NUCLEAR MAGNETICRESONANCE (NMR)

SPECTROSCOPY

Characterization of Organic Compounds

O

O

OH

OMe

OMe

2

NMRUVIRMS

I = 0, Nuclei is NOT NMR ACTIVEI ≠ 0, Nuclei are NMR ACTIVE

Like electrons protons and neutrons have spins and they pair.

16O,

Number of spin states = 2I + 1

(NOT NMR ACTIVE)

(NMR ACTIVE)

(NMR ACTIVE)

1H NMR SPECTROSCOPY

m (magnetic dipole)

A very small magnetic field is associated with the nucleii.e. nuclei with spin behave as tiny bar magnets

Influence of External Magnetic Field (B0)

parallel antiparallel

Bo

Precessional motion

Precessional orbit

The frequency of this motion is called precessional frequency (n)

Nuclear magnetic dipole

Further Influence of External Magnetic Field

parallel anti-parallelApplied magnetic field

DE = hgB0

is the magnetogyric ratio (a proportionality constant, differing for each nucleus which essentially measuresthe strength of nuclear magnets).

B0 = is the strength of the applied magnetic field.h = Plank’s constant

DE = hn

Precessional frequency (n) is directly proportional to Bo

2p

n = gB02p

Bo = 1.4T, n = 60 MHz

Bo = 14T, n = 600 MHz

Bo = 4.7T, v = 200 MHz

= hgB02p

The basic NMRequation

For transition from the lower state to the upper state, the energy falls within the radio frequency range

i.e. when the precessional frequency = radio frequency

When the appropriate radio frequency is supplied,

Absorption will occur and the nucleus and the radio frequency are said to be in resonance and hence the term NUCLEAR MAGNETIC RESONANCE

DE = hB0/2p

the frequency is also called resonance frequency

n = B0/2p

DE = hB0/2p

NMR is not sensitive method of analysis

A small excess (0.00003%; Bolzmann’s excess) in the lower energy level is responsible for absorption)

SENSITIVITY OF NMR SPECTROSCOPY

DE = hB0/2p

Sensitivity can be enhanced by the use of strongerMagnetic field

2.35T, 4.7T, 7.05T, 9.4T, 11.74T, 14.10T, 16.2, 18.4T

Increasing order of sensitivity

• Na (population in lower level) and Nb (population in upper level) approach equality.

• No further net absorption of energy.• Resonance signals will fade out and this is

called saturation of the signal.

When a proton absorb the correct radiofrequency, nuclei in the lower energy state undergo transition to the higher energy state.

Such energy loss is called relaxation(the process through which the Boltezmann’s excess

is restored)There are two types of relaxation:1) Spin-lattice relaxation:-

The transfer of energy to the lattice (entire framework or aggregate of neighbours i.e. solvent molecules, or other atoms in the molecule).

2) Spin-spin relaxation:-Transfer of energy to a neighbouring nucleus having identical precessional frequency.

High energy nuclei lose energy by radiationless process.

The secondary magnetic field (B) generated by the electron circulation shields the nucleus from the external magnetic field

d is shielding factor

0

From this equation, can NMR beuseful in structure elucidation?

Thus NMR equation has to be modified as

n = B0/2p

n = B0(1-d)/2p

n will vary with electron environment

Electron donatingElectron donating groups increase the electron density around nearby hydrogen atoms resulting in increased increased shielding, shielding, Electron withdrawingElectron withdrawing groups decrease the electron density around nearby hydrogen atoms resulting indedecreased creased shieldingshielding, (, (dedeshieldingshielding).).

CH3F, CH3O, CH3N, CH3C, CH3Si Decreasing order of electronegaitivity

Increasing order of shielding

Tetramethylsilane (TMS) (CH3)4Sihas among the most shielded nuclei and is used as a reference point in NMR

NMR is useful to distinguish nuclei in differentElectronic environment

Why is it necessary to have a reference point in NMR?Because: the magnetic field strength is not constant.The change in field strength will affect both the sampleand the reference compound in the same way.Relative absorption positions are generated in NMR.

CHEMICAL SHIFTThe difference in the absorption position (in Hz) of a nucleus (eg. 1H) from the absorption position of a Reference (usually TMS).

