CHAPTER 14 Molecular Spectroscopy 3: Magnetic Resonance
I. Magnetic Resonance Spectroscopy - General Idea.
A. Excite Nuclear and electron spin state transitions. (NMR, ESR)
B. In absence of magnetic field, spin states are degenerate.
States split by constant magnetic field. (Stark field)
-the energy of a magnetic moment µ in a magnetic field B; is :
E = −µ • B
C. A second oscillating magnetic field (sweep field) can then excite spin flips, which lead to absorption of the secondary field energy. This field oscillation frequency is varied (scanned). € D. Splitting depends not only on Stark field strength, but on presence of other electrons and nuclei near one of interest. So frequency at which transition occurs reflects “chemical environment” in neighborhood of nucleus undergoing transition.
Probe of structure of molecules.
II. Effect of Magnetic Fields on Electrons and Nuclei
A. Electrons
1. The orbital angular momentum of the electron produces a magnetic moment that interacts with the external field B and gives energies:
E =m µ B m B o
m =magnetic quantum number µB = Bohr magneton
Bo =magnetic field magnitude
e −24 µB = −γe = = 9.724x10 J/T 2me so :
Em = −γem Bo
CHAPTER 14 1 € 2. The spin angular momentum also interacts with the magnetic field with energy:
E g m B ms = e sµB o
ms = spin quantum number(±1/2) g g-value = 2.0002319 e = so : E = −g γ m B ms e e s o
3. These equations are useful in interpreting ESR spectra, which observes electron spin state changes for unpaired electrons only. € B. Nuclei.
1. The spin quantum number I of a nucleus is a fixed property depending on number of protons and neutrons in the nucleus:
Table 14A.1 Nuclear constitution and the nuclear spin quantum number
2. Spin angular momentum of the nucleus is, as usual:
I(I+1)
3. For a given I there are spins states mI = -I, …. +I representing orientation of the nuclear magnetic moment relative to the external field. € 4. Energy of interaction of nuclear magnetic moment with the field B is:
CHAPTER 14 2 E B m = −µz o I nuclear magnetic moment along field direction µz = E m B m = −γ I! o I m spin orientation quantum number( 1/2 for proton) I = ± γ = magnetogyric ratio for specified nucleus E g m B m = − µN I o I g = nuclear g-factor
e! 27 µ = = 5.051 ×10− J / T N 2m proton (nuclear magneton constant) For proton: gµ B E = ± N o 2
Table 14A.2 Nuclear spin properties
CHAPTER 14 3
III. NMR - nuclear magnetic resonance spectroscopy.
A. Field strengths and frequencies involved.
1. An external magnetic field of strength B (typically magnitude ~ 10 Tesla (T)) produces a splitting of the spin states (for a spin 1/2 nucleus) given by 1 1 E E E g B ( ) Δ = −1/2 − +1/2 = − µN o − − 2 2 E g B Δ = µN o
2. So now, what sweep field frequency is needed to cause spin state changes from +1/2 to -1/2 in this diagram?
hν = ΔE = gµNBo
ν = gµNBo /h
3. Work it out, for proton NMR; for C-13 NMR:
ν = 425 MHz for protons; 107 MHz for C-13 nuclei
€ These are the resonance frequencies for an isolated nucleus in a field of 10 Tesla.
CHAPTER 14 4 B. The Chemical Shift.
1. Nuclei interact not just with the external field Bo, but with the local magnetic field B.
2. Local field B is combination of external field Bo plus field induced in the neighborhood Bind.
3. External field induces electronic orbital angular momenta giving rise to a field of its own. (field due to circulation of electronic currents)
4. Induced field Bind = -σBo
where σ = shielding constant Bo = external field magnitude
5. So local field felt by nuclei is:
Bloc = Bo + Bind =(1-σ) Bo
6. The chemical shift δ of a particular nucleus is then defined:
ν − ν° δ = ×106ppm ν°
where ν = actual resonance freq ν° = resonance freq in a reference standard compound (TMS) €
7. Relation of δ to σ.
gµN gµN ν = B = (1 − σ)Bo h loc h
gµN gµN νo = Bloc(TMS) = (1 − σo)Bo h h € gµN (1 − σ)Bo − (1 − σo)Bo h { } 6 € δ = ×10 gµN Bo h
6 δ = (σo − σ) ×10 ppm € where σo is the shielding constant in TMS σ is the shielding constant in actual environment € CHAPTER 14 5
Notice δ is independent of magnitude of external field Bo.
8. Deshielding:
As shielding σ gets smaller, chemical shift δ gets larger.
CHAPTER 14 6
9. Typical chemical shifts δ in proton and C-13 NMR.
CHAPTER 14 7
10. Contributions to shielding.
σ = σ(local) + σ(neighbor) + σ(solvent)
e- on e- in other same atom parts of molecule
σd + σp
diamagnetic paramagnetic contrib contrib
1 σd ∝ r
r = e- nuclear distance; depends on e- density close to nucleus
€ - σp only contributes if unpaired e present on the atom, or if molecule has low-lying excited states.
σ(neighbor) - arises from field-generated e- currents in neighboring groups of atoms.
1 ∝ where r = distance to neighbor atoms r3
Particularly strong around aromatic groups because B induces a ring current. €
CHAPTER 14 8
C. The Fine Structure. (spin-spin coupling)
1. Splitting of NMR lines due to effect of the other nuclei.
2. Strength of interaction expressed as scalar coupling constant J (Hz).
3. J is independent of field B.
4. For two interacting nuclei A...X with coupling J, each is split by the other by J Hz.
1 J ν = νo ± 2
in absence of neighbor nucleus
€ One line is split into two for each nucleus.
5. For nuclei A interacting with two equivalent nuclei X, in other words A...X2, the equivalent X nuclei resonate like a single nucleus but with double the intensity.
So the lines due to X are a doublet still.
CHAPTER 14 9
6. The nucleus A however is split into a 1:2:1 triplet.
7. For a nucleus A interacting with 3 equivalent nuclei A...X3 the spectrum of A is a 1:3:3:1 quartet.
CHAPTER 14 10
8. Now interpret ethanol splitting pattern.
CHAPTER 14 11
9. Magnitudes of coupling constants J.
1J means only 1 bond separates the two interacting nuclei.
13 1 1 e.g. C - H is JCH (~120 to 250 Hz)
2J means 2 bonds separate.
13 1 2 e.g. C - C - H is JCH (~10 to 20 Hz)
3J means 3 bonds separate. 1 1 H H e.g. 3 C— C is J HH (~1 to 15 Hz)
NJ for N>4 are negligible.
10. The Karplus Equation = three-bond coupling constant depends on dihedral angle as:
3J = A + B cos φ + C cos 2φ
φ
3 Important for JHH coupling:
1H 1H
C— C
CHAPTER 14 12