INTRODUCTION TO SPECTROSCOPIC METHODS IN ORGANIC CHEMISTRY

Organic spectroscopy is the study of how photons are absorbed by organic molecules. A study of the
relationship between the molecular structure and the kinds of photons absorbed by the molecule helps us
to predict structures and obtain information which can explain the chemistry of the molecule. The passage
of electromagnetic radiation through a molecule can increase one or more of the quantized energy levels
in the molecule. The increase takes place only when radiation of thecorrect frequency (and therefore
possessing the correct energy) is absorbed.

ELECTROMAGNETIC SPECTRUM

The electromagnetic spectrum consists of radiation of wavelength (𝜆) from 10-10 to 104 cm. High
wavelength radiation (i.e. low frequency radiation as 𝜆 = 𝑐/𝜐, where 𝜐=frequency and c=velocity of
light) is of low energy, as the Einstein-Planck relationship tells us that each photon of any beam of
radiation carries an energy E=h 𝜐, where h=6.62 × 10-34 Joule sec (Planck constant).

Types of radiation and their properties:

Radio TV Radar Microwave Far Near Visible UV X-ray γ-ray

IR IR
𝜆 (𝑐𝑚) 104- 102 1 10-1 10-2 10-4 10-5 10-6 10-8 10-10
102
𝜐 (𝐻𝑧) 3×105 3×108 3×1010 3×1011 3×1012 3×1014 5×1014 1015 3×1018 1020
E (eV) 10-10 10-6 10-4 10-3 10-2 1.24 1.55-3 4 104 106
Transitions

ROTATIONAL VIBRATIONAL
NUCLEAR SPIN ELECTRONIC Nuclear
(Molecular transitions)

Transitions possible in a molecule are electronic,vibrational, and rotational, while nuclei can undergo spin
transition.

EElectronic>> E Vibrational >> E Rotational >>>> E Nuclear spin

Long wavelength radiation (Radiowaves) affect nuclear spin transitions and rotational transitions
(Microwave) require less energy than electronic transitions (UV or visible).
 ULTRAVIOLET AND VISIBLE SPECTROSCOPY

Radiation used: The UV and visible range consists of

a. Far UV 4-200 nm: less useful as vacuum equipment needed for study
b. Near UV 200-400 nm: Colourless UV active compounds absorb in this region. Quartz
cells are needed to study the 200-300 nm region since glass absorbs below 300 nm.
c. Visible 400 (violet)-800 (red) nm: coloured compounds absorb in this region

Beer Lambert’s law

The fraction of incident radiation absorbed is proportional to the number of absorbing molecules in its
path.

𝑰𝟎
𝑰 = 𝑰𝟎 × 𝟏𝟎−𝜺𝒄𝒍 𝒐𝒓 𝒍𝒐𝒈 = 𝜺𝒄𝒍 I0 = Intensity of incident radiation
𝑰
I = Intensity of transmitted radiation
I/I0 = T =Transmittance
log I0/I = A = absorbance
𝜀 = Molar extinction coefficient (a measure of the
intensity of the absorption of the molecule at that 𝜆)
c = molar concentration of solute in moles/litre
l = Internal length of cell in cm

Energy transitions (for non-conjugated systems)

Only 𝑛 → 𝜋 ∗ transitions cause useful absorptions in nonconjugated molecules. e.g. CH3CHO (𝑛 →

𝜋 ∗ absorbs at 294 nm,𝜋 → 𝜋 ∗ at 198 nm). Conjugation causes tremendous shifts in 𝜆max.e.g.
CH3CH=CHCHO 𝜆max= 320 and 218 nm)

Chromophore: An isolated functional group capable of absorbing UV or visible radiation

e.g. benzene ring or carbonyl group.

Auxochrome: Functional groups which do not absorb UV or visible radiation, but when attached to a
chromophore causes a shift to longer wavelength and increases degree of absorbance.
e.g. -OHgroup in Phenol shifts UV absorption of benzene although OH alone (e.g. ethanol) does not
absorb.
Effect of Auxochromes: Substituents like -OH, -NH2 have n electrons making 𝑛 → 𝜋 ∗ transitions
possible. The effect of this is to shift the 𝜆max to longer wavelengths (bathochromic shifts).

e.g. Phenol has 𝜆maxat 210.5 ( 6700) and 270 ( 1450) nm. In base, it shifts to 235 nm ( 9400) and 287
( 2600) nm. In the phenolate anion, n electrons are more available than in phenol due to the negative
charge on oxygen.

Effect of Chromophores: These substituents contain π electrons, leading to extended conjugation with
the aromatic ring.
e.g. -COR group in C6H5COR, 𝜆maxshifts to 238 ( 13 000) and 278 ( 890) nm due to the possibility of
resonance structures with positively charged benzene ring.

Information from UV: If many bands are present, some even in visible region, long chain conjugated or
polycyclic aromatic chromophore present. If compound is coloured, at least 4-5 conjugated chromophores
and auxochromes are present. (Exceptions: Some nitrogen containing compounds, e.g. nitro, diazo,
nitroso compounds, iodoform, 𝛼-diketones).

max of 10000-20000 is usually due to 𝛼, 𝛽-unsaturated ketones or dienes and max of 1000-10000 is due to
the aromatic system. An max of less than 100 suggests 𝑛 → 𝜋 ∗ transitions and if this is in the 270-350 nm
region, an unconjugated chromophore containing n electrons is possible.

Woodward-Fieser Rules can be used to predict 𝜆maxfor dienes and carbonyl compounds.

𝛼, 𝛽-Unsaturated Dienes

𝜆max basic value for unsubstituted conjugated acyclic or heteroannular 214 nm

dienes

Increments (to be added to basic value)

For homoannular component +39 nm

Extended conjugation (each extra C=C) +30 nm
Exocyclic double bond +5 nm
Substituents on vinyl carbons -alkyl groups, halogens, etc (each) +5 nm
Worked example 1:

Calculated 𝜆max

214 nm (Basic value) + 39 nm (Homoannular) + 30 nm (Extended

cong.) + 5 nm (1 exocyclic double bond) + 7 × 5 = 35 nm (7 vinyl
substituents) = 323 nm

Carbonyl compounds

𝜆maxBasic value for unsubstituted acyclic or 215 nm

6-membered ring enone

Increments (to be added to basic value)

For homoannular component + 39 nm

Extended conjugation (each extra C=C) + 30 nm
Exocyclic double bond + 5 nm
Carbonyl group present as aldehyde - 5 nm
Enone system in 5-membered ring - 10 nm
Alkyl substituent 𝛼 = +10, 𝛽 = +12, 𝛾 = +18 nm
OH substituent 𝛼 = +35, 𝛽 = +30, 𝛾 = +50 nm
Cl substituent at 𝛼- position +15 nm

Note: Where 𝜆max can be calculated for two different chromophores or by two different methods, the higher 𝜆max
obtained is the one observed.

Worked example 2:

Calculatedλmax

215 nm (Basic value) + 5 nm (1 exocyclic double bond)– 5 nm (Aldehyde)

+ 0 nm (0𝛼-substituents) + 12× 2 = 24 nm (2 β-substituents) =239 nm

Polyenes
The above rules hold well only for compounds with up to 4 conjugated double bonds.

Aromatic molecules
A low intensity band at 255 nm is usually present in all aromatic compounds, both benzenoid and non-
benzenoid.
Benzene 𝜆max 184 nm (=47000) (𝜋 → 𝜋 ∗ ), 202 nm (=7400) (𝜋 → 𝜋 ∗ ) and 255 nm (=730). Band at 255
nm is seen as a fine structure band with a number of peaks in a non-polar solvent. Effect of substituents
on benzene rings cannot be predicted qualitatively unlike in dienes and enones.

INFRARED SPECTROSCOPY

Infrared (IR) radiation extends from the visible to the microwave region of the electromagnetic spectrum.
It can cause transitions between vibrational and rotational energy levels of molecules. The IR region
extends from 750 nm (0.75 µ) to 830 µ, but only the Fundamental region, 2.5-15.4 µ is usually used in
organic chemistry. Organic chemists usually express IR absorption in wave number,

𝜈 = 1 𝜆 = 𝜆 𝑐 𝑐𝑚−1

Sample:If the sample is a solid, it can be mixed with solid KBr and pressed into a transparent disc or
mixed with nujol (parrafin hydrocarbon mixture) to form a suspension or nujol mull. The mull and also
neat liquids are pressed between NaCl plates and the IR of the liquid film is obtained. Solids and liquids
could also be determined in solutions. (Solvents CCl4, CHCl3 and CS2) in NaCl cells. Special cells are
used for gasses.