CH3

Ois 120 Hz at 1.4T (60 MHz)

is 200 Hz at 2.3T (100 MHz)is 1200 Hz at 14T (600 MHz)

n = gB0(1-d)/2pn = gB02p

B0 = 1.4T, n = 60 MHzResonance frequencyin the neighbourhoodof 60 MHz

Chemical shift = nnucleus - nTMS

Chemical shift (d) = nnucleus - nTMSz

Z operational frequency in MHz

Unit of d will be: HzMHz

Hz106 Hz

= = 1106

= ppm

Chemical shifts are not given in frequency (Hz)Since it is dependent on the field strength

Field independent parameter can be obtained ifChemical shift in frequency is divided by the operational frequency (Z)

CH3

O is 120 Hz at 1.4T (60 MHz)

is 200 Hz at 2.3T (100 MHz)

is 1200 Hz at 14T (600 MHz)

dCH3 = nsignal-nTMS

z

12060

= 200100

= 1200600

= = 2ppm

Position of resonance signal is in a unitIndependent of the field strength

At 60 MHz, 1ppm = 60 HzAt 100 MHz, 1ppm = 100 HzAt 600 MHz, 1ppm = 600 Hz

Operational frequency has implications on resolution

High frequency Low frequency

High fieldLow field

1) Effect of electronegativitySi-(CH3)4 dH = 0.00 ppm

d-d+

H

XC

C in X issp or sp2

sp3

XX

sp3sp3

sp3

2) ANISOTROPYSecondary magnetic field generated as the result ofcirculation of bonding electrons in a p-bond(also in s-bond)

Paramagnetic anisotropy -(deshielded = downfield)Diamagnetic anisotropy -(shielded = upfield)

d 7.15 ppmH

H

HH

H

H

[18][18]--ANNULENEANNULENE

HH

d -3.00 d 9.30

If aromatic it should have (4n+2)p electrons Is this aromatic?

[18]-Annulene corresponds to (4X4+2)p electrons.

1H NMR evidence for aromaticity

It should be aromatic.

How about[16]-annulene?

- Deshielding zone => nuclei will be downfield shifted

+ Shielding zone => nuclei will be upfield shifted

Anisotropy in other p-systems

CH3

CH3

H H

H H

Single Bond AnisotropyH H

Hax

Heq

1

3

56

2 4

1.1 ppm

1.6 ppm

Deshieleded Occurs in rigid cyclohexane ring or in cyclohexanone

Hax

Heq1

3

56

2

4 Hax

Heq1

3

5

6

24ring

flipThe axial and equatorial protons will have the same value

HHd 0.22 ppm

CH3

CH3

H H

H H

s-bond anisotropic effectshields the protons

Free rotation

‘aromatric like’ property

C C

a

X

b

H

X dHa dHbH 5.28 5.28OCH3 6.38 3.85COCH3 5.85 6.68

C C

a

OCH3

b

....

C C

a

O+

CH3

b

-

..

C C

ab

CH3

O..

.. C C

ab

CH3

O..

....

+

dHa by Inductive effect and dHb by resonance effect

3) Substituent effect on p-system (Resonance)

HH

HH

H

H

d 7.15 ppm

H

HH

OMe..

:H

HH

OMe

-

:+

6.81 ppmH

HH

OMe

-

+:

6.86 ppm

7.17 ppm

H

HH

NO2

8.25 ppm

7.66 ppm7.45 ppm

=

Resonance, is deshielding for ortho and para protonsInductive effect, is deshielding for ortho > meta > para protonsAnisotropy, is deshielding for ortho protons

H

HH

N+O O

H

HH

ON

+ O

+H

HH

ON

+ O

+

4) Hybridization of the carbon atomChemical shift is dependent on electronegativity,Anisotropy, resonance and hybridization

s electrons are closer tocarbon than p electrons

Carbon has more controlOn s-electrons thanp-electrons

How about in hybrid orbitals sp3, sp2 and sp?

