NMR SPECTROSCOPY AND SOME PROBLEMS BASED ON IT

This presentation gives a brief overview on the theoretical aspects of NMR (proton & carbon), Instrumentation and some problems based on NMR

1. Nuclear Magnetic
Resonance (NMR) Spectroscopy
By
Dr. Kalam Sirisha,
Associate Professor & Head,
Department of Pharmaceutical Chemistry,
Vaagdevi College of Pharmacy, Ramnagar, Warangal, Telangana
E-mail: [email protected]

2. 2
Introduction
•Nuclear magnetic resonance (NMR) is a physical phenomenon in
which nuclei in a magnetic field absorb and re-emit
electromagnetic radiation.
•NMR spectroscopy is the most powerful tool available for organic
structure determination.
•It is used to study a wide variety of nuclei (
1
H,
13
C,
15
N,
19
F,
31
P etc).
•Most common types of NMR: proton and carbon-13
•
1
H NMR (PMR): To determine the type and number of H atoms
and spatial arrangement of them in a molecule.

•
13
C NMR (CMR): To determine the type and number of carbon
atoms and spatial arrangement of them in a molecule.
•The source of energy in NMR is radio waves which have long
wavelengths (75-0.5m), and thus low energy and frequency (4-
600MHz).
=>

3. 3
Nuclear Spin
•A nucleus with an odd atomic number (no. of protons or
electrons) or an odd mass number (no. of neutrons and
protons in the nucleus) has a nuclear spin.
•If the number of neutrons plus the number of protons is odd,
then the nucleus has a half-integer spin (i.e. 1/2, 3/2, 5/2)
•If the number of neutrons and the number of protons are both
odd, then the nucleus has an integer spin (i.e. 1, 2, 3)
•A nucleus of spin I will have 2I + 1 possible orientations.
•A nucleus with spin 1/2 (
1
H &
13
C) will have 2 possible
orientations.

4. •The spinning charged nucleus
generates a magnetic field.
4
=>

5. 5
External Magnetic Field
When placed in an external field, spinning
protons act like bar magnets.
=>

6. The magnetic fields of the spinning nuclei will align
either with the external field, or against the field.
6

7. 7
Two Energy States
A photon with the right amount of energy can be
absorbed and cause the spinning proton to flip.

8. 8
DE and Magnet Strength
•Energy difference is proportional to the magnetic field
strength.
DE = hn = g h B
0

2p
•Gyromagnetic ratio, g, is a
constant for each nucleus.
g=µ/p µ-dipole moment
p-angular momentum
•In a 14,092 gauss field, a 60 MHz
photon is required to flip a proton.
•A nucleus is in resonance when it
absorbs RF radiation and “spin flips” to a higher energy
state.
=>

9. Larmor Precession
Spinning particle precesses about the external magnetic
field axis at an angular frequency known as Larmor
frequency.
9
fo = γ Bo
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10. 10
The NMR Spectrometer
=>
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11. 11
1
H NMR
Magnetic Shielding:
•If all protons absorb the same amount
of energy in a given magnetic field, not
much information can be obtained.
•But protons are surrounded by
electrons that shield them from the
external field.
•Circulating electrons create an induced
magnetic field that opposes the external
magnetic field. =>

12. 12
Shielded Protons
Magnetic field strength must be increased for
a shielded proton to flip at the same
frequency.
=>

13. 13
Protons in a Molecule
Depending on their chemical environment,
protons in a molecule are shielded by
different amounts.
=>

14. 14
NMR Signals
•The number of signals shows how many
different kinds of protons are present.
•The location of the signals shows how
shielded or deshielded the proton is.
•The intensity of the signal shows the
number of protons of that type.
•Signal splitting shows the number of
protons on adjacent atoms. =>

15. 15
Number of Signals
•Protons within a compound experience different magnetic
environments, which give a separate signal in the NMR spectrum
•Equivalent: Protons that reside in the same magnetic environment are
termed chemically equivalent. They have the same chemical shift.
Equivalent Non-equivalent
=>

