H NMR

1. Introduction to NMR
Types of NMR
Source of NMR
Theory of NMR
Principle of NMR
Chemical shift
Acquisition of spectra
1H NMR
NMR set-up
Problems
Summary
1
Nuclear Magnetic Resonance
Spectroscopy

2. Nuclear magnetic resonance involves the interaction
between an oscillating magnetic field of
electromagnetic radiation and the magnetic energy of
certain nuclei when they are placed in an external static
magnetic field.
NMR spectroscopy is a powerful analytical technique
used to characterize organic molecules by identifying
carbon-hydrogen frameworks within molecules.
It is a research technique that exploits the magnetic
properties of certain atomic nuclei.
It determines the physical and chemical properties of
atoms or the molecules in which they are contained. 2

3. 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 of 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).
NMR - Basic principles

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

5.  Two common types of NMR spectroscopy are
used to characterize organic structure:
 1HNMR:- Used to determine the type and number of H
atoms in a molecule
 13CNMR:- Used to determine the type of carbon atoms
in the molecule
5

6. • The source of energy in NMR is radio waves which
have long wavelengths having more than 107nm,
and thus low energy and frequency.
• When low-energy radio waves interact with a
molecule, they can change the nuclear spins of
some elements, including 1Hand13C.
6
radio frequency range 60–1000 MHz

7. In a magnetic field, there are now two energy states for a proton:
a lower energy state with the nucleus aligned in the same
direction as Bo, and a higher energy state in which the nucleus
aligned against Bo.
 When an
difference
external energy source that
between these two states is
matches the energy
applied, energy is
absorbed, causing the nucleus to “spin flip” from one orientation
to another.
The energy difference between these two nuclear spin states
corresponds to the low frequency RF region of the
electromagnetic spectrum.
7

8.  When a charged particle such as a proton spins on its axis, it
creates a magnetic field. Thus, the nucleus can be considered to
be a tiny bar magnet.
Normally, these tiny bar magnets are randomly oriented in space.
However, in the presence of a magnetic field B0, they are oriented
with or against this applied field.
More nuclei are oriented with the applied field because this
arrangement is lower in energy.
 The energy difference between these two states is very small (<0.1 cal).
8

9. 9

10.  Thus, two variables characterize NMR: an applied magnetic
field B0, the strength of which is measured in tesla (T), and
the frequency  of radiation used for resonance, measured in
megahertz (MHz).
1
A nucleus is in resonance when it absorbs RF radiation and ‘spin
flips’ to a higher energy state

11.  The frequency needed for resonance and the
applied magnetic field strength are proportionally
related:
  BO
 The stronger the magnetic field, the larger energy
difference between two nuclear spin states and
higher the  needed for the resonance.
11
2πʋ = γB0

12.  Both liquid and solid type of samples can be used in NMR
spectroscopy.
 For liquid sample, conventional solution-state NMR spectroscopy is
used for analysing where as for solid type sample, solid-state
spectroscopy NMR is used.
 In solid-phase media, samples like crystals, microcrystalline
powders, gels, anisotropic solutions, proteins, protein fibrils or all
kinds of polymers etc. can be used.
 In liquid phase, different types of liquid solutions, nucleic acid,
protein, carbohydrates etc. can be used.
12

13.  The sample is dissolved in a solvent, usually
CDCl3(deutero-chloroform), and placed in a
magnetic field.
 A radiofrequency generator then irradiates the
sample with a short pulse of radiation, causing
resonance.
 When the nuclei fall back to their lower energy
state, the detector measures the energy released
and a spectrum is recorded.
13
Principle of NMR

14. Electromagnet
14

15. Protons in different environments absorb at
slightly different frequencies, so they are
distinguishable by NMR.
The frequency at which a particular proton
absorbs is determined by its electronic
environment.
The size of the magnetic field generated by
the electrons around a proton determines
where it absorbs. 15

16. 15
The spin state of a nucleus is affected by an applied magnetic
field….
Modern NMR spectrometers use a constant magnetic field
strength B0, and then a narrow range of frequencies is
applied to achieve the resonance of all protons.
Only nuclei that contain odd mass numbers (such as 1H,
13C, 19F and 31P) or odd atomic numbers (such as 2H and
14N) give rise to NMR signals.

