Basics of NMR. Suitable for UG and PG courses. Includes principle, instrumentation, solvents. chemical shift and factors affecting it. Some problems. resolving agents, coupling constant and much more

1. NMR - NUCLEAR MAGNETIC
RESONANCE SPECTROSCOPY

2. WHAT WE KNOW…
 Tool for structure determination.
 Used for both organic or inorganic
compounds..?
 Easily available..?
 Expensive or cheap..? Affordable..?
 Quantitative analysis

3. REFERENCES

6. INTERNET REFERENCES
 http://www.youtube.com/watch?v=7aRKAXD4dAg
- Type ‘Magritek’ in youtube
 http://www.youtube.com/watch?v=jRxgX-7FO8g
 http://www.freelance-teacher.com/videos.htm
- Type ‘freelanceteach’ in youtube
• http://www.youtube.com/watch?v=uNM801B9Y84

8. VECTORS
 Quantities which have both magnitude and
direction.
 Examples for vectors are Force, velocity, angular
momentum, spin etc
 When two vectors (P and Q) interact with each
other, the resultant vector (R) has a different
direction and magnitude, which is influenced by
the properties of the interacting vectors (P and Q)
R = (P2+푸ퟐ + ퟐ푷푸풄풐풔θ)0.5

9. SPIN
 Form of angular momentum. This is an
intrinsic property shown by elementary
particles and some atomic nuclei.
 This is a vector quantity, has both magnitude
and direction.
 Particles which posses spin and have charge,
interact with magnetic fields, since they
generate their own magnetic field

10. Bose – Einstein
Statistics
Fermi – Dirac
Stats
• Indistinguishable
particles
• One particle per
energy level
• Particles have half
integral spin
• Examples for
Fermions are
protons, electrons,
NO, 2He3
• Indistinguishable
particles
• Any number of particles
can occupy a given
energy level
• Particles have integral
spin.
• Examples for Bosons are
N2, H2, photons.

11. NUCLEAR SPIN
 Nuclei consist of nucleons, each having spin quantum
number of ‘Half’ (1/2)
 When these nucleons combine to give the nucleus of
an atom, the nucleus possesses ‘Spin’
 For a nucleus, the number of allowed spin states is
quantized and determined by it’s ‘Nuclear Spin
Quantum Number’ (Symbol : I)
 For each nucleus, there are (2I+1) allowed spin states,
with integral differences ranging from +I to –I
 + I, (I-1), (I-2),…., (-I+1), (-I)

12. MORNING EXERCISE
 Exercise: Calculate the allowed spin states for
a. Hydrogen (I= ½)
b. Chlorine (I= 3/2).
 You should get 2 spin states for Hydrogen i.e
(+1/2 and -1/2)
 4 spin states for Cl i.e. (-3/2, -1/2, +1/2, +3/2)

13. NMR ACTIVE NUCLEI
 All nuclei carry charge, they possess spin angular
momentum
 Only nuclei having spin quantum number I > 0 are
NMR active.
Mass no Atomic no. Spin quantum
no.
Examples NMR Activity
Odd Odd or Even Half Integral ½(H),
3/2 (13C),
5/2 (17O)
Active
Even Even Zero 0 (12C, 16O) Inactive
Even Odd Integral 1(2H, 14N), 3(10B) Active

14. OF ENERGY AND POPULATION
 In the absence of an applied magnetic field, all
spin states of a nucleus are degenerate (have
equal energy)
 In a collection of atoms, all spin states should be
equally populated.
 This situation changes, when an external
magnetic field is applied.
 Degeneracy is lost, the spin states have different
energies.
 Also, the population of atoms occupying each
state changes.

15. ALIGN AND OPPOSE
 For a Hydrogen nucleus, there are two spin
states, (+1/2) and (-1/2)
 Under an applied field, each nucleus can adopt
only one of these states.
 (+1/2) is of lower energy since it is aligned to
the field. This is called α state
 (-1/2) is of higher energy since it is opposed to
the field. This is called the β state

16. NUCLEAR MAGNETIC RESONANCE
 NMR occurs when nuclei aligned to applied
field, absorb energy and oppose it.
 This is a Quantized process , and energy
absorbed should be equal to difference in
energy between the two spin states.
Eabs = E(1/2) – E(-1/2) = hν
 Energy difference is also a function of applied
magnetic field.