CH3-CH3

sp3, 109.5º25% S character

CH2=CH2

sp2, 120º33% S character

sp, 180º50% S character

CH CH

d 0.9 ppm d 5.8 ppm d 2.35 ppm

An increase in %S character will result:-1) In an increase in bond angle2) (Order of electronegativity: sp > sp2 > sp3)

s electrons will be closer to C than HHence, H will be de-shielded, higher d value

Electronegativity & Anisotropic effects are:additive (deshielding) in alkenesand subtractive in alkyenes,Anisotropy (shielding), hybridization (deshielding)

d+d-RO H +

d+d-RO H R

O HR

O Hd+d- d- d+

Hydrogen bonding

5) Hydrogen bonding

Hydrogen atoms involved in hydrogen bondingare even more deshieldedbecause the O-H bond is more polarized with lower electron density at hydrogen atom.R

O H + D3CO D

RO D + D3C

O H

Exchangeable protons arenot observable in ‘protic solvents’

solvent

a) Intermolecular H-bonding

O-H, N-H, S-H De-shielded compared to C-H

RO H

HO H

‘Protic solvent’sample

These are also temperature dependent,

These are conc. Dependent,higher conc. results in stronger H-bonding,hence, stronger deshielding effect

Higher temp. results in weaker H-bondingHence, lower d values (weaker de-shielding effect)

RO H +

DO D

RO D + D

O H

D2O exchange OH Signal will disappear

OR

OH

O

RO

H

dimer even atdilute solution

b) Intramolecular H-bonding

OOH

O-Hydroxyacetophenone

d 10 – 14 ppm

R R

O O

b-DiketoneR R

O OH

d 11 – 16 ppm

Concentration andtemperature have little effect on d value

O Oconcentration andtemperature have NO effect on d value

6) Solvent effectsCDCl3, (CD3)2O , CD3OD, (CD3)2S=O, D2O

Change of solvent results in some change in d values

Some Typical 1H Chemical Shifts (δ values) in Selected Solvents SolventCompound

CDCl3 C6D6CD3COC

D3

CD3SOCD3

CD3C≡N D2O

(CH3)3C–O–CH3C–CH3O–CH3

1.193.22

1.073.04

1.133.13

1.113.03

1.143.13

1.213.22

(CH3)3C–O–HC–CH3O–H

1.261.65

1.051.55

1.183.10

1.114.19

1.162.18

------

C6H5CH3CH3C6H5

2.367.15-7.20

2.117.00-7.10

2.327.10-7.20

2.307.10-7.15

2.337.15-7.30

------

(CH3)2C=O 2.17 1.55 2.09 2.09 2.08 2.22

R R

OOH

d 10 – 18 ppm

A

B HAHB

What is the cause of splitting?

SPIN-SPIN COUPLING

Proton-Proton Coupling

Because HA ‘sees’ HB in two spin states and vice versa

Two proton System

Is the splitting of NMR signals into doublet, triplet,quartet etc., as the result of spin-spin interaction,through bonding electrons, between/among neighbouring NMR active nuclei.

HA in isolation

HB

Reason: The presence of HB makes HA to come to resonance twice

Similarly, the presence of HA makes HB to come to resonance twice

HA in the presence of HB

B0 1 1

Two proton system

A

BB’

Three proton system

What is the splitting of HA?

HA HBHB’

B0

B B'

B B'

B B' B B'

‘SEES’

HA

HBHB’

Free rotationHB and HB’ are identical,splitting is not observedbetween HB and HB’

1 1

triplet

Multiplicity = 2nI+1

n = Number of equivalent nuclei causing signal splittingI = spin quantum number of the nuclei causing the splitting

For 1H, I = ½ , multiplicity = n+1

n01

234

Is the distance between lineswithin a set of NMR signals in Hz

5 lines (pentet), 6 (sextet), 7 (heptet), 8 (octet),

Integral ratio, 1:1

Integral ratio, 2:1

Integral ratio, 1:2Integral ratio, 1:1

Integral is the relative peak area under a set of NMR signal(s) and is dependent on the numberof nuclei contributing to the signal(s)

Integral ratio 1:3 Integral ratio 2:3

1H NMR Spectrum - C4H8O2

2H3H3H

IHD= [(2X4)+2]-8 = 12

CH3

CH3

CH2-CH2-CH3

C=O -O- 2

33

triplet

O

O

Cl H

H

H

H

H

H

HHH

HH

Propionic acid 2-chloro-propyl ester

J depends mainly on: a) Number of bonds separating the nucleib) Dihedral angle for vicinal coupling (3J)But it is independent of field strength.