17. absorb E
-spin states -spin states
release E
Signals detected by NMR
17

18.  Chemical shift is the resonant frequency of a nucleus relative
to a standard in a magnetic field.
 NMR spectra show applied field strength increasing from left to right.
 Left part is downfield, the right is upfield.
 Nuclei that absorb on upfield side are strongly shielded where
nuclei that absorb on downfield side is weakly shielded.
 Chart calibrated versus a reference point, set as 0, tetramethylsilane
[TMS].
18
Si
CH3
CH3
CH3
H3C

19. Theelectrons surrounding a nucleus affect theeffective
magnetic field sensed by the nucleus.
19

20. Measuring Chemical Shift
 Measured in parts per million.
 Ratio of shift downfield from TMS (Hz) to total spectrometer frequency (Hz).
 Same value for 60, 100, or 300 MHz machine.
 Called the delta scale.
Absorptions normally occur downfield of TMS, to the left on the chart.
τ = 10-δ

21.  Deshielded nuclei have a much higher energy difference between the -
and -spin states and these resonate at a much higher frequency.
21
 Deshielded nuclei have a much higher energy difference between the -
and -spin states and these resonate at a much higher frequency.
Shielded nuclei do not ‘sense’ as large a magnetic field as deshielded
nuclei do. As a result, the energy difference between the - and -
spin states is much lower in energy for shielded nuclei and resonate
at a lower frequency.

22. 22
Factors Affecting Chemical Shift
1.Electron Density – Shielding /Deshielding
2.Electronegativity
3.Hydrogen Bonding – low field

23.  The received nuclear magnetic resonance response is very weak in
signal and requires a sensitive radio receiver to pick up.
 A Fourier transform is done to extract the frequency-domain
spectrum from the raw time-domain spectrum.
 Good 1H NMR spectra can be acquired with 16 repeats, which
takes only minutes.
 However, for heavier elements than hydrogen, acquisition of
quantitative heavy-element spectra can be time-consuming, taking
tens of minutes to hours.
 Then a average of all the acquired spectrum will be generated and
displayed through the graph.
23

24.  Proton NMR is much more sensitive than 13C and the active
nucleus (1H) is nearly 100% of the natural abundance.
 Shows how many kinds of nonequivalent hydrogens are in a
compound.
 Theoretical equivalence can be predicted by seeing if replacing
each H with “X” gives the same or different outcome.
 Equivalent H’s have the same signal while nonequivalent are
“different” and as such may cause additional splitting
(diastereotopic effect).
24

25.  Replacement of each H with “X” gives a different
constitutional isomer.
 Then the H’s are in constitutionally heterotopic
environments and will have different chemical shifts –
they are nonequivalent under all circumstances.
25

26. 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.
NMR Signals

27. 0 ppm4681012 2
X H
X = N, O, S
O
HR
O
OR
H
R C
H
H
H
C C
H
H
C C H
H
X C
H
H
H
X = N, O, S, halogen
Common NMR Shifts

28. Chapter 13 28
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. =>

29. Chapter 13 29
1,1,2-Tribromoethane
Nonequivalent protons on adjacent carbons.
=>

30. Chapter 13 30
Doublet: 1 Adjacent Proton
=>

31. Chapter 13 31
Triplet: 2 Adjacent Protons
=>

32. Chapter 13 32
The N + 1 Rule
If a signal is split by N equivalent protons,
it is split into N + 1 peaks.
=>

33. 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.

34. The coupling constant (J) is the distance between two adjacent peaks of a split
NMR signal in hertz (Hz)
Coupled protons have the same coupling constant
Not dependent on strength of the external field
Coupling Constants

35. NMR Spectrum of Ethanol

36. 1H NMR of Neopentyl Bromide

37. O
OH3C
CH2
CH3
3
2
3
1H NMR of Ethyl Acetate
What are these strange signals?
Integral
ratios

38. Identify the compound having molecular formula C4H8O2

39. Relative intensities of X,Y and Z is 3:2:3 since number of protons giving the signal are in that
order
C4H8O2

40. Let us compare the spectra of ethyl ethanoate and methyl propanoate

41. 41
The signal for proton Hb is split into
triplet by the two hydrogens (Ha) on the
adjacent carbon
The Ha is split into a quintet by the four
equivalent Hb protons on its adjacent
carbons because the two set of protons
are equivalent
Splitting of the signals by two adjacent groups of equivalent protons

42. 42
split into
doublet by the
He protons
split into septet
by the Hb
protons
split into triplet
by the Hc
protons
split into triplet
by the Hc
protons
Shows five
signals
The signal for Hc protons will split by both the Ha and Hd protons. Because Ha
and Hd are not equivalent, the n+1 rule has to be applied separately to each
set. Thus the Hc proton will split into
(na+1)(nd+1)=(4)(3) = 12 It is called multiplet in this case
Splitting of the signals by two adjacent groups of nonequivalent protons

43. 43
Protons in Substituted Aromatic Benzene Ring (Beyond syllabus)
a
c
b
d
cd
e
Suppose to
have five
signals
split into triplet by
the Hb protons
split into quartet
by the Ha protons
split into doublet
by the Hd protons
split into triplet by
the two Hd
protons
Hc and He protons are not
equivalent protons.
Hd protons will split into
doublet by He proton and
then each of the doublet peak
split into another doublet by
Hc protons forming a doublet
of doublet (split into four
signals of equal intensities
Since the electronic effect (electron-
donating ability) of an ethyl
substituent is not sufficiently large to
cause a great effect on the
environment of He, Hc, and Hd, the
signals of these protons are
overlapped to form a multiplet signal
We can generally say that the
protons in mono-substituted
benzene ring show a multiplet
signal
doublet of doublet

44. 1- How many signals would you expect to see in the 1H-NMR spectrum of each of the
following compounds?
2 1 3 3
Least shieldedmost shielded
Least shieldedmost shielded
Least shieldedmost shielded
2- Which set of protons in EACH of following compounds is the least shielded?
Which set of protons in the above compound s are the most shielded?