16. Position of signals (chemical shift)
•The position on the horizontal frequency scale at which
the equivalent proton signals occur is called a chemical
shift.
•The chemical shift depends on the varying local
magnetic fields from the neighboring protons.
16

17. 17
Chemical Shift
•Measured in parts per million (ppm), called the delta
(δ) scale.
•Ratio of shift downfield from TMS (Hz) to total
spectrometer frequency (MHz).
•Same value for 60, 100, or 300 MHz machine.
•Tau () value=10-
Ƭ
δ

=>

18. 18
Delta Scale
=>

19. 19
Tetramethylsilane (TMS)
•TMS is added to the sample.
•Since silicon is less electronegative
than carbon, TMS protons are highly
shielded. Signal defined as zero.
•Organic protons absorb downfield (to
the left) of the TMS signal.
=>
Si
CH3
CH
3
CH
3
H
3C

20. 20
Location of Signals
•More electronegative
atoms deshield more and
give larger shift values.
•Effect decreases with
distance.
•Additional electronegative
atoms cause increase in
chemical shift.
=>

21. 21
=>

22. 22
Intensity of Signals
•The area under each peak is proportional to the
number of protons.
•Shown by integral trace.

23. 23
Integration
When the molecular formula is known, each integral rise
can be assigned to a particular number of hydrogens.
=>

24. 24
Spin-Spin Splitting
•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 and splitting
patterns (Multiplicity) results.
=>

25. Common NMR peaks
25

26. 26
N + 1 Rule
If a signal is split by N equivalent protons, it is split
into N + 1 peaks.
=>

27. 27

28. 28
1,1,2-Tribromoethane
Nonequivalent protons on adjacent carbons.
=>

29. 29
Doublet: 1 Adjacent Proton
=>

30. 30
Triplet: 2 Adjacent Protons
=>

31. 31

32. 32
Range of Magnetic
Coupling
•Equivalent protons do not split each other.
•Protons bonded to the same carbon will
split each other only if they are not
equivalent.
•Protons on adjacent carbons normally will
couple.
•Protons separated by four or more bonds
will not couple.
=>

33. 33
Splitting for Ethyl Groups
=>

34. 34
Splitting for
Isopropyl Groups
=>

35. 35
Coupling Constants
•Distance between the peaks of multiplet
•Measured in Hz
•Not dependent on strength of the external
field
•Multiplets with the same coupling
constants may come from adjacent groups
of protons that split each other.
=>

36. 36
Values for
Coupling Constants
=>

37. 37
O-H and N-H Signals
•Chemical shift depends on concentration and solvent.
•Hydrogen bonding in concentrated solutions deshield the
protons, so signal is around d3.5 for N-H and d4.5 for O-H.
•Proton exchanges between the molecules may broaden the
peak.

=>
=>

38. 38
Identifying the O-H
or N-H Peak
•To verify that a particular peak is due to O-H or N-H, shake
the sample with D
2
O (heavy water).
•Deuterium will exchange with the O-H or N-H protons.
•On a second NMR spectrum the peak will be absent, or
much less intense.

=>

39. 39
Time Dependence
•Molecules are tumbling relative to the
magnetic field, so NMR is an averaged
spectrum of all the orientations.
•Axial and equatorial protons on
cyclohexane interconvert so rapidly that
they give a single signal.
•Proton transfers for OH and NH may occur
so quickly that the proton is not split by
adjacent protons in the molecule.
=>

40. 40
NMR of Paracetamol.mp4

41. 41
Carbon-13
•
13
C has a magnetic spin, but is only 1% of the carbon in a sample.
•It is more difficult to obtain a
13
C-NMR spectrum than a
1
H-NMR
spectrum. The difficulty stems from two sources:
•1.the natural abundance of
13
C is low, so there are fewer NMR-
active nuclei per mole of compound to absorb energy.
•2.the inherent signal intensity per nucleus is less for
13
C than
for
1
H. For equal numbers of
1
H and
13
C nuclei, the signal intensity
for
13
C is roughly 1/4 that of
1
H. When combined with the fact that
the natural abundance of
13
C is roughly 1% of that of
1
H, this means
that the signal intensity of
1
H is over 400 times greater than that of
13
C.
Hundreds of spectra are taken, averaged.
=>