17. Inaugural Derivation..!
Concentrate on the Board..!

18. VISUALISE

19. PRECESSION AND RESONANCE

20. COMPARE…
Source : Google Images

21. SPINNING TOPS AND NUCLEI
 The nucleus behaves as a spinning top under the
influence of an external field.
 The nucleus undergoes precession around its
own axis with an angular frequency ω.
 Stronger the magnetic field, higher the ω.
 Precession generates an oscillating electric field,
of same frequency
 If radio waves of same frequency are supplied
externally, absorption occurs.

22. PROTON FLIPPING
 Absorption occurs since the electric field
generated by the nucleus and the electric field
of the incoming radiation undergo coupling.
 Energy can be transferred from the incoming
radiation to the nucleus.
 Spin change occurs from (+1/2) to (-1/2).
 This condition is called Resonance.

23. ANATOMY
Source : Google Images

24. LABELLING THE PARTS
 There are primarily 4 parts :
 - The Magnet : With a controller to produce a
specified magnetic Field
 - Radio Frequency Source
 - Detector : To detect the output signal
 - Recorder : To record the output and plot it against
the magnetic field.

25. FURTHER DETAILS
 Sample is taken in a glass tube, placed between
2 poles of a magnet.
 Sample is exposed to Radio Waves, frequency
kept constant.
 The tube is spun around at a steady rate, such
that all parts of the sample get equal exposure
to the radiation and the magnetic field.
 Using electromagnets, the magnetic field
strength is varied.

26. SOLVENTS
 As the field strength is varies, the precessional
frequency of the proton matches the frequency
of incident radiation.
 Small amount of the compound is taken in
0.5-1 ml of solvent.
 Solvents used are Deuterated. Common ones
are D2O, CDCl3, DMSO.
 These solvents do not contain protons.
 Deuterium though being NMR active, does not
resonate along with a proton.

27. STRUCTURES
- Pure CDCl3 and DMSO do not
show any peaks in 1H NMR.
- However, commercial
samples are not 100% pure
and hence show peaks.
E.g: DMSO – 2.5ppm,
CDCl3 – 7.32,
D2O – 4.79*
Source : Wikipedia
Ref: Fulmer, Miller et al : NMR Chemical Shifts of Trace Impurities: Common
Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the
Organometallic Chemist, Organometallics 2010, 29, 2176–2179

28. FT NMR
 The older NMR spectrometers used to excite
nuclei of one type at a time.
 Resonances of individual nuclei are recorded
separately and individually.
 Modern instruments use short (1-10 μs) bursts
of energy (radiation) called Pulses.
 A Pulse contains a range of frequencies, which
is enough to excite all the types of protons at
the same time.

29.  Each type of proton, emits radiation of different
frequencies. This emission is called a Free
Induction Decay. (FID)
 The intensity of this FID decreases over time as
the nuclei lose their excitation.
 The FID signal is a superimposed combination of
all the frequencies emitted.
 A mathematical method called ‘Fourier
Transform’ is used to extract individual
frequencies.

30.  The FT breaks the FID into individual sine or
cosine wave components.
 From these components the individual
frequencies can be determined.
 The computer collects intensity vs time data
and converts it into intensity vs frequency data.
 The final spectrum obtained (I vs ν) is called the
FT NMR Spectrum.

31. Left : Free Induction Decay Signal
Below: The FID Signal broken into a
sine wave using FT.

33. ADVANTAGES OF FTNMR
 Entire spectrum recorded, computerized and
transformed in few seconds
 400 spectra accumulated in approx. 13mins
 Samples at very low conc. Can be measured.
 Studies on nuclei with low natural abundance,
and small magnetic moments ( 13C, 15N & 17O)
 Improved spectra for sparingly soluble
compounds.
 Many spectra for same sample : Averaging

34. DEFLECTION

35. SHIELDING
 Each proton is in a slightly different chemical
environment within the molecule and hence not
all protons resonate at the same frequency.
 Electrons surrounding the protons shield (or
protect) the protons from the applied magnetic
field.
 The electrons circulate and form a current.
 This current generates a secondary magnetic field
which opposes the applied field. This is called
‘Diamagnetic Sheilding’

36. QUANTITATIVE ASPECTS
 Greater the electron density around the proton,
greater the induced counter field.
 Proton experiences a lower magnetic field, and
hence the precessional frequency of the proton
decreases.
 Due to this the proton undergoes absorption at
a lower frequency.
 Since each proton is in a different environment,
each has a different resonance frequency.