45. 3- The following 1H-NMR spectrum corresponds to one of the compounds
shown below. Which compound is responsible for this spectrum?
Hint: This question can be solved by considering the integration of each
peak

46. 46
O
ethyl methyl ketone
HO
isopropyl alcohol
a- C4H8O:  1.0 ppm (triplet, 3H),  2.1 ppm (singlet, 3H),  2.5 ppm (quartet,
2H).
b- C3H8O:  1.2 ppm (doublet, 6H),  2.0 ppm (singlet, 1H),  4.0 ppm (septet,
1H).The signal at  2.0 ppm disappears upon shaking with D2O.
7- Deduce the structure that is consistent with the following NMR data:
Notice that the hydroxyl
protons (amine protons)
appear as singlet in the
NMR spectrum (not split
by neighboring H)

47. 8- Identify the compound from its molecular formula and 1H-NMR spectrum.
Propyl benzene
3H
2H2H
Note: 3 signals in the spectrum are zoomed
and shown in boxes for more clarity

48. Applications
1. MRI: Magnetic Resonance Imaging is a diagnostic technique that uses magnetic
fields and radio waves to produce a detailed image of the body’s soft tissues and
bones with the help of a computer. This technique helps in diagnosing various
diseases such as tumor, cardio-vascular complications, cirrhosis and also helps in
identifying sports injuries. (details is given in the ppt titled MRI)
2. Determination of structure of organic compounds: Based on the number,
location, intensity and splitting pattern of the spectrum structure of unknown
compounds can be established. However, structure elucidation requires help
from other spectral techniques also.
3. Differentiate between isomers (evident from the spectra of C4H8O2 given
earlier): The NMR spectra of isomers can be differentiated based on location,
intensity and splitting pattern seen in the spectra.
4. Determination of water content in liquid N2O4 : If water is present its
protons will show a singlet at about δ=4.7ppm. By comparing the intensity of
the peak for the test sample with that of known standard it is possible to
determine the water content in N2O4.

49. 5. For studying hydrogen bonding: Hydrogen bonding causes deshielding
and hence the signal for OH group protons appear at higher δ value. Inter
molecular and intra molecular hydrogen bonding can be differentiated by NMR
since inter molecular hydrogen bonding is affected by concentration changes
whereas intra molecular hydrogen bonding is unaffected by changes in
concentration.
6. Sample purity determination: Impurities present in the sample will show
its characteristic signals in the NMR spectrum. Hence by comparing with
standard spectrum it is possible to determine the purity of the sample.
7. To study reaction kinetics: By monitoring the spectrum of reactant or
product at definite time intervals it is possible to study the reaction kinetics.

50. 14- Predict the structure of the compound that show the NMR data shown below:
i. C9H10O  1.2 (triplet, 3H),  3.0 (quartet, 2H),  7.4-8.0 (multiplet, 5H)
ii. C10H14  1.3 (singlet, 9H),  7.2 (multiplet, 5H)
iii. C10H12O2  2.0 (singlet, 3H),  2.9 (triplet, 2H),  4.2 (triplet, 2H),
 7.3 (multiplet, 5H)
iv. C8H7N  3.7 (singlet, 2H),  7.2 (multiplet, 5H)
v. C4H6Cl2O2  1.4 (triplet, 3H),  4.3 (quartet, 2H),  5.9 (singlet, 1H)
vi. C7H14O  0.9 (triplet, 6H),  1.6 (sextet, 4H),  2.4 (triplet, 4H)
vii. C3H6Br2  2.4 (quintet, 2H),  3.5 (triplet, 4H)
viii. C4H8O2  1.4 (doublet, 3H),  2.2 (singlet, 3H),  3.7 (singlet, 1H),
 4.3 (quartet, 1H)
The singlet signal at  3.7 is a broad one that disappears upon
shaking with D2O
ix. C10H14  1.2 (doublet, 6H),  2.3 (singlet, 3H),  2.9 (septet, 1H),
 7.0 (multiplet, 4H),

51. 51
i- ii-
iii-
iv- v- vi-
vii- viii-
Ethyl phenyl ketone
ix-
Answer of Q 14:

52.  Nuclear magnetic resonance spectroscopy basically
provides the detailed information about the structure,
dynamics, reaction state, and chemical environment
of molecules.
 It has various applications in food industries, food
science, chemical analysis of different products,
pharmaceutical approach etc.
 To analyse the carbon-hydrogen framework in the
molecule is the basic work of NMR technique.
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