B.Sc (Hons.

VIth Semester)
Paper: CHB 605 B (Application of Spectroscopy)
Topic: NMR Spectroscopy
by
Dr. M. K. Bharty
Asstt. Professor
Department of Chemistry
Institute of Science
Banaras Hindu University
NMR spectroscopy is one of the most powerful tools or structural elucidation. Various
applications, that is the various aspects such as such as classification of chemical shifts into
different region, effect of structure on spin-spin coupling, hydrogen bonding and different
types of π electron circulation with several example have been already discussed. Nuclear
magnetic resonance (NMR) spectroscopy, as is implied in the name, involved the change of
the spin state of a nuclear magnetic moment when the nucleus absorbs electromagnetic
radiation in a strong magnetic field. Two types of NMR spectroscopy in common use today
are notably 1H (proton, i.e. PMR) and 13C (carbon, i.e. CMR). To study the radio-frequency
radiation of nuclei is called nuclear magnetic resonance. The method was first discovered by
E.M. Purchell and Felix Bloch in 1946. Three broad principles for the nuclear spins:

1) If the sum of protons and neutrons is odd, then I is half integral (1/2, 3/2, 5/2, …). For
example: 1H, 13C => I = 1/2; 11B, 35Cl => I = 3/2. (I = spin quantum number)
2) If the sum of protons and neutrons is even, then I is zero or integral (0, 1, 2, 3 …).
3) If both protons and neutrons are even in numbered, then I is zero. 12C and 16O fall this
category and give No NMR signal

In fact magnetic properties occur with those nuclei which have…

a) Odd atomic number and odd mass number, e.g. 1H1, 15N7, etc.
b) Odd atomic number and even mass number, e.g. 2H1, 14N7, etc.
c) Even atomic number and odd mass number, e.g. 13C6, etc.

The NMR spectroscopy is most often concerned with I = 1/2, the best example of which
is proton 1H1 with a spin of ½, so, the NMR spectroscopy sometimes also known as proton
magnetic resonance (PMR) spectroscopy.

Number of Signals:- (Depends upon equivalent and non-equivalent protons).

Compounds No of 1H signals Compounds No of 1H signals

CH2 CH2 One CH3 CH2 CH3 Two
Br Br
1,2-dibromoethane

1
 H3C H Three H3C H Two
C C C C
Br H H3C H
2-Bromopropane
O NHNH2
Two C
Four
CH3-CH2OCH2CH3
Diethyl ether

N
Isonicotinic acid hydrazide
CH3 O NHNH2
Four C
Six
H H
H

N
H H
Methyl Cyclopropane Nicotinic acid hydrazide
CH3 O
Two
CH2 C O CH2CH3 Three

CH2 C O CH2CH3

CH3 O
CH3 CH3
Two
Four

H3C CH3
Mesitylene C OCH3

O
Four CH3 Two
O
H3C C NH2
H3C CH2 O CH2 C CH3
Ethoxy acetic acid CH3
t-Butylamine
CH3 CH Br Two H3C CH3 One
CH3 C O C CH3
Br
1,2-dibromoethane H3C CH3
2-tert-Butoxy-2-ethyl-propane
b a c
H2C CH3
Three OCH3
Three

a
c
b

Cl
Environmental effects on NMR spectra:
1- Chemical Shift: The observed NMR frequency of nuclei in atoms or molecules is slightly
differ from the theoretical value. This is chemical shift or shielding effects. It arises from the
magnetic field generated due to the movement of electron around a nucleus when it is placed
in magnetic field, opposes the applied field. Thus the effective field observed by a nucleus is
not equal to the external magnetic field but less by a factor equal to induced magnetic field.
Heff = Happlied – Hinduced
Chemical shift described the dependence of nuclear magnetic energy levels on the electronic
environment in a molecule.

2
The separation in the position of the spectral signal of hydrogen atom in different chemical
environment from that of some arbitrarily chosen standard is called the chemical shift. The
difference is usually reported as the chemical shift δ, described as

It is also reported as τ value.

Electron surrounding a nucleus, under the influence of magnetic field will circulate and in
doing so they generate their own magnetic field. This is opposed to be applied field at the
nucleus as shown for a proton of a C-H bond (see above figure).

Fig.: Chemical shift.