37. DESHIELDING
 In some cases, like aromatic nuclei, closed loops
of pi-electrons are found.
 These electrons cause strong diamagnetic
shielding around the aromatic nucleus, but the
protons align themselves to the magnetic field
and hence are deshielded.
 In other words, when protons align themselves
to the direction of the applied field, they are
said to be deshielded.

38. MEASUREMENTS AND REFERENCES
 The difference in absorption frequencies is very
low (50-100 Hz in field of 60 MHz)
 The exact resonance frequencies are not easily
measurable, and hence a reference compound
TMS (Tetramethylsilane) is used.
 TMS protons are most shielded. TMS gives one
sharp peak, which is taken as zero.
 All other proton resonance frequencies are
measured relative to TMS

39. NUMERICALS
 Calculate the resonance frequency of a proton
at
a.) 1.41 Tesla.
b.) 2.35 Tesla.
c.) Find the ratio of the given field strengths
and the calculated resonance frequencies
 From this we infer that, for a given proton, the
shift in frequency from TMS is field dependent.

40. CHEMICAL SHIFT
 Protons resonate at different frequencies
depending on the strength of the applied field
and the radiation used.
 This can be confusing if chemists have
spectrometers of different magnetic fields.
 To avoid this confusion, a new parameter which
is independent of the applied field is defined.
 This parameter is called ‘Chemical Shift’

41. MATHEMATICALLY SPEAKING
• Values of δ for a given proton are same
irrespective of the spectrometer frequency.

42. MORE NUMERICALS
 For example : Bromomethane protons resonate at
162Hz in 60 MHz NMR, but in 100 MHz NMR they
resonate at 270 Hz. Calculate δ for both.
 Calculate the frequency and energy corresponding
to 1 ppm shift from TMS in
a.) 600 MHz NMR Spectrometer
b.) 300 MHz NMR Spectrometer

44. SUMMARISING…
 Chemical shift of frequency.
independent
 Shielding : Opposing the applied magnetic field,
peaks at low δ values. E.g : Aliphatic, vinylic,
benzylic protons etc
 Deshielding : Alignment towards applied field. Peaks
at higher δ values. E.g : Aromatic protons and
protons of aldehydes and carboxylic acids.
 So tell me where do ketone proton peaks appear..?

45. FACTORS AFFECTING CHEMICAL SHIFT
 Inductive or Electronegativity effects
 Hybridization effects
 Hydrogen Bonding
 Solvent effects
 Van der Waal’s deshielding
 Geminal and Vicinal Coupling
 Proton exchange

46. ELECTRONEGATIVITY EFFECTS
 If electronegative elements are attached to the
same carbon of interest, the absorption shifts
downfield.
 More electronegative the atom, more
downfield is the absorption.
 Multiple substituents have a stronger effect
 Effect decreases over distance
 This idea can be used to measure
electronegativity of elements

47. ELECTRONEGATIVITY EFFECTS

48. HYBRIDIZATION EFFECTS
 Higher the s character of the bond, closer the
electrons are to the nucleus. Proton resonates
downfield. Other effects can affect this order.
 For sp3 hybridised C atoms
 3o > 2o > 1o > strained ring
δ Values: 2 1 0

49. Sp2 and sp Hybrid Carbons
 The s character is greater in protons attached to
sp2 carbons. Appear at 4.5 – 7 ppm.
 Aromatic Protons appear at 7-8 ppm, and
aldehydic protons at 9-10 ppm. This high shift is
caused by another effect called anisotropy.
 For acetylic hydrogens, anisotropy results in
shift at 2-3 ppm, though it has greater
s character.

50. MAGNETIC ANISOTROPY
 Protons attached to benzene rings are influenced by
3 magnetic fields.
 i. External Magnetic field
ii. Magnetic field generated by circulating π
electrons.
Iii. Magnetic field generated by valence electrons
around the protons.
 These fields interact with each other and caused
anisotropy. (non uniformity).
 Benzene protons lie in the deshielded region and
hence they appear downfield.

51. MAGNETIC ANISOTROPY
All groups with π electrons cause anisotropy. In
acetylene, the geometry of the generated field is such
that the protons get shielded. Hence they appear
upfield.

52. FINDING AROMATICITY

54. Hydrogen Bonding
 In samples that can undergo intermolecular H
bonds, the δ value is highly concentration
dependent. Intramolecular H bonding is very
slightly concentration dependent.

