2.60 2.50 2.40 2.30 2.20 2.10 2.00 1.90 1.80 1.70 1.60 1.50 1.40 1.30 1.20 1.10 1.00 0.90 0.80

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5 CH3

O

CH3

Nuclear Magnetic Resonance Spectroscopy

Basic Principles

Dilip D. DhavaleDepartment of ChemistryUniversity of Pune, Pune – 411 007ddd@chem.unipune.ernet.in

Spectroscopy

Study of the interaction of light with matter

Light: Visible part of a large range of Electromagnetic Radiationhaving corpuscular as well as wavelike properties.

Energy E = h ( h = Planck’s constant 6.62 x 10 –27 erg/sec)

Ultra-Violet

Visible Infra-Red

Near Far

Microwave RadioCosmic andGamma Rays

A0 10-4 1000 2000 4000 8000 104 107 1014

NuclearTransition

Electronic Transitions

MolecularVibrationalTransitions

MolecularRotationalTransitions

Nuclear SpinTransitions

Energy cal/mole 1014 – 1010 100 – 35 K ~28.5 K 10-2 – 10-6

Sample atEquilibrium

RadiationExcited State Spectrum

Relaxation

Observation

Spectroscopy

UV-Visible: Presence of chromophoric system/conjugation in the molecules

IR Spectroscopy: Presence of Functional Groups in the molecules

1H NMR Spectroscopy: The number of different types of Hydrogens in the moleculesThe relative numbers of different types of Hydrogens in the moleculesThe electronic environment of different types of Hydrogens in the moleculesThe “neighbours and a neighbours” of a functional group

These spectroscopic techniques are mutually complimentary and a combination of these three-along with a Mass Spectroscopy form a powerful device in the determination of structures of organic molecules.

History of NMR Spectroscopy

In 1945 - The phenomenon was simultaneously discovered by two groups: Purcell, Torrey and Pound at Harvard University: Paraffin Bloch, Hansen and Packard at Stanford University: Water

In 1950 – The first NMR spectra of ethyl alcohol was recorded by Arnold, Dharmatti and Packard.

In 1952 – The two discoverers (Bloch and Purcell) were awarded the First Nobel Prize in NMR (Physics).

In 1991 – The second Nobel Prize in NMR (Chemistry) was awarded to Richard Ernst for discoveries of advanced methodologies.

In 2002 – The third Nobel Prize in NMR (Chemistry) was awarded to Kurt Wuthrich for determination of structures of Biomolecules in solution by NMR. Nobel Prize awaited to Paul Lauterbur for Magnetic Resonance Imaging

Behavior of Magnetic Nuclei

Randomly oriented nuclearspins of equal energy in the absence of any magnetic field

12

_

12

+

EP

12

_

12

+

ElectromagneticRadiation in R F range with energy E = Ep

Ho

Precisely oriented nuclear spinsin the presence of Magnetic field

For nuclei with spin I = ½Two possible orientations as per equation 2I + 1

In NMR, we are measuring the energy required for the flipping of the nucleus

Nuclear Spin• A nucleus with an odd atomic number or an

odd mass number has a nuclear spin.

• The spinning charged nucleus generates a magnetic field.

=>

l

Nucleii spin + charge

l

Nucleii spin + charge

ll

Nucleii spin + charge

Spinning charged particle is a magnetSpinning charged particle is a magnet

Magnetic Properties of Nuclei

The spinning of positively charged particle produces:

(1) Spin angular momentum or Spin quantum number (I)

(2) Magnetic moment () along the axis of spin

(3) Electric quadrupole moment (Q)

(as a result of non-spherical distribution of nuclear charge)

The angular momentum of spinning nucleus is described in terms of spin quantum no.I

The spin quantum no. I is a characteristic constant of a nucleus, and is dependent on the number of protons and neutrons.

1) Nuclei with odd mass number and odd or even no. of protons havehalf – integral spin such as 1/2, 3/2, 5/2 etc.

2) Nuclei with even mass number and odd no. of protons have integral spin such as 1, 2, 3

3) Nuclei with even mass number and even no.of protons always have zero spin (Due to pairing of oppositely directed spins in the nucleus)

In general three rules apply to the nuclear spins.

Nucleus No. of

Proton

No. of Neutron

Mass

No.