Excitation: Molecules which are made up of similar or dissimilar atoms held by chemical
bondsapproximate to point masses connected by small springs. It has associated with it a number of
vibrational motions of the atoms against one another. These vibrations are excited by a change in the
dipole moment induced by IR radiation, giving IR spectra. A similar effect on vibrational energy is
achieved when there is a change in polarization caused by visible radiation -- called Raman spectra.

Raman spectra: Raman spectra is observed when irradiation with intense visible radiation (usually blue
mercury light) leads to a change in polarization of molecules of the sample. Incident radiation (E=h𝜈)
may:

a) Be re-emitted without loss of energy and scattered in all directions (Rayleigh Scattering)
b) Be absorbed if energy levels correspond with vibrational and rotational levels within the
molecule, but almost all the absorbed radiation may be released in all directions when molecules
revert to ground state (Fluorescence) or
c) If E<<Eelectronic but E>>Evibrational, give rise to Raman scattering (Radiation in visible range
required).

The energy relations are given by;

ℎ𝜈 𝑝ℎ𝑜𝑡𝑜𝑛 + 𝐸1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 → 𝐸2 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 ∗ + 𝑝ℎ𝑜𝑡𝑜𝑛 𝑜𝑓 𝐸 = ℎ𝜈 ± ∆𝐸

where molecule refers to the molecule in different quantum state of energy level, E 2> or <E1 and E
represents the difference in energy between the two rotational-vibrational energy levels E1 and E2.
Measurements of ∆𝜈 gives the difference between various vibrational and rotational levels. Radiation at
right angles to incident beam is observed in Raman spectroscopy. Lines representing wavenumbers of
incident radiation are seen in the spectrum together with new lines of higher and lower wavenumbers
separated by ∆𝜈. The line due to incident radiation is strong while those on either side (Stokes and anti-
Stokes lines) are weaker. The shifts of the weaker lines from the strong lines (±∆𝜈) gives information
about the frequencies of molecular vibrations.
Raman and IR absorption: Atoms have electrical properties and the molecule may or may not possess a
dipole moment. A beam of radiation can act as an external field and electrically polarize the atoms.
During molecular vibrations both the dipole moment and polarizability may change. If dipole moment
changes, a stationary alternating electric field is produced. This field interacts with the moving electric
field of electromagnetic radiation. When 𝜈of electromagnetic radiation is equal to that of the alternating
field produced by changes in dipole moment, vibrational motions are excited, and radiation is absorbed. If
there is no change in dipole moment, radiation is not absorbed, and the compound is IR inactive. e.g.
CH2=CH2 does not show C=C stretching absorption at 𝜈 = 1600 cm-1. For Raman activity, no change in
dipole moment is necessary, but for vibrational motions to be excited, a change in polarizability, relative
to ground state, is required. Thus, in molecules with a centre of symmetry, IR inactive vibrations are
Raman active while Raman inactive vibrations are IR active.

Vibrations of a diatomic molecule: The frequency of vibration motion, is given by the Law of simple
harmonic motion as
𝜈 = frequency
1 𝑘 k = force constant
𝜈= 𝑚 .𝑚
2𝜋 𝑚𝑟 𝑚𝑟 = 𝑚 1+𝑚2 m1 and m2 are the masses of the two atoms
1 2

Although this equation enables us to calculate the frequency of vibration of a diatomic molecule, quantum
restrictions permit vibrational energy levels to assume only values given by,

1
𝐸= 𝜈+ ℎ𝜈
2
Where 𝜈 is an integer 0.1,2 etc. The vibrational energy levels, 𝜈0 , 𝜈1 , 𝜈2 are therefore separated from one
another by almost equal amounts. If radiation of frequency 𝜈 causes transition 𝜈0 → 𝜈1 , radiation 2𝜈
should cause transition 𝜈0 → 𝜈2 (𝜈=fundamental, 2 𝜈=1st overtone). Mathematical calculation of
frequencies of IR absorptions are only possible in the simplest of molecules. IR spectra of polyatomic
molecules contain in addition to fundamental and overtone absorptions of several different vibrations and
rotations, bands due to combination, coupling and Fermi interactions and are very complex. The
interpretation of IR spectra is usually made on the basis of empirical data (i.e. using previously obtained
IR data of compounds whose structures and functional groups are known).

Factors governing IR band frequency: Most important are,

a) Bond elasticity represented by force constant, k
b) Relative masses of bonded atoms, mr
Other effects like electronegativity, electrical effects, size of neighbouring atoms, hydrogen
bonding, phase changes and steric effects also influence absorption frequency.

Types of vibrations:
 + + + -

Frequencies of vibrations depend largely on atomic masses and the bond order of the bonds
joining the atoms. The lighter the atoms, the higher 𝜈 becomes, while the higher the bond order, the
higher the 𝜈 as force constant is higher.

Complex molecules: The important IR stretching absorptions (𝜈max) obtained empirically are as follows:

𝜈(𝑐𝑚−1 ) 𝜈(𝑐𝑚−1 ) 𝜈(𝑐𝑚−1 )

-C-H 2850-3300 -S-H 2550-2600 -C≡C- 2100-2360

-N-N 2080-2170 -O-H 2500-3650 -N-H 3030-3500
-C≡N 2010-2260 -C=O 1660-1850 -C=C 1600-1670
C-O- 1050-1230 -C=N- 1690-1630

It is seen that most recognizable functional groups absorb above 1350 cm-1. The region 650-1350 cm-1 is
called the “fingerprint region” and includes usually bending and skeletal vibrations. This region can be
used to compare compounds as usually no two compounds have the same pattern in this region. If two
samples have the identical IR pattern in the “fingerprint region”, it is very likely that the two are samples
of the same compound.

The actual 𝜈max of a particular functional group will depend on the other groups and atoms in its
immediate environment. C-H, C=C, O-H, C-O, C=O and H-H stretching absorptions can vary as shown:

𝜈 (𝑐𝑚−1 ) 𝜈 (𝑐𝑚−1 )
ALIPHATIC CARBONYL GROUPS
C-C-H 3010-3090 Aldehyde C=O 1740-1715
C=C 1650-1670 Acyclic ketone C=O 1730-1710
C=C-H 3310-3200 Six membered ring C=O 1730-1710
C≡C-H 2300-2050 Five membered ketone C=O 1730-1760
AROMATIC Anhydride C=O 1820-1760
C=C-H 3080-3030 Acid C=O (dimeric) 1724-1698
C=C 1600-1620 Amide C=O 1678 & shoulder at 1760
ry
C-O (I alcohol) 1085-1058 Acyl chloride C=O 1812-1790
ry
C-O (II alcohol) 1125-1087 Ester C=O 1765-1735
ry
C-O (III alcohol) 1205-1125 N-H
ry
C-O (phenol) 1230-1140 I amine 3550-3350 & 3450-3250
ry
C-O (Ester) 1300-1180 II amine 3550-3350
 ry
HYDROXYL GROUPS I amide 3540-3480 & 3420-3380
ry
O-H (free) 3670-3580 I amide (H-bonded) 3360-3320 &3220-3180
ry
O-H (intramol H-bond) 3590-3420 II amide 3460-3420
ry
O-H (carboxylic acid) 3300-2500 II amide (H-bonded) 3330-3170

Carbonyl groups of saturated acyclic ketones absorb at 1705-1725 cm-1, but those of α,β-unsaturated
acyclic ketones absorb at 1665-1685 cm-1. Generally speaking, conjugation shifts C=O frequency to a
lower 𝜈 by 25-30 cm-1 due to partial loss of C=O double bond character (+C-C≡C-O-). Intermolecular
forces affect the 𝜈max of absorption and so IR should ideally be measured in the gaseous state. However,
this is not technically feasible. Dilute solutions are usually reliable.

The exact position of O-H and N-H stretching bands in the 3700-3400 cm-1 region depends on the extent
of H-bonding. In chelated compounds, a very broad band may appear extending to as low as 2500 cm-1.
Hydrogen bonding broadens the peaks due to O-H stretching and shift to lower 𝜈.

Substitution positions around a benzene ring can often be assigned by the combination and overtone
bands in the 2000-1650 cm-1 region. These bands although weak and observed only in concentrated
solutions consist of 2-8 peaks depending on the ring substitution.

NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

Atomic nuclei are positively charged and nucleonic particles spin on their own axis. They possess
quantized spin angular momentum or spin quantum number, I. Individual protons and neutrons have spin
quantum numbers of +1/2and -1/2 only. Due to angular momentum of nucleons, most nuclei possess spin.
 Nuclei with odd mass number like 1H, 9F and 13C have half integral spin1/2, 3/5, 5/2 etc.
 Nuclei with odd number of protons and neutrons and therefore even mass number like2H, and 14N
have integral spins 1,2,3 etc.
 Nuclei with even number of protons and neutrons have zero spin.

