NMR: Nucelar Magnetic Resonance

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UNIVERSITY OF PÉCS MEDICAL SCHOOL

www.medschool.pte.hu

NMR: Nucelar

Magnetic Resonance

Miklós Nyitrai, April 6, 2016

What is the medical importance of NMR?

(see lecture No. 13. in this semeter!)

www.medschool.pte.hu www.medschool.pte.hu

Physical bases, spins and nuclei

Protons and neutrons (and also the electrons) are particles having a spin angular momentum of 1/2.

If the number of protons or neutrons is add in the nucleus, then the total nuclear spin will not be zero!

Spins and nuclei

Properties of nuclear spin

Nucleus: proton and neutron

Two magnetic moments (spin) pointing in opposite directions zero net magnetic moment!

nuclei with even Z and even N have nuclear spin I=0 (isotopes are important)

If there are unpairedprotons or neutrons in a nucleus, then the net nuclear spin (the intrinsic nuclear magnetic

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Atoms for NMR

1_H,2_H,3_H,3_He,4_He,12_C,13_C,14_C,14_N,15_N16_O,17_O,19_F,23_Na,31_{P, etc.}

The most frequently applied nuclei: 1_H,13_C,15_N,17_O,19_F,31_P

Signal depends on:

-Magnitudeof the magnetic moment -Concentrationof the isotope

The split of the energy states (Zeemann)

N

S

Spinning nucleus with it’s magnetic field alligned with the external field.

Spinning nucleus with it’s magnetic field alligned against the external field.

α-spin state, favourable, lower energy N S N S β-spin state, unfavourable, higher energy

The energy difference between the spin states does depend on the

strength of the external magnetic field.

www.medschool.pte.hu m: magnetic moment of the individual atom

Orientation and reorientation on the

microscopic level

µ

H=0; M=0

µ

_µ

µ

H>0; M>0

M=

Σ

µ

i www.medschool.pte.hu

Graphical representation

Remember, the energy is proportional to the frequency!

H

E

at no magnetic field there

is no difference between these states α-spin state β-spin state ∆E1

<

∆E2 H1

<

H2 N S N S S N N S

An example: CH

4

H

E

α-spin state β-spin state h x 300MHz 0 T 7,05 T 11,75 T h x 500MHz

We can probe the energy difference of theα- andβ- state of the protons by irradiating them with EM radiation of just the right energy. In a magnet of 7.05 Tesla, it takes EM radiation of about 300 MHz (radio waves). So, if we bombard the molecule with 300 MHz radio waves, the protons will absorb that energy and we can measure that absorbance. In a magnet of 11.75 Tesla, it takes EM radiation of about 500 MHz (stronger magnet means greater energy difference between theα -andβ- state of the protons).

What is the problem with this concept?

It is difficult to compare the data from two instruments with different strenghts of magnetic fields.

Let’s use the chemical shift!

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Chemical shift or

δ

We need a reference sample to standardize our insturments!

1. Measure the absorbance frequency of the standard: fr;

2. Measure the frequency for your own sample: fs;

3. Let’s calculate the difference between the two: ∆f = fr-fs;

4. Normalise the difference with the ferquency of the reference:δ =∆f / fr;

Theδ,i.e. the chemical shift will be characteristic for your sample, but will not depend on the parameters of the

instrument used for the experiments!

Why? Let’s take an example!

Imagine that we have a magnet where our standard absorbs at 300,000,000 Hz (300 megahertz), and our sample absorbs at 300,000,300 Hz. The difference is 300 Hz, so we take

300/300,000,000 = 1/1,000,000 and call that 1 part per million (or 1 PPM).

Now lets examine the same sample in a stronger magnetic field where the reference comes at 450,000,000 Hz, or 450 megahertz. The frequency of our sample will increase proportionally, and will come at 450,000,450 Hz. The difference is now 450 Hz, but we divide by 450,000,000 (450/450,000,000 = 1/1,000,000, = 1 PPM).

(We do not have to calculate all these, the NMR machine does it for us!)

An example: chemical

shift

The absorption is proportional to the concentrationof the corresponding nuclei.

The NMR spectrumis the energy absorbed by the system as the function of the freqeuncy (f) of the excitation energy (ΔE) or the magnetic field (H, B). Due to local effects and fields the excitation energy (and

thus frequency) is different for different nuclei.

