Engineering chemistry SPECTROSCOPY 4.pdf

Chemistry

1. SPECTROSCOPY
Spectroscopy deals with the interaction of matter with
electromagnetic radiation
There are two types of spectra
Absorption spectrum and Emission spectrum
Absorption spectrum is obtained when a molecule is
excited from a lower energy level to a higher energy
level by the absorption of energy
Emission spectrum is obtained when a molecule falls
from a higher energy level to a lower energy level by
the emission of energy.

2. •Atoms give rise to atomic spectra when transition
occurs between their atomic energy levels
•Molecules give rise to molecular spectra when
transition occurs between their molecular energy
levels
Electromagnetic spectra is the entire range
overwhich electromagnetic radiation exists

4. Molecular energy levels
Molecule has different quantum mechanical energy states
1.Electronic energy level : It is associated with the electrons
present in the molecule which are present in various
molecular energy levels which can be bonding or anti bonding.
The gap between electronic energy levels falls in the UV-Vis
region of the electromagnetic spectrum
2.Vibrational energy level: Atoms in a molecule can vibrate by
keeping the position of centre of mass a constant. The energy
associated with this motion is called vibrational energy
The energy gap between vibrational energy levels falls in the
infrared region

5. 3. Rotational energy level: If a molecule rotates in
space about an axis passing through its centre of mass
it is said to possess rotational energy
Rotational energy is possessed by molecules of gas and
liquid.
The energy gap between rotational energy levels falls
in the microwave region.
4. Translational energy level: If the position of centre
of mass changes with time the molecule is said to
possess translational energy. It is not quantized

7. E
total = E
tran +E
rot+ E
vib +E
el
•E
el>>E
vib>>E
rot>>E
tran
Translational energy is negligible
∴ E
total = E
rot+ E
vib +E
el

8. Beer-Lamberts Law
•When a parallel beam of monochromatic
electromagnetic radiation is passed through a solution
of an absorbing substance of concentration ‘c’ the
rate of decrease in intensity (-dI) of radiation with
thickness of the absorbing medium (d ) is
proportional to the intensity of radiation (I) at that
point and also the concentration of the solution.
•-dI/d ∝ Ic
•-dI/I ∝ cd ; -dI/I = kcd --------------(1)
•Where k is the constant of proportionality

10. •Integrating equation (1)
ln(I
0 /I) = kc
2.303 logI
0/I = kc ;
logI
0/I = A = εcl; ε = k/2.303 →molar extinction coefficient or
molar absorptivity
A is the absorbance
Transmittance (T) is the ratio of the intensities of the
transmitted light to incident light (T = I/I
0)

11. Beer Lamberts Law is useful
•Identification of a substance
•Determination of concentration of a solution

12. Limitations of Beer Lambert’s Law
•Law is valid only in dilute solutions. In concentrated
solutions, due to electrostatic interactions between
molecules in close proximity law is invalid.
•Turbid solutions cannot be measured due to scattering
of light between particulates.
•Only monochromatic light can be employed
•Deviations from the law are observed during
association, dissociation or reaction of the absorbing
species with the solvent or if there is fluorescence or
phosphorescence in the sample

13. UV-Vis Spectroscopy (200-800 nm)
•Electrons of the molecule are excited from a lower energy
level to a higher energy level and vice versa
•Energy levels correspond to molecular orbitals. Transitions
are not obtained as discrete lines, because electronic
transitions are accompanied by large change in vibrational
energy which lead to the occurrence of a group of bands
•Vibrational energy changes are accompanied by large
change in rotational energy of the molecule which leads to
the formation of a series of lines in each bands.

14. Types of Electronic Transitions
Major electronic transitions are σ→σ*, π → π*,
n→ π* and n→σ*
σ→σ* transitions
It is a high energy process. Saturated hydrocarbons can
give only σ→σ* transitions of c-c σ bond and
C-H σ bond. This type of transition occurs below 150 nm.
The ordinary UV machines take spectra only from
200-700 nm and hence saturated hydrocarbons cannot
be detected

15. π → π* Transitions
This occurs in the unsaturated centres in the molecule
(double bonds, triple bonds, aromatics
Eg. of ethylene gives transition at 169 nm,
conjugated alkenes 170-190 nm, carbonyl compounds
180 nm
Alkenes , alkynes, carbonyl compounds, cyanides, azo
compounds etc. show π → π* transition.
In the case of 1,3-butadiene π → π* transition is
observed at 213 nm.

