2 Nuclear Magnetic Resonance

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NMR - Basic principles

Subatomic particles like electrons, protons and neutrons are associated with spin - a

fundamental property like charge or mass. In the case of nuclei with even number of

protons and neutrons, individual spins are paired and the overall spin becomes zero.However, there are many cases such as 1H and 13C, where the nuclei possess a net spin ,

which is important in Nuclear Magnetic Resonance (NMR) Spectroscopy . A set of rules

to determine the overall spin a nucleus is given below.

When there are even number of protons and even number of neutrons in the

nucleus, the net spin is equal to zero.

When there are odd number of neutrons and odd number of protons in the

nucleus, it will have an integer spin (i.e. 1, 2, 3)

If the sum of the number of neutrons and the number of protons is odd number,

the nucleus will have a half-integer spin (i.e. 1/2, 3/2, 5/2).

These rules can be summarized in terms of atomic mass and atomic number as shown

below.

12 C (0)EvenEvenZero

2H (1)OddEvenInteger

13 C (1/2)EvenOddHalf-integer

1H (1/2)OddOddHalf-integer

ExamplesAtomicNumber

AtomicMass

I

12 C (0)EvenEvenZero

2H (1)OddEvenInteger

13 C (1/2)EvenOddHalf-integer

1H (1/2)OddOddHalf-integer

ExamplesAtomicNumber

AtomicMass

I

NMR active

Not NMRactive

Nuclei are charged and those with a net spin would generate a magnetic dipole

along the spin axis. The magnitude of this dipole is given by the nuclear magnetic

moment , which is give by

= I h / 2 ..................................(1)

where I is the spin quantum number (with values , 1, 3/2 etc.) and


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= Gyromagnetic ratio, which is a characteristic constant for individual

nuclei.

In Nuclear Magnetic Resonance Spectroscopy, we study the behavior of magnetic nuclei

in presence of an external magnetic field. Quantum mechanics tells us that a nucleus of

spin = I can have 2I+1 orientations with respect to an external magnetic field. Since the

main aim of this chapter is to illustrate the use of NMR in the study of aromaticity, we

will focus on the magnetic properties of 1H and see what special information does 1H

NMR spectroscopy provide when this nuclei ( 1H) is part of an aromatic system.

A nuclei such as 1H with spin = , can orient in two ways (2 x + 1 = 2) with

respect to the external field. Of these, the spin state represented as +1/2 (or state) is of

lower energy where as the one represented as -1/2 (or state) is of higher energy. The

former reinforces the applied field and the latter opposes it. The energy difference

between the two spin states is proportional to the applied field and can be written as

E = h = h B B0 /2 ..................................(2)

where, h = Planks constant,

= Gyromagnetic ratio which is a characteristic constant for individual

nuclei,

B B0 = strength of the magnetic field at the nucleus

From equation (2), it is clear that a radiation of frequency = B B0 /2 possess the

right amount of energy to effect a transition from lower energy state to higher energy

state. Absorption of energy takes place only with a certain combinations of field strengths

and radio frequencies during which the system is said to be in a state of resonance .

For proton, = 2.675x10 8 T-1s-1

If the NMR spectrometer is equipped with a magnet with a field of 7.046T,2.675 x 10 8 T -1 S -1 x 7.046 T

2 x 3.1416= 300 x 10 6 Hz = 300 MHz = B 0 /2 =


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This can be pictorially represented as shown below:

I = 1/2

Spin state I = -1/2 ( )

Spin state I = +1/2 ( )

Field strength B 0

Energy

NMR experiments can be performed in one of the following ways,

i) keep the external magnetic field strength constant and vary the frequency of

radiation to see an absorption

ii) keep the radio frequency constant and slowly vary the field strength until the

splitting of spin states corresponds to the energy of radio wave.

In new versions of instruments, instead of Continuous Wave (RF) sweeping, an intense

radiofrequency pulse is used to excite all nuclei simultaneously and their individual

absorptions determined using Fourier transform methods

Shielding and Deshielding

So far we were considering the magnetic moments generated by the spinning

nucleus and its interaction with an external field. However, in realty, nuclei are

surrounded by electrons which also generate small local magnetic fields (B loc) as they

circulate. These local magnetic fields can either oppose or augment the external magnetic

field. If the field created by the electron oppose the external field, nuclei experience an

effective field which is smaller than the external field and it is said to be SHIELDED. If

the field created by the electron augments the external field, nuclei experience aneffective field which is larger than the external field. It is said to be DE-SHIELDED.

i.e. B Beff = B o - B loc

This can be represented as B eff = B o( 1 - )


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magnetic shielding or screening constant; depends on electron density.

Equation (2) can now be written as

E = h = h Bo( 1 - )/2 ..................................(3)

Equation (3) shows that magnetic field felt by individual nuclei varies depending upon

their chemical environment; E and hence the energy of the radiation required to excite

them differ consequently.

