NMRNMR NNuclear M Magnetic RResonance

ProtonProton NMR (and other nuclei):NMR (and other nuclei):Conformational exchangeConformational exchange

Index NMR-basicsNMR-basics H-NMRH-NMR NMR-SymmetryNMR-Symmetry Heteronuclear-NMRHeteronuclear-NMR

Dynamic-NMRDynamic-NMR

Chemical Equivalence by InterconversionChemical Equivalence by Interconversion

1. Keto-Enol Interconversion2. Restricted Rotation (amides)3. Rings Interconversion

Axial HAxial H -> equatorial Hequatorial HRapid at room temperature, Slow at –60C

Equivalence in Shift Depends on Temperature, Solvent, Catalyst…Equivalence in Shift Depends on Temperature, Solvent, Catalyst…

Ha

HeHe

Ha

Keto-Enol Interconversion

15 10 5

3.002.80 0.890.70 0.15

43

O

52

6

CH31

O

78

7.5

43

O

52

6

CH31

O

78

H

KetoKetoCHCH22

CHCH33

=CH=CH

Keto-Enol Interconversion: slow on NMR time

OHOHenolenol

CHCH22 =CH=CH

200 150 100 50

24.62

25.82

54.62

76.62

77.04

77.46

96.65

126.94

128.56

128.77

132.23

133.75

134.80

183.26

193.71

43

O

52

6

CH31

O

78

H

43

O

52

6

CH31

O

78

Keto-Enol Interconversion: 13C

KetoKetoCHCH22

CHCH33

=CH=CH

KetoKeto enolenol

CHCH22 =CH=CH

Restricted Restricted RotationRotation

Restricted Restricted RotationRotation

Examine processing data

Simulating spectra

Ring Ring InterconversionInterconversion

C-NMRC-NMR3030ooCC

-60-60ooCC

Cis-1,4-dimethylcyclohexane

Ring InterconversionRing Interconversion

Ring InterconversionRing Interconversion

Exchange : DNMR – Dynamic NMRExchange : DNMR – Dynamic NMRNMR is a convenient way to study rate of reactions – provided that the lifetime of participating species are comparable to NMR time scale (1010-5-5 s s)

H

H

H

H

H

GeMe3

At low temperature, hydrogens form an A2B2XA2B2X spin system

At higher temperature germaniumgermanium hop from one C to the next

Chem-805Chem-805

Fluxional OrganometallicsFluxional OrganometallicsFrom:From:

http://www.ilpi.com/organomet/fluxional.html

Chemical exchange

• A molecule can undergo fluxional process by interchanging two or more sites

• If the rate of exchange is faster than the NMR time scale, the two different groups will appear at an average shift

• As temperature is cooled the rate become slower and separate shift can be obtained.

HHaa

HHee

HHee

HHaa

k1

k-1

Spectra of Cyclohexane-d11:

k at coalescence (61C): kc = /2

Chemical exchange

• Cp rings appear as separate signals in the 1H or 13C NMR spectrum. Slow exchange at room temperature

• Above room temperature the 2 Cp coalesce into a singlet. • The Cp rings themselves freely rotate about the Ti-ring with a very low energy barrier.

Therefore, all five protons on the same ring are chemically equivalent at all temperatures. • From line separation at RT, we can calculate rate constant rate constant kkcc

(kc = kc = //22) at coalescence temperature TTcc for the ring interconversion process.

• Then we can calculate Gc≠

GGcc≠≠ = 4.57 10 = 4.57 10-3-3 TTcc (log (log TTcc – log – log kkcc + 10.32) kcal/mol + 10.32) kcal/mol

Ti

SS

S

SS Ti

SS

SS

S

Interpreting Dynamic NMR dataInterpreting Dynamic NMR data

• Start at the low-TStart at the low-T: find the static structures that are consistent with the data. Use your knowledge of the system, integrals, multiplicities, coupling constants, decoupling data, and chemical shifts

• Look at the high-TLook at the high-T. to understand which groups are interconverting on NMR time scale

• Find a chemically reasonable pathway for the interconversionFind a chemically reasonable pathway for the interconversion

Pathways for interconversionPathways for interconversion• Dissociation and re-coordination of a ligand. Dissociation and re-coordination of a ligand.

