Nuclear Magnetic Resonance (NMR) Spectroscopy cont... Recommended Reading:

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Nuclear Magnetic Resonance Nuclear Magnetic Resonance

(NMR) Spectroscopy (NMR) Spectroscopy

cont...

cont...

Applied

Spectroscopy

Recommended Reading:

Banwell and McCash Chapter 7 Skoog, Holler Nieman Chapter 19 Atkins, Chapter 18

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Relaxation processes Relaxation processes

We need a mechanism for the spins to relax back to the ground state.

To avoid saturation the relaxation rate must be at least as great as or greater than the rate of absorption of energy from the RF field.

In optical transitions the excited state relaxes by emitting radiation at a frequency corresponding to the energy difference between the two

states ⇒ fluorescence or radiative decay.

Rate of spontaneous emission ∝ ν3 (see, Einstein Coefficients)

⇒ rate of spontaneous emission is very small for RF transitions.

⇒ non-radiative mechanisms are very important for NMR transitions.

To reduce saturation and increase absorption signal, relaxation should occur as fast as possible ⇒ lifetime of excited state should be as small as possible.

BUT

From Uncertainty Principle, ΔEΔt ≥ h/4π ⇒ if Δt is small (short lifetime) then ΔE will be large ⇒ Broad spectral lines, and this prevents high

resolution spectroscopy. ⇒ Tradeoff between resolution and lifetime Optimum lifetime for NMR: 0.1 to 10 sec

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Relaxation Mechanisms Relaxation Mechanisms

Two important relaxation mechanisms in NMR spectroscopy.

(1) Spin-Lattice (longitudinal) relaxation (2) Spin-Spin (transverse) relaxation

1) Spin-Lattice relaxation

Absorbing nuclei are part of a larger collection of atoms. The entire

collection of atoms in the sample is called the Lattice (even if it is a liquid or gas). Atoms and molecules in sample are in thermal motion, this

creates a complex, time varying field about each nucleus (Lattice field) some of the magnetic components of this field will be varying at the

Larmor frequency and will interact with the nucleus of interest to cause it to relax from upper to lower state. The nucleus gives up its spin energy to the thermal motion of the lattice (⇒ tiny increase in sample

temperature) .

Spin Lattice relaxation is a first order decay process and is characterised by an exponential decay in the population of the excited state

characterised by a spin-lattice relaxation time (or longitudinal relaxation time ) T1.

T1 depends on magnetogyric ratio and mobility of atoms in the lattice.

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2) Spin-Spin relaxation

All other effects that cause relaxation from upper to lower level are collected together and are called spin-spin or transverse relaxation effects.

Two nuclei close enough together in the lattice such that one relaxes to the ground state and gives its energy to excite the second nucleus.

This does not change the population of the excited state BUT it decreases the average relaxation time of the excited state ⇒ Lifetime broadening.

Again, this is a 1st order effect with an exponential decay and an

associated lifetime. The spin-spin (or transverse) lifetime denoted by T2. T2 very small in solids (~ 10-4 sec) and dominates relaxation process ⇒ difficult to carry out high resolution NMR spectroscopy on solid

samples.

Solids: T2 ~ 10-4 sec << T1. ⇒ Broad Lines, low resolution experiments.

Liquids: T2 ≈ T1. Sharp lines, high resolution experiments.

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Chemical Shift Chemical Shift

So far we have only considered an isolated nucleus in the presence of an applied magnetic field.

This is not realistic since all nuclei are associated with electrons in atoms and molecules.

When placed in a magnetic field the surrounding electron cloud tends to produce a field that opposes the applied field. So the total field

experienced by the nucleus is

B - B

= B

effective applied induced

And since the magnitude of the induced field is proportional to the applied field

B

induced

= σ B

applied

where σ is a constant. So we have

( 1 - σ ) B

= B

effective 0

where B0 is the external applied magnetic field.