Spin-spin coupling: The actual field observed by a proton (or nucleus) not only depends on
the surrounding electrons but also on the influence made by neighboring magnetically active
nuclei. This is due to spin-spin interaction. A proton can have either α spin (MI = +1/2) or β
spin (MI = -1/2). Now let us consider the proton A is present in the neighboring of the other
proton B. The field observed by proton A will be dependent on spin of proton B. Example:
(n+ 1) = 1+1 = 2
(doublet) O
O H3C
O
CH
H3C C CH3 (doublet) NH2
CH2 O C H3C (septate)
H2 (singlet)
(triplet) (quartet)
(n+ 1) = 6+1 = 7
(n+ 1) = 2+1 = 3 (n+ 1) = 3+1 = 4 (septate)
(triplet) (quartet)
(i) (ii)
n= number of neighboring protons (neighboring magneticalyy active nucli)

3
In real protons are surrounded by a cloud of charge due to adjacent bonds and atoms. In an
applied magnetic field (Bo) electron circulate and produce an induced field (Bi).
In all the PMR spectra presented so far, one has seen almost undistorted multiplets, with the
multiplicity obeying the (n+1) rule. The splitting patterns are rather easy to recognize, since
in each case, one has only two sets of non-equivalent protons. A modification of the n+1 rule.
Generally, when a proton has neighboring sets na, nb, nc and so on of chemically equivalent
protons, then its signal is split into (na+1) (nb+1) (nc+1) and so on. This leads to complex
multiplet patterns, and their analysis is of great help in structure determination.
Factors influencing chemical shift:
1: Inductive effect (Electronegativity): In an organic compound a proton is covalently
bonded to carbon, nitrogen, oxygen or other atoms by sigma bond. When placed in a strong
magnetic field the electrons of the sigma bond circulate to generate a small magnetic field
which opposes the applied field. The chemical shift for the protons of halomethane is directly
related to the electronegativity of the halogen atom. F is most effective at drawing electron
density from the methyl group. Therefore it will be more deshielded than other halogen.

CH3F CH3Cl CH3Br CH3I CH4

4.26 3.0 2.80 2.16 0.9 (ppm)

downfield upfield

2: Hydrogen Bonding: If a proton is hydrogen bonded, it causes downfield shift relative to

the unbounded state. For example, the paramagnetic shifts associated with hydrogen bonding
in hydrogen bonded protons of phenol and carboxylic acids appear at δ values greater than 10
(negative τ values). Other factors which affect chemical shifts are the deshielding effect
caused by anisotropy of carbonyl group and van der Waals deshielding effect

Approximate chemical shift δ (ppm) for certain protons:

Group δ (ppm) Group δ (ppm)

R-CHO 9.0-10.0 >C-H 1.5
Carboxylic acid 10.0-12.0 >CH=CH2 4.5-5.0
Acid H-C-COOH 2.0-2.6 -CH=CH= 5.0-5.5
Enolic C=C-OH 15.0-17.0 -CH=CH-(cyclic) 5.5-6.0
Ar-OH 4.0-7.0 -C≡CH- 2.0-3.0
Aliphatic OH 2.0-5.0 Aromatic CH=CH 6.5-9.0
Aromatic C-H 6.0-8.5 CHCl3 7.25
Benzylic Ar-CH 2.2-3.0 CH2Cl2 5.28
R-SH 1.0-2.0 CH3Cl 3.0
CH3-C=O 2.0-3.0 CH4 0.9
-CH3 0.8-0.9 H-C-F 4.0-4.5
-CH2 1.2-1.5 H-C-Cl 3.0-4.0
H-C-OR 3.3-4.0 Cyclopropane 0.2-0.3
H-C-COOR 2.0-2.2

4
Magnetic Anisotropy:

It means different magnetic properties at different points in space due to the application of
magnetic field. It is observed that proton attached to C=C in alkene is more deshielded than
alkane and their peaks at higher δ value.
H
H
C C (4-8 ppm) Ar-H (6-9 ppm) C (9.5-10.0 ppm)
H
O

But for alkyne (C≡C-H) it is very low (1.5-3.5) but it is expected to come at larger δ value.
This is because of the induced magnetic field generated due to the movement of π electrons in
>C≡C< is parallel to the applied field. Alkyne molecule is linear and symmetrical distribution
of π electrons that’s why the alkyne protons are shielded.