To probe for such behavior, add some of the free ligand to the solution. If the fluxional process involves dissociation of that species, the chemical shift of the free ligand and bound species will come at the weighted average

• Rotation about a hindered bond Rotation about a hindered bond (for large groups)• Opening/closing of bridgesOpening/closing of bridges.

In dinuclear systems, a carbonyl or alkoxide ligand can switch between a bridging and terminal position

• Monomer-dimer or dimer-tetramer equilibriumMonomer-dimer or dimer-tetramer equilibrium. To probe such equilibria, try decreasing or increasing the concentration. This should modify the ratio of the equilibrating species.similarly, high T should favor the dissociated form and low T the more associated form

• 5-coordinate systems are quite notorious for fluxional behavior 5-coordinate systems are quite notorious for fluxional behavior as the energy barrier between trigonal bipyramid (TBP) and square pyramidal (SP) geometries is often quite low. Such interconversions can occur through a Berry pseudorotation or turnstile mechanism.

ExampleExample

• 1H NMR spectra of (tetramethylallene)Fe(CO)(tetramethylallene)Fe(CO)44 at -60 oC and 30 oC (tetramethylallene is Me2C=C=CMe2Me2C=C=CMe2)

• low temperature spectrum: the integrated ratios of the peaks are 1:1:21:1:2. • When tetramethylallene is combined with the complex, the 30 oC 1H NMR

spectrum of the mixture "consists of two singlets"

Explain the spectraand the behavior

Example (con’t)Example (con’t)

2. Draw the structure of the molecule at the low temperature limit

Explain the spectra

1.1. What is the molecular structure of What is the molecular structure of (Tetramethylallene)Fe(CO)(Tetramethylallene)Fe(CO)44

In Tetramethylallene. the two CMe2 planes are 90 degrees apart

Fe COOC

OC CO

AA

BBCCCC

it is not plausible for the allene to coordinate only via its central carbon atom because that carbon has no orbitals available to do this

Two of methyl groups (C) are chemically equivalent Two of methyl groups (C) are chemically equivalent the other two (A and B) are (A and B) are notnot equivalent: one points equivalent: one points at the metal and one awayat the metal and one away.

Example (con’t)Example (con’t)

• Is it an Is it an intermolecularintermolecular process? ( process? (the allene becomes free and recoordinatethe allene becomes free and recoordinate))• Is it an Is it an intramolecularintramolecular process? (the process? (the allene remain bound to the iron but intramolecular allene remain bound to the iron but intramolecular

process exchange the methyl positionprocess exchange the methyl position))

What is going on at elevated temperature?

Fe COOC

OC CO

AA

BBCCCC

Both process would explain room temperature spectra.

When free allene is added to the complex, a different singlet is observedfree allene is added to the complex, a different singlet is observed. If an intermolecularintermolecular process, only one allene shift should be observed at an average shiftonly one allene shift should be observed at an average shift

As this is not what is observed it is an intramolecular processAs this is not what is observed it is an intramolecular process

Example (con’t)What is going on at elevated temperature?

Fe COOC

OC CO

AA

BBCCCC

• intramolecular process: The Allene is "hopping" from one pi face of the intramolecular process: The Allene is "hopping" from one pi face of the allene to anotherallene to another

Example 2Example 2

• At +80 oC, In 1H NMR (31P-decoupled) Equimolar amount of the Osmium complex with ethylene shows a single sharp line at 6.0 ppm single sharp line at 6.0 ppm (ignoring the PMe3 and PPh3 resonances).

• Cooling this solution to 0 Cooling this solution to 0 ooC C results in the splitting of this splitting of this resonance into a single line at 4.9 ppm (about where free resonance into a single line at 4.9 ppm (about where free ethylene is observed) and two doublets at 7.5 (J = 2 Hz) and ethylene is observed) and two doublets at 7.5 (J = 2 Hz) and 6.7 (J = 2 Hz) ppm6.7 (J = 2 Hz) ppm.

• Upon cooling to -80 Upon cooling to -80 ooCC, the two doublets split further into a two doublets split further into a complex multipletcomplex multiplet.

