Background to organic spectroscopy€¦ · a quantized spin angular momentum and a magnetic moment. Thus, proton and carbon-13 have the spin quantum number I= ½ and has two allowed

Introduction

Electric Quadrupole and Nuclear magnetic moments

NMR experiments

Energy consideration

Chemical shift scale

11/07/2012Organic Spectroscopy 1

Lecture 6 & 7

011th November 2012

Nuclear magnetic resonance is an importantspectroscopic technique which uses thelonger wavelengths (radio frequency)absorption to give information about numberof each type of nuclei (hydrogen/carbon) alsothe nature of its chemical environment.

An absorption spectrum is recorded byvarying the energy of radiation incident on asample and monitoring how much radiation isabsorbed at each energy level.


Radiation can be absorbed when its energy

corresponds to the difference in energy

between energy levels in a molecule.

Energy transition in NMR technique; for

electron spin, ∆E corresponds to energy in

microwave region and for nuclear spin, ∆E

corresponds to the energy of radio waves.


Any atomic nucleus which possesses eitherodd mass or odd atomic number or both hasa quantized spin angular momentum and amagnetic moment.

Thus, proton and carbon-13 have the spinquantum number I = ½ and has two allowedspin states. i.e +½ and –½.

Recall that I is a physical constant for eachnucleus, and there are 2I + 1 allowed spinstate which range with integral differencesfrom +I to -I


i.e -I (-I + 1), …, (I –1), I;

For proton, Allowed spin state =2I + 1

Then, 2(½) + 1 = 2

Allowed spin states for proton is 2.

So, an applied magnetic field nuclei spins in

+½ (a-state) or -½ (b- state).

Whereas, the spin state of any given nucleus

are degenerate and equally populated in

absence of applied magnetic field



Nuclear spin

(I)

Half integer

I = ½, 3/2 , 5/2,

Have

Odd

Atomic mass

Odd or Even

Atomic number

Integer

I = 1

Have

Even

Atomic mass

Odd

Atomic number

For zero

I = 0

Have

Even

Atomic mass

Even

Atomic number

I =1/2 : 1H, 13C, 19F

I= 3/2: 11B

I= 5/2: 17O

I =1: 2H, 14NI = 0: 12C, 16O

Spin state of some isotopic elements with respect to

atomic number and atomic masses

NMR involve absorption of Radio frequencies

on EMS.

A signal in an NMR spectrum is referred to as

resonance.

1H and 13CNMR spectroscopic experiment

gives information about number, type and

connectivity of hydrogen and carbon atoms in

a molecule


NMR data are commonly acquired from 1D & 2D

experiments

The NMR spectra give information about the nature

of the chemical environment of each magnetically

active nucleus in the molecule


H3C

C

O

O

H2C

C

CHH3C

H2C

O

H

3 types of H atomsNumber of H atoms= 6

5 types of C atomsNumberof C atoms = 5




Nucleus may have spherical chargedistribution or non-spherical (prolate oroblate) charge distribution.

That is, all nucleus with I = 0 or I = ½ haveapproximately spherical charge distributionwithin their nuclei hence small electricquadrupole moment.

Those with I > ½ have non-spherical(ellipsoidal) charge distribution within theirnucleus hence large electric quadrupolemoment as shown below.


Intrinsic magnitude of the generated dipole is

expressed in nuclear magnetic moment, μ.

The magnetic moment is generated by the charge

and spin of a charged particle.

Many moving charge generates a magnetic field

of its own and a weak secondary field (Bsec) due

movement of electron around it that opposes

applied field that shields the proton nucleus.


All nuclei with non-zero spin (I >0) have

magnetic moment, μ, but the non-spherical

nuclei also have eQ.


1H 12C 16O 19F 2H < 14N << 35Cl, 37Cl << 79Br,81Br 127I

eQ negligible Increasing eQ (absolute value)

15

When the frequency of the oscillating electric

field component of the incoming radiation

just matches the frequency of the electric

field generated by the precessing nucleus the

two fields can couple, and energy can be

transformed from the incoming radiation to

the nucleus thus causing spin change. This

situation is called Resonance

The strong applied magnetic field (Bo) induces

the electrons to circulate around the nucleus

(left hand rule)


e-

Bo

The induced circulation of electrons sets up a

secondary (induced) magnetic field (Bi) that

opposes the applied field (Bo) at the nucleus

(right hand rule)

Therefore, the effective field (Beff) is the difference

between Bo and Bsec . That is, Beff = Bo - Bsec


e-

Bi

Bo

Hydrogen nucleus may have +½ or –½ spin

with μ. These two cases are in opposite

In an applied magnetic field, Bo all protons

will have their μ either aligned with the field

or opposed to it.