H

(10 Hz) (16 Hz)

4J and above are called long range coupling

(8 Hz) (2 Hz) (less than 1 Hz)

Long range coupling is observed in: aromatics, olefins, acetylenesand strained aliphatic ring systems

p-electrons are more effective in transmitting coupling information.

CH3

H4J = 0.6-0.9 Hz

broad signal

O CH3

O

sharp signal

Allylic coupling

Homoallylic coupling

HH

HPh

5J = 9 Hz

H

H O

H

4J = 1.1 Hz

Zigzag arrangement is necessaryi.e. ‘W’ or ‘M’ arrangement of atomsIt is also called ‘W’ coupling

H

H

In strained bicyclic system the coupling constantMay be large

Extensive overlap of the rear lobes is responsible for the coupling interaction

4J = 4 Hz

H

H

H

H

stronger overlap

larger J value

4J = 0 Hz

Hax

Hax

HeqHeq

13

56

2

4

Rigid system

Φ H6eq

H6ax

C5

H1eq

H1ax

C2

6

Newman projection

Φ is dihedral angle

Dihedral relationship (Φ)

Axial-axial (180°)Axial-equatorial (60°)

Equatorial-equatorial (60°)

Cycloalkanes

J (Hz)

8-122-5

2-5

Dependence of Vicinal Coupling on Dihedral Angle

H

average value

Geminal coupling:-

Geminal coupling is common in rigid ring systemHax

Hax

HeqHeq

13

56

2

4

2J = -12 Hz

Coupling between nuclei/protons attached to the same carbon atom (2J) and is negative usually

Vicinal coupling:-Coupling between nuclei/protons attached to adjacent carbon atoms (3J)

HH

XY2J = - 4 Hz

2J = decreases with an increase in bond angle

Effect of Electronegativity of Substituents

ClCl

Cl

HHH

3J = 6.0 Hz

Electronegative elements directly attached to the same carbon atom as vicinaly coupled protons, reduce coupling constant, and electropositive elements raise it.

For freely rotating chains, the effect is small,In cyclic systems the effect is larger

HCl

H

HHH

3J = 6.5 Hz

HLi

H

HHH

3J = 8.4 Hz

Hax

Hax

OHHeq

3

56

2

4

Jee = 4.5 Hz Jaa = 10.0 Hz

When an electronegative element is anti-periplanarwith respect to one of the coupling protons, the effect of reducing J value is larger

Rigid Systems OH

Hax

HeqHeq

13

56

2

4

Jae = 4.5 HzJea = 2.5 Hz

Chemical Equivalence of NucleiTwo or more nuclei are said to be chemically equivalentwhen they are in identical chemical environment,Then, they will have identical chemical shift values.

Chemically equivalent nuclei are interchangeable by dynamic process or symmetry operation.

Symmetry Elements Symmetry OperationAxis of symmetry (C2) rotationPlane of symmetry (s) reflection

Hax

Heq1

3

56

2

4 Hax

Heq1

3

5

6

24ring

flip

None

*

stereocentre

Enantiotopic diastereotopic

Magnetic Equivalence of Nuclei• Magnetically equivalent nuclei are those

chemically equivalent nuclei, which also have the same coupling constant with every other nucleus in the molecule.

CH3 CH3

HH7.2 Hz7.2 Hz

triplet

septet

H, H are magnetically equivalent nucleiCH3, CH3 are magnetically equivalent nuclei

OHH

HH

H

H 12

34

5

6

J2,3 is different from J3,6 OR J5,6 is different from J2,5

(8 Hz) (2 Hz) (less than 1 Hz)

H2 and H6, H3 and H5are chemically equivalentbut they are not magnetically equivalent

• In situations when a proton is coupled to neighbouring protons with different coupling constants:

• Multiplicity = [(na+1)(nb+1)(nc+1)…]• Where na, nb, nc .. are sets of magnetically

equivalent protons.