• Explain

Consider the pseudo-octahedral ethylene complex of osmium

Example 2 (Con’t)Example 2 (Con’t)

• One ethylene ligand is coordinatedOne ethylene ligand is coordinated, the other ethylene is free. In the coordinated ethylenecoordinated ethylene, all 4 protons are not equivalent all 4 protons are not equivalent because the molecule has no symmetry elements

• the axial ligands axial ligands are different and there are three different kinds of three different kinds of equatorial ligandsequatorial ligands. Therefore, each of the four hydrogens four hydrogens on the coordinated ethylene are non equivalent non equivalent and coupled to the other threecoupled to the other three. These coupling constants would differ, making each individual resonance a doublet of doublet of doublets (dddddd). In addition, two sets of hydrogen are in similar environments so we would expect that the chemical shifts of these would be similar. Therefore, we'd expect to see two sets of overlapping ddd, or a pair of "complex multiplets".

Step 1: explain low tempStep 1: explain low temp. -80 -80 ooCC, the two doublets split further into a complex multiplettwo doublets split further into a complex multiplet(single line at 4.9 ppm do not change)(single line at 4.9 ppm do not change)

Example 2 (Con’t)Example 2 (Con’t)

• Observed data is consistent with exchange with free ethylene with exchange with free ethylene and rapid rotation in the complexrapid rotation in the complex. We observe an averaged peak at 6.0 an averaged peak at 6.0 ppmppm.

There is an equimolar amount of equimolar amount of complexcomplex andand free ethylene free ethylene:

Complex: 2 H at 7.5 ppm, 2H at 6.7 ppm Free ethylene: 4 H at 4.9 ppm.

The weighted average of these is The weighted average of these is [(2)(7.57.5) + 2(6.76.7) + 4(4.94.9)]/8 = 6.0 ppm, exactly where we see our peak6.0 ppm, exactly where we see our peak.

Step 2: explain high tempStep 2: explain high temp.single sharp line at 6.0 ppmsingle sharp line at 6.0 ppm

Example 2 (Con’t)Example 2 (Con’t)

The peak at 4.9 is free ethylene. For the complex choices are:1. The doublets are from an ethylene coordinated parallel to the axial ligands2. The doublets are from an ethylene coordinated in the equatorial plane3. The doublets are from an ethylene rotating rapidly on the NMR time scale

In choices 1 and 2, regardless of how the coordinated ethylene is fix, all four all four hydrogens are chemically inequivalenthydrogens are chemically inequivalent. This cannot be a fixed system. There is rapid rotation.

The rotation exchange C1 and C2 of the ethylenerotation exchange C1 and C2 of the ethylene.

The geminal protons on a given carbon are still non-equivalent geminal protons on a given carbon are still non-equivalent because of the lack of symmetry in the molecule. They form an AX pattern with a geminal an AX pattern with a geminal coupling constant between them (2 Hz)coupling constant between them (2 Hz)

Step 3: explain spectrum at 0 Step 3: explain spectrum at 0 00CCsingle line at 4.9 ppm (about where free ethylene is observed) and single line at 4.9 ppm (about where free ethylene is observed) and two doublets at 7.5 (J = 2 Hz) and 6.7 (J = 2 Hz) ppmtwo doublets at 7.5 (J = 2 Hz) and 6.7 (J = 2 Hz) ppm

Example 3

• The compound shown above exhibits fluxional behavior in the 31P{H} NMR.

• As the sample is cooled, the singlet resonance broadens and decoalesces between -80 and -100 oC (the T depends on the nature of group X). Below the coalescence point a pattern that appears to be a quartet is observed.

• Addition of excess PEt3 to this reaction mixture has no effect on the observed NMR behavior.

• Explain what is being observed and show how this is consistent with the NMR observations.

Example 3 (con’t)

• start at the low temperature limitstart at the low temperature limit. At low T limit what symmetry elements does this molecule possess?

• This molecule has no symmetry elements. The monohapto Cp has an implicit hydrogen that is not drawn. That means this ligand is not planar as the drawing misleadingly suggests, but goes either in front of or behind the plane of the screen!

• This render the 2 Phosphorous non-equivalent2 Phosphorous non-equivalent. One is close to the CpOne is close to the Cp, the other one is close to the hydrogenother one is close to the hydrogen.

• The 2 Phosphorous are coupled together forming and AB patternAB pattern. The shift difference is such that it resemble a quartet.

Example 3 (con’t)

• explains the coalescence to a singlet as the T is raised? Is it because:explains the coalescence to a singlet as the T is raised? Is it because:

1. Rotation about the phenyl-Pd bond is rapid2. Rotation about the Cp-Pd bond is rapid3. Change from a square planar to tetrahedral geometry4. Dissociation/reassociation of a phosphine ligand.