17. •The molecule has 4 π molecular orbitals (ψ
1,ψ
2,ψ
3,
ψ
4) which are formed by the linear combination of
four atomic orbitals.
•ψ
1,ψ
2 are BMOs and ψ
3, ψ
4 are
ABMOs
•Ψ
2 is the HOMO and ψ
3 is the LUMO. Transition
takes place from HOMO to LUMO

20. n → π* Transitions
•This is observed in aldehydes and ketones where
carbonyl group which has both π electrons and
non bonding electrons.
•These transitions require
The least amount of energy
•These transitions are observed with least intensity

21. n → σ* Transitions
•This transition occurs in saturated compounds
containing at least one heteroatom (Eg. S, N, O) with
unshared pair of electrons. (saturated halides,
alcohols, ethers, amines)
•These are also forbidden and occurs with low intensity
Water absorbs at 167 nm,
methanol at 174 nm,
CH
3Cl at 169 nm

22. •A chromophore is any isolated covalently bonded
group that shows characteristic absorption in the UV-
Vis region.
▪All the above groups contain at least a double or triple
bond or lone pair of electrons which are either loosely
bound or unbound which can be excited by absorbing a
small quantity of energy corresponding to visible
region or close to UV region

23. Auxochrome
Any group that cannot act as a chromophore but
whose presence brings about a shift of the
absorption band towards longer wavelength.
They extend the conjugation of the chromophore by
the sharing of the non bonding electrons
Example: -OH, -NH
2, -NHR, -NR
2, -SH, -X etc

24. •Benzene absorbs at 250 nm and aniline absorbs at
280 nm. Hence amino group is an auxochrome.
•

25. Instrumentation of UV
•Commercial instruments for UV has two light
sources a deuterium or hydrogen discharge tube
for the region 200-370 nm and a tungsten filament
lamp for the region 325-750 nm
•Ordinary spectrometers cover a range of 220-800
nm.
•The monochromator is a quartz prism or
diffraction grating

26. •Light coming from the source is dispersed with the
help of a prism and then selected by slits. The
selected beam is monochromatic.
•Most of the spectrometers are double beam
instruments.
•The selected beam is divided into two beams of
equal intensity

27. UV-Vis Spectrophotometer

28. •The two beams of exactly equal intensities are
made to pass through the sample cell and the
reference cells (containing the solvent).
•The two beams emerging from the sample and the
reference cells are lead to the detector system.
•The detector transmits the signal to a recorder
which gives the output in the form of a graph.
•The absorbance is recorded as a function of
wavelength and the graph is called absorption
spectrum.

29. Applications of UV-Visible spectroscopy
•Detection of aromatic compounds, conjugated dienes
•Characterization of dyes and colorants
•Determination of unknown concentration. This is done
by comparing the absorbance with absorbance of
standard solutions
•Study of kinetics of chemical reactions
•Quantitative estimation of blood sugar, cholesterol etc
•Detection of impurities (benzene is a common impurity
in cyclohexane which can be detected by the
absorption at 255 nm)

30. Q1. A dye solution of concentration 0.04 M shows absorbance of 0.045 at 530 nm,
while a test solution of the same dye shows absorbance 0.022 under the same
conditions. Find the concentration of the test solution.
•A = εcl
0.045 = ε x0.04 x l0.022 = ε xc xl
∴C = 0.0195 M
Q2. A solution of thickness 2 cm transmits 40 % incident light. Calculate
the concentration of the solution if ε = 6000 litre mol
-1
cm
-1
Sol: A = log(I
o/I) = εcl
Log100/40 = 6000 x c x 2
∴c = 0.3979/12000 = 3.316 x 10
-5
moll
-1

31. Q3. A monochromatic radiation is incident on a solution of 0.05 M concentration of an
absorbing substance. The intensity of the radiation is reduced to one fourth of the initial
value after passing through 10 cm length of the solution. Calculate the molar absorption
coefficient of the substance
•