The nucleus is shielded from the applied field B0 by the diamagnetic contribution from the electrons. σ is called the shielding constant. The size of the shielding will depend on the electron density around the atom in a particular position in the molecule.

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We can generalise this by writing

( 1 - ) B

= B

i 0

σ

i

where Bi is the field experienced by nucleus i whose shielding constant is σi .

Example: Oxygen is more electronegative than carbon, So the electron density around the hydrogen nucleus in a C - H bond should be much higher than that around a hydrogen nucleus in a O - H bond. So we would expect the shielding constant for the hydrogen in C - H to be greater than the shielding for the hydrogen in an O- H bond

OH

CH

> σ

and hence

σ

( 1 - ) < B = B ( 1 - ) B

= B

CH 0

σ

CH OH 0

σ

OH

⇒ Field experienced by H in O-H is greater than field experienced by H in C-H bond.

⇒ For a given externally applied field the C-H hydrogen nucleus will have a smaller Larmor frequency than that of the O-H hydrogen.

Therefore the C-H hydrogen will come into resonance at a lower radio- frequency than the O-H hydrogen. Conversely, to come into resonance at a given frequency, the C-H hydrogen requires a bigger magnetic field.

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Effect of an applied magnetic field on the methyl (C-H) and hydroxy (O-H) nuclei of

methanol.

For a given applied radio- frequency:

The OH protons are shielded less, so they experience a bigger

magnetic field ⇒ bigger splitting ⇒ they come into resonance before the C-H protons.

Note that in CH3OH there are three CH3 and only one OH protons so the ratio of the areas (absorption intensities) of the NMR peaks is 3:1

12C and 16O nuclei have no spin, so they have no NMR signal

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Previous figure illustrates two important points

1) Identical nuclei (e.g. 1H) give rise to absorption peaks at different positions when they are in different chemical environments. For this reason the separation between the absorption peaks is usually referred to as their chemical shift.

2) The area of an absorption peak is proportional to the number of

equivalent nuclei (I.e nuclei with the same chemical shift position) giving rise to the absorption.

There are several ways to quantify the chemical shift difference Δ between the O-H and C-H signals.

1) we assume spectrum to be recorded by varying the applied magnetic field, so we could attach a tesla (micro-tesla) scale to the spectrum. In this case Δ = 3.26 μT (see scale on previous figure)

2) we could also have recorded the spectrum by holding the magnetic field constant and varying the frequency of the electromagnetic

radiation. In this case 2.3487 T = 100 MHz.

So Δ = 3.26 μT = 139Hz

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Neither of these measurements of the chemical shift is satisfactory.

( ) ( 1 - )

2

= B -

h 1 B

= g h

B

= g h

= E

L i N i N 0 i 0

σ

i

π σ γ

β β

ν Δ

And so in Hertz

(

CH OH

)

0

(

CH OH

)

0 N

2

= B h

B

= g

σ σ

π γ

β σ - σ -

Δ

The shielding constant σi is independent of the applied field or

frequency, so the measured chemical shift separation is proportional to the applied magnetic field B0. This is very inconvenient, since the

measured shifts will vary with the operating conditions of the

instrument, particularly since different spectrometers use fields varying between 0.6 and 15 Tesla. Measurements will vary from lab to lab. How do we overcome this problem?

Ans: Give chemical shifts as a fraction of the applied field or frequency.

Example: for methanol: Δ = 139 Hz at 100 MHz ⇒ Δ =1.39 × 10-6 = 1.39 p.p.m or, in terms of the field, Δ = 3.26 × 10-6 T at 2.3487 T = 1.39 p.p.m.

Even if we double the field or frequency the separation would also double so the chemical shift would still be 1.39 ppm.

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In the previous example we have looked at the shift of the C-H proton relative to the O-H proton in methanol.

Ideally we would like to reference all shifts to the resonance position of the unshielded proton for which σ = 0. BUT this is an impracticable

standard - cannot have a sample in which we have a bare proton.

So we choose some reference substance as a secondary standard and measure the shift in the resonance position of other hydrogen atoms from the resonance position for hydrogen in this standard in p.p.m.