42. 42
13
C-NMR: Number of Signals
Number of
13
C-NMR signals reveals equivalent carbons
•One signal per unique carbon type
•Reveals molecular symmetry
Examples
CH
3
CH
2
CH
2
CH
2
OH CH
3
CH
2
OCH
2
CH
3
Two
13
C-NMR signals
2 x CH
3
equivalent
2 x CH
2
equivalent
No equivalent carbons
Four
13
C-NMR signals
Symmetry exists when # of
13
C-NMR signals < # of carbons in formula

43. 43
13
C-NMR: Position of Signals
•Position of signal relative to reference = chemical shift
•
13
C-NMR reference = TMS = 0.00 ppm
•
13
C-NMR chemical shift range = 0 - 250 ppm
•Downfield shifts caused by electronegative atoms and pi electron clouds
OH does not have carbon
¯
no
13
C-NMR OH signal
Example: HOCH
2
CH
2
CH
2
CH
3

44. 44
13
C-NMR: Position of Signals
Trends
•RCH
3
< R
2
CH
2
< R
3
CH
•EN atoms cause downfield shift
•Pi bonds cause downfield shift
•C=O 160-210 ppm
Characteristic Carbon NMR Chemical Shifts (ppm)
(CH3)4Si = TMS = 0.00 ppm (singlet) CDCl3 (solvent) = 77.0 ppm (triplet)
RCH3 0 – 40 RCH2Cl 35 – 80 benzene ring 110 – 160
RCH2R 15 – 55 R3COH 40 – 80 C=O ester 160 – 180
R3CH 20 – 60 R3COR 40 - 80 C=O amide 165 – 180
RCH2I 0 – 40 RCºCR 65 – 85 C=O carboxylic acid 175 – 185
RCH2Br25 - 65R2C=CR2100 - 150C=O aldehyde, ketone180 – 210

45. 45
13
C-NMR: Integration
1
H-NMR: Integration reveals relative number of hydrogens per signal
13
C-NMR: Integration reveals relative number of carbons per signal
•Rarely useful due to slow relaxation time for
13
C
time for nucleus to relax from
excited spin state to ground state

46. 46
Hydrogen and Carbon
Chemical Shifts
=>

47. 47
Combined
13
C
and
1
H Spectra
=>

48. 48
Differences in
13
C Technique
•Resonance frequency is ~ one-fourth,
15.1 MHz instead of 60 MHz.
•Peak areas are not proportional to
number of carbons.
•Carbon atoms with more hydrogens
absorb more strongly.
=>

49. 49
Spin-Spin Splitting
•It is unlikely that a
13
C would be adjacent
to another
13
C, so splitting by carbon is
negligible.
•
13
C will magnetically couple with
attached protons and adjacent protons.
•These complex splitting patterns are
difficult to interpret.
=>

50. 50
13
C-NMR: Spin-Spin Coupling
•Spin-spin coupling of nuclei causes splitting of NMR signal
•Only nuclei with I ¹ 0 can couple
•Examples:
1
H with
1
H,
1
H with
13
C,
13
C with
13
C
•
1
H NMR: splitting reveals number of H neighbors
•
13
C-NMR: limited to nuclei separated by just one sigma bond; no pi bond “free spacers”
Conclusions
•Carbon signal split by attached hydrogens (one bond coupling)
1
H
13
C
13
C
12
C
Coupling observed
Coupling occurs but signal very weak:
low probability for two adjacent
13
C
1.1% x 1.1% = 0.012%
No coupling: too far apart
No coupling:
12
C has I = 0

