fourlines, and the chemical shift changes are small, with one exception. One ring is now trans to chlorine ligand
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instead of to nitrogen, and so one of resonances associated with this ring has been shifted much more than
the others.
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Example 3 Double resonance
Reaction between [Ir(CO)H(PPh3)3] and SiBrH3leads to the elimination of PPh3 and the formation of the
complex shown in Figure 4, whose31
P-{1H} NMR spectrum shows the presence of two equivalent
31P
nuclei. The proton NMR spectrum (excluding the aromatic region) is shown below. How would you use
double resonance to help to account for the splitting pattern observed in the spectrum for Ht, at 10.03?
Figure 41H NMR spectra of the products of the reaction between [Ir(CO)H(PPh3)3] and SiBrH3.
The pattern observed for Htis a wide doublet of narrower doublets of triplets. The wide doublet can be put
down to coupling to the trans-31
P nucleus and the narrower to coupling to the cis-31
P nucleus; this may be
confirmed by single-frequency 31P-{1H} double resonance. The triplet coupling is more puzzling. We wouldexpect to observe coupling between Htand Hc, but this should lead to a doublet and not a triplet. There are
three other proton resonances in the spectrum shown: one of these is at low frequency, and must be due to
Hc, and the other two must be due to the SiH protons, which are made non-equivalent by the chirality.
Irradiation at Hcreduces the triplet splitting in Htto a doublet splitting; irradiation at 4.89 has no effect,
but irradiation at 5.68 also reduces the triplets to doublets. This shows that the triplet splitting in Htis due
to two doublet couplings that happen by chance to be equal; one is to Hcand the other to one of the SiH2
protons. The two SiH resonances look very complicated, but can be analyzed easily in terms of a series of
doublet splittings; there are five nuclei with spin one-half that are involved in the spin system, and at high
resolution all 32 lines expected in the group of resonances centered at 5.68 can be resolved.[3]
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Example 4 Structure assignment by 2D NMR
Figure 5 below shows 1D (proton decoupled) and 2D COSY11
B NMR spectra of B9H11NH. Assuming that
coupling is only observed between adjacent boron nuclei, deduce the structure of the boron cage.
Figure 5
1D (proton decoupled) and 2D COSY11
B NMR spectra of B9H11NH
Three of the resonances in the 1D NMR spectrum are twice as intense as the others, so they represent two
boron nuclei each. The pairs of nuclei B, C and D couple with one another, so they must be arranged as in
structure (a) in Figure 6.. Nucleus F couples to the pairs Band C, and must therefore cap that face of the
prism, while E couples to the C and D pairs, and thus lies over that face [Figure 6, structure (b)]. The
remaining nucleus, A, shows coupling to the unique E and the pair of D nuclei, and so should be placedover the triangular face DDE. The NH group is in fact linked to the boron atoms CCF, and if we add a bond
between the two B atoms we complete the cage ([Figure 6, structure (c)]. Each boron atom carries a
terminal hydrogen atom, and there are also hydrogen atoms bridging the two AD bonds.
Figure 6
Development of the structure of B9H11NH. Taken, with permission, from [4]. Copyright 1988 The American Chemical
Society.
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Example 5 Multinuclear NMR spectroscopy
The compound [(CH3)2SnS]3, which has a structure similar to I, shown below, exists in two crystalline
forms. In the monoclinic form there is no molecular symmetry, while there is a two-fold axis through the
molecules in the tetragonal form. Describe the13
C and119
Sn NMR spectra expected for each form.
I
The13
C NMR spectrum of the monoclinic form will have six resonances, each with117
Sn and119
Sn
satellites, as the methyl groups are all inequivalent. In fact, only five resonances are observed, as two of
them overlap. The three tin nuclei are also inequivalent, so there will be three resonances. Each of these will
have eight satellite lines, due to coupling with117
Sn and119
Sn at each of the other tin sites.
In the tetragonal form none of the methyl groups lies on the two-fold axis, so there are three pairs of
equivalent carbon atoms, and three13
C resonances, with the usual satellites due to coupling with the
adjacent tin atoms. Two of the tin atoms are symmetry-related, while the third one is unique, so the119
Sn
NMR spectrum has two resonances, one twice as intense as the other (Figure 7). The satellite pattern,
shown in the spectrum below, is quite complex. The resonance due to the unique tin, which is at the high-
frequency end of the spectrum, has satellites due to119
Sn and117
Sn in the other position. However, the119
Sn
satellites are part of a second-order [AB] pattern, and are therefore not of equal intensity, nor centered on
the main resonance. The same applies to their counterpart satellites in the low-frequency half of spectrum.