NMR spectrum

Structureof organic molecules and subtances;

Interactionsof organic molecules; Structure of macromolecules(proteins, nucleic acids);

Biological and artificialmembranes; MRI: Magnetic Resonance Imaging.

Applications of NMR

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SUMMARY

Jablonski diagram

S0 S1 Ground-state S0– S1 Excited-state

Vibrational relaxation –heat(10-12_s)

excitation (10-15_s) hν T1 T1→S0(10-3– 10-1s ) vibrational levels 0 1 2 3 4 5 S1→T1: inter-system crossing (10-10_{– 10}-8_s) S1→S0(10-9s ) A → kt D www.medschool.pte.hu substance

OD = A = - log (I / I

0

) =

ε

(

λ

) c x

I

0

I

„optical density”

_{I = I}

0

10

-ε(λ) c x

The definition of absoprtion

light source monochromator sample detector

How to measure absorption?

A scheme of a photometer.

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Kasha’s-rule

The emission of the fluorescence light is allways

starting from the lowest vibrational level of the

first excited level (S

1

).

Michael Kasha December 6, 1920 -June 12, 2013 www.medschool.pte.hu

S

T

S

In the ns range In the > ms range

Definition of fluorescence and

phosphorescence

Basic fluorescence parameters

Fluorescence spectrum, intensity;

Quantum efficiency

Fluorescence lifetime

Polarisation

Stokes-shift

The difference (measured innm) between the peak of the excitation and the emission spectrum (energy loss).

flu o re sc en ce em is sio n sp . fl u o re sc en ce ex ci ta ti o n sp . wavelength (nm) Stokes www.medschool.pte.hu

Scheme of a spectrofluorometer

Non linear arrangement !!!

F M S

M

D

90

o

light source excitation monochromator emission monochromator detection sample holder www.medschool.pte.hu Er

Photoselection

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Scheme of a spectrofluorometer

F M S M D

90

o

light source excitation monochromator emission monochromator detection sample holder

Polarisers in the light paths!!!

IVV IHV IVH IHH P P

Vertical or horizontal polarisers:

Emission anisotropy

r = (I

_V_V

-

G

I

_V_H

) / (I

_V_V

+ 2

G

I

_V_H

)

G = IHV/ IHH • dimensionless

• depends on rotational motion of the fluorophore

• additive! Vertically aligned polarizer on the excitationside Horizontally aligned polarizer on the emissionside Isum= IZ+ IX+ IY Isum= IVV+ IVH+ IVH Isum= IVV+ 2IVH Remember! www.medschool.pte.hu

Conditions of FRET

• A fluorescent donor molecule.

• The appropriate orientation of the donor and • Overlap between the donor emission and acceptor absorption spectra.

• Distance range of 2-10 nm!

FRET efficiency

E = 1 – (F

DA

/ F

D

)

where

FDA: donor intensity with the acceptor FD : a donor intensity without the acceptor. Can also be calculated with lifetimes!

E = 1 – (

τ

DA

/

τ

D

)

Distance dependence of the FRET

efficiency

6

0

6

0 R

R

E

+

=

Determination of distances! Molecular ruler!

Applications of FRET

• The determination of FRET distances

→ To study the establishment of interactions between molecules; → To study intra-molecular structural changes.

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Vibration modes of water

molecules

http://www1.lsbu.ac.uk/water/vibrat.html

spring = bond

Model for molecular vibrations

if matom↑ then f ↓

(C-H: 3000 cm-1_{, C-Cl: 700 cm}-1₎

if bond strength ↑ then f ↑ (C=H: 1650 cm-1_{, C≡C: 2200 cm}-1₎

mass = atom

f ~ D / m

Infra and Raman spectroscopy

IR abs. Rayleigh

scatter Raman scatter

S0 S1 Vibrational levels hf0 h(f0-f1) h(f0+f1) Vibrational levels elastic vs. non-elastic www.medschool.pte.hu

Example: regions of an IR spectrum

Double bonds (2000-1500) Triple bonds (2500-2000) ”Fingerprint” region (1500 - 600) Bonds to H (X–H) (4000-2500 cm-1) www.medschool.pte.hu m: magnetic moment of the individual atom