Diamagnetic and paramagnetic anisotropy

Two effects that originate from electronic delocalization in aromatic systems are

1) External field induces a flow (current) of electrons in system ring current effect

2) Ring current induces a local magnetic field with shielding (decreased chemical shift)

and deshielding (increased chemical shifts) zones . This is the basis of diamagnetic

anisotropy in aromatic systems which can be diagrammatically represented as shown

below.

In the case of compounds with unpaired electrons, paramagnetism associated with

the net spin overrides the diamagnetic effects and lead to a different type of magnetic

effect (vide infra)

Chemical shift and scale

As mentioned, effective magnetic field at individual nuclei vary depending upontheir chemical environment. Radiofrequency required to excite them also will be

different under different external field strengths. This means that if we take NMR spectra

of a compound using instruments of different field strengths, peaks appear at different

positions and it would be difficult to compare them without applying corrections for


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differences in absorptions. To avoid this problem, a different system, based on chemical

shift is often used. In this, we use a reference compound in our experiments. It is

generally tetramethylsilane, which is unreactive with other organic compounds, volatile,

and gives an absorption which is relatively shielded in comparison with most of the

protons important to organic chemists. Solvent used in NMR experiments should not

contain 1H due to interference. Deuterated solvents such as CDCl 3, DMSO-D 6 etc. are

generally used in modern instruments.

Now let us see what are chemical shift and scale .

At first, the absorbance frequencies of the standard and our sample are measured.

The latter is then subtracted from the former and then divided by the frequency of the

standard. This gives a number called the chemical shift ( ). Let us assume that in a given

magnetic field (external field), the standard absorbs at 300,000,000 Hz (300

megahertz), and our sample absorbs at 300,000,300 Hz. The difference between our

sample and the standard is 300 Hz, and we take 300/300,000,000 = 1/1,000,000 and call

it 1 part per million (or 1 ppm). If we examine the same sample in a stronger magnetic

field where the reference comes at 500,000,000 Hz (or 500 megahertz), the frequency of

our sample will increase proportionally, and will come at 500,000,500 Hz. The

difference in this case would be 500 Hz. But if we divide this difference by 500,000,000

ie., 500/500,000,000, we will get 1/1,000,000, = 1 ppm! So, chemical shift values remainthe same irrespective of the instruments (field strengths). Although we did these

calculations manually, all these are done automatically in the computer associated with

the NMR spectrometer. A correlation of radiofrequencies and values and their relative

positions with respect to TMS (standard) is presented below.


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A representative NMR spectra is given below

Shielding and deshielding zones in Aromatic systems

Due to the magnetic field generated by the circulating electrons, hydrogens which

lie in the plane of the ring experiences a deshielding effect as shown below ( A). At the

same time those situated above and below the plane experience shielding , as the

magnetic lines of force are opposite in direction with respect to the applied field. Similar

effect, but to a lesser extent, can be seen in simple olefins ( B), aldehydes etc.

H H

B0

Appliedmagneticfield

7-8 ppm 5-7 ppm

C C

H

H

H Br

H

H

C C

H

H

H Br

H

H

A B

Illustrative examples:Chemical shift positions of hydrogen atoms which are placed in the shielding anddeshielding zones of aromatic systems are given below.


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H H

Bridge head protons at -0.5

Outer protons at 6.9-7.3

HH H

H

Inner protons 0.00

Outer protons 7.6

H

H

[14]-annulene

H3CCH3

Outer ring protons at 8.14-8.67

CH3 protons at -4.25

H

H HH

H HH

[18]-annulene

H HH

H

H

HHH

H

H

H

HOuter protons at ~ 9

Inner protons at ~ -3

H aH b

H1H7

23

4

5 6

Ha at -0.3 Hb at 5.1 H1 & H7 at 6.4 H2-H6 at 8.5

Antiaromatic systems

Antiaromatic systems are paratropic. That is, they are able to sustain a

paramagnetic ring current, which lead to shielding of outer ring protons and deshielding

of inner protons (opposite to that of aromatic compounds which show diamagnetic

effect). Examples presented below demonstrate the behavior of magnetic nuclei situatedin an antiaromatic environment.

1) At -170 oC, inner protons of [12]-annulene comes at ~8 ppm and outer protons comes

at ~6 ppm which is characteristic of antiaromaticity. Above -150 oC, all protons are

magnetically equivalent showing conformational flexibility. Above -50 oC, it rearranges

to a bicyclic system as shown below.

[12]-annulene

above -50 oC


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2) At -130 oC [16]-annulene is paratropic with four central protons at 10.56 , and

twelve outer protons at 5.35 . Above -50 oC, all protons are magnetically equivalent

showing conformational flexibility

in solution

Nonaromatic[16]-annulene

H H

H H

3) As discussed previously, the locked form of [14]-annulene show significant

aromatic character, with outer protons resonating at 8.14-8.67 and CH 3 protons coming

at -4.25 . However, the dianion of this compound is antiaromatic which is evident fromthe paramagnetic anisotropic effect seen in its NMR.

H3CCH3

Outer ring protons at -3

CH3 protons at 21

H

H3CCH3

Outer ring protons at 8.14-8.67

CH3 protons at -4.25

H

reduction

2 Nuclear Magnetic Resonance

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