In this part can also be seen two pairs of117
Sn satellites, one arising from coupling to 117
Sn at the unique
position, and the other from coupling between117
Sn and119
Sn in the two symmetry-related sites.
Figure 7
The119
Sn NMR spectrum of [(CH3)2SnS]3. Taken, with permission, from [5]. Copyright 1989 Elsevier Inc.
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Example 6 Compound identification
Reaction between trans-[Ir(CO)Cl(PMe3)2] and SF4in CCl2H2gives a single product whose19
F spectrum at
180 K is shown in Figure 8 below. Three groups of resonances are observed: a doublet of triplets of
doublets at 68 ppm (integrated intensity 2), a triplet of doublets of narrow triplets at 61 ppm (intensity 1),
and a triplet of quartets at 383 ppm (intensity 1). At 230 K the two resonances at higher frequencies
broaden and lose detail, and at 300 K they have coalesced to a single broad peak; the lowest frequency set
of resonances is unchanged. Account for these observations, and predict the31
P NMR spectrum of the
product.
Figure 819
F NMR spectrum of the product of the reaction between trans-[Ir(CO)Cl(PMe3)2] and SF4.
There is a resonance of intensity 1 to very low frequency, where peaks due to fluorine bound to a transition
metal appear. The peaks in the high frequency region, where we would expect to see SF resonances, are of
total relative intensity 3. Moreover, the low frequency resonance is split into a triplet of quartets. All this
suggests that SF4has reacted with trans-[Ir(CO)Cl(PMe3)2] by oxidative addition of SF to Ir, giving the
complex [Ir(Co)ClF(PMe3)2(SF3)] with F transto CO and transto SF3.
Sulfur has a lone pair of electrons, and so coordination is likely to be based on a trigonal bipyramid,
with the lone pair and the metal atom in the equatorial plane. With easy rotation about the IrS bond the
axial fluorine atoms bound to sulfur become magnetically equivalent. The spin system then becomes
[AM2QX2], with the high frequency resonance assigned to the two axial fluorine atoms, the middle-
frequency resonance assigned to the equatorial SF nucleus, and the low frequency resonance to the IrF
nucleus. All three SFnuclei couple more or less equally to the IrF nucleus.
As the system is warmed, pseudorotation becomes faster on the NMR timescale. This leads to blurring
of the distinction between axial and equatorial fluorine sites, and at 300 K the spectrum shows that
exchange of position is beginning to lead to effective equivalence. The interchange of SFsites is intra- and
not inter-molecular. The P{H} NMR spectrum at low temperature appears as a doublet of triplets of
narrow doublets, but on warming this changes to a doublet of quartets as the SFnuclei become equivalent.
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Example 7 Intramolecular exchange
An intramolecular exchange process between two isomers of a ruthenium-tungsten cluster is implied by the
spectra shown in Figure 9. At the lowest temperature the major isomer (II, 87%) has two hydride
resonances, with intensity ratio 2:1, while the minor isomer (III) has three equal resonances. The strongest
line in the spectrum shows183
W satellites. On warming all the lines broaden, coalescing at 220 K and
finally giving one sharp resonance with 183W satellites by the time room temperature has been reached. As
only one signal remains, the hydrogen atoms must be exchanging rapidly between all five sites, so the two
isomers are interconverting, and this provides a mechanism for all the hydrogen atoms to move between all
the sites. The retention of coupling to183
W demonstrates that the process must be intramolecular.
Thus NMR spectroscopy provides a convenient and extremely important way of studying both
mechanisms and rates of exchange reactions, provided the lifetimes of the participating species are
comparable with the NMR timescale, within two or three orders of magnitude of 105
s. Reactions which
are somewhat slower than this can also be studied, by using the technique of saturation transfer. Suppose
nuclei are exchanging between sites A and B, with lifetimes long compared with the NMR timescale, but
short compared with their relaxation times, say of the order of one second. They will then give distinct
resonances, but if one frequency is strongly radiated so that saturation occurs, the exchange process willthen transfer saturation to the other signal, which will diminish in intensity. From the intensity change and
the mean relaxation time of the nuc1ei in sites A and B the exchange rate can be calculated. Thus, the
method fills a gap in determining exchange rates, between those measurable by conventional NMR
methods, and slower ones which can be measured by following the course of a reaction toward equilibrium,
over a period of minutes or longer.
Figure 91H NMR spectra of [Ru3W(C5H5)(CO)11H3], see complexes VIII and IX, at temperatures between 185 K and 294 K.