Spin No. (I)

Natural

% Abundance1H 1 0 1 1/2 99.982H 1 1 2 1 00.015611B 5 6 11 3/2 81.1712C 6 6 12 0 98.8013C 6 7 13 1/2 01.10814N 7 7 14 1 99.63515N 7 8 15 1/2 00.36516O 8 8 16 0 99.9517O 8 9 17 5/2 00.03719F 9 10 19 1/2 100.0029Si 14 15 29 1/2 04.7031P 15 16 31 1/2 100.00

Nuclear Properties of Important Nuclei - I

Requirements of nuclei to be NMR active

Three important characteristics: Nuclei should have Spin no. I > 0 and magnetic momemtum > 0 Nuclei should have even charge distribution that is nucleus should be spherical in shape so as Q = 0. Nuclei should have high % of natural abundance

1H, 13C, 19 F and 31 P nuclei have I = 1/2 and > 0 These nuclei are spherical in shape (even charge distribution) and Q = 0So observed by NMR technique.1H, 19F and 31P have high % abundance

12C and 16O nuclei are also spherical in shape Q = 0; but I = 0 and = 0So non-magnetic and not observed by NMR

2H, and 14N nuclei have I > 0; > 0 and Q > 0 Nuclei are ellipsoidal prolate in shape, so no even charge distribution. So difficult to study by NMR

Ellipsoidal oblate35Cl

Ellipsoidal prolate

Basic NMR Equation

For proton spin no. I = ½. Therefore, there are (2I + 1) two possible orientations.The energy of orientation is a product of magnetic moment and strength of the appliedField Ho (E = Ho).

At resonance:h = 2 Ho

Ho /hThe equation is rewritten as

Ho /2

Where = 2. / h.IIt is a proportionality constantbetween and I.Also called as Gyro magnetic ratio.It is constant for a particular nucleibut different for different nuclei.

HO

E2 = + HO

E1 = HO

Aligned with the fieldLow energy orientation

Aligned against the fieldHigh energy orientation

E = 2 HO

E = h

Precessional orbit

Nuclear magnet

HO

H

The behaviour of a nuclear magnet in a magnetic field

o

Two ways of doing NMR experiment

Sweeping – change magnetic field

Keep constant frequency

Change the frequency

Keep magnetic field constant

= Ho / 2

External Magnetic FieldWhen placed in an external field, spinning protons act like

bar magnets.

=>

Nucleus Spin quantum number

Magnetic moment

()

Gyromagnetic ratio ()

Resonance frequency (MHz at a

Field of 1 T)

Relative sensitivity

at constant field

Natural abundance

(%)

Quaderpole moment, Q

1H 1/2 2.792 2.675 42.577 1.000 99.98 -

2H 1 0.857 0.411 6.536 0.009 0.0156 0.003

10B 3 1.8007 0.288 4.575 0.02 18.83 0.111

11B 3/2 2.6880 0.858 13.660 0.165 81.17 0.036

13C 1/2 0.702 0.673 10.705 0.016 1.108 -

14N 1 0.403 0.193 3.076 0.001 99.635 0.02

15N 1/2 -0.282 -0.271 4.315 0.001 0.365 -

17O 5/2 -1.8930 -0.363 5.772 0.029 0.037 -0.004

19F 1/2 2.627 2.517 40.055 0.834 100.0 -

29Si 1/2 -0.5549 -0.531 8.460 0.079 4.70 -

31P 1/2 1.131 1.083 17.235 0.066 100.0 -

Nuclear Properties of Important Nuclei -II

The NMR Spectrometer

=>

The NMR Graph

=>

Interpretation of NMR Spectrum

NMR Spectra is analysed on the basis of following parameters Integration Chemical shift Coupling constant Rate processes

CH3 OEt

O

4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.50.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

9.0

9.5

NMR Signals

• The number of signals show how many different kinds of protons are present.

• The location of the signals show how shielded or deshielded the proton are.

• The intensity of the signal shows the number of protons of that type.

• Signal splitting shows the number of protons on adjacent atoms.

Integration

The process of excitation in NMR involves the flipping of the nucleus. This process of transition and the probability of the transition is same for all the protons, irrespective of the electronic environment. As a result, the area under the NMR resonance is proportional to the number of hydrogens which that resonance represents.

In this way, by integrating the different NMR resonance, information regarding the relative numbers of chemically distinct hydrogens can be found.

The integrals will appear as a line over the NMR spectrum. Integration only gives information on the relative number of different hydrogens, not the absolute

number.

2.5 2.0 1.5 1.0 0.50.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

9.0

9.5

0.91

0.93

0.94

2.08

2.40

2.422.43

2.45

H3CCH3

O

4 mm

6 mm

6 mm

Intensity of Signals• The area under each peak is proportional to

the number of protons.