Fig.: Deshielding of alkene and aldehydic protons.

Fig.: Shielding of acetylene protons.

A proton may experience shielding effects caused by electronic circulations that
originate in other parts of the molecule. Effects resulting from electronic currents in other
parts of the molecule will not contribute to the shielding of a proton because such effects will
be averaged to zero by rapid thermal motions. But in a relatively rigid molecule such currents
can affect either shielding or deshielding of a proton. These effects depend on the orientation
of the proton relative to the induced magnetic currents and are called anisotropy effects.

5
 A ring current is similar to current in a wire loop that is a secondary induced field that
acts in opposition the applied field. This induced field however exerts a magnetic effect on
the protons attached to the ring in the direction of the field as shown in below figure.

Fig.: Ring current effect in Benzene.

NMR spectrum of [18]-annulene

Fig.: [18]-annulene

(i) 6 inside hydrogen [Ha]: 1.9 δ, i.e. highly shielded due to inside ring current
(ii) 12 outside hydrogen [Hb]: 8.8 δ, i.e. deshielded due to being outside ring current
Certain large ring compounds obey Huckel’s rule i.e. they contain (4n+2)π electrons and have
some protons that are shielded and others that are deshielded. A graphic example of shielding
and deshielding by ring currents is found in some annulenes. In [18] annulene, the 12
peripheral protons i.e. the protons which are outside the ring are strongly deshielded δ = 8.9,
while the inside protons are so strongly shielded that they absorb at field strength (δ = 1.8)
greater than that used for the reference point (TMS = 0).

6
*An external magnetic field indices a proportional current which in turn generates a (smaller)
magnetic field whose direction is opposed to that of the external magnetic field on the outside
of the conducting ring and parallel on the inside. For the antiaromatic dianion, the situation is
exactly reversed [14] Annulene.

CH3 ring (  ppm) = 8.14-8.64

CH3  ppm) = 4.25

CH3

Dihydrodimethylpyrene

NMR spectrum of Benzaldehyde: The adehyde proton is shifted down field both by
anisotropic effects and by electron withdrawal by the carbonyl oxygen. The combination of
these effects results in absorption that is for down field (9-10 ppm).

Fig.: NMR spectrum of Benzaldehyde

NMR spectrum of Salicylaldehyde:
Acidic protons (-OH, NH, SH) are rapidly exchanged through intermolecular hydrogen
bonding. Two different shifts are involved in this exchange process (free and hydrogen
bonded). Intramolecular hydrogen bonding (salicylaldehyde) is usually strong enough to
suppress the exchange and we observe only one species with a shift that is largely
independent of the concentration and temperature.

7
 Figs.: NMR spectrum of salicylaldehyde (2-hydroxybenzaldehyde)

NMR spectrum of methyl salicylale:

Figs.: NMR spectrum of methyl salicylale

Ignoring the meta and para couplings which are small and may or may not appear then we
observe:

Chemical shift Pattern Coupling constant Integration Assignment Hydrogen

10.79 Singlet - 1 Hydroxy OH
7.82 Doublet 8.0 1 ArH H or H6
3

7.44 Triplet 7.8, 7.8 1 ArH H4 or H5

6.98 Doublet 8.4 1 ArH H3 or H6
6.87 Triplet 7.6, 7.6 1 ArH H4 or H5
3.92 Singlet - 3 Methyl CH3

Hydrogen bonding effects: At lower concentration the intermolecular hydrogen bonding is

diminished in simple OH, NH & SH compounds. Since hydrogen bonding involves electron
cloud transfer from the hydrogen atoms to a neighboring electronegative atom (O, H, S), the
hydrogen experiences a net deshielding effect when hydrogen bonding is strong and is less
deshielded when hydrogen bonding is diminished. Thus, at high concentration (strong

8
hydrogen bonding, strong deshielding) OH, NH and SH protons appear at higher δ value than
in dilute solution. Increased temperature also reduces intermolecular hydrogen bonding. So,
the resonance positions for these protons are temperature dependent (at higher temperature δ
value low).
Intermolecular hydrogen bonding is unchanged by dilution and the NMR spectrum
from such systems is virtually unaltered by varying concentration. Salicylates and enols of β-
dicarbonyl compounds are examples of such systems: chelates such as salicylates, show the
OH resonance at very high (δ = 10-12) and enol OH appears even higher (δ = 11-16).
Carboxylic acids are a special case of hydrogen bonding because of their stable
dimeric association, which persists even in very dilute solution. Carboxylic OH appears
between δ = 10 and δ = 13, usually nearer δ = 11-12.
O H O
C
(carboxylic acid dimer)
O H O

Intermolecular hydrogen bonding: δ values largely unaffected by concentration change.

