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Demystifying the Black Box: The Mechanics Behind Nuclear Magnetic Resonance Spectroscopy

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### Demystifying the Black Box:The Mechanics Behind Nuclear Magnetic Resonance Spectroscopy

Kelsie Betsch

Chem 381

Spring 2004

What is NMR good for?

- Spectroscopic method widely used by chemists
- Provides information about:
- The number of magnetically distinct atoms of the type being studied
- The immediate environment surrounding each type of nuclei

Overview

- NMR involves transitions of the orientations of nuclear spins in magnetic fields
- Examine the quantum-mechanical states of nuclear spins interacting with magnetic fields
- Specific focus on hydrogen

Spin

- Electron has intrinsic spin angular momentum
- z component of ±ħ/2
- Spin of ½
- Nuclei also have intrinsic spin angular momenta, I
- Spins not restricted to 1/2

Spin eigenvalue equations

- Nuclear spin eigenvalue equations for protons:

Î2=½(½ +1) ħ2 Î2=½ (½ +1) ħ2

Îz=½ħ Îz =½ħ

() and () are spin functions

- is a spin variable
- ↔ Iz= ħ/2 and ↔ Iz= -ħ/2
- and are orthonormal

Ah, physics

- Motion of an electric charge around closed loop produces a magnetic dipole:

μ = iA

i= current (amperes)

A= area of loop (m2)

- Substitution of i=qv/2πr and A=πr2

μ = qrv/2

- Noncircular orbit

μ= q(r×v)/2

Physics

- Express μin terms of angular momentum, L
- L = r×p and p =mv

μ = (q/2m)L

- Replace classical angular momentum with spin angular momentum, I

μ = gN(q/2mN)I = gNβNI = I

gN= nuclear g factor, βN= nuclear magneton, mN= mass of nucleus, = gN βN=magnetogyric ratio

Physics

- Magnetic dipole wants to align itself with magnetic field
- Potential energy, V, for the process

V = -μ•B

where F = q(v×B)

- Take magnetic field to be in the z direction:

V = -μzBz = -γBzIz

Dipping into Quantum Mechanics…

- Replace Iz by its operator equivalent, Îz
- Can now write the spin Hamiltonian

Ĥ = -γBzÎz

- Corresponding Schrödinger equation

Ĥ = -γBzÎz = E

- Wave functions are the spin eigenfunctions

Îz 1 = - ħγm1Bz

E = - ħγm1Bz

Energy differences

- Interested in transitions between alignment with the field (m1= ½) and against the magnetic field (m1= -½)
- Energy difference

E = E(m1= -½) –E(m1= ½) = ħγBz

- Note that E depends upon strength of magnetic field

Condition for resonance

- Sample is irradiated with electromagnetic radiation
- When E matches the energy of the radiation:
- The proton will make a transition from the lower energy state to the higher energy state,
- The sample will absorb and give the NMR spectrum
- Condition for resonance/absorption

E = ħγBz= hν

Shielding

- Frequency of associated transition:

ν = γBz/2π

Bz = magnetic field experienced by nucleus

- Seems all protons would absorb at the same frequency
- Account for magnetic field induced by moving electrons
- Total magnetic field = sum of applied field and shielding field

B0 = (2)/((1-))

Resonance Frequency and Chemical Shift

- Resonance frequency

H = ((γB0)/(2))(1- H)

- Chemical shift

H = ((H - TMS)/spectrometer) 106

- Degree of shielding with electron density
- Greater electron density = smaller chemical shift
- Deshielded – left, downfield, weak field
- Well-shielded – right, upfield, strong field

Why does splitting occur?

- Any given hydrogen is also acted upon by the magnetic field due to the magnetic dipoles of neighboring hydrogen nuclei
- Effect is to split the signal of the given hydrogen nuclei into multiplets

A quantitative approach: Step 1

- Hamiltonian that accounts for spin-spin interaction

Ĥ = -γB0(1- 1)Îz1- γB0(1- 2)Îz2+ (hJ12/ħ) Î1Î2

J12 = spin-spin coupling constant

Step 2: Perturbation theory

- Assume first-order perturbation theory is adequate
- Unperturbed Hamiltonian

Ĥ(0) = -γB0(1- 1)Îz1- γB0(1- 2)Îz2

- Perturbation term

Ĥ(1) = (hJ12/ħ) Î1Î2

Step 3: Solve Schrödinger Eqn

- Unperturbed wave function

1 = (1)(2) 2 = β (1)(2)

3 = (1)β(2) 4 = β(1)β(2)

- Energy equation through first order

Ej = Ej(0) + d1d2 j*Ĥ(1) j

- Solve unperturbed and perturbed portions separately

Step 3: Solve Schrödinger Eqn

- For unperturbed part, recall

Ĥ(0) j = Ej(0) j

- For first-order corrections

Hii(1) i = (hJ12/ħ)d1d2i*Î1Î2 i

- Turns out that only z components contribute to first-order energies

Energies and selection rules

- Only one type of nucleus at a time can undergo a transition

First-order Spectra

- Resonance frequencies
- Occur as a pair of two closely spaced lines doublet
- Condition for use of first-order perturbation theory

J12 << 01- 2

- Leads to two separated doublets, which is called a first-order spectrum

The Case of Equivalent Protons

- Similar calculations
- Two shielding constants are equal
- Equivalent, indistinguishable nulcei wave functions are combinations
- Spin-spin coupling constant effect cancels in the transition frequencies due to selection rules
- Single proton resonance observed

Visiting the Variational Method

- Second-order spectra can be calculated exactly
- Same Hamiltonian
- Linear combination of possible wave functions as trial function

= c11+c2 2+c3 3+c4 4

First- or second-order?

- Observed spectra depend upon the relative values of 01- 2 and J
- J = 0 two separate singlets; two distinct hydrogen nuclei with no coupling
- 1 = 2 two chemically equivalent protons with one signal
- Cases between these conditions, the spectrum can varies
- This is a second-orderspectrum

First- and Second-order Examples

- Spectrum depends upon field strength, B, because depends upon B

Conclusion

- Classical physics behind Nuclear Magnetic Resonance Spectroscopy
- Chemical shifts
- Quantum mechanical methods used to determine spectra
- Splitting patterns

References

- D.A. McQuarrie, J.D. Simon, Physical Chemistry: A Molecular Approach, University Science books, CA. 1997.
- D.L. Pavia, G.M. Lampman, G.S. Kriz, Introduction to Spectroscopy, 3rd ed. Thomson Learning, Inc. 2001.
- F.L. Pilar, Elementary Quantum Chemsitry, 2nd Ed. Dover Publications, Inc. NY. 1990.

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