• Shown by integral trace.

=>

10 56

Integration: Determination of keto – enol ratio

H1

electron current

Induced magnetic moment opposing the applied magnetic field

Chemical Shift: Shielding of a proton in a magnetic field

Basic equation of NMR is = Ho/2 wherein resonance frequency is simply a function of applied magnetic field and gyro magnetic ratio. If so, then all the protons in a molecule should resonate at one place- NMR of no value to organic chemists!!!!! What is true?

= Ho/2 Where, Ho = strength of the applied magnetic field experienced by the nucleus.And Ho = H1 (1 ) where H1 is the actual strength of the applied field and is the shielding constant.More the electron density – more is the shieldingLess the electron density – more is the deshielding

The resonance frequencyis, to a small extent,dependent on its Electronic EnvironmentMaking NMR Useful to Structure Determination

The difference between the positions of absorption of reference standard and that of a particular proton or a group of protons

Chemical Shift

012345

Reference peak is kept at zero of the chartWhat is reference ?

CH3-CH2-OH

Requirements for a reference standard in NMR

A good standard should meet the following requirements:• It should be chemically inert (non-reactive).• It should give a single sharp line.• It should be magnetically isotropic.• It should have unique line position.• It should be Miscible with organic solvents.• It should be readily volatile to allow recovery of the compound.

TetraMethylSilane

Silicon is more electro-positive than carbon. Therefore,pushes electron density towards carbon and thus to hydrogen-making methyl protons strongly shielded.

H3CSi

CH3

C

CH3

HH

H

Units of Chemical shift Chemical shifts are expressed in Hz.

Chemical shifts expressed in Hz are proportional to the applied magnetic field HO/oscillator frequency.

For example, In CH3CH2OH 60 MHz 100 MHz 300 MHz CH3 60 Hz 100 Hz 300 Hz CH2 216 Hz 360 Hz 1050 Hz OH 300 Hz 500 Hz 1500 Hz This can be calculated by the equation: Frequency of X proton (Hz) in A instrument X Osc. frequency of B in Hz Osc. frequency of A in Hz (major difficulty in calculating chemical shift as variety of instruments)

Chemical shift are also expressed in ppm)

is dimensionless and is NOT PROPORTIONAL to Ho or Osc.

frequency.

value is same in all the different instruments

Universal scale in NMR

Higher the value of chemical shift in Hz or deshielded is the proton.

Lower the value chemical shift in Hz or shielded is the proton.

Units of Chemical shift

Shift in frequency from TMS (Hz) Frequency of spectrometer (Hz)

X 106

General regions of chemical shift

012345678910

Aldehyde

Aromatic and heteroaromatic

Alkene

-Disubstituted aliphatic

-Monosubstituted aliphatic

Alkyne

-Substituted aliphatic

Aliphatic alicyclic

High fieldLow field

Deshielding Shielding

Solvents for NMR

D3COOD 2.02, 11.53

D3CCOCD3 2.05

D3CCN 1.95

C6D6 7.20

CDCl3 7.25

CD2Cl2 5.35

D3C-SO-CD3 2.50

F3C.COOD 10-11

A satisfactory solvent should be It should be chemically inert It should not contain protons It should be non – polar and should have low – boiling point It should have low viscosity

Chemical Equivalence Chemically equivalent protons

H3C CH3

OO

H3C CH3 H3CO OCH3

O

X Y Z

X, Y, and Z have one set of equivalent protons.

Chemically non-equivalent protons

CH3CH2Cl CH3CH2OCH2CH3 CH3CH2OH

2 signals 2 signals 3 signals

Number of SignalsEquivalent hydrogens have the same chemical

shift.

=>

Parameters that affect the Chemical shifts

Electron Withdrawing Inductive Effect

Diamagnetic anisotropic shielding and deshielding effect

Electron donating and withdrawing mesomeric (resonance) effect

R CH3 0.9

R CH2 R 1.3

R3 CH 1.7

R CH2 I 3.2

R CH2 Br 3.5

R CH2 Cl 3.7

C-CH3 0.9

N-CH3 2.3

O-CH3 3.3

CH3 CR 2.1

R C CR2CH3

1.6

O

CH3 2.3

CH3 OH 3.3

CH3 OCOCH3 3.6

CH3 OR 3.0

R2C CH R 5.3

R C C H 2.5

H7.25

RC-H 9.7

O

Chemical shift: Electron Withdrawing Inductive effect

Electron withdrawing inductive effect is one of the parameter that affects the chemical shiftStronger the electron withdrawing group - more is the deshielding