The peak marked with an asterisk is due to an impurity. Taken, with permission, from [6]. Copyright 1989 Elsevier
Inc.
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II III
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Example 8 The application of 13C13C COSY and NOESY
In Figure 10 the13
C COSY and NOESY spectra for [Os3H2(CO)10] (complex IV) 75% labeled with13
C are
shown. As both spectra are symmetrical about the diagonal line for homonuclear experiments it is possible
to compose the two sets of results into a single 2D graphic. The intensity of the peak at 175.2 ppm in the 1D
spectrum identifies it as being due to CO(7), while a
13
C NMR spectrum with
1
H coupling retained indicatesthat the CO(6) resonance is at 176.3 ppm. The COSY spectrum shows that these two couple with one
another, while the peak in the NOESY spectrum is in this case indicative of chemical exchange between the
CO(6) and CO(7) sites (see Section 4.17.2). The COSY spectrum also shows coupling between13
C nuclei
of CO(4) and CO(5) groups as expected, although the assignment of these two resonances is uncertain.
However, the spectrum indicates one further coupling, between C(7) and either C(4) or C(5). As the
couplings C(7)C(4) and C(6)C(5) both involve a trans relationship at one Os atom and a cis relationship
at another, we might expect that if we see one such coupling we should also see the other. So the most
likely explanation is that the observed coupling is C(7)C(5), with cis relationships at both Os atoms.
Figure 10
Homonuclear 2D13
C scalar coupling (COSY) and chemical exchange (NOESY) spectra for [Os3H2(CO)10], complex
IV, obtained in two separate experiments. The one-dimensional13
C NMR spectrum is shown at the bottom.
Reproduced from [7]. Copyright 1985 The Royal Society of Chemistry.
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IV
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Example 9 Intramolecular exchange
The spectra presented in Figure 11 indicate that two separate exchange processes can occur as theruthenium-silver complex V is warmed from 180 K to room temperature. The Ag(PEtPh2) group is bound
to one of two equivalent Cl3faces of the octahedron associated with one of the ruthenium atoms, and as the
molecule therefore has no plane of symmetry, all five phosphorus nuclei are inequivalent. The31
P NMR
spectrum shows two [AB] patterns arising from the pairs of phosphine ligands bound to ruthenium, and the
phosphorus bound to silver appears as two doublets, the splitting being caused by coupling to the two spin-
l/2 isotopes of silver. On warming to 220 K the lines in the [AB] patterns collapse to a single resonance,
while the silver phosphine pair of doublets is unchanged. The coalescence of the ruthenium phosphine reso-
nances can be attributed to the Ag(PPh2Et) group switching between four equivalent sites, two on each
ruthenium atom, so that all four phosphines become equivalent. On further warming, all the resonances
coalesce, so we can deduce that there is additional exchange of phosphines between coordination sites on
ruthenium and silver.
Ru
Cl
Ru PPh
2EtEtPh
2P
PPh2Et
Cl
EtPh2P
Cl
Ag
Cl
Cl
PPh2Et
Ru
Cl
Ru PPh
2EtEtPh
2P
PPh2Et
Cl
EtPh2P
Cl
Ag
Cl
Cl
PPh2Et
4.XXXII
Figure23
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Figure 111H decoupled
31P NMR spectra of [RuCl5(PPh2Et)4Ag(PPh2Et)] (a) at 180 K, (b) at 220 K, and (c) at 300 K. The lines
marked and in spectrum (a) indicate the two separate [AB] spin systems associated with phosphines bound to
ruthenium.
A two-dimensional NMR technique called EXSY (for exchange spectroscopy) also provides rate
information for slow exchange reactions. The method is the same as that used to evaluate Nuclear
Overhauser Effects, and the proximity of spinning nuclei to one another (NOESY, Section 4.13.3). Figure
12 shows an EXSY spectrum for the methyl group protons of the platinum complex XI. Here there are four
resonances due to methyl groups, and the off-diagonal peaks clearly indicate that group B is exchanging
with C, and A with D. Analysis of the cross-peak intensities yields the rate constants for the exchangeprocesses, and from the constants measured at a series of temperatures it is a simple matter to derive the
activation energy and entropy for the exchange process, which evolves the free alkene group displacing the
coordinated one.[8]
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Figure 12
EXSY 2D1H NMR spectrum of the platinum complex VI. The off-diagonal peaks show that the methyl group giving
the resonance B is exchanging with that giving resonance C, and groups A and D are also exchanging. Reproduced
from [8]. Copyright 1989 The Royal Society of Chemistry.