Thus, hydrogen nuclei can adopt either +½ or –½ with respect to the applied field


Consider hydrogen nuclei acting like a tiny bar magnet (having North pole (N) and South pole (S) when placed in magnetic field, like-poles repeals each other and opposite poles attracts each other.


Proton spinning in an

external magnetic field,

where the magnetic axis

of the proton precesses

about z axis of the

stationary magnetic field

Bo.


protons in a molecule are shielded hence resonate at differentfrequencies.

Alignment of proton spin states when an external magnetic field, Bo is applied.


Aplied magnetic

field, Bo

spin + ½

aligned to Bo

spin - ½

oppose Bo

Spinning proton has electron which also do spin and both generatemagnetic field.


There is excess spin state in a-state because of lower inenergy.

The difference in energy, DE betweenthe two states (+½ and –½) dependson strength of the applied field, Bo.


24

A nucleus is in resonance when it absorbs radio frequency radiations and spin flip to a high energy state

Energy absorbed is a quantized process and most equal the energy different between two state involved

Eabsorbed = (E -1/2 state – E +1/2 state) = h

In practice DE = f (bo)‏

The stronger the applied magnetic fields, the greater the energy difference between possible spin.

i.e the magnitude of the energy level separation depend on the

particular nuclear involved

25

Magnetogyric ratio, is constant for eachnucleus and determines the energydependence on the MF

DE = f (bo) = h

bx

D =l

where l = 1/2

bx

The charge distribution which is slightly excess ina state (a lower energy level) is explain byBoltzmann charge distribution, N α > Nβ

Nα / Nβ = e-ΔE/kT

where k is Boltmann constant and T is temp (K) and change in energy is given by;

∆E = (hγ/2π)βo Where h - Planck’s constant, γ – nuclear constant (magnetogyric ratio).


Spin states of proton in absence and in presence of Bo


Spin states of chlorine in absence and in presence of Bo


Once two levels for the proton have beenestablished energy in form of radiofrequency(rf or v1) can be introduced to effect thetransition between energy levels in astationary magnetic field of a given strengthBo;

Then, this equation (∆E = (hγ/2π)βo ) have tobe modified due to secondary magnetic field

by the nucleus i.e ν = βo (1- d)/2π; d isshielding factor



DE =h

2_

Bo1 =

2_

Bo

Obtainable is the fundamental NMR equation correctingthe applied radiofrequency, 1 (MHz) with the Bo,

since DE = hν.


1. NMR spectroscopy uses radiofrequency absorption to giveinformation about number of each type of nuclei and theconnectivity

2. Proton and carbon-13 have the spin quantum number I = ½and has two allowed spin states. i.e +½ (α sate) and –½ (βstate) which important for NMR techniques

3. The spin state of any given nucleus are degenerate andequally populated in absence of applied magnetic field

4. A signal in an NMR spectrum is referred to as resonance

5. The NMR spectra give information about the nature of thechemical environment of each magnetically active nucleus inthe molecule

6. All nuclei with non-zero spin (I >0) have magnetic moment,μ, but the non-spherical nuclei also have electricalquadrupole moment.

7. Spinning proton has electron which also do spin and both generate magnetic field The strong Bo induces the electrons to circulate around the nucleus (left hand rule)

which sets up secondary MF (Bi) that opposes Bo (Right hand rule)

8. The difference in energy, DE between the two states (+½ and –½) depends on strength of the applied field, Bo

9. Protons in a molecule are shielded hence resonate at different frequencies

10. There is excess spin state in a-state because of lower in energy

11. ∆E = (hγ/2π)βo ; since ∆E = hv

v = (γ/2π)βo


Background to organic spectroscopy€¦ · a quantized spin angular momentum and a magnetic moment. Thus, proton and carbon-13 have the spin quantum number I= ½ and has two allowed

Documents