R2NH

ROH

ArOH

RCO2H

2 - 4

1 - 6

6 - 8

10 - 12

Protons attached to O, N, S are resonating anywhere between 1 to 15 . The position depends on a) substrate b) solvent c) concentration d) temperature.The best method to detect these protons is to re-run the spectra on addition of drop of D2O, wherein these protons will change their position

Carboxylic Acid Proton, 10+

=>

Magnetic Anisotropic EffectCarbon-Carbon double bond and triple bond

(The influence on induced magnetic moments of neighbouring bonds)

H

R1

R

H

ShieldedRegion

ShieldedRegion

DeshiededRegion

+

+

DeshiededRegion

Olefinic Protons at 4.5-6.0

H

R

ShieldedRegion

ShieldedRegion

+

+

DeshiededRegion

DeshiededRegion

Acetylenic Protons at 2.5-3.0

Ho

Circulation of electronsperpendicular to Ho

Induced Magnetic Field

Magnetic Anisotropic Effect

Ring Current Effect in Benzene Ring

Benzene protons at 7.25

Magnetic Anisotropic Effect

Ho

Ho

Ho

(Induced by magnetic moments of neighboring bonds)

HC

H

OHO

H

H H

CH3

CH3

H

H

HH

HH

H H

H

H

H

H

HH

H

H

H

H

H OH HO HH H

OO O

O

OCH3

H3CO H

H3CO

H

OO O

O

OCH3

H3CO H

H

H3CO

8.99 6.42 5.48

(Ring) 8.14 - 8.64(CH) 4.25

3.53 3.75

(Ring) 7.27; 6.95(CH) 0.51

0.42 .42

(H outer) 9.28 (H inner) 2.99

Chemical Shift: Resonance effect

Electron donating resonance effect increases electron density at the carbon and in turn to hydrogen – Shielding of proton.Electron withdrawing resonance effect decreases electron density at carbon and in turn to hydrogen – Deshielding of proton.

OCH3H

H

H

-H-deshielded due to -I effect-H-shielded due to +M effect

OCH3H

H

H

CH

H

H

O

CH3C

H

H

H

O

CH3

-H-deshielded due to small -I effect-H-deshielded due to strong -M effect

R

H

H

H

5.28

5.40

6.38

3.85

5.85

6.40

ED ED ED EW EW EW

OCH3 - 0.43 - 0.09 - 0.37

OH - 0.5 - 0.14 - 0.40

OAc - 0.2 - 0.02 -

NH2 - 0.75 - 0.24 - 0.63

NMe2- 0.64 - 0.10 - 0.60

Substituent Ortho Meta Para Substituent Ortho Meta Para

NO2 + 0.95 + 0.17 + 0.23

CHO + 0.58 + 0.21 + 0.27

CN + 0.27 + 0.11 + 0.30

Chemical Shift: Resonance effect

2.5 2.0 1.5 1.0 0.50.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

9.0

9.5

0.91

0.93

0.94

2.08

2.40

2.422.43

2.45

Spin-spin Coupling

• Three groups of non-equivalent hydrogens, therefore three signals are expected.• These signals are split into peaks due to the spin-spin interactions of non-equivalent protons.• Spin-spin interaction occurs between adjacent non-equivalent protons only.

H3CCH2

CH3

O

Coupling constant J is always expressed in HzJ is Independent of Operating frequency.Multiplets are symmetric about the mid point =

Mutually coupled protons have identical coupling constant J

JJ

Spin-Spin Coupling• Nonequivalent protons on adjacent carbons

have magnetic fields that may align with or oppose the external field.

• This magnetic coupling causes the proton to absorb slightly downfield when the external field is reinforced and slightly upfield when the external field is opposed.

• All possibilities exist, so signal is split.

Types of Coupling

Three types of coupling in NMR

Geminal Coupling [1,3 coupling]: Coupling between nuclei on the same carbon (Two bond coupling)

Vicinal Coupling [1,4 coupling]: Coupling between nuclei on the adjacent carbon atom (Three bond coupling)

Long range Coupling [1,5; 1,6- coupling] Coupling between nuclei separated by more than three carbon atoms (Coupling between nuclei separated by more than three bonds)

C C

HH1

2

3

4

Coupling constant ‘J’ always expressed in Hz It is independent of applied magnetic field/Operating frequency

C C

H

C

1

24

H

3

5

C

H

H

21

3

Rules of coupling (First Order Analysis)

Three rules govern the number and nature of multiplets in NMR

Equivalent protons do not couple with each other.