J1,2 = 2 Hz J1,3 = 14 Hz

J2,3 = 10 Hz

H

H1

H3

OO

H2

Multiplicity and Magnetic Equivalence

What will be the multiplicities for H1, H2 and H3?

Multiplicity = [(na+1)(nb+1)(nc+1)…]

Multiplicity of H1 = [(1+1)(1+1)] = 4

Multiplicity of H2 = [(1+1)(1+1)] = 4

Multiplicity of H3 = [(1+1)(1+1)] = 4

na = 1, nb = 1

H

H1

H3

HO2C

2

J1,2 = 2 Hz J1,3 = 14 Hz

J2,3 = 10 Hz

H1

‘Coupling tree’

J1,3 = 14 Hz

J1,2 = 2 Hz

Double doublet (dd) for H1H2 and H3 are also double doublets, each

H

H1

H3

HO2C

2

(dd)

(ddd)

(s)

(d)

H

C-H

H

H

16 Hz

10 Hz

7 Hz

16 Hz

7 Hz

triplet of a doublet (td)

m of H1= [(na+1)(nb+1)]

m of H1 = [(1+1)(2+1)]

(na=1, nb=2)

m of H1 = 6 lines

16 Hz

7.2 Hz

quartet of a doublet (qd)

m = [(na+1)(nb+1)]

m = [(1+1)(3+1)]

(na=1, nb=3)

m = 8 lines

16 Hz

10 Hz

7.2 Hz

triplet of a double doublet (tdd)

1st and 2nd Order SpectraDn/J >10 [the difference between the absorption positions (in Hz)between coupling protons divided by the coupling constant is more than 10], spectrum is 1st order

Dn/J ≤ 10 the spectrum is 2nd order

1st Order 2nd OrderMultiplicity determined by 2In+1 rule (n+1 for 1H)

The 2In+1 rule is not valid. Increased multiplicity

Relative intensity determined by Pascal’s triangle

There is no definitive rule to determine intensity

Spectra are simple Spectra are complex

Dn(2.8)/J(7) = 0.4

Dn(7)/J(7) =1.0

Dn(28)/J(7) =4

Dn(105)/J(7) =15 1st ORDER

2ND ORDER

105 Hz

J =7 Hz J =7 Hz

28 Hz

Spin SystemSpin system- a designation which indicatesthe number of chemically and/or magneticallyequivalent protons which are mutually coupled.

AnXm or AnBmA, X and B: relative chemical shift positions(A for highest and X and above for the lowest d)n and m: number of magnetically equivalent protonsdepending on the chemical shift (in Hz) difference

Dn/J >10 the spectrum is 1st order and spin system designation will be AnXm

Dn/J ≤10 the spectrum is 2nd order and spin system designation will be AnBm

Dn/J = 0.4

Dn/J =1.0

Dn/J =4

Dn/J =15 1st Order

2nd order

AX

AB

Roofing effect

Two proton System

CH3

ClO

H H

A2

Dn (0)/J(7) = 0

The two methylene protons couple giving two doublets.However the outer lines are too small to be detectedand the inner lines have completely merged.

AX or AB A2X or A2B

AX2 or AB2 A2X2 or A2B2

First or second order depends on1) Difference in electronic environment2) Strength of magnetic field used for measurement

H

H

H

NC 1

23

AMX (fully first order)

OHH

HI

H

H

AA’XX’ or AA’BB’ABX, AXY(mixture of 1st and 2nd order)

A b s o r b a n c e

012345678 d, ppm

A b s o r b a n c e

01234 d, ppm

TMS

TMS

60 MHz

100 MHz

60 MHz

100 MHz

Chemical Shift (d, ppm) =Observed chemical shift from TMS (Hz)

Sptectrometer frequencey (MHz) = ppm

Dn/J >10

Dn/J ≤10

1st order at 2.35 T

2nd order at 1.4 T

Resolutionincreased

A2X3 or A2B3

Five Proton System

HH

XH

HH

Problems set A, Q1

Problems set A, Q2

Problems set A, Q3

Problems set A, Q4

C5H11Br

Problems set A, Q5

C8H9O2Br

1H1H 1H

6H

Problems set A, Q6

Problems set A, Q7