32. Q5. A 0.01 M solution absorbs 10 % of an incident monochromatic light in a path
of 1 cm length. What would be the concentration of its solution if it is to absorb 90
% of the same radiation , also the same path length.
A = εcl C1 = 0.01 mol/litre
I
o = 100; I = 100-10 = 90
A
1 = log100/90 = 0.01 x ε x l
A
2 = log100/10 = c x ε x l = 1
∴c = 0.01/0.045 = 0.222

33. Q6. A solution of an organic dye in water absorb 20 % of an incident radiation in a
path length of 2 cm. What percentage of the same incident light would be
absorbed if the concentration of the solution is double for the same path length.
•Sol. A = log100/80 = ε x cx 2-------------(1)
•log100/x = ε x 2c x 2 -----------(2)
Solving equations 1 and 2,
log100/x = 2 x log100/80 = 0.1938
•100/x = Antilog(0.1938)
•X= 64
Percentage of light absorbed = 100-64 = 36

34. Infra red Spectroscopy
•This involves transition between vibrational energy levels
of a molecule.
•These transitions are brought by absorbing IR radiations in
the range 500-4000 cm
-1
.
•Vibrations are of two types-stretching and bending
vibrations
•Stretching vibration is a vibration along the bond axis such
that the distance between the two atoms is decreased or
increased
•Bending vibration deformation) brings change in bond
angles.

35. For interaction to be possible between the IR radiation
and the bonding system of the molecule
1The dipole moment of the bond must vary during
the vibration
2The frequency of the incident radiation must exactly
correspond to the frequency of the particular
vibrational mode
NO absorption results from stretching vibrations in a
homonuclear diatomic molecule
Or heteronuclear diatomic molecule like HCl stretching is
IR active as there is change in dm with change in
internuclear distance

36. Number of Vibrational modes
•For describing the position of atom in a molecule, it
requires 3 degrees of freedom corresponding to 3
cartesian coordinates. The number of coordinates
required to specify the position of all atoms ina
molecule is called the the number of degrees of
freedom. Hence if the number of atoms ina molecule is
‘n’, the no of degrees of freedom is 3n
Non-linear molecule
•The no of translational degrees of freedom = 3
•No. of rotational degrees of freedom = 3

37. ∴The no. of vibrational degrees of freedom = (3n-6)
Eg: H
2O, SO
2, NO
2 are non-linear triatomic molecules.
Hence no.of vibrational modes=(3X3)-6= 3
Linear Molecule
•The no of translational degrees of freedom = 3
No. of rotational degrees of freedom = 2 ( (There is
no rotation along the bond axis)
∴The no. of vibrational degrees of freedom = (3n-5)

38. Vibrational modes of carbondioxide
CO
2 is a linear molecule.
No. of Vibrational degrees of freedom of CO
2 is 3n-5 = 4

39. •Symmetric stretching is IR inactive but asymmetric stretching is
IR active
•The two bending modes are IR active as they involve a change
in dipole moment. The two bending modes are equivalent or
degenerate.
•So only two absorption frequencies are observed in the case of
CO2 (at 2349 cm
-1
for asymmetric stretching and 667 cm
-1
for
bending vibration)

40. Vibrational modes of water
•This is a non- linear molecule and so the number of vibrational
modes is equal to (3n-6) = 3x3-6 = 3
•These are symmetric stretching, asymmetric stretching and
bending

42. •All the three vibrational modes of water are
infrared active

43. A simple harmonic oscillator can be a
model for a vibrating diatomic molecule
•

44. •

45. •ΔE = h ν = (v’- v) hν
o
ΔV = ±1; But for absorption spectroscopy it should be +1
h ν = h νo ; ν = ν
o
The frequency of absorbed IR radiation ν is equal
to the fundamental vibrational frequency of the
molecule ν
o .
There should be only a single line in the vibrational spectra of diatomic molecule.
But
a
band is observed actually. This is due to the anharmonicity at high quantum
levels.