Maximum No. of Lines = (2na.Ia + 1) x (2nb. Ib + 1) x (2nc. Ic + 1)……. na, nb, nc = number of neighboring non-equivalent nuclei Ia,Ib, Ic = spin numbers of the respective nuclei. In case of 1H, 19F and 31P nuclei, spin no. = ½. Therefore, Maximum no. of lines = (na + 1) x (nb + 1)….

When the multiplicity is produced by a group of equivalent nuclei with I = ½, the intensities of the different lines are given by the coefficient of

expression (X + 1)n where n = number of interacting nuclei.

These rules are valid if J > 6 (where is a frequency difference between coupled protons)

C C

HA

No adjacent proton

C C

HA HB

One adjacent non-equivalent proton

C C

HA HB

HB Two adjacent nonequivalent protons

C C

HA HB

HB

HB

Three adjacent non-equivalent protons

Structure Spin States Signals

Pascal’s Triangle

Relative intensities of first order multipletes

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1

n = number of equivalent nuclei Relative Intensity

0

1

2

3

4

5

6

7

8

Coupling of proton with neighboring nuclei

1,1,2-TribromoethaneNonequivalent protons on adjacent carbons.

=>

Doublet: 1 Adjacent Proton

=>

Triplet: 2 Adjacent Protons

=>

Splitting for Isopropyl Groups

=>

Values for Coupling Constants

=>

Spectrum for Styrene

=>

HO

HA

HA

HA

HA

HA

HA

HB C2 C1 HA HA C1 C2 HB

HB C2 C1 HA HA C1 C2 HB

E E2E1

HA in the absence of HB (s)

HA in the presence of HB (d)

Line due to of HB Line due to HB

8

4 4Jab

+1/2 J -1/2 J

Nuclear Spin

Electron Spin

Mechanism of Spin-Spin Coupling

E – E1 = E2 - E

J Vicinal always +ve

Geminal Coupling

HA HB

HBHA

E2

HO

If HB is aligned to Ho-drop in ground state and increase in exited state energy This effect is opposite to that observed in vicinal couplingGeminal coupling constant is always negative

HA C HB

Low field

High field

Due to of HB

Due toof HA

Splitting arising from spin-spin coupling with the protons of the methylene group

Spin arrangements of the methylene protons

Unperturbed signal

HO-CH2-CH3

Total Spin Orientations -1 1

0 2

+1 1

The splitting of the signal from the methyl protons in ethanol by spin-spin interaction with the protons of the methylene group

+1J -1J

206 200 194

1

2

1In ethanol, methyl appears at 1as a triplet with J = 6 Hz.

Line positions in 200 MHz NMR

The splitting of the signal from the methylene group protons in ethanol by spin-spin interactions with the protons of the methyl group

Splitting arising from spin-spin coupling with the protons of the methyl group

Spin arrangements of the methyl protons

Unperturbed signal

Total spin Orientations

-3/2 1

-1/2 3

+1/2 3

+3/2 1

In ethanol, methylene protons appear at 3.5 as a quartet with J = 6 Hz

HO-CH2-CH3

709 703 697 691700

+3/2 J -3/2 J

+1/2

-1/2

1

3 3

1

Lines in 200 MHz

16

8 8

4 4 4 4

12

1 12

1 12

1 12

1

15 Hz

10Hz 10Hz

4Hz4Hz

4Hz 4Hz 4Hz 4Hz4Hz

Splitting due to Hb

Splitting due to Hc

SplittingDue to Hd

Jab

Jac

Jad

V1 –V2 = JadV1 – V4 = JabV1 – V6 = Jab

12 lines for Ha First Order Analysis (ddt)

Hc

Hb

X

HdHd

Ha

First Order Multiplet Analysis

Multiplets are reported starting with the largest coupling first, e.g. td J = 8.0 and 3.0 Hz implies a triplet with a J = 8.0 Hz and a doublet with J = 3.0 Hz, which is very different from a doublet, J = 8.0 Hz and a triplet J = 3.0 Hz.

X

H

X

H

H

H

X

X

H

H

H

H

3 Hz

Intermediate analysis: 3 < J Second order analysis: J < 3

Spin System If /J ratio is large (greater than 8), the interacting nuclei are weakly coupled. They are well separated, Designated as AM or AX.

