NMR for inorganic chemistry
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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 Inorganic Chemistry Part 2: NMR Dr. Chris, Feb. 2016
- 2. Rotating Charge produces a magnetic moment ฮผ 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 I of 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 = ยฝ For bigger I there is a quadrupole moment which affects the broadness of the peaks
- 4. In a strong magnetic Field B0 the spins can have the values + or โ ยฝ. The energy difference between these states is:
- 6. Energy States for I = ยฝ
- 7. RF Pulse Experiment The sum 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 ฮด The magn.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.
- 12. Example
- 14. 13C NMR 13C has I = ยฝ; its natural abundance is 1.1% โข 13C sensitivity is only 1/5700 that of 1H โข 13C experiments require higher concentrations and more scans/time โข S/N increases with square root of # of scans
- 17. Exercise How would the spectrum look like for 1H NMR 19F NMR 31P NMR
- 21. Compare CH4 and BH4 - 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
- 23. C contains 1.1% of 13C -> coupling between the H would result in a doublet 13C โ 1H coupling And a main peak in the middle from 1H alone
- 24. B has I = 3/2 => possible mL are 3/2, ยฝ, -ยฝ and -3/2 E B0 4 possible spin orientations ๏ 3 possible transitions (in a 11B NMR spectrum !) ๏ ฮmL = +/- 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 11B NMR of BH4 - ? The interactions of the B-spin with the 4 H-spins
- 26. The 4 protons can have 5 different total spins: ยฝ ยฝ ยฝ ยฝ Spin 2 -> 1 way ยฝ ยฝ ยฝ -ยฝ Spin 1 -> 4 ways ยฝ ยฝ -ยฝ -ยฝ Spin 0 -> 6 ways ยฝ -ยฝ -ยฝ -ยฝ Spin -1 -> 4 ways -ยฝ -ยฝ -ยฝ -ยฝ Spin -2 -> 1 way
- 27. Splitting pattern by spin ยฝ 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 ยฝ x 4 +1 = 5 ) โPascal triangleโ
- 28. Example 2: Diborane Consider only the 2 H in the bridge ๏ What is the 1H NMR spectrum ๏ What is the 11B NMR spectrum ?
- 29. 11B spectrum 2 equivalent B atoms couple with 2 equivalent protons ๏ 2 (1/2) 2 + 1 = 3 signals Coupling by spin ยฝ nuclei -> use the Pascal triangle
- 30. 1H NMR spectrum Number of 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 ยฝ and ยฝ 3/2 I = 2 2 ways ยฝ ยฝ and 3/2 -1/2 and - ยฝ 3/2 I = 1 3 ways ยฝ -ยฝ and -ยฝ ยฝ and 3/2 -3/2 and -3/2 3/2 I = 0 4 ways
- 31. How would the spectra change if we had 3 bridging Hydrogens ?
- 32. 19F NMR example (spin ยฝ)
- 33. To calculate the coupling constant J, we have to convert the difference in ฮด to Hz ฮฮด = 0.07 The measuring frequency was 282 MHz J = 0.07 1/106 x 282 MHz = 20 Hz
- 35. We refer this simply to be a multiplet
- 36. Read the spectrum from left to right
- 37. DYNAMIC NMR
- 38. Principle How would the spectrum look like if the groups A and B and โfrozenโ and if they can quickly interchange ?
- 47. equivalent F
- 48. t
- 49. Which F-NMR signals would we expect ? Replace by C6H5 here
- 50. a a b c a b We would expect a broadened triplet for each of the equatorial flourines
- 54. c a b d Protons c and d are not equivalent, so we would expect 2 different doublets