NMR spectrum of Phenol:
At ordinary concentration, absorb in the range δ 7.7-6.0 ppm. At infinite dilution they absorb
in the range 4-5 δ ppm. Some O-substituted phenols are intramolecularly H bonded so that
the change on dilution is not so appreciable. Phenol with very strong intramolecular hydrogen
bonds absorb in the region 12.5 to 10.5 δ ppm and do not show a great shift on dilution (For
example: methyl salicylate, -0.58τ, salicylatdehyde -0.95τ, o-Hydroxyacetophenone -2.05τ.

A = 7.24
B = 6.93
C= 6.83
D = 5.35

1
H NMR spectrum of Phenol
NMR spectrum of o-Hydroxyacetophenone:

Chemical shift (δ ppm)

A = 12.25
B = 7.71
C = 7.44
D = 6.96
E = 6.89
F = 2.609

9
NMR spectrum of acetylacetone:
H
O O 5.5 ppm 2.0 ppm
2.0 ppm C
H3C C C CH3
C
H3C CH2 CH3
2.13 ppm 3.57 ppm O
2.13 ppm O
H
14.92 ppm
Fig: Keto & enol form of acetylacetone
Hydroxylic absorption and enol occurs at very low field and is usually unaffected by dilution
or solvent extractions. The hydroxyl group of the enol from of acetylacetone absorb at 14.92
δppm. The NMR spectra of enols show absorptions of both keto and enol froms. Thus,
interchange between these two forms on the NMR time scale is very low. Integration of the
spectrum of acetylacetone indicates the presence of 84 % of the enol form and 16 % of the
keto form in the pure liquid at 40°C.

NMR spectrum of acetaldehyde:

NMR spectrum of methyl p-toluate:

NMR spectrum of Limonene:

1.65 ppm 1.96 ppm
1.70 ppm
H3C 2.33ppm 1.63 ppm
C CH3

H2C
4.63 ppm 1.96 ppm H
5.32 ppm
Limonene
Methyl groups attached to isolated carbon-carbon double bonds absorb in the range 1.8-1.6 δ
ppm (example: α-pinene and limonene). Conjugation of the double bond with some
unsaturated group causes the methyl absorption to be shifted to lower field (2.1-1.8 δ ppm)
(down field). The magnitude of this shift depends on the geometrical isomerism about the
double bond (a diamagnetic anisotropic effect).

10
NMR spectrum of n-propyl iodide: The protons of propyl iodide form an A2B2C2 system.
The A protons (adjacent to iodine) are coupled to the B protons and the methyl C protons are
coupled to B protons. Examination of an expanded spectrum of propyl iodide reveals that J AB
= 6.8 cps, JBC = 7.3 cps. If these values for the coupling constants are used, the appearance of
the pattern of the B protons may be calculated. The unperturbed line for the B protons (a) is
split into a quartet (JBC = 7.3 cps) by the methyl C protons, (b) each component of this quartet
is then split into triplet (JAB = 6.8 cps) by the methylene protons A (C).

i = split in to 4
ii = each split into 3
b
= 12 lines

C B A
CH3CH2CH2I {(n+1) (n+1) = (3+1) (2+1) = 12 lines
(Due to overlapping or merged of peaks in the spectrum only sextet peak observed for B protons)

NMR spectrum of phenyl acetic acid

Carboxylic acid absorbs in the region 10.5-12.0 δ ppm. This absorption does not shift
appreciably on dilution of inert solvents because of the strong hydrogen bonded dimer
structure of this group. Hydroxyl and carboxyl groups of phenyl acetic acid that contain both
groups usually undergo rapid chemical exchange, the average absorption position that results
(-COOH and -OH) is concentration.
3.53 ppm
H2
C
COOH

7.21 ppm 12.0 ppm

1
H NMR spectrum of o-Chloroaniline
Amines ordinarily give rise to single, sharp absorption lines, a behavior (similar to alcohol
groups) that indicates rapid rapid chemical exchange of the amino hydrogen atoms. Aliphatic
amines absorb in the region -2.2-0.3 δ ppm and aromatic amines 4.7-2.6 δ ppm.