55. EXCHANGEABLE PROTONS
Protons which can undergo exchange show
variable and broad absorptions.

56. VARIABLE ABSORPTIONS
Acids RCOOH 10.5-12.0 ppm
Phenols ArOH 4.0- 7.0 ppm
Alcohols ROH 0.5-5.0
Amines RNH2 0.5-5.0
Amides RCONH2 5.0- 8.0
Enols CH=CH-OH > 15

57. Functional Group Proton
Chemical Shift
(ppm) Reason
Ethers R-O-CH 3.2-3.8 ppm Electronegativity of
oxygens
Haloalkanes -CH-I 2.0-4.0 ppm
Electronegativity of
halogen atom
-CH-Br 2.7-4.1
-CH-Cl 3.1-4.1
-CH-F 4.2-4.8
Nitriles -CH-C≡N 2.1-3.0 α- Protons deshielded
by C≡N group
Amines R-N-H
CH-N-Ar-
N-H
0.5-4.0
2.2-2.9
3.0-5.0
Protons undergo
exchange
Resonance removes
electron density from
nitrogen

58. Functional Group Proton Chemical
Shift (ppm)
Reason
Ketones R-CH-C=O 2.1-2.4 Anisoptropy of C=O
group
Esters 2.1-2.5
3.5-4.8
Anisoptropy of C=O
group
Electronegativity of
Oxygen
Carboxylic
acids
R-COOH
-CH-COOH
11.0-12.0
2.1-2.5
Electronegativity of
Oxygen
Signal usually broad.
Exchangeable
Amides R(CO)-N-H-
-CH-CONH-R(
CO)-N-CH-
5.0-9.0
2.1-2.5
2.2-2.9
Variable absorption
Deshielded by carbonyl
group
- Nitrogen’s
electronegativity
C
H2
O
O
CH2

59. INTEGRALS AND INTEGRATION
 Apart from finding the types of hydrogens
present, NMR can also be used to find the
relative number of hydrogens.
 This can be done by calculating the area under
each peak.
 This is done by tracing a vertically rising line
over each peak, this line is called the integral.
 The integral rises by an amount proportional to
the area under the peak.

60. INTEGRALS AND INTEGRATION

62. CALCULATING THE HYDROGENS
 CONCENTRATION ON THE BOARD..!
WORD FILE..!

63. ALCOHOLS
 C-OH – 0.5-5.0 ppm
 CH-OH – 3.2- 3.8 ppm
 No spin-spin coupling if exchange is rapid.
 Exchange promoted by increased temperature,
small amounts of acidic impurities and also water
content.

64. DEUTERIUM SHAKE
 Used to identify –OH absorption.
 Add D2O to the sample, and shake. Hydrogen
gets replaced by D atom, and peak disappears.

65. A compound with molecular formula C8H10 has
in its 1HNMR spectrum two signals δ 2.3(6H)
and 7.1(4H) deduce the structure
p-xylene

66. Deduce the structure of an organic compound(C7H8O
having following spectral data
IR : 3300cm-1( br), 3050cm-1 (m) , 2950cm-1(str)
NMR: δ7.1(m,5H) , δ4.4(s,2H) and 3.7(s,1H)
C6H5CH2-OH
Benzyl alcohol

67. Deduce the structure of an organic compound
(C8H9NO) having following spectral data IR :
3430cm-1, 1710cm-1 ,
NMR: δ7.3(m,5H) , δ2.2(s,3H) and 8.3(br,1H)
C6H5NH-CO-CH3
acetanilide

68. A compound with molecular formula
C8H8O shows following NMR peaks
Deduce the structure of the compound δ
2.8 (2H, d): δ 7.28 (5H,m) δ 9.78(1H,t)
C6H5CH2CH=O
Phenyl acetaldehyde

69. COUPLING CONSTANT (J)
 Distance between peaks of a simple multiplets
 Geminal Coupling 2J : Coupling between 2 protons
on same carbon, this carbon can be C-13 .

70. COUPLING CONSTANTS OF SOME GROUPS
ALKANES:
 Three types of hydrogens :
- Methyl Hydrogens (-CH3): δ 0.7- 1.3 ppm
- Methylene hydrogens (-CH2-): δ 1.2 – 1.4 ppm
- Methine Hydrogens (-CH--): δ 1.4-1.7 ppm
Coupling Constant : 3J = 7-8 Hz

71. ALKENES
 Vinylic hydrogens : (C=C-H) δ 4.5 – 6.5 ppm
 Allylic Hydrogens : (C=C-C-H) δ 1.6 – 2.6 ppm
 3Jtrans = 11 – 18 Hz
 3Jcis = 6 -15 Hz
 2J = 0-3 Hz
 4J = 0-3 Hz

