Nuclear Magnetic Resonance Spectroscopyesilrch1.esi.umontreal.ca/~syguschj/cours/BCM6200/BCM...

Nuclear Magnetic Resonance Spectroscopy!

Sample

Data

Spectrum!

Detection

Fourier!

transformation

Storage

Response

Magnetization

Magnet

NMR Spectroscopy!

Nuclear Magnetism!

NMR is a manifestation of nuclear spin angular momentum (P).!

P = h[ I (I + 1)]1/2!

I = angular momentum quantum number!

h= Planck’s constant/2!!

!

Possible Values of I:!

1.! I = 0, mass number (A) and atomic number (Z) are even!

2.! I = half-integral value, A is odd.!

3.! I = integral value, A is even and Z is odd.!

!

Nuclear Magnetism!

For NMR:

1.! I = 0 are NMR-inactive

2.! I > 1/2, nuclei possess electric quadrupole moment due to

non-spherical nuclear charge distribution. The lifetime

of the magnetic states for quadrupoles in solution are

much shorter than for I = 1/ 2 This results in line

broadening and the can be more difficult to study.

3.! I = 1/2 include 1H, 13C, 15N and 31P.

!

Nuclear Magnetism!

Due to restrictions of quantum mechanics only one of the

three Cartesian coordinates can be specified.

Pz = hm

m = (-I, -I + 1, …., I – 1, I) – magnetic quantum number.

!

Nuclear Magnetism!

Iz has 2I + 1 values of m. This behavior is called

directional quantization.

In the case I=1/2 In the case I=1!

Nuclear Magnetism!

Nuclei with non-zero spin angular momentum also possess nuclear magnetic moment:

! = " P

-" in part determines the receptivity of a nucleus in NMR spectroscopy. It is a constant for a given nuclide and it is referred to as the gyromagnetic ratio.

Therefore:

µz = m"h

- No simple model can predict or explain the actual magnetic moments of nuclei.

Nuclear Magnetism!

More Quantitatively:

µ = g ehI 4!Mc

M = mass

e = charge uniformly spread over the surface

c = speed of light.

Properties of some nuclei of importance in NMR

Nuclear Magnetism!

•! In the absence of an external magnetic field:

2I + 1 states of m are equivalent

•! In the presence of an external magnetic field, the

spin states of a nucleus have energies given by:

E = µz B0 = - Iz" hB0

B0 = static magnetic field strength.

Nuclear Magnetism!

The projection of angular momentum of the nuclei

onto the z-axis of the laboratory frame results in 2I +

1 equally spaced energy levels, which are referred to

as the Zeeman levels or Zeeman States.

In the case I=1/2 In the case I=1!

Nuclear Magnetism!

For Iz = 1/2 we have m = +1/2 and m = -1/2 in which

µz is either parallel or anti-parallel to the field

direction.

B0!

Nuclear Magnetism!

In quantum mechanics:

m = + 1/2 = spin function #$

m = - 1/2 = spin function %$

**Since we are restricted to single quantum

transitions:

$ $ $&m = ± 1

Nuclear Magnetism!

!

!

!

!

!

!

!

&E = "hB0!

Nuclear Magnetism!

What Effects &E and why is this important?

•!Gyromagnetic ratio

•!Strength of static magnetic field

At equilibrium the energy states are not equally populated and the relative populations is given by the Boltzmann distribution.

In our case, Em = -mh"B0

!

!"=

##$

%&&'

( "##$

%&&'

( "=

I

Im B

m

B

mm

Tk

E

Tk

E

N

Nexpexp

!!"

#$$%

&!!"

#$$%

&= '

(=Tk

Bhm

Tk

B

B

I

ImB

00 exphm

exp))

Nuclear Magnetism!

Since m!"B0 << kbT the exponential can be expanded to first order using Taylor series.

!

!

!

!

!

!

!

•! exp (-E/k T) " 1 – E/k T

•! denominator sums to " 2I + 1

( )121

11

0

00

+!!"

#$$%

&+'

!!"

#$$%

&+!!

"

#$$%

&+' (

)=

ITk

Bhm

Tk

Bhm

Tk

Bhm

B

I

Im BB

*

**

Nuclear Magnetism!

Here is an example for I = 1/2

k: Boltzmann constant (1.3805 x 10-16 erg K-1)

T: temperature (K)

h: Plank constant (1.0546 x 10-27 erg S)

B0: magnetic field strength

(Tesla, T; 100 MHz ! 2.35 T)

": gyromagnetic ratio (107 Hz T-1)

TkBhTkEeN

N

bb

TkE b

011 !

"

#$=%$&=

%$

Nuclear Magnetism!

Example:

At T = 300 K and B0 = 5.875 T (250 MHz),

1H:

300103805.1

875.5100546.1107519.261

16

277

!!

!!!!"=

"

"

#

$

N

N

( ) !!" NNN 99996.000004.01 =#$

!" NN 99987.0#

For B0 = 18.8 T (800MHz),

Nuclear Magnetism!

