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NMR for inorganic chemistry
1) NMR spectroscopy provides information about atomic nuclei by analyzing their magnetic properties when placed in a strong magnetic field. 2) For nuclei with spin 1/2 like 1H and 13C, NMR spectra show peaks representing different spin states that are split by spin-spin coupling with neighboring nuclei. 3) Nuclei with higher spin like 11B display more complex splitting patterns due to interactions between their multiple spin states and coupled nuclei.
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NMR for inorganic chemistry
- 1. Spectroscopic Methods in InorganicChemistry Part 2: NMR Dr. Chris, Feb. 2016
- 2. Rotating Charge producesa magnetic moment μ For ELECTRONS that means that molecules become dia- or paramagnetic. For the NUCLEUS it means that it gets a magnetic moment that can be adjusted to an outer magn. Field.
- 3. The spin Iof particles is QUANTIZISED, means it can only have certain values For the nucleus of elements, the spin depends on the number of protons and neutrons: Most important for NMR are elements with I = ½ For bigger I there is a quadrupole moment which affects the broadness of the peaks
- 4. In a strongmagnetic Field B0 the spins can have the values + or – ½. The energy difference between these states is:
- 6. Energy States forI = ½
- 7. RF Pulse Experiment Thesum of all magn.moments is a vector that rotates around B0 B1 as magn.field perpendicular to B0 causes the magn.moments to move towards the y-axis
- 10. Chemical Shift δ Themagn.Field B0 also affects the electrons moving around the nucleus. They will align with the B0 field, creating a local field opposing B0 for the nucleus. The nucleus is shielded by the electrons.
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- 14. 13C NMR 13C hasI = ½; its natural abundance is 1.1% • 13C sensitivity is only 1/5700 that of 1H • 13C experiments require higher concentrations and more scans/time • S/N increases with square root of # of scans
- 17. Exercise How would thespectrum look like for 1H NMR 19F NMR 31P NMR
- 21. Compare CH4 andBH4 - How would the 1H NMR spectra look like ? Consider that C has an NMR-active isotope 13C B has a nuclear spin of 3/2
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- 23. C contains 1.1%of 13C -> coupling between the H would result in a doublet 13C – 1H coupling And a main peak in the middle from 1H alone
- 24. B has I= 3/2 => possible mL are 3/2, ½, -½ and -3/2 E B0 4 possible spin orientations 3 possible transitions (in a 11B NMR spectrum !) ΔmL = +/- 1 -3/2 +3/2 +1/2 -1/2 The 4 protons can interact with 4 different B spin states => quadrupel signal
- 25. How about 11BNMR of BH4 - ? The interactions of the B-spin with the 4 H-spins
- 26. The 4 protonscan have 5 different total spins: ½ ½ ½ ½ Spin 2 -> 1 way ½ ½ ½ -½ Spin 1 -> 4 ways ½ ½ -½ -½ Spin 0 -> 6 ways ½ -½ -½ -½ Spin -1 -> 4 ways -½ -½ -½ -½ Spin -2 -> 1 way
- 27. Splitting pattern byspin ½ nuclei 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Number of peaks: 2 x I x n + 1 ( = 2 x ½ x 4 +1 = 5 ) “Pascal triangle”
- 28. Example 2: Diborane Consideronly the 2 H in the bridge What is the 1H NMR spectrum What is the 11B NMR spectrum ?
- 29. 11B spectrum 2 equivalentB atoms couple with 2 equivalent protons 2 (1/2) 2 + 1 = 3 signals Coupling by spin ½ nuclei -> use the Pascal triangle
- 30. 1H NMR spectrum Numberof peaks: 2 I n + 1 = 2 (3/2) 2 + 1 = 7 The proton interacts with 2 B spins -> no Pascal triangle ! which can have the following orientations: 3/2 3/2 I = 3 1 way 3/2 ½ and ½ 3/2 I = 2 2 ways ½ ½ and 3/2 -1/2 and - ½ 3/2 I = 1 3 ways ½ -½ and -½ ½ and 3/2 -3/2 and -3/2 3/2 I = 0 4 ways
- 31. How would thespectra change if we had 3 bridging Hydrogens ?
- 32. 19F NMR example(spin ½)
- 33. To calculate thecoupling constant J, we have to convert the difference in δ to Hz Δδ = 0.07 The measuring frequency was 282 MHz J = 0.07 1/106 x 282 MHz = 20 Hz
- 35. We refer thissimply to be a multiplet
- 36. Read the spectrumfrom left to right
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- 38. Principle How would thespectrum look like if the groups A and B and “frozen” and if they can quickly interchange ?
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- 49. Which F-NMR signalswould we expect ? Replace by C6H5 here
- 50. a a b c a b We would expecta broadened triplet for each of the equatorial flourines
- 54. c a b d Protons c andd are not equivalent, so we would expect 2 different doublets
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