VI
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Example 10 Polymer analysis
As an example of polymer analysis we show here that the structure of hydrocarbon polymers can be
analyzed by13
C NMR. Polypropylene exists in various forms, which can be idealized as isotactic,
sydiotactic and atactic [Figure 13(a-c)]. If methyl groups are placed on opposing sides of the chain, this is
called a meso pair (m), if placed at the same side, it is called racemo (r). Different combinations of mand r
(called pentades) give distinguishable resonances in the13
C NMR spectrum and can be assigned by
comparison with spectra of samples of known structure. The spectrum shown in Figure 13 is a13
C NMR of
atactic polypropylene, which contains small contributions of isotactic (mmmm, 2%) and syndiotactic (rrrr,
7.8%) areas. The ratios are obtained by integration. For comparison there is also a spectrum of a reference
sample of polypropylene with an isotactic purity of 98%. This procedure is a very helpful and widely used
tool in polymer analysis and can be used for many different polymers, including truly inorganic ones, once
their NMR/structure relationship has been firmly established.
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Figure 13
Structures of (a) isotactic, (b) syndiotactic and (c) atactic polypropylene and 13C NMR spectrum of an atactic sample
of polypropylene (solid line, sample obtained by catalytic polymerization with an hydroxylaminato-titanium complex
[9]) with small isotactic, and syndiotactic contributions, as well as a reference sample spectrum (dotted line) of an 98%
isotactic polypropylene. (Note the small peaks at 21.30, 20.80 and 19.75 ppm indicating some small contributions of
mmmr, mmrrand mrrmpentades). Figure courtesy of Dr Alexander Willner (Bielefeld).
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Example 11 2D INADEQUATE
A further solution of the problem of the low abundance of neighboring
183
W-
183
W pairs is the use of 2DINADEQUATE spectra (Incredible Natural Abundance Double Quantum Transfer Experiment). This
experiment is rarely used, and then most often to detect13
C13
C couplings in organic molecules, in order to
establish the connectivity pattern of bigger systems. The technique selects only the transitions that have a
coupling between the nuclei under observation (13
C or183
W in our case). In this way, the main signals (for
the majority isotopes) are suppressed. As there is no polarization transfer from sensitive and highly
abundant nuclei involved, the method is inherently insensitive and long acquisition times are also necessary.
As an example, consider the183
W-INADEQUATE spectra for [PPbW11O39]5
and [SiZnW11O39]6
[12,13] shown in Figure 14. The compounds comprise eleven WO6octahedra arranged as shown in Figure
15. In the heteropolytungstates the twelfth octahedral site (dotted in Figure 15), linking at corners to 1, 1', 4
and 4', is occupied by a different metal ion, and there is a silicon or phosphorus atom at the center of the
ion. Assigning the six observed183
W resonances in the 1D-183
W NMR spectra for each compound is
difficult, even making use of the observation that183
W183
W coupling is only significant for nuclei in
adjacent octahedral (2J). The spectra in Figure 14 show the double quantum frequencies in one dimension
and the183
W chemical shifts in the other. Double quantum transitions are only possible for molecules which
have two nuclei coupling to each other, which in this case means in adjacent octahedra. The spectrum
consists of pairs of signals, equidistant from the diagonal line (any signals lying on the line arise from
imperfections in the experiment), and each pair, which lies on one horizontal line, indicates that the nuclei
with those chemical shifts are coupled. For example, in each of the spectra of Figure 14 there is one
tungsten coupling to four others, and this must be W(3). The unique tungsten atom, W(6), can be
recognized in the 1D183
W NMR spectra as its resonances are half the intensity of the others, and it can be
seen to be coupled to W(2) and W(5), and so on. Thus, the whole of the skeleton of each ion can be
deduced from a single spectrum, with the exception of the connection between W(4) and W(5) in
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[PPbW11O39]5
. The full analysis shows that the two ions have similar structures, and reveals a large
difference between the W(4) chemical shifts, which probably arises from different placements of the Pb and
Zn atoms in the ions.
Figure 14183
W 2D INADQUATE spectra for (a) [PPbW11O39]5and (b) [SiZnW11O39]
6. The pairs of off-diagonal peaks indicate
connections between the associated WO6 octahedra. Normal183
W NMR spectra are shown on top the 2D spectra.
Taken, with permission, from [10]. Copyright 1983 The American Chemical Society.
Figure 15
Relationship between the WO6 octahedra in the structure of Na5PPbW11O39. Taken, with permission, from [10].
Copyright 1983 The American Chemical Society.
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