As addition of excess PEt3 to this reaction mixture has no effect we can rule out 4As addition of excess PEt3 to this reaction mixture has no effect we can rule out 4

Rotation about the Cp-Pd bond render the 2 Phosphorous equivalentRotation about the Cp-Pd bond render the 2 Phosphorous equivalent.

The Cp ring is closer to each phosphine 50% of the time, effectively making the two phosphine ligands equivalent. Therefore, we'd see a singlet.

IndexNMR-basicsNMR-basicsH-NMRH-NMR

NMR-SymmetryNMR-SymmetryHeteronuclear-NMRHeteronuclear-NMRDynamic-NMRDynamic-NMR

NMR and Organometallic compoundsNMR and Organometallic compounds

NMR Special topicsNMR Special topics

Multinuclear NMR

• There are at least four other factors we must consider• Isotopic Abundance. Some nuclei such as 19F and 31P are 100% abundant (1H is

99.985%), but others such as 17O have such a low abundance (0.037%). Consider: 13C is only 1.1% abundant -- that's one of the reasons we need to use a lot more sample and take more scans to obtain a 13C spectrum versus a 1H spectrum.

• Sensitivity goes with the cube of the frequency. 103Rh (100% abundant but only 0.000031 sensitivity): obtaining a spectrum for the nucleus is generally impractical. However, the nucleus can still couple to other spin-active nuclei and provide useful information provided it has good abundance. In the case of rhodium, 103Rh coupling is easily observed in the 1H and 13C spectra and the JRhX can often be used to assign structures

• Nuclear quadrupole. For spins greater than 1/2, the nuclear quadrupole moment is usually larger and the line widths may become excessively large. This can sometimes be overcome by running the sample at low temperature

• Relaxation time

Notable nuclei• 19F: spin ½, abundance 100%, sensitivity (H=1.0) : 0.83

2JH-F = 45 Hz, 3JH-F trans = 17 Hz, 3JH-F Cis = 6 Hz 2JF-F = 300 Hz, 3JF-F = - 27 Hz

• 29Si: spin ½, abundance 4.7%, sensitivity (H=1.0) : 0.0078The inductive effect of Si typically moves 1H NMR aliphatic resonances upfield to approximately 0 to 0.5 ppm, making assignment of Si-containing groups rather easy. In addition, both carbon and proton spectra display Si satellites comprising 4.7% of the signal intensity.

• 31P: spin ½, abundance 100%, sensitivity (H=1.0) : 0.07 1JH-P = 200 Hz, 2JH-P ~2-20 Hz, 1JP-P = 110 Hz, 2JF-P ~ 1200-1400 Hz, 3JP-P = 1-27 Hzthe chemical shift range is not as diagnostic as with other nuclei, the magnitude of the X-P coupling constants is terrific for the assignment of structuresKarplus angle relationship works quite well

Notable nuclei• 31P: spin ½, abundance 100%, sensitivity (H=1.0) : 0.07

1JH-P = 200 Hz, 2JH-P ~2-20 Hz, 1JP-P = 110 Hz, 2JF-P ~ 1200-1400 Hz, 3JP-P = 1-27 Hzthe chemical shift range is not as diagnostic as with other nuclei, the magnitude of the X-P coupling constants is terrific for the assignment of structuresKarplus angle relationship works quite well

2JH-P is 153.5 Hz for the phosphine trans to the hydride, but only 19.8 Hz to the (chemically equivalent) cis phosphines.

See Selnau, H. E.; Merola, J. S. Organometallics, 1993, 5, 1583-1591.

Notable nuclei• 103Rh: spin ½, abundance 100%, sensitivity (H=1.0) : 0.000031

1JRh-C = 40-100 Hz, 1JRh-C(Cp) = 4 Hz,

For example, in the 13C NMR spectrum of this linked Cp, tricarbonyl Rh dimer at 240K (the dimer undergoes fluxional bridge-terminal exchange at higher temperatures), the bridging carbonyl is observed at d232.53 and is a triplet with 1JRh-C = 46 Hz. The equivalent terminal carbonyls occur as a doublet at d190.18 with 1JRh-C = 84 Hz:

See Bitterwolf, T. E., Gambaro, A., Gottardi, F., Valle G Organometallics, 1991, 6, 1416-1420.