The spinning of positively charged nuclei gives rise to the equivalent of a current flowing in a circular
path, producing a magnetic field with a magnetic moment directed along the axis of spin. Nuclei with
Nuclear Spin Quantum number I>0 like 1H and 13C behave like tiny bar magnets.

Nuclear Precession

The effect of the applied magnetic field on the nuclear spin is to make the nucleus precess. The angular
frequency ωL of precession depends on the strength of the applied magnetic field and the nature of the
nucleus.
ω𝐿 = 𝐻0 . γ
ωL:Larmour angular frequency
H0 : applied field
γ : gyromagnetic ratio
(ratio of magnetic moment, μ of the
nucleus to angular momentum of
nucleus)
The Lamour linear frequency of precession, νL is given by,
𝜔𝐿 = 2𝜋𝜐𝐿

Therefore,
2π𝜈 𝐿 𝛾𝐻0
γ= or 𝜈𝐿 =
𝐻0 2𝜋

ℎ𝛾𝐻0
∆𝐸 = ℎ𝜈𝐿 = Where, h = Planck’s constant
2𝜋

In terms of quantum theory, a precessing nucleus can have only certain orientations with respect to H0.
Number of possible orientations for a nucleus with spin quantum number, I = 2I + 1

Nuclei with I=1/2 like 1H, 13C and 9F can have two orientations and 2H and 14N with I=1, three possible
orientations.

The two orientations of nuclei like 1H, 13C and 9F with I= 1/2 are,alignment with and alignmentagainst the
field. Alignment against the field is of higher energy, the difference in energy for these nuclei being given
by hγH0/2π.

In general, for a nucleus with spin quantum No. = I,

ℎ𝛾𝐻0 𝐼
∆𝐸 =
𝜋
For 1H nuclei precessing in an applied magnetic field, radiation providing energy of 2μH0 will cause
transition from the lower energy state to higher energy state. The energy required is only about 10 -10 eV
and can be supplied by low energy Radiofrequency (low frequency 3 × 105 Hz, high wavelength 102 - 104
cm) range of the electromagnetic spectrum. If a field due to a radiofrequency radiation is oscillating in a
plane at right angles to H0, and if the frequency of radiation ν of this field is the same as the frequency of
precession of the nucleus, transfer of energy from the radiation to the nucleus can occur. This transfer of
energy is called resonance.

At resonance, ∆E = hν = hγH0/2π or ν = γH0/2π, for nuclei like 1H with I=1/2

In general, ν = γH0I/π. Since π is a constant and H0 determined by the user, ν ∝ γI

Since the energy difference ∆E between the orientations is small [e.g. in a 300 MHz (Rf used = 300 MHz)
NMR spectrometer, ∆E=0.12 kJ/mole for a proton], there are almost as many nuclei in the upper energy
state as in the lower energy state.

No. in upper energy state

= e−∆E RT
No. in lowerenergy state

For ∆E = 0.12 kJ/mole, this ratio is roughly = e–(5×10-5) as RT = approximately 2500 kJ/mole. Therefore, of
every 2,000,050 protons, there will be 1,000,050 in the lowest state and 1,000,000 in the upper state. Even
in very powerful NMR instrument, the lower energy state has an excess of less than one in every 20,000
protons. We can therefore place a magnetic nucleus in a powerful magnetic field H0 making it precess and
expose it to radio waves whose frequency is changed until it matches the precession frequency of the
nucleus. At this stage, there will be an exchange of energy between the radiofrequency (Rf) field and the
precessing nucleus, whose magnetic moment will change from being oriented with H0 to being oriented
against H0. The frequency of the Rf field at resonance will give us information on the nucleus magnetic
properties like γ and I. Theoretically, for a particular H0, nuclei of one type like 1H or 13C should achieve
resonance at the same Rf. If this were so, NMR would not be very helpful in structural studies.
Fortunately, other effects, which we shall study about, help us to distinguish between nuclei of the same
type in different environments. In this form of NMR spectroscopy called continuous wave NMR either
the magnetic field or the Rf frequency can be kept fixed and the other changed. Most modern NMR
spectroscopy however are Fourier Transform NMRs which acquire and manipulate data using more
sensitive methods about which we learn later.

When irradiation takes place there will be a net absorption of energy since upward transition will be
slightly greater than downward transition. Nuclei in the upper energy state are able to return to the lower
energy state through a process called relaxation. The energy can be lost to the surrounding environment
called the “lattice”- spin-lattice relation or exchanged with a nucleus in the lower energy state-spin-spin
relaxation. In the latter case, there is no net change in the lower and upper energy states.

The NMR spectrum:

The NMR spectrometer consists essentially of a powerful magnet and a radiofrequency source. The
magnet used can vary in strength from 1.4 T (Tesla) in the older machines and should be maintained at
the temperature of liquid helium. Most NMR machines in use today have superconducting magnets.
Protons achieve resonance in our old 1.4 T magnet spectrometer (Varian T 60) in our department at an Rf
of about 60 MHz and in the newer 7.2 T magnet spectrometer (Mercury 300) at 300 MHz. These
machines are referred to as 60 MHz NMR and 300 MHz NMR spectrometer; although other nuclei
achieve resonance at different Rfs with these magnets since their gyromagnetic ratios are different. 13C
nuclei achieves resonance at 75 MHz in an NMR Spectrometer with a 7.2 T magnet, which is still called a
300 MHz spectrometer as 1H nuclei achieve resonance at 300 MHz with this magnet. If it is necessary to
run spectra for a particular nucleus, say 19F or 13C, the spectrometer should be able to provide and detect
Rf in the range at which it achieves resonance and this capability is provided by the “probe” used, e.g. a
19
F probe.

The spectrum is a plot of resonance signal against

magnetic field. The more powerful the NMR
Rf RF
oscillator detector spectrometer the better the spectrum and less
compound is needed for good spectra. The compound
Sample

is usually dissolved in a deuterated solvent (solvent

in which H is replaced by D) like CDCl3, CD3COCD3
or CD3OD. Since it is not possible to exactly measure
Magnet Magnet
magnetic fields at which resonance occur, as internal
standard, usually tetramethylsilane (TMS) (CH3)4Si
is used and the positions (chemical shifts) of the
signals are measured from TMS. TMS is a good
Sweep
generator internal standard as it gives a sharp signal at one end
Recorder of spectrum with most proton signals on its left, does
not react with organic compounds and is volatile and
NMR spectrometer (Continuous wave) so can easily be removed. The units used are ppm
(parts per million) and chemical shift, δ is given in
ppm, 1 ppm being one millionth of the applied field (τ was also used by the English originally but it is
rarely used today τ = 10 - δ). The chemical shift from TMS in Hz will depend on the external field. Thus,
a signal for a particular proton in our 300 MHz machine will appear five times the distance in Hz from
TMS when compared with the same signal on a 60 MHz machine. Since the unit of chemical shift, δ ppm
is obtained as a fraction of the applied field, the chemical shift in δ ppm is obtained as fraction of the
applied field, the chemical shift in δ shown by both instruments will be the same.
𝜐 𝐻 −𝜐 𝑇𝑀𝑆
𝛿 𝑝𝑝𝑚 =
𝑆𝑝𝑒𝑐𝑡𝑟𝑜𝑚𝑒𝑡𝑒𝑟 𝜐 (𝑖𝑛 𝑀𝐻𝑧 )

The areas of the different signals in the spectrum are proportional to the number of resonating nuclei
responsible for the signal. Their areas are electronically calculated by the machine.

Proton NMR:

The positions of the signals depend on the chemical environment of the particular proton. The 1H NMR
spectrum of methyl acetate CH3COOH shows two lines of equal intensity at δ 2.20 (CH3CO) and 3.50
(OCH3) ppm. Environments and appear at different chemical shifts but all three protons of each methyl
group are in the same environment and appear at the same chemical shift. Since the signals are both due
7 6 5 4 3 2 1 TMS
to three protons each, they are of the same intensity, i.e. they are equal in area.
 Sweep
generator
Recorder

NMR spectrometer (Continuous wave)

7 6 5 4 3 2 1 TMS
Why do signals appear at different positions? We have so far considered the proton is isolation. However,
the proton is surrounded by electrons, which are also electrically charged. Electrons normally rotate in a
random manner. When an external magnetic field is applied, the electrons create a small magnetic field to
oppose the applied field, H0. Since the electron is a negatively charged body, rotation in a plane at right
angles to H0 can create a field opposing H0 at the nucleus-called a diamagnetic field as it opposes the
external field. In effect, this small field create by the rotation of the electrons, Hs, “shields” the nucleus
(diamagnetic shielding). The extent of shielding depends on the electron density around the nucleus. The
magnetic field experience by the nucleus will therefore differ slightly from each other even though the
same field is applied on them. They therefore achieve resonance at slightly different Rf frequencies and
show signals at different chemical shifts. In p-xylene, for example, two types of protons are present, and
two signals appear. The six protons of the two CH3 groups appear at δ 2.3 ppm and the four benzene ring
protons at δ 6.95 ppm, their areas being in the ratio of the number of protons contributing to each signal,
3:2. The total area or integral of the signal ∝ No of absorbing nuclei. The area is determined
electronically or integrated by the NMR spectrometer. The H NMR chemical shift of some important
protons are shown below. For a more detailed list, see Table 1.