O O

Ha

Hb

6.45

7.72

In 100 MHz/J = 772 – 645 = 127 HzJ = 10 Hz/J = 127/10 = 12.7

If /J ratio is small (less than 6), the interacting nuclei are strongly coupled. Designated as AB

S Br

Hb

Cl

Ha 5.96.0 In 100 MHz/J = 600 –590 = 10 HzJ = 4 Hz/J = 10/4 = 2.5

Such a collection of set insulated from further coupling form a spin system

AB Multiplets: Effect of / J ratio

AB Multiplet Analysis

Non – equivalence due to restricted rotation

19 F spectrum of BrCF2-CCl2Br

J. D. Robbert and P. M. Nair 1959

Five lines at 120 °C

Br

FF

Cl

Br

Cl

Both F nuclei equivalentSinglet

Br

FbFa

Cl

Cl

BrBr

FaFb

Br

Cl

Cl

Both are mirror images

Both F nuclei are non equivalentFive lines

Only one line at 25 °C

Three possible conformers

Non – equivalence due to restricted rotation

In case of compounds wherein methylene group is attached to the carbonhaving three different groups- the two methylene protons are non – equivalent.

C

R1

R2

R3

C R

Ha

Hb

Three possible conformers

Ha

RHb

R3

R2

R1

Ha

RHb

R2

R1

R3

Ha

RHb

R1

R3

R2

Some Nonequivalent Protons

C CH

H

Ha

b

cOH

H

H

H

a

b

c

d

CH3

H Cl

H H

Cl

a b =>

AB Multiplet Analysis

Some approximate J values

J = 7- 9 Hz Ortho

J = 1 - 2.5 Hz Meta

J = very low Para

H

H

H

H

H

H

Aromatics

H

H

H

H

H

HC

H

C

H

J = 0 - 10 Hz Ja,a = 7 - 10 Hz Je,e = 0 - 5 Hz Ja ,e = 0 - 7 Hz

Alkanes and Cycloalkanes

H

H

H H H

H

J = 11 - 18 Hz Trans

J = 5 - 14 Hz Cis

J = 8 - 11 Hz

H

HH

HJ = 0 - 3 Hz SP2 Geminal

J = 11 - 18 Hz SP3 Geminal

Alkenes

Factors affecting Vicinal Coupling Constant (3J)

The magnitude of 3J (sign is always positive) depends in essence upon four factors

The dihedral angle , between the C-H bonds under consideration (a).

The C,C bond length, R(b).

The H-C-C valence angles, and ’ (c)

The electronegativity of the substituent R on the H-C-C-H

HH

a

H H

R,b

H H '

c

H H

Rd

Karplus Correlation

Relationship between dihedral angle () and coupling constant for vicinal protons

Axial - axial

Axial - equatorial

Equatorial - equatorial

Dihedral Angle Calculated J (Hz) Observed J (Hz)

180°

60°

60°

9

1.8

1.8

8 -14

1 - 7

1 - 7

Rate Processes

NMR of CH3CH2OH ( with trace of acid/base impurity)

CH3-CH2-O-HH+

CH3-CH2-OH

HCH3-CH2-O

H

H

H+H+

NMR of pure ethanol

Effect of High Oscillator Frequency

Chemical shift in Hz is proportional to the oscillator frequency.Coupling constant J in Hz is not proportional to the oscillator frequency.As a result, NMR spectra are more resolved at high operating frequency.For example:The NMR of compound BrCH2CH2CH2 shows: 3.70 (t, J = 7 Hz), 3.55 (t, J = 7 Hz), 2.27 (quin, J = 7 Hz)

As /J ratio increases - NMR is more resolved Interpreted by first order analysis

3.70 = 1110 Hz

1 2 1

1117

1110

1103

1 2 1

1072

1065

1058

3.55 = 1065 Hz

1117

1110

1103 1072

1065

1058

300 MHz instrument3.70 = 222 Hz

1 2 1

229

222

215

1 2 1

220

213

205

3.55 = 213 Hz

229

222

220215

213

205

60 MHz instument

/ J = 9/7 =1.3

/ J = 45/7 =6.5

AMX Spin System

ADVANCES IN MRI

Magnetic Resonance Imaging (MRI) has reached a high level of maturity and has established itself as the diagnostic modality of choice in almost all neurological system disorders, joint diseases, Mediastinal and heart pathologies, work-up of abdominal and pelvic malignancies and evaluation of vascular system of the body.