72. H
H
a n g le = 1 0 9 o
2 J = 1 2 -1 8 H z
H
H
a n g le = 1 1 8 o
2J = 5 H z
H
H
a n g le = 1 2 0 o
2J = 0 -3 H z
H
H
C H 3
H 3C
C H 3
O
a n g le = 1 0 7 o
2 J = 1 7 .5 H z

73. Karplus Relationship
 Relates coupling constant with dihedral angle.
 variation of Jvic is given by Karplus’s equation
3JH,H = A + B cosф + C cos2ф
 Modified as :
 If 0o˂ ф˂90o , Jvic = 8.5cos2 ф-o.28
 If 90o˂ ф˂180o , Jvic = 9.5cos2 ф-o.28

75. Types of NMR spectrum
 First order spectra:
 Spectra shows multiplets obeying n+1 rule and
intensities following Pascal’s triangle
 n+1 rule is applied only if 1) ratio of difference
in chemical shift (ΔνHz) of interacting nuclei to
their coupling constant J(Hz) is more than 10,
ΔνHz/JHz≥10
 Values of J of all protons in neighboring groups
with protons of interest must be identical

76. First Order Spectra:
 J does not depend on field strength, chemical
shift value depends on field strength
 At field strength ΔνHz/JHz≥10 possible
 Spectra becomes first order at higher field
strength .

77. Second Order Spectra
 If ratio ΔνHz/JHz ˂ 10 l 1H NMR becomes
complicated
 Splitting not obey n+1 rule
 Intensities of peaks not belong Pascals tringle
 Also known as non-first order spectrum

78. Limitations of the (n+1) rule
 This is valid only when vicinal inter-proton coupling
constants are exactly the same for every successive
pair of carbons.
H 2 C
 JAB = JBC = JCD = JDE ….
H 2
C
CH
2
H 2
C
CH
2
A
B D
C
E

79. H
16
8 8
4 4 4 4
2 2 4 4 2 2
1 13 3 3 3 1 1
1 4 6 4 1
H
H
H
C
C
C
H
H
In the above
molecule, count
the number of 3J
H-H couplings, for
the 2nd carbon.
You will find 4 of
them.
Hence, 24 = 16
H

80. Rules are meant to be broken..?
 Use 1,1,2-trichlorethane – predict spectrum.
This obeys (n+1) rule.
 Styrene oxide does not…
 All protons of the side chain
appear as quartets.
HA
HC
H
 Reasons: Coupling constants not equal for
vicinal protons.
 Restricted rotation across the ring.
H
OH
HC

81. J1 J2
CH CH2
J1
J2
J2
CH2
O
J1 J2
CH2
CH CH
J1
J2 J2

82. Hindered Rotation
 N,N- Dimethyl Formamide shows 2 distinct peaks for
the methyl protons even though they are chemically
equivalent.
 Free rotation of the methyls around the nitrogen is
possible.
 This is due to resonance, the molecule is forced to
adopt a planar geometry making the methyl
magnetically non equivalent.
 At higher temperatures, the rate of rotation increases
and a single peak is equivalent.

83. Spin System Notations – Pople Notation
 Each chemically different proton is given a capital
letter A, B, C so on…
 If there are more than one proton of the same type
then numerical subscripts are used A2 , B2 etc
 Protons with widely different chemical shift are
denoted with letters far off in the alphabet.
E.g AB protons are more closely related than AX or
AY protons.
 If three protons with different chemical shifts exists
then notations like AMX, AMY etc are used.

84.  For protons to give distinct peaks, the ratio of
Δν/J (Both measured in Hertz) should be more than 10.
 For AB protons this ratio is very low compared to that of AX
protons.
 Δν (Hz) is dependent on spectrometer frequency, but J (Hz) is
independent of frequency.
 Hence to get larger Δν/J spectrometer of higher frequencies have
to be used.
 In other words AB protons behave as AX protons in higher
frequency spectrometer.
 Which means that at higher frequencies, second order spectra
become first order spectra.

86. Some Examples
 1,1,2-trichloroethane coupling constant is 7Hz
 Difference btw CH & CH2 protons is 1.8 δ (5.7-3.9 δ)
corresponding to 108Hz (60MHz spectrum)
 Hence ΔνHz/JHz ˃ 10
 If Ha & Hb become more alike relative intensities of
peaks do not follow Pascal’s triangle
 Spectrum deviates from first order
 During distortion of signals inner signal intensity
increases outer signal intensity decreases

88. McConnell Equation