3000 Hz Chemical shifts from TMS in Hz at 300 MHz 0 Hz

(10 ppm)

300,003,000 Hz Rf in Hz at 7.2 T 300,000,000 Hz

Saturated CH2
H-bonded
Ar-OH

MeC=O
RCOOH

C=C-H

O-Me

Me-C
C≡CH
Ar-H

TMS
CHO

14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
Deshielded Highly shielded
Downfield Upfield (Strong field)

Table 1: 1H NMR chemical shifts (𝛿 H in ppm)

C-NH 0.5-3.0 Ar.CH 2.5-3.0 O-CH2-C=O 4.0-5.5
C-OH 0.5-5.0 CH2N & CHN 2.5-3.3 O-CH2-O 4.0-5.5
CH3-C 0.7-2.0 CH.CO 2.5-3.6 Ar-CH2-Cl 4.5-5.2
CH2-C 1.3-2.3 CH3NC=O 2.7-3.8 Ar-OH 4.5-7.0
CH-C 1.5-2.3 OC.CH2.CO 3.0-4.5 Ar-CH2.O 4.7-5.7
CH3C=C 1.6-2.3 OCH3 3.2-4.1 NH.C=O 5.0-9.0
CH2C=C 1.7-2.0 NCH2.CO 3.2-4.5 Ar.H 6.0-9.0
CH3C=O 1.7-2.0 ArCH2C=O 3.5-4.2 H-bond OH 7.0-14.0
CH3-Ar 2.0-3.0 Ar.NH 3.5-6.5 H-bond NH 7.0-14.0
CH3-N 2.1-3.1 C=C-H 3.5-8.5 O.CHO 8.0-8.3
CH2.CO 2.1-3.1 CH-N-C=O 3.5-4.2 CHO 9.5-10.0
Ar.CH2 2.4-3.2 CH2-O.CO 4.0-4.3 COOH 9.5-13.0

Fourier Transform NMR:

The continuous wave NMR method described above is rarely used in modern NMR spectrometers. In
Fourier Transform NMRs,the Rf signal is applied as a single powerful pulse, which effectively covers the
whole frequency range and lasts for atime, t usually a few microseconds. This pulse generates an
oscillating magnetic field (H1) along the x axis. Because of the slight excess of nuclei in the lower energy
state, the sample has a net magnetization (M) aligned in the direction of the applied field. The effect of
the pulse is to tip the magnetization through an angle, θ = γH1t, the time t usually being chosen so that
θ=90, and such pulsesare therefore called π /2 pulses.
z z

x B0 x B0
M
π/2 pulse
0 y y
Relaxation T1 Precession

x x

FT
0

-1

The magnetization, disturbed from its orientation along the z axis, precesses in the xy plane. A receiver
coil is placed to detect magnetization oriented along the y axis. Thus after the pulse has been applied, the
detected signals starts along +y (positive signal), precesses to the x axis (zero signal), then to -y (negative
signal), and so on. The signal is detected typically for times of the order of a second. The frequency of
oscillation detected is the difference between the NMR resonance frequency and the excitation frequency.
The magnetization in the xy plane due to excitation of the resonance frequency differing by νHz from the
excitation frequency oscillates from positive to negative (and back to positive) νtimes per second. As it
does so, it decays exponentially as relaxation gradually allows the signal to return to the signal to return to
the equatorial position along the z axis. What is obtained is an exponentially decaying cosine of frequency
νHz called free induction decay (FID). These decaying cosine waves (in the time domain) can be
converted into the frequency domain by a mathematical manipulation called Fourier transformation
(FT).The FT spectrum is xxxxxxxx exactly the same form as a CW spectrum and will show the signal for
the excited nucleus. A NMR resonance frequency differing from the excitation frequency by a higher
frequency than νsimilarly gives a decaying signal which oscillates and relaxes at different frequencies
from that for ν . This results in a different FID, which on FT gives a spectrum with a signal at a different
chemical shift. If the two NMR frequencies are simultaneously excited, then the individual FIDs of the
two nuclei interfere, but the information necessary to extract both line frequencies is still present in the
resulting interferogram. Interferogram contains information about all the nuclei excited by the pulse and
can be very complicated since spectra typically contain many NMR resonances over a larger range of
higher frequency differences. In the CW method, the resonances are excited sequentially, whereas in the
FT method they are excited simultaneously. Hence, FT NMR is much more sensitive and faster than CW
NMR and give better signal-to-noise ratios. Further in FT NMR, pulses can be successively applied, the
results acquired after each pulse and the FIDs in digital form can be added together in a computer and
finally subjected to Fourier transformation to give a much better spectrum with very little sample. It is
possible to obtain good NMR spectra for nuclei such as 13C, which is a low sensitive nucleus present as
only 0. 1 % in natural carbon, only by the use of high field machines using FT.

Splitting:

So far, we have seen that protons in different nuclear environments give signals at different chemical
shifts and that the integral of each signal depends on the number of protons responsible for the resonance.
We would expect the NMR spectrum of 1,1,2,2-tetrachloroethane, Cl2HC-CHCl2 to have only one signal
as both protons are in the same environment. In 1,1-dibromo-2,2-dichloroethane, Br2HC-CHCl2, the two
protons are in different environments and should show two signals. But its NMR spectrum shows four
lines, two lines for each proton. The signal due to each proton is split into two forming two doublets. Why
does this occur?

We have studied that just over half the protons are in the lower and the remainder
in the higher energy states. In half the molecules, Hx will be aligned with the
applied magnetic field, H0 and half against the field. The adjacent proton HA will
experience a small field due to Hx, which is only three bonds away. This field will
be in the same direction as H0, in half the molecules and in the opposite direction in
the other half. If this field is denoted by H0, will experience a field of H0 – HS +HC in half of the
molecules HA and a field of H0 – HS– HC in the other half. So, in half the molecules HAwill achieve
resonance at a slightly different chemical shift from the other half and the signal for HA will appear as two
lines (doublet) of equal intensity. Similarly Hx will appear as a doublet because of the effect of the two
orientations of HA .

The protons are said to couple with each other or

split each other. Both doublets are separated by the
same distance as the effect of Hx on HA will be the
same as the effect of HA on HX. The distance is
called the coupling constant, J and is about 7 Hz
for this particular coupling. Coupling constants for
important couplings are shown in table 2.

Across bonds with completely free rotation,

observed coupling constants are the average of
those for all the possible confirmations. For
protons across such bonds, the averaged coupling
constant, J is about 7 Hz. The coupling constant is
independent ofHo unlike shielding and chemical
shift differences in Hz. Coupling is usually
important only across three or less bonds, although
long range coupling is observed if a double bond
separates the protons. Coupling across two bonds
is called geminal coupling and coupling across
three bonds, victual coupling. Geminal protons are
usually equivalent due to free rotation. If they are
chemically and magnetically non-equivalent, J can
be 0-30 Hz(usually12-15 Hz). For geminal protons
on an sp2 carbon, J is about 0.3 Hz. Vicinal
coupling constants in cyclohexane rings are higher
for axial-axial coupling than for axia1-equatorial and equatorial-equatorial.

It has been shown that coupling constants for protons on vicinal carbon atoms depends on the dihedral
angle (∅). This is shown by the Karplus curve. The coupling constant, J=0 when ∅=90 and J=12 when ∅=
180. The protons of the two methyl groups in CH3COOCH3 are far separated (five bonds) and do not split
each other and the NMR spectrum as we learnt consists of two singlets of equal area ratio.
z z

x B0 x B0
M
π/2 pulse
0 y Karplus curve y
Relaxation T1 Precession

x x

π
Spin system:
1
Protons close to each other in chemical shift are labeled with letters next to each other like A, B or C,
those far from them with X, Y or Z and those intermediate with M or N when describing the spin system
FT of nuclei coupling with each other. Cl CH.CHCl will therefore be
in a0compound. A spin system consists 2 2
an A2 system, Br2CH.CHCl2 an AM system and CH3COOCH3 has two A3 systems, since they do not split
each other.
-1
Chemical and Magnetic Equivalence:

Magnetic nuclei in identical environments are said to be chemically equivalent

and appear usually at the identical chemical shift and show the same signal, e.g.
CH3COCH3 and (CH3)3CC(CH3) 3 each show only a single signal for all the
6 2 methyl protons in their 1H NMR spectra. The signals of chemically equivalent
protons will not split each other, e.g. CH3CH3 shows only one singlet in its
5 3 spectrum although there are 3 protons on the adjacent carbon atom. Chemically
equivalent nuclei may sometimes be magnetically non-equivalent and therefore
Ç Ç show separate signals and split each other. Magnetically equivalent nuclei should
√ √ not only be in the same environment, i.e. be chemically equivalent but also their
p-Bromoanisole coupling constant with other chemically equivalent nuclei in the spin system
CHEM3013 -2015 NMR SPECTROSCOPY
should be identical, e.g. in p-bromoanisole(4-bromomethoxybenxene), 2- and 6-Hs are chemically
equivalent and 3- and 5-Hs are chemically equivalent but they are not magnetically equivalent as the
coupling constant of 2-H with 5-H (J=1 Hz) is not the same as that of 6-H with 5-H (J = 9 Hz). The spin
system is referred to as AA'BB' system implying that A &A' and B & B’ are chemically equivalent but
magnetically non-equivalent.