11
 6.57 ppm
NH2 6.27 ppm
7.08 ppm

6.75 ppm Cl
7.40 ppm

Fig.: 1H NMR spectrum of o-Chloroaniline

These absorption values are shifted to higher field on dilution with inert solvents. Because the
nitrogen nucleus has a spin of unity (I = 1), the absorption of a proton attached to nitrogen
will in theory be split into a triplet. This behavior is observed for amines in acid solution.
NMR spectrum of N-methyl acetamide
O
1.92 ppm
H3C C
NH 8.13 ppm

CH3
2.70 ppm

Aliphatic amine = 7.8-9.7 τ

Aromatic amine = 5.3-7.4 τ
Amide NH groups rise to somewhat broad absorption bands in the region 8.5-5.0 δ
ppm. Protons attached to amide nitrogen do not undergo rapid chemical exchange, because
spin-spin coupling with alkyl groups also attached to the same nitrogen can be observed.
Because the position absorption of the amide protons is dependent to a small extent on
concentration, it is possible that show chemical exchange does exist.

12
 13
C NMR
13
The magnetic resonance of C is much weaker. Thus only 1% of carbon atoms are a simple
being the magnetic isotope. The sensitivity of 13C NMR is decreased by a factor of 100.
Moreover due to gyromagnetic ratio (the ratio between the nuclear magnetic moment µ and
the nuclear angular momentum I, it is denoted by γ) of 13C being only one fourth that of a
proton, the 13C resonance frequency is only one fourth of that for PMR (at a given magnetic
field). Due to these factors CMR is less sensitive that PMR and therefore the weak signals to
be observed 13C spectra are scanned repetitively and stored in a computer. Common range of
13
C NMR is 0.0 to 210 ppm. Thus lower peaks overlaps in 13C NMR. The magnetic moment
of 13C is about one-quarter that of proton 1H. So that signals are inherently weaker, but the
overwhelming problem is that the natural abundance of 13C is only 1.1%.

where, ν = exact frequency

Bo = applied field
γ = magnetic ratio
Compounds No of 1H signals Compounds No of 1H
signals
Three Two

CHO O NHNH2
Six C Four

N
OCH2CH3
Isonicotinic acid hydrazide
O NHNH2
Three C

Six
N
N

Nicotinic acid hydrazide

CH3
Two
One

CH3
CH3
Eight
COCH3
Seven
OH

C OCH3

Approximate chemical shift δ (ppm) for certain protons:

Group δ (ppm) C≡C 50-70
Keton (C=O) 205-210 C-N 100-140
Aldehyde 200-210 C=S 190-210
Carboxylic acid 175-185 Alkane 0-50
1o amide 170-180
2o amide 160-170

13
Multiplicity: 13C and 1H NMR have I = 1/2, so that we should expect to see coupling in the
spectrum [a] 13C-13C, [b] 13C-1H. The probability of two 13C atoms being together in the same
13
molecule is so low that C-13C coupling are not usually observed. Coupling from 13
C-1H
interactions should be observed in the 13C NMR spectra.
13
C NMR spectra of 2-butenol and 1-butenal

OH

H3C CH CH2 CH3

H3C CH2 CH2 CH2OH
22.4 69.0 32.3 10.3  ppm)
13.9 19.4 35.3 61.7  ppm)
Fig.: 13C NMR spectrum of 1-butanol (chemical shift (δ Fig.: 13C NMR spectrum of 2-butanol (chemical
ppm): 61.7, 35.3, 19.4, 13.9) shift (δppm): 69.28, 32.07, 22.83, 10.01)

13
C NMR spectrum of 2-butanol, the carbons in electron dense environments produce
up field signals and the carbons close to electron withdrawing groups produce down field
signals. The signals are not normally split by neighboring carbons because there is little
likelihood of an adjacent carbon being 13C. The probability of two 13C carbons being next to
12
each other is 1.11% × 1.11% (about 1 in 10000). (Since C does not have a magnetic
moment, it cannot affect the position of a signal.)
The signals in a 13C NMR spectrum can be split by nearby hydrogens. However, this
splitting is not usually observed, because the spectra are recorded using spin-decoupling. In
other words, the carbon proton interactions are decoupled. Thus all the signals are singlets in
an ordinary 13C NMR spectrum. If the spectrometer is run in spin-coupled mode, the signals
show spin-spin coupling. The splitting is not caused by adjacent carbons by protons bonded
to the carbon that produces the signal. Protons on other carbons do not cause splitting. The
multiplicity of the signal is determined by n+1 Rule.