In classical representation the nuclear dipole process about the

magnetic field direction with Larmor Frequency:

'o = "B0 (Radians/sec)

( = '0/2! = "B0 /2! (Hertz or cycles/sec)

The z-component is given by the magnetic moment:

µz = "Pz = m"h

Nuclear Magnetism!

When m = ± 1/2; the nuclear dipoles precess around a

double cone with a half-angle of the cone is 54.7°.

Due to the differences in populations of the energy states

there is macroscopic magnetization along the magnetic field

direction.

Nuclear Magnetism!

For a macroscopic sample:

The x and y components are random and sum to zero so:

But we know that :

Doing the math:

!"=

=

I

Im

mmNhM #

0

( ) ( )!= TkBhmTkBhmNNbbm 00 expexp ""

( ) ( )TkIIBhNMb

310

22

0+= !

Nuclear Magnetism!

So the macroscopic magnetization (M0) depends on:

•! The External Magnetic Field- B0

•! The angular momentum quantum number- I

•! The gyromagnetic ratio- "2

•! The temperature- 1/T

Note: I and " are specific to the nuclei of interest.

Nuclear Magnetic Resonance (NMR)!

What happens in NMR:

To induce NMR transitions a radiofrequency pulse is

applied to the sample. The pulse is generated by a linear

alternating electromagnetic field (B1) along a transverse

axis. This results in M0 being rotated into the XY plane.

Resonance Condition:

h(1 = &E

Figure 1-6

Nuclear Magnetic Resonance!

Larmor frequency : (1 = (L = |" /2!| B0

At 11.7 T : for 1H" = 2.675 x 108 (0 = 500 MHz

for 13C" = 0.6726 x 108 (0 = 125 MHz


Pulsed NMR:

- All nuclei are excited simultaneously by a radiofrequency pulse.

- Pulse angle is proportional to the pulse width (duration) and pulse power (Magnitude of B1 field).

-The coil is arranged so that this field is perpendicular to the applied field (in the x-y plane).


The oscillating magnetic field is equivalent to two counter-

rotating magnetization vectors:

We can use this pair of counter-rotating vectors as an

equivalent representation of the rf signal.


We will represent the rf along the x-direction as two vectors with the same magnitude (B1).

•! one vector rotates

clockwise B1 (r).

•! one vector rotates

counter-clockwise B1 (l)


Only one of the two components is capable of interacting with the processing nuclear dipoles (B1).

The sample magnetization is

static along the z-axis.

What happens when the rf

magnetization (which is moving)

interacts with the sample

magnetization?


The Rotating Frame:

The experimentally detected signal is subtracted from the

carrier frequency. We chose a set of coordinates that rotate

along with the nuclear precession.


!

!

ROTATING FRAME


The Rotating Frame:

How can we use this to look at rf pulses (B1)?

- In the rotating frame B1 is at right angles to M.

- The net result is to produce a torque acting around B1 at

a speed depending on the field strength.


The Rotating Frame:

How does this look in our two frames of reference?


The Rotating Frame:

What happens to M as we increase the length of the pulse?


The Rotating Frame:

How des this effect the NMR signal?

Our receiver is oriented with its axis along the y-direction.


)!= pulse angle or pulse flip-angle.

Four common pulses and their phases (*):




Pictorial representation of phase coherence:

Free Induction Decay (FID) and Fourier

Transformation (FT)!

Time

E

FID T2

AQ

+

+

+

FT

X

Y

YO



Fourier Transformation (FT):!

- dispersion component is removed by phase cycling.

( )

tite

dtetfwg

ti

ti

!!!

!

sincos

)(

+=

=

"

#

#"

$

real part imaginary part

(Absorption) (Dispersion)



FIGURE 1.15

90MHz 1H NMR spectrum of methyl iodide CH3I(1);

One pulse, spectral width 1200 Hz, 8 K data points,

acquisition time 0.8s. A: time domain spectrum (FID);

the generator frequency is almost exactly equal to the

Resonance frequency of the sample; B: frequency

domain spectrum obtained by Fourier transformation of

A.

FIGURE 1.16

22.63 MHz 13C NMR spectrum of methanol 13CH3OH (2);

Solvent: D2O, 17 pulses, spectral width 1000 Hz, 8 Kne

pulse, spcectral width 1200 Hz, 8 K data points, A: Time

domain spectrum (FID); B: frequency domain spectrum

obtained by Fourier transformation of A. This consists of a

quarter, as the 13C nucleus is coupled to the three protors of

the methyl group.

Signal Detection!

Sensitivity of the NMR Signal$

+!= electromagnetic induction force in a detection coil

(Faradays Law of Induction).

Procession of a bulk magnetic moment about a static field

yields a time varying magnetic field. This produces an

induced electromotive force.

Signal Detection!

