Spectroscopic Methods in Inorganic Chemistry

Part 2: NMR

Dr. Chris, Feb. 2016

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.

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

In a strong magnetic Field B0 the spins can have the values + or – ½. The energy difference between these states is:

Energy States for I = ½

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

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.

Spin Coupling

http://orgchem.colorado.edu/Spectroscopy/nmrtheory/splitting.html

Example

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

Exercise

How would the spectrum look like for 1H NMR 19F NMR 31P NMR

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

http://users.auth.gr/akrivos/Metaptyxiaka/Fysikes%20Methodoi/008-P-C-NMR-complexes_Coord-Chem-Rev_2008.pdf

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

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

How about 11B NMR of BH4- ?

The interactions of the B-spin with the 4 H-spins

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

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”

Example 2: Diborane

Consider only the 2 H in the bridge What is the 1H NMR spectrum What is the 11B NMR spectrum ?

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

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

How would the spectra change if we had 3 bridging Hydrogens ?

19F NMR example (spin ½)

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

We refer this simply to be a multiplet

Read the spectrum from left to right

DYNAMIC NMR

Principle

How would the spectrum look like if the groups A and B and “frozen” and if they can quickly interchange ?

Example: Methanol

Example: Re Complex

http://www.rsc.org/images/18_Solving_Inorganic_Spectroscopic_Problems_tcm18-29994.pdf

equivalent F

t

Which F-NMR signals would we expect ?

Replace by C6H5 here

a

a

b

c

a

b

We would expect a broadened triplet for each of the equatorial flourines

c a

b

d Protons c and d are not equivalent, so we would expect 2 different doublets

c

b

Ca. 1-3 Hz

14Hz