nmr for inorganic chemistry
TRANSCRIPT
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Spectroscopic Methods in Inorganic Chemistry
Part 2: NMR
Dr. Chris, Feb. 2016
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Rotating Charge produces a 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.
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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 = ½ For bigger I there is a quadrupole moment which affects the broadness of the peaks
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In a strong magnetic Field B0 the spins can have the values + or – ½. The energy difference between these states is:
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Energy States for I = ½
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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
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Chemical Shift δ
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.
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Spin Coupling
http://orgchem.colorado.edu/Spectroscopy/nmrtheory/splitting.html
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Example
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13C NMR 13C has I = 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
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Exercise
How would the spectrum look like for 1H NMR 19F NMR 31P NMR
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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
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http://users.auth.gr/akrivos/Metaptyxiaka/Fysikes%20Methodoi/008-P-C-NMR-complexes_Coord-Chem-Rev_2008.pdf
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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
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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
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How about 11B NMR of BH4- ?
The interactions of the B-spin with the 4 H-spins
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The 4 protons can 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
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Splitting pattern by spin 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”
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Example 2: Diborane
Consider only the 2 H in the bridge What is the 1H NMR spectrum What is the 11B NMR spectrum ?
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11B spectrum
2 equivalent B atoms couple with 2 equivalent protons 2 (1/2) 2 + 1 = 3 signals Coupling by spin ½ nuclei -> use the Pascal triangle
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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 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
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How would the spectra change if we had 3 bridging Hydrogens ?
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19F NMR example (spin ½)
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To calculate the coupling 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
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We refer this simply to be a multiplet
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Read the spectrum from left to right
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DYNAMIC NMR
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Principle
How would the spectrum look like if the groups A and B and “frozen” and if they can quickly interchange ?
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Example: Methanol
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Example: Re Complex
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http://www.rsc.org/images/18_Solving_Inorganic_Spectroscopic_Problems_tcm18-29994.pdf
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equivalent F
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t
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Which F-NMR signals would we expect ?
Replace by C6H5 here
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a
a
b
c
a
b
We would expect a broadened triplet for each of the equatorial flourines
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c a
b
d Protons c and d are not equivalent, so we would expect 2 different doublets
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c
b
Ca. 1-3 Hz
14Hz