51. 51
Proton Spin Decoupling
•To simplify the spectrum, protons are
continuously irradiated with “noise,” so
they are rapidly flipping.
•The carbon nuclei see an average of all
the possible proton spin states.
•Thus, each different kind of carbon
gives a single, unsplit peak.
=>

52. 52
Off-Resonance Decoupling
•
13
C nuclei are split only by the protons
attached directly to them.
•The N + 1 rule applies: a carbon with N
number of protons gives a signal with
N + 1 peaks.
=>

53. 53
Interpreting
13
C NMR
•The number of different signals indicates
the number of different kinds of carbon.
•The location (chemical shift) indicates the
type of functional group.
•The peak area indicates the numbers of
carbons (if integrated).
•The splitting pattern of off-resonance
decoupled spectrum indicates the
number of protons attached to the
carbon. =>

54. 54
Two
13
C NMR Spectra
=>

55. NMR Problems
55

56. If molecular formula is given, calculate the double
bond equivalence to know the level of unsaturation
in the molecule
56

57. 57
1) Predict the number of peaks that you would expect in the PMR spectrum of
the following compounds by assigning equivalent and non-equivalent protons.

CHOOCH
3
i) ii)
CH
3
COOCH
2
CH
3
iii)
CH
3CCH
2CH
3
CH
3
OH
iv)
CH
3CCH
2Cl
CH
3
CH
3 v)
NH
2
COOCH
2
CH
3
2) Propose the structure corresponding to the following spectrum

Mol. Formula : C
3
H
7
Cl
δ ppm splitting Integration
3.7 t 2.0
1.75 m 2.0
1.0 t 3.0

58. 58
3) What splitting pattern you would expect in the PMR of the protons in the
following compounds:
CH
CH
3
CH
3
a) b) c)
d) e)
CH
3
O
CH
2
N
CH
3
CH
3
Br
H
H
NO
2
H
H
CH
2
CH
2
CH
2
O
CH
3
C
CH
3
CH
3 OCH
2CH
3

4) Give a structure or structures consistent with the following CMR data:
Mol. Formula : C
3
H
8
O
CMR data: a) δ 14.7 (q)
b) δ 57.6 (q)
c) δ 67.9 (t)

59. 59
5) Molecular Formula: C
4
H
8
O
2
IR (cm
-1
): 3012 (m),1725 (s), 1423 (m), 1262 (m), 1058 (m).
1H NMR (δ ppm): 0.98 (t, J=6.7 Hz, 3H), 2.67 (q, J=6.7 Hz, 2H),
3.52 (s, 3H).
6) Molecular Formula: C
10
H
12
O
IR (cm
-1
): 3032 (m),1718 (s), 1424 (m), 876(m), 728 (m) & 647 (m)
1H NMR (δ ppm): 1.02 (t, J=7.2 Hz, 3H), 1.21(sext, J=7.2 Hz, 2H),
2.54 (t, J=7.2 Hz, 2H), 7.23-7.32 (m, 5H).

60. 7) Molecular Formula: C
8
H
8
O
3
IR (cm
-1
): 3521 (s), 3018 (m),1738 (s), 1605 (m), 1262, 1051 (m),
842(m), 767 (m) & 668 (m)
1H NMR (δ ppm): 3.52 (s, 3H), 7.28 (t, J=5.2 Hz, 1H), 7.32 (d,
J=5.2Hz, 1H), 7.38 (d, J=5.2 Hz, 1H), 7.45 (s, 1H), 11.57 (s, 1H).
8) Molecular Formula: C
8H
12O
UV: 281 nm (λmax)
IR (cm
-1
): 3018 (m), 1710 (s), 1608 (m), 1424 (m), 1350(m)
1H NMR (δ ppm): 1.9 (s, 6H), 2.5 (s, 3H), 4.2 (d, J=5.4 Hz, 1H), 4.3
(t, J=5.4Hz, 1H), 4.5 (d, J=5.4 Hz, 1H).

61. 9)

62. 62
10)

63. 63
11)

64. 64
12)

65. 65