We learnt that different nuclei like 1H, 13C, 19F and 14N will achieve resonance at very different Rf in an
NMR instrument with a magnet of a particular fixed strength and if we run a 1H spectrum we will not see
the signals of the other magnetic nuclei in the compound. However, other magnetic nuclei in close
proximity should be able to split 1H nuclei and be split by 1H. We do not however see the splitting of 1H
signals by 13C nuclei because only 0.1% of carbon is 13C, while 99% the protons are 1H. Most of the
carbon nuclei close to 1H are non-magnetic 12C nuclei. However, in a compound like 1,l-dibromo-2,2-
dichloro-l-fluoroethane, Br2FC-CHCl2, the 1H NMR spectrum will show only one signal, a doublet due to
the proton which is split by the F. If we run a 19F NMR spectrum of the compound, that too will show
only one doublet for the F split by the proton.

In 1,1,2-trichloroethane, there are two chemically non-equivalent protons HA

and Hx. HAon the carbon with two Cl atoms will be more downfield (δ 5.75)
than Hx on a carbon with only one Cl atom ( δ 3.90) making it a AX2 spin
system. The resonance of the fix protons appear as a doublet because in half
the molecules HA will be oriented with the field and in the other half HA
against the field and their effect on Hx would be to make Hx in half of the
molecules to achieveresonance at a slightly different chemical shift from the other half. H x signal
therefore appears as two lines of equal intensity (doublets) as we saw in Br2CH.CHCl2.

There are two HX, protons and they can be oriented in three ways. Both Hx protons could be aligned with
the field(↑↑), both aligned against the field (↓↓) or one Hx could be aligned with and one against the field
(↑↓). In every 100 molecules of the compound, Hx will be aligned with the field (↑) in 50 molecules and
against the field (↓) in 50 molecules. Of the 50 in which the Hx is aligned with the field (↑), 25 will have
the other Hx also aligned with the field (↑), i.e. Hx in (↑↑) combination and 25of the other Hx aligned
against the field (↓),i.e Hx in (↑↓) combination. Similarly in the 50 in which Hx is aligned against the field
(↓), 25 will have the other HX also aligned against the field (↓), i.e. (↓↓) combination and 25, the other Hx
aligned with the field (↑), i.e. (↓↑) combination. So, altogether half the molecules will have the two Hxs in
opposite orientations (aligned with andagainst the field,( ↑↓)), quarter both Hx aligned with the field (↑↑)
and quarter both aligned against the field (↓↓).Inhalf the molecules with the two Hxs in opposite
orientations, the effect on HA of one Hx would be cancelled by the other and the signal appears at the
chemical shift it would have if there was no coupling. In the quarter of the molecules with both H x
oriented against the field and the quarter with both Hx aligned with the field the HA signal would be
shifted to a slightly higher and a slightly lower chemical shift. The signal for HA will appear as atriplet (t)
or 3 lines in the ratio 1:2:1, while that of Hx will be a doublet (d) of equal intensity. The area ratio of the
two signals HA:Hx:1:2. The NMR spectrum can be written as δ 5.75 (t, J=7 Hz, 1H), 3.90 (d, J=7 Hz, 2H).

In ethyl acetate, CH3COOCH2CH3, we have an A3 system and an A2X3 system. CH3CO will appear as a
singlet at δ 1.95 while CH2 appears as a quartet (q) at δ 4.05 and
CH3 CH3 as a triplet at δ 1.2.
1
The CH3 protons are affected by three different CH2 orientations
like the HA proton in1,1,2-trichloroethane. Both Hs can be aligned
1/2 J=7 Hz 1/2 with the field (↑↑) with total spin +1,(+1/2+1/2), contribution 1, Hs

1/4 J=7 Hz J=7 Hz 1/4

1/4+1/4
1 : 2 : 1
 CH3
1

can be aligned in opposite directions with total spin 0,(+1/2-1/2), contribution 2, and both HS can be
1/2 J=7 Hz 1/2
aligned against the field (↓↓) with totalspin -1.,(-1/2-1/2), contribution 1. The CH3 signal will be split as
shown into a triplet.
1/4 J=7 Hz J=7 Hz 1/4
1/4+1/4
1 : 2 : 1

CH2
The CH2 proton is affected by four different CH3 orientations. All three
Hs canbe aligned with the field (↑↑↑) with total spin +3/2 1
(+1/2+1/2+1/2), contribution1, two Hs aligned with and one H aligned
against the field (↑↑↓), total spin +1/2 (+1/2+1/2-1/2), contribution 3, 1/2 J=7 Hz 1/2
one H aligned with and two HS aligned against the field (↑↓↓), total spin
-1/2 (+1/2-1/2-1/2), contribution 3 and all three Hs aligned against the
1/4 J=7 Hz J=7 Hz 1/4
field (↓↓↓) with total spin -3/2, (-1/2-1/2-1/2), contribution 1.The effect
of the CH3 protons would be to split the CH2 signal into a quartet (q) as
shown, the four lines being in the ratio 1:3:3:1. 1/2
1/8 1/8
1/8 + 1/4 + 1/4 + 1/8
+
1 : 3 : 3 : 1
In general, if N is the number of equivalent protons in adjacent atoms,
the multiplicity of a group is given by N + 1

The relative areas of a multiplet are symmetric about the midpoint and will follow the coefficients of the
binomial distribution. (d 1:1 , t 1:2:1, q 3:3:1, qn 1:4:6:1 etc ) This is called the N +1 rule.The N+1 rule
applies only for chemically and magnetically equivalent nuclei which are not close to each other in
resonance frequency, i.e: ∆ν/J> 6.The protons (N protons) involved in the coupling should therefore be
chemically equivalent, the coupling constants for coupling between the two types of protons should be the
same for every coupling and they should belong to an AM or AX system and not an AB (∆ν/J< 6) system.
Spectra in which all the couplings satisfy these conditions are called first order spectra and are easy to
interprit. If coupling involves signals where Δν/J< 6, the spectra are second order and may be distorted
from those predicted from first order treatment. Since Δν will increase and is not affected by increase in
spectrometer magnetic strength it is possible to simplify spectra by using a spectrometer with a more
powerful magnet and thereby obtain first order spectra for compounds showing second order spectra with
less power spectrometers.

The N+1 rule can be extended to systems involving magnetically active nuclei at three or more chemical
shifts. In these cases, a modified N+1 rule can be used provided all coupling constants of each type are
equal, e.g. ClHAC=CHXHM, has an AMX system, the three protons HA,HX and HM absorb at different
chemical shifts. According to the extended N +1 rule the resonance of HA is split into a doublet by
coupling with Hx protons and split again by HM giving two doublets or a double doublet (dd). Hx and HM
will also appear as double doublets, making total of three double doublets for the AMX system (Total 12
lines).

Double Resonance and Spin Decoupling:

It is possible to simplify NMR spectra and identify spin-spin coupling in compounds using a double
irradiation technique called double resonance or spin decoupling. The nucleus is observed at the
radiofrequency at which it achieves resonance while it is at the same time being irradiated by a second Rf.
The second Rf is one at which the nucleus with which the observed nucleus is coupled achieves
resonance. In e.g. ethyl acetate, the CH2CH3 group protons make up an A2X3 spin system and appear at δ
3.95 (q) and 1.20 (t). During spin decoupling the quartet is irradiated and observed while it is also
irradiated with the frequency at which the CH3 protons achieve resonance. Irradiation at the second
frequency makes the methyl protons undergo rapid transition between its two orientations reducing the
lifetime in any spin state. The CH2 protons do not any longer "see" two orientations of the CH3 group. The
result is that the quartet at δ 3.95 collapses into a singlet. Similar observation and irradiation of the CH 3
signal while irradiating the CH2 frequency will result in the collapse of the CH3 signal into a singlet. Spin
decoupling can be used to simplify complex and overlapping signals and to confirm coupling relationship.