The material usually chosen as a standard for 1H and 13C NMR spectroscopy is trimethylsilane Si(CH3)4 or TMS.

Advantages of TMS as a reference

1) All 12 hydrogen nuclei are equivalent ( i.e. have the same chemical environment) and so absorb at the same position – TMS resonance is sharp and intense.

2) The resonance position of protons in TMS is to the high field side of almost all of the other hydrogen resonances in organic molecules, so it can be easily recognised.

3) Low boiling point so it can be easily removed from samples.

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10

6

×

=

o

o

ν ν - δ ν

where νo is the resonance frequency of the protons in TMS.

The relationship between the shielding constant σ, and the chemical shift δ, is then

( ) ( )

( ) × 10

6

= × 10

6

~ ( ) × 10

6

B

B

= B σ - σ

σ - 1

σ - σ σ

- 1

σ - 1 - σ -

δ 1

o o

o o

o

We can now define the chemical shift as

where we have used the fact that σo << 1 for the protons in TMS.

Note that as the shielding σ gets smaller the chemical shift δ gets bigger

nuclei with large chemical shifts are strongly deshielded

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B

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Example: the 1H

Example: the 1H - - NMR NMR spectrum of ethanol spectrum of ethanol

CH3 CH2

OH

Ratio of areas under peaks is 1:2:3 (indicated by red lines)

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NMR Experiment NMR Experiment

The early years of NMR spectroscopy (1945-1970) were dominated by continuous wave (CW-NMR) methods, using a weak, fixed-amplitude radiofrequency field.

Spectra were obtained either by

(i) keeping the electromagnetic frequency fixed, while slowly sweeping the magnetic field strength, or

(ii) Keeping magnetic field fixed and sweeping the radio frequency

Either method will bring spins with different chemical shifts sequentially into resonance, recording the NMR signal all the while.

This approach has now been almost completely superseded by pulse techniques employing short, intense bursts of radiofrequency radiation, this is Fourier Transform NMR spectroscopy FT-NMR (described later).

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Continuous Wave NMR (CW

Continuous Wave NMR (CW - - NMR) NMR)

Transmitter and

receiver coils at 90

O

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NMR spectrometer consists of:

1) a magnet that can produce a uniform, intense magnetic field.

Nowadays superconducting magnets are usually used, older systems used water cooled electromagnets.

The magnet poles are 20-30 cm in diameter, and the gap between them is only some 2-3 cm.

When recording a spectrum in the field-sweep mode, the radio-frequency oscillator (the source) bathes the sample in radiation of, say, 100 MHz by means of the transmitter coil placed near the sample in a vertical plane.

The magnetic field is set to about 2.5 T and this is smoothly raised (swept) by means of a current produced in the sweep generator fed to auxiliary coils round the magnet poles.

As each nucleus is brought to resonance it absorbs energy from the oscillator and then, when it reverts to the ground state, the emitted energy is collected by the detector coil wound round the sample, amplified, and passed to the recorder.

The detector coil is in the horizontal plane so it is perpendicular to the transmitter coil and therefore does not interact with it.

Continuous Wave NMR (CW

Continuous Wave NMR (CW - - NMR) NMR)

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The swept magnetic field method is by far the simplest technically, since it is easy to change a magnetic field smoothly, and fairly easy to maintain a precise frequency by using a carefully controlled crystal oscillator.

The alternative frequency-sweep mode, wherein the magnetic field is maintained constant and the radiofrequency oscillator swept through a frequency range, is more difficult, but has advantages, particularly for some double-resonance experiments, and is sometimes employed.

Both of these sweep techniques have proved successful for an adequate quantity of a sample containing either hydrogen, fluorine, or phosphorus nuclei.

For example, 20 mg of sample dissolved in 0.5 ml of a solvent enables a good spectrum to be obtained in about 5 minutes: 10mg samples take longer, typically about 20 minutes.

Figure

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References

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