14
Both n-butanol (1-butanol and 2-butanol) will give four signals each. In t-butanol alcohol due
to equivalence of three methyl groups only two signals will be found, While iso-butyl alcohol
i.e. 2-methyl-1-propanol will give three signals.

Example: t-Butyl alcohol (2-methyl-2-butanol)

CH3

H3C C OH

CH3 (Two signals)

Iso-butyl alcohol (2-methyl-1-butanol)

H3C
CH CH2 OH

H3C (Three Signals)

13
C NMR spectrum of Methyl benzoate

13
C NMR spectrum of Phenyl acetate

13
C NMR spectrum of Fullerene: The 13C NMR spectrum of purified C60, reported by Kroto
et al. (Taylor 1990). The NMR spectrum of C60 fullerene contained a single peak at δ = 142.7
ppm as expected for the highly symmetrical truncated icosahedron structure in which all the
15
carbons are identical. This result eliminated plannar graphite fragments and fullerenes of
lower symmetry as possible structure for C60.

Fig.: Structure of C60 and C70 fullerenes.

13
The C NMR of purified C70 was reported by Kroto and as expected, it contained five
peaks. The proposed football shaped C70 fullerene possesses five sets of inequivalent carbon
atoms in ratio of 10:10:20:20:10. This is precisely the ratio of the line intensities observed in
the 13C NMR spectrum.

Fig.: 13C NMR spectra of C60 and C70 fullerenes.

Lanthanide shift or chemical shift reagent

The lower spectrum is recorded in normal CDCl3. The upper spectrum was recorded after
addition of a soluble Europium(III) complex to the solution and the spectrum is pulled out
over a much wider range of frequencies so that it is simplified almost to first order. The
paramagnetic Europium(III) ion complexes with the quinoline and induces enormous shift to
higher frequency in the quinoline resonance. The use of europium and other lanthanide
derivatives as chemical shift reagents or lanthanide shift reagents has extended the
applicability and detailed NMR studies to very complex molecules.

16
Fig.: Effect of lanthanide shift reagent Eu(DPM)3 on the 1H NMR spectrum of 6-methyl
quinoline (ring protons only 60 MHz in CDCl3) [Eu(DPM)3; DPM = dipivaloylmethane].

Question: Suggest the structure for the following data:

1) M.W. = C8H14O4 (3 signal) triplet = 1.26, quartet = 4.18 and singlet = 2.6 ppm
I.R. = 3000-2860, 1750, 1455, 1285, cm-1
Ans:
O

CH2 C O CH2CH3

CH2 C O CH2CH3

2) M.W. = 150
I.R. = 3031(V), 2941(V), 1725, 1608, 1504, 1060, 830 cm-1
NMR = 3.35 (singlet, 3H), 3.82 (singlet, 3H) and unsymmetrical pattern 7.85-7.20 (4H)
Ans:
CH3

C OCH3

3) M.W. = C8H18O, IR = 2960-2851(m), 1342(w) and 1075 cm-1

NMR = 1.25 (singlet)
Ans:
H3C CH3

CH3 C O C CH3

H3C CH3

4) M.W. = 130

17
 I.R. = 3082-2860(m), 1285(s), 1755(m), 1455(m) cm-1
NMR = 8.7 τ (triplet), 7.8 (quartet)
Ans:
O O

H3C CH2 C O C CH2 CH3

5) M.W. = 108
I.R. = 3404, 3065(w), 1499(w), 1455 cm-1
NMR = 7.26 (4H), 4.6 (singlet, 1H), 3.9 (singlet, 3H)
Ans:
OH

CH3

6) M.W. = C4H9NO
I.R. = 3500(m), 3402(m), 2460(w), 1682(s), 1610(s) cm-1
NMR = 1.05 (doublet), 2.1 (septet), 8.12 (singlet)
O
H3C

CH
NH2
H3C

18