Two-dimensional NMR:

The FT NMR technique permitted the development of a number of complex NMR experiments from
which a range of structural information could be obtained. It is possible for instance to apply a multi-pulse
sequence on different axes and also introduce a time delay between pulses. The output is called a two-
dimensional spectrum and today, more than fifty different 2-D NMR experiments are known. We will
only with the most commonly used 2-D NMR techniques, COSY and here too only with the interpretation
of the spectrum which results from a COSY experiment.

All 2-D NMR techniques involve the application of a pulse, which changes the magnetization from its
equilibrium state in a specific manner ("preparation phase"), allows the magnetization to evolve as a
function of time ("evolution phase”) and permits the interaction of nuclei with each other ("mixing
phase"). The resultingmagnetization is detected by acquiring FIDs and subjected to Fourier
Transformation to obtain a 2-D spectrum in which two parameters are plotted on two axes. The 2-D
spectrum obtained is actually three-dimensional and would appear as stacked plots if the intensities of the
signals are shown. It is however usual to draw it in a two-dimensional form by showing peak intensities
as contours like the heights of mountains are shown in a map. In COSY or more properly 1H-1H
COrrelationSpectroscopY, the spin-spin coupled protons are allowed to affect each other's behaviour, i.e.
magnetization is transferred between them in the mixing phase. The COSY spectrum of m-dinitrobenzene
is shown below.
 In this spectrum, two
f2 essentially identical chemical shift
axes, frequently labelled f1 and f2, are plotted. A one-
dimensional (normal) spectrum of the compound, m-
dinitrobenzene, is plotted on one or both of the chemical
shift axes. The diagonal of the 2-D NMR spectrum
shows all the peaks due to the protons of m-
dinitrobenzene and is essentially a one-dimensional
spectrum of the compound. In addition, the spectrum
shows cross-peakssymmetrically placed about the
diagonal when peaks are spin-spin coupled with each
other. Since H2 is coupled to H4/H6, there are cross-
f1
peaks that occur at the chemical shift of H2 in one
dimension and at the chemical shift of H4/H6 (shown by
dashed lines).Similarly, a cross-peak connecting H4/H6
with H5 tells us that they are coupled to each other.

The cross-peaks themselves show the coupling constants, but not the full multiplicity. Although H5 signal
on the diagonal is a triplet, the cross-peaks do not have the centre-line of the triplet in either direction.
The same loss of the center-line is found with quintets, but doublets and quartets appear as doublets and
quartets in the cross-peaks. The planes used to define contours have to be chosen carefully if all cross-
peaks are to be displayed. m-Dinitrobenzene provides a simple example but COSY is a powerful
structural tool for the analysis of complex spectra containing many overlapping signals.
Chemical shift and molecular Structure:

We have learnt that when electronegative groups are close to a proton, it is deshielded and its signal
appears relatively downfield. If we consider methyl(CH3),methylene (CH2) and methine (CH) protons in
the same environment, it is found that generally CH3 appears more upfield than CH2 and CH2more upfield
than CH in the same envirenmente.g.CH3-C δ 0.8-1.1, -C-CH2-C δ 1.2-1.5 and C-CH-C. δ 1.4-1.65. The
alkyl group appears to have a deshielding effect compared with hydrogen. It is also found that C≡C-H
appears at δ 2.6-3,1 while C=C-H appears at δ 5.0-5.7, although we would expect the C≡C-H to be
more deshieided than C=C-H as electronegativities decrease in the order sp> sp2> sp3. Similarly, benzene
protons (δ 6.6-8.0) appear to be much more deshielded than olefinic protons. To understand the chemical
shifts, we must examine the effect of currents induced by the applied magnetic field.An isolated H atom
has spherical symmetry and the H(shielding) at the nucleus will always remain the same no matter how the
nucleus orients itself with respect to the applied field. But at a point outside the atom, the induced field
depends on the orientation of the nucleus. In most organic compounds, the large bulk of nuclei are able to
have random orientations in the usual solution state used in NMR determinations. The induced field at a
point outside the atom due to the random orientations will be in different directions. If all the electrons in
the atom have equal freedom to circulate in all directions, the induced field at a point outside should
average to zero. Such an electronic environment is said to be isotropic (symmetrical). For such molecules
the field arising from local diamagnetic circulation about one nucleus has no effect on the amount of
shielding of a neighboring nucleus (Remember in splitting the effect of spin orientation of the adjacent
atom is considered, while here we are considering effects of induced field due to the electron cloud).

In molecules where electron moment is not completely free, i.e. when the circulation of the charge
induced by the applied field has some preferred orientation with respect to the field, the secondary field
induced by circulation of electrons at a point outside the nucleus may not average to zero but can increase
or reduce the applied field at that point. The electronic environment is then said to be anisotropic
(unsymmetrical) and the induced circulation of electrons around the nucleus may cause additional
shielding or deshielding of neighbouring nuclei.

Acetylenic protons:
The acetylene C≡C (and also the C≡N) is linear and the molecular orbitals constituting
the triple bond are symmetrical about the molecular axis. The π electrons are able to
circulate around this axis within a cylindrical area between the carbon atoms. When
placed in an external field, H0, the π electrons can circulate in a plane at right angles to
the applied field and induce a diamagnetic field to oppose H0 only if the C≡C bond is
aligned with H0. The molecules therefore prefer to orient themselves so that the axis
of the C≡C is parallel to H0as the electron current then flows in a direction
perpendicular to H0. The circulation of electrons produces a diamagnetic field parallel
to the bond axis as shown and an unexpectedly high shielding it protons attached to
carbon atom is observed.

Ethelynic protons:
In olefins and other compounds containing functional groups with double bonds like
C=O, C=N, C=S, the molecular orbitals permit circulation of electrons above and
below the plane of the double bond. The molecule orients itself to the applied field,
H0, so that the electrons circulate perpendicular to H0. The electrons circulate as
shown and induce a magnetic field opposed to the applied field. The induced field is
perpendicular to the bond axis unlike in the acetylenes and paramagnetic deshielding
is experienced at the proton attached to the vinyl carbon atom. These protons are
deshielded to greater extent than expected from the inductive effect of the sp 2
carbon.

Benzene and aromatic nuclei:

The electrons are delocalized in aromatic systems and ring currents are readily
developed. Electrons circulate in ring planes above and below the molecule and the
ring orients itself perpendicular to H0. The ring currents produce stronger
deshielding effects at the ring protons than those produced in simple ethylenic
compounds. Therefore aromatic protons appear in NMR much more downfield than
expected.

Chemical Exchange and Splitting:

Protons attached to oxygen, nitrogen and sulphur do not usually split adjacent protons or become split by
them. e.g. in CH3CH2OH, the OH proton appears as a singlet in 1H NMR, although the proton is separated
from the CH2 protons by three bonds and we would expect vicinal coupling to take place. This is because
the slight traces of acids or bases usually present in the NMR sample result in a rapid exchange of
hydroxyl protons between ethanol molecules. If Ha and Hb are of different orientations, the effect of these
spin orientations on CH2 will cancel out.
Since the populations of the two orientations are equal, exchange takes place rapidly from one orientation
to the other and the Hcoupling experienced by CH2 protons due to OH proton cancels out to zero. The proton
is therefore not split by the OH proton and similarly, the OH proton is not split by the CH2 proton. If
exchange takes place rapidly, OH (similarly, NH and SH) will appear as it sharp singlet. If exchange does
not take place rapidly enough for complete decoupling, the signal appears as a broad singlet. The NMR
spectrum of ethanol usually consists of two signals at δ 1.10 (t) and 3.62 (q) due to CH3 and CH2 protons
and a broad singlet for OH around δ 1.8-2.0 but its chemical shift can vary depending on the extent of
hydrogen bonding. If ethanol is purified carefully to exclude acid or base, CH2signal (δ 3.62) will appear
as a double quartet (8 lines) and the OH proton as a triplet. If a drop of D2O is added to a compound with
OH or NH protons, the OH/NH proton signal will disappear due to exchange between the OH/NH proton
and D. OH/NH is converted to OD/ND which will not appear in 1H NMR.

13
Carbon NMR:
13
Carbon, we know, has a spin quantum number, I = 1/2 and 13C nucleispectra like protons. The main
difficulty in 13C NMR is that unlike 1H, 13C is of low abundance (1.1%) in natural carbon and has a low
gyromagnetic ratio. Until the general use of superconducting magnets and the Pulse Fourier transform
(PFT) technique, routine 13CNMR spectra could not be run. The signals in 13CNMR are usually seen to be
split by the protons on the carbon but not by neighbouring carbon nuclei. This is because the chances of a
13
C being next to a second 13C is extremely low. 13C-1H coupling constants are large and these coupling
can result in complex overlapped signals. Since I= ½, the number of lines for each signal is given by N+1
where N is the number of protons on the carbon atom. The signals of quaternary, methine, methylene and
methyl carbons appear respectively as singlets, doublets, triplets and quartets.