Higher field = higher signal (lose some to noise)

For two nuclei with same value of I their sensitivity is

related by :

a = 13C b = 1H

( ) ( )

( ) ( )TkIIBhNBM

TkIIBhNM

BMdttdM

b

b

31

31

)(

20

230

022

0

0

+=

+=

=!

""

"

"#

( ) ( ) 0159.041 13C vs1Hfor 33=!

ba""

Properties of some nuclei of importance in NMR

Relaxation!T1: spin-lattice relaxation time (longitudinal)!

T2: spin-spin relaxation time (transverse)!

!

-! After we remove the pulse, the magnetization vector

returns to equilibrium.!

-! Bloch assumed that the processes were first-order and

defined them by T1 and T2:!

dMz/dt = - Mz – Mo/T1!

so!

Mz = M0(1-e-t/T1)!

dMx,/dt = -Mx,/T2!

dMy,/dt = -My,/T2!

Relaxation!

!

!

!

!

On Resonance:!

T1 Relaxation!

Experimental determination:

T1 Relaxation!

1H:

T1 Relaxation!

13C:

T1 Relaxation!

13C:

T1 Relaxation!

How do we determine the T1 value?!

! !M0 - Mz = Ae--/T1!

After a 180° pulse during the inversion recovery experiment M0 = -Mz so A = 2M0:!

!M0 - Mz = 2M0e--/T1!

!ln (M0 - Mz) = ln2M0 - -/T1!

!

Substituting I for M where I is intensity of signal:!

ln (I0-Iz) = ln2I0 - -/T1!

Tzero = T1 ln2!

T1 values can vary greatly from one nuclei to the next in the same molecule.!

T1 Relaxation!

Need to wait 5T1 periods for system to return to

equilibrium so:!

! !(5 T1 - 180°x' - - - 90°x' - FID)n !

!

Why measure T1?!

! !/2 pulse elicits maximal NMR signal at thermal

equilibrium.!

- for repetitive signal averaging this is not always the best

pulse angle.!

- need to wait about 5T1 for z-magnetization to return to

equilibrium.!

- more efficient to repeat sooner.!

T1 Relaxation!

Ernst Angle:!

! !Cos#E = e -Tr/T1!

- Tr = the delay between pulses.!

- best value of # actually depends on offsets of signals in

spectrum but this is an average value.!

If stick with !/2 pulses then:!

- the optimum repetition rate is Tr = 1.27 T1!

- in this case you are about 80% efficient.!

T1 Relaxation!

Influence of protons on 13C T1 values:!

!

- main contribution is via dipole-dipole

interactions, so directly bonded protons can have

large effects.!

- increasing the number of protons shortens the

T1 although the correlation is not always direct.!

Influence of molecular size on T1:!

- T1 values decrease as the size of the molecules. !

T2 Relaxation!

After 90°x' pulse Mz = 0 and net magnetization is along the

y-axis. The population of N# and N% are equal . !

Phase Coherence: a small proportion of nuclear dipoles are

bunched along the y-axis.!

T2 - determines how fast Mx' and My' return to zero.!

T2 Relaxation!

T2 Relaxation!

How do we view this?:!

-Nuclei change from one energy state to the other and we lose the phase coherence.!

- this is not the main contribution.!

Magnetic Field inhomogeneities: there is a small &B0 throughout the sample which leads to fanning out of signals.!

-fanning out process causes signals that are chemically equivalent to process with slightly different Larmor frequencies.!

- the net result is an increase in the line width of the various signals.!

By definition though T1 . T2!

T2 Relaxation!

Experimental Determination of:!

- We want to measure a true value which is independent of the

&B0 fanning out process.!

Spin-echo experiment:!

After initial 90°x' pulse you apply a 180°x' pulse at set interval -.!

At !2- - resultant transverse magnetization is -y' direction.!

!4- - magnetization is y' direction.!

!6- - magnetization is -y' direction.!

Etc……we get an echo with alternating phase at intervals of 2-.!

However, the net magnetization is decreasing with each echo

due to true spin-spin relaxation.!

T2 Relaxation!

T2 Relaxation!

!

-! the 180°x' pulse turns the Ay' component into the (-y') direction but has no effect on the Ax'. The angle between vector A and the y'-axis is unchanged if we disregard sign.!

- the effect of a 180°x' pulse is that vectors A and B undergo reflections in the x', z' plane.!

!

T2 Relaxation!

Quantitatively:!

! ! !My' = A e -t/T2!

!

At t=0 , A=M0!

!

! ! !ln My' = lnMo - t/T2!

!

since I # My'!

! ! !ln I(t) = ln I0 - t/T2!

!

Plot I(t) vs t where I(t) is intensities at each echo. Slope of line is -1/T2.!

!

T2 Relaxation!

For the most part we do not concern ourselves with T2

measurement.!

- Of more consequence is T2* which takes into account the

inhomogeneity of the magnetic field.!

Nuclear Magnetic Resonance Spectroscopyesilrch1.esi.umontreal.ca/~syguschj/cours/BCM6200/BCM...

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