Proton broad band decoupled spectra:

Proton broad band decoupled spectra are usually run to simplify 13C spectra. In these spectra, all 13C-1H
multiplets are decoupled while the spectrum is being run by irradiating the sample with a strong spin
decoupling field covering the entire range of frequencies within which the protons in the molecule come
into resonance. The spectrum consists of uncoupled singlets for each non-equivalent carbon atom.
Decoupling affects the intensity of signals differently as the protons being decoupled are involved in the
relaxation of the carbon nuclei. Intensities of signals in proton decoupled spectra therefore give no
indication of the number of carbon nuclei responsible for the signal. Since quaternary carbon atoms have
no protons on them, their intensity is relatively low in proton decoupled spectra and can usually be
recognized as such. 13C chemical shifts are in the range of 0-240 ppm relative to TMS and the important
chemical shifts are shown in Table 3.Generally carbonyl carbons appear at 155-240 ppm, vinyl and
aromatic carbons at 80-160 ppm, carbons attached to oxygen at 50-110 ppm and saturated carbons not
having heteroatom substituents below 70 ppm.
13
Table 3: C NMR chemical shifts (in ppm)

CH3 0-30 C-CH2-O 55-90

-CH2- 10-70 C-(C-)CH-O 65-100
CH3COX (X=C, O, N) 15-35 C-(C-)C(-C)-O (quaternary) 70-110
C-CH(-C)-C 15-70 C-CH2=C and aromatic C-H 80-140
C≡C 20-100 C-(C-)C=C(C-)-C 80-160
CH3N 25-50 Aromatic C-X (X=any substituent) 90-160
-CH2COX (X=C, O, N) 25-60 -O-C-O 90-160
C-(C-)C(-C)-C(quaternary carbon) 25-80 𝛼, 𝛽-unsat. -C=O(X) X=O,OH,N,Cl 155-175
C-CH(-C)-X (X=S,N,C=O) 25-90 Saturated -C=O(X) X=O,OH,N,Cl 165-185
CH3O- 50-65 𝛼, 𝛽-unsaturated -CHO 185-195
C-CH2N 35-75 -CHO, 𝛼, 𝛽-unsaturated -C=O 190-210
C-(C-)C(-C)-N (quaternary) 30-100 -C=O 200-220

Proton Off resonance Decoupled Spectra:

In broad hand decoupled spectra, one cannot distinguish between primary, secondary or tertiary carbons
although quaternary carbons can be recognized because they are much weaker proton off-resonance
decoupled (OFRD) spectra can be used to obtain information on the multiplicity of the 13C signals. When
the spin decoupling field used for irradiation in broad band decoupling is of a frequency close to but not
coinciding with the frequency of the protons, the carbon signals appear as multiplets but with much less
than the normal JC-H. The multiplets are in effect narrowed but not completely removed as in broad band
decoupled spectra. With increasing offset, J increases to approach normal J C-H coupling constants. By
using an offset between 0.5-1 kHz, it is possible to obtain an OFRD spectrum in which coupling is seen
but signals do not overlap.

HETCOR Spectra:

The OFRD spectrum is rarely used today as information on carbon proton correlation can be obtained
conveniently from 2D NMR using a technique in which the delay in the pulse sequence is set to ½ JC-H.
The technique called 1H-13C HETeronuclearCORrelation Spectroscopy or HETCOR correlates 13C
carbon signals with signals of the protons to which they are attached. The spectrum has the 1H spectrum
plotted on one axis and the 13C spectrum on the other. Peaks will be observed at the chemical shift of the
13
C in line with the chemical shifts of the Proton(s) attached to it. Quaternary carbons show no peaks in
HETCOR, while methyl carbons and methane carbons correlate to only one proton signal. Methylene
carbons are correlated to one proton signal unless the two methylene protons are chemically non-
equivalent and appear at different chemical shifts, in which case, they will be correlated to two proton
signals.

MASS SPECTROMETRY
Substances lose electrons and form cations when subjected to high voltage electric current in gaseous
state. These cations can be accelerated by applying a potential field and deflected by magnetic or electric
fields. The extent of deflection depends on mass, charge and velocity. If charge, velocity and deflecting
force are constant, deflection is less for heavy particles and more for light ones.

Electron-impact is usually used to produce the cations. The vapourised sample is introduced under
vacuum into a tube called the ion source and the entering molecules are bombarded with a beam of
electrons from a hot tungsten or rhenium wire. If the energy of bombarding electron is less than the
ionization potential of molecules (about 10 eV), no change occurs. But as energy is increased above the
ionization potential, ionization is induced.

M + e  2e + M+e.g. CH3OH + e 2e + CHOH+ (radical cation)

M+ is called the molecular ion. If the energy of electrons is increased to 50 – 70eV, the molecular ion
formed with an excess of energy and can break down into a mixture of neutral compounds, radicals and
positive ions, the latter being capable of further breakdown or fragmentation.

e.g. CH3OH+CH3+ + OH orCH3OH+ CH2OH+ + H

CH2OH+H2 + CHO+

These ions are accelerated through s slit system by a potential (V) of1000-2000 volts. The neutral
compounds and radicals formed are not accelerated and do not appear in the mass spectrum. The fast
moving ionic beam after leaving the ion source enters the analyzer tube, where it is subjected to a uniform
magnetic field, H (500-10,000 gauss) in low resolution mass spectroscopy. This field is perpendicular to
the direction of the ionic beam and the ions are deflected along a path of radius, r given by,
𝑚𝑣
𝑟=
𝑧𝐻
where m = mass, z charge, v =accelerated velocity of the ions
𝑚2 𝑣 2 𝑟 2 𝑧 2 𝐻2
𝑟 2 = 2 2 𝑜𝑟 𝑣 2 =
𝑧 𝐻 𝑚2
1
But, 2 𝑚𝑣 2 = 𝑧𝑉 as kinetic energy of accelerated ions (1/2mv2) must be equal to the electrostatic energy
(zV) acquired by passing through the applied voltage, V, since the kinetic energy before acceleration is
negligible.

2𝑧𝑉 𝑟 2𝑧 2𝐻2 𝑟 2 𝑧𝐻 2 𝑚 𝐻2𝑟 2

Then,𝑣 2 = 𝑚
= 𝑛2
= 2𝑉 = 𝑚
𝑜𝑟 𝑧
= 2𝑉

𝑚
𝑧
∝ 𝑟 2 if H and V are constant.

If V and H are fixed, the radius of the circular path of an ion in the mass spectrometer will depend on m/z.
Since most ions are singly charged (z=1), ions of different mass will follow different paths and be
recorded at different mass values. The mass spectrum which is a plot of ion beam intensifies against m/z,
is characteristic of a compound provided the identical conditions are used. The largest peak is called the
base peak (100% peak) and other peaks are expressed as % of base peak. Every peak in a mass spectrum
cannot be interpreted as ions may undergo rearrangement, form doubly charged ions, etc, in the mass
spectrometer.
However, certain basic fragmentations can be recognized.
In a simple electron impact mass spectrometer (EIMS), the sample is placed on a probe usually with a
ceramic tip like the one shown above and introduced into the ion source region. The probe tip is placed
near the heated filament. Electrons are accelerated from the hot filament to an anode as shown above
usually through the a potential difference of 70V. The 70 eV electron converts the sample molecules into
molecular ions which are cation radicals. The molecular ions may fragment extensively if a 70eV electron
source is used. The mixture of molecular ions and fragment ions are expelled from the ionization chamber
by a positive voltage applied to the repeller plate shown above and enter the ion analyzer. The ion
analyzer found in a double focusing mass spectrometer described earlier is shown. The ions pass through
a magnetic sector and then through an electric sector. The EIMS described is suitable for mass spectral
analysis of low molecular weight non-polar compounds. Several techniques like Field Desorption (FD),
Fast Atom Bombardment (FAB) and Electrospray Ionization (ESI) have been developed for the analysis
of high molecular weight compounds like proteins and polysaccharides and polar non-volatile compounds
like phenolics compounds. We will not deal with these techniques in the present course.

Molecular ion:

The peak at highest m/z in a mass spectrum is usually the molecular ion (M +). Molecular ions of some
compounds e.g. alcohols are unstable under electron bombardment and are weak or even absent. The peak
of highest m/z will not then be the molecular ion. In organic compounds, an odd molecular ion implies
that an odd number of N atom is present. Since there are is a small percentage of isotopes with higher
atomic weights present, a small M++1 and a smaller M++2 peak can be observed e.g. benzene will show
M+ at 78 and also peaks at m/z 79 and 80, 6.5% and 0.2% as intense as M+ due to 12C513CH6 and
12
C413C2H6. These peaks are of low intensity as 13C is found only to extent of 1.1% in natural carbon. But
isotope peaks are important for Br and C1.79Br and 81Br make up 50% each of natural bromine while 35Cl
is 75% and 37C1 25% of natural chlorine. CH3Br will therefore show M- peaks at m/z 94 and 96 (equal
intensity) and peaks, about 1% as intense, occur at m/z 95 and 97 due to 13C Isotopes.

Chemical Ionization:

The absence of a molecular ion in some ELMS spectra is a serious disadvantage as it would be difficult to
analyze fragmentations if M+is not known. Further, weak molecular ions are difficult to recognize as they
may be mistaken for MS peaks of an impurity. Chemical Ionization Mass Spectrometry (CIMS) can be
used to determine the M+ of compounds showing weak or absent molecular ion. In this technique, a
reagent gas usually methane, isobutene or ammonia is allowed to pass under pressure into the ion
chamber and ionized at about 300 eV. The molecular ion formed e.g.𝐶𝐻4+∙. in methane can undergo some
fragmentation and give 𝐶𝐻3+ and𝐻 ∙ . The ions produced at the high pressures can unlike in EI combine
with neutral counterparts. g.𝐶𝐻4+∙ reacts for example with methane to give g.𝐶𝐻5+ and g.𝐶𝐻3∙ . The net
result is that the reagent gas has neutral molecules with trapped𝐻 + like 𝐶𝐻5+and these act as strong acids
and can protonate sample molecules (M) to give MH+ ions. In the formation of these ions much of the
energy is used up and the resultant MH+ ions will show very little further fragmentation. The CIMS
spectrum will show a prominent MH+ peak and a few weak fragmentation peaks. It is therefore possible
to determine the m/z of the molecular ion of the sample from the CIMS experiment.

Fragmentation:

In hydrocarbons, cleavage takes place preferentially at branched carbon as carbocation stability decreases,
in the order IIIry>IIry>Iry.

The base peak of 2,2,4,4-tetramethylpentane, MeC(Me)2CH2C(Me)2CMe at m/z = 57 is due to C4H9+ (i.e.

MeC+Me2) ions formed by cleavage at both branched carbons.

Bonds ꞵ to double bonds, aromatic rings or heteroatoms are readily cleaved as the ions formed are
stabilized by resonance. Allyl carbonium ions are often formed.

e.g.

Cleavage takes place at benzyl carbon in aromatic compounds, as benzyl ion is resonance stabilized.
Compounds with a C6H5CH2 fragment in their structures always show an intense peak at m/z = 91 due to
the benzyl ion initially formed rearranging into the stable aromatic tropylium ion, which has 4n+2
electrons (Hückel's Rule).

Molecules containing heteroatoms like O, S, N are cleaved at the bonds 𝛽to these atoms. The parent ion
thus formed is also able to eliminate stable molecule like alkenes. H2O,CO, NH3, H2S HCN, alcohols, etc.
Similar eliminations of stable molecules can occur in olefins, alkylbenzenes, ethers and carbonyl
compounds.
e.g.

Phenols and benzyl alcohols

lose CO or CHO to give M-28 and
M-29 peaks due to the formation of ions which have 4n +2 electrons and are therefore aromatic.
e.g.

Aldehydes and ketones like other oxygen compounds lose one of their lone pairs of electrons to form a
molecular ion, which then undergoes α-cleavage to give an oxonium ion.
ꞵ-Cleavage can occur if a γ-hydrogen is present through the McLaferty rearrangement. Aldehydes show
intense M-1 peaks due to (R-C≡O+) and often a base peak at m/z 29 due to H-C≡O+. In carboxylic acids
and esters, both α and ꞵ cleavage can occur. ꞵ-cleavage occurs if a γ-hydrogen is present while α-cleavage
can give 4 different peaks as shown below.

Some of the characteristic mass losses caused by fragmentation and the formulae of some significant ions
are given in the following tables. An M-15 peak suggests the loss of a methyl from a branched position
while M-17 or M-18 peaks suggest the loss of water from an alcohol or a carboxylic acid. The loss of an
acetic acid residue gives an M-60 peak in the mass spectra of acetates and monocarboxylic acids. When a
carbonyl group is present, M-28 peaks are often produced while M-31 peaks are seen for the loss of OMe
from ethers and esters and for the loss of CH2OH from primary alcohols.
 Table 4: Characteristic mass losses through fragmentation from molecular ions

Ion Fragment lost Structure of fragmentation types indicated

M-1 H Aldehydes (some ethers and amines)

M-15 CH3 Methyl substituents
M-17, M-18 H2O Alcohols
M-28 C2H4, CO, N2 C2H4, McLafferty rearrangement, CO extrusion
from cyclic ketones
M-29 CHO, C2H5 Aldehydes, Ethyl substituents
M-31 OCH3, CH2OH Esters, ethers, primary alcohols
M-34 H2S Thiols
M-35, M-36 Cl, HCl Chlorides
M-43 CH3CO, C3H7 Methyl ketones, propyl substituents
M-45 COOH Carboxylic acids
M-60 CH3COOH Acetates
 TUTORIAL QUESTIONS

1. Given velocity of light = 3 × 1010 cm/sec, calculate energy of

i) UV radiation of wavelength 3 × 10-5 cm
ii) IR radiation of frequency 6×1013 Hz

2. Why are UV bands broad, unlike IR or NMR peaks?

3. Aniline has 𝜆max230 (E 8600) and 280 (E 1430) nm. In alkali, it shifts to 203 (E7500) and 254 (E
160). Explain.
 4. Calculate 𝜆maxfor following.

5. Usually 𝜆maxof trans-polyenes are greater than those of cis-polyenes. Why is this?

6. C6H5COCH2CONHC6H5 shows 𝜆max at 245 band 308 nm. In alkali, the 308 nm peak shifts to 323
nm. Explain.

7. 𝜆max for o-methylacetophenone are at 237 nm (E 10 700) and 245 nm (E 8360), m-

methylacetophenone at 241 nm (E 12400) and 248 (E 9400) and p-methylacetophenone at 246 nm
(E 18 000). Comment.
8. -C-O and -C-H vibrates at 𝜈=2100 and 2900 cm-1, while -C≡C- and -C=C- vibrates at 1600 and
2200 cm-1. Assign their frequencies.

9. Stretching vibrations require more energy than bending vibrations. Will they occur at higher or
lower wavenumbers in IR?

10. Two phenolic esters, C8H8O3 showed broad IR bands at 3300 cm-1 in CCl4. Dilution shifted the
bands to 3600 cm-1 and made it sharp in one case but had no effect in the other. Explain.

11. A 60 MHz NMR signal appears as a doublet at 𝛿 1.0 ppm (J=7 Hz). If a spectrum was run at 100
MHz, what will be the 𝛿 and J of the signal?
12. How many signals will there be in the 1H NMR spectra of each of the following? What will be their
relative positions from TMS and area ratio?

13. How would the NMR spectrum of CHCl2CH2CHCl2 be look like?

14. The proton NMR spectrum of the following compounds consist of 2 singlets. Predict the relative
intensities of the 2 peaks?

15. What will be the relative positions from TMS of the signals in the NMR spectra of the following.
Predict the relative intensities and multiplicities of the signals.

16. Predict the multiplicity, relative intensity and relative positions from TMS of the signals in,
17. Give the relative intensities, multiplicities, and relative positions from TMS of signals in,

18. Sketch the 1H NMR spectrum of

If JAM = 9 Hz, JAX = 5 Hz and JMX = 2 Hz, draw the splitting pattern of each signal.
What would be the NMR spectrum of the compound when double irradiation at frequency of a) Hx
and b) HM is carried out.

19. The aromatic bridge compound shown has signals for two protons at 𝛿 0.3 in its 1H NMR spectrum.
What is the signal due to? Explain.

20. What would be the NMR spectrum of very pure ethanol be like?

21. What is the structure of the compound, C6H12O which shows the following OFRD 13C NMR
spectrum: 𝛿22.6 (q), 42.8 (t), 56.0 (q), 68.1 (d), 115(d) and 131.8 (t)? Draw the 1H NMR and COSY
spectra you would expect for it.

22. Sketch a) the broad band decoupled and OFRD 13C NMR spectra and b)1H NMR and COSY spectra
you would expect for lactone A.
23. What is the structure of the hydrocarbon, which shows a mass spectrum: m/z 98 (M+, 4%), 83(8), 57
(70) and 41 (100).

24. Show the fragmentation responsible for the MS peak at m/z 57, 44 and 29 in 3-methylbutanal. What
would be the corresponding peaks in the MS of 2-methylbutanal.