46. Applications of IR spectroscopy
Determination of force constant of diatomic molecules
Identification of functional groups in organic molecules
(from the stretching vibration of bonds present in the
functional group. For Eg.
Absorption range of in saturated ketone is
1700-1725 cm
-1
.
O-H group in alcohol is at 3300-3500 cm
-1
,
NH
2 group is at 3200-3300 cm
-1
,
C C in alkene is 1620-1650cm
-1
,
H-C in organic compounds is at 2900-3000 cm
-1
.

47. •Determination of purity. Impurities will give rise to
extra absorption bands
•To distinguish between inter and intra
molecular hydrogen boding.
The O-H stretching frequency (3300-3500 cm
-1
)
should not change on dilution in the case of
intramolecular hydrogen boding (o-hydroxyphenol)

48. •In parahydroxyphenol there exists inter molecular
hydrogen bonding. On dilution, the molecules get
separated, intermolecular hydrogen bonding
weakens and there is shift in O-H absorption
frequency

49. Q1. The CO molecule absorbs IR frequency of 2140 cm
-1
. Calculate the
force constant of the chemical bond, given that the atomic masses of C =
12 amu and O = 16 amu
•

50. Q2. If the fundamental vibrational frequency of HCl is 8.667 x 10
13
s
-1
,
calculate the force constant of HCl bond [H = 1.008; Cl = 35.45]
μ = (1.008 x 1.66 x 10
-27
kg x 35.45 x 1.66 x 10
-27
)
(1.008 x 1.66 x 10
-27
kg + 35.45 x 1.66 x 10
-27
)
= 1.627 x 10
-27
Kg
ν = 8.667 x 10
13
s
-1
K = 4π
2
ν
2
μ
k= 4 x 3.14²x(8.667x10¹
3
s⁻¹)
2
x(1.673 x10
⁻27
kg)= 482.086 kg s
⁻2
=
482.086kgms⁻
2
m⁻¹ = 482.086 Nm⁻¹

51. Q3. Calculate the frequency in hertz and cm
-1
of O-H bond if the force
constant and reduced mass of the atom pair are 770 Nm
-1
and 1.563
x 10
-27
kg respectively
•

52. Q4. The vibrational frequency of HCl molecule is 2886 cm
-1
. Calculate the
force constant of the molecule. Reduced mass of HCl is 1.63 x 10
-27
kg
•

53. •

54. Q7. The fundamental vibrational frequency of
12
C
16
O is 2140 cm
-1
. Without
calculating K find the fundamental frequency of
13
C
17
O in m
-1
.
• ῦ
1
/
ῦ
2
ῦ1

55. NMR Spectroscopy
All electrons and some nuclei possesses a property called spin. Let us
discuss the sort of spectra the spin can give rise to. All electrons have
a spin of ½.
1.Nuclei with both p and n even have zero spin
Eg:,
2He
4
,
6C
12
,
8O
16
etc
2. Nuclei with both p and n odd (charge odd but mass (p+n)
even have
integral spin
Eg:
2
H,
14
N (spin = 1),
10
B (spin = 3)
3. Nuclei with odd mass have half integral spin
Eg.
1
H,
15
N,
19
F,
31
P,
17
O,
13
C

56. •The spin of the nucleus (the spin quantum number) is given the
symbol I
•I can take values 0, ½, 1, 3/2, ----
•For proton I = ½
•The nuclear spin angular momentum quantum number can have
(2I+1) different orientations
•I, I-1, -----(-I+1), -I
I = ½ m
I = +1/2, -1/2
I = 1 m
I = +1, 0, -1
In the case of proton with I = ½, there can be two orientations ,
alignment with the field (+1/2) or alignment against the field (-1/2)

57. In the absence of a magnetic field all orientations of
the nuclear moment are degenerate. But in the
presence of an external magnetic field this
degeneracy will be destroyed

58. m
I = +1/2 (alignment with the field) will be lower in
energy. The energy difference ∆E is not very large
compared to thermal energies (kT)
The proton spins about its own axis. A charged
particle spinning about an axis constitutes a
circular electric field which in turn produces a
small magnetic moment and thus behaves like a
tiny bar magnet. When such a spinning proton is
placed in an external field the nucleus is in the
lower energy level.
The axis of the spin will precess around the
magnetic field just like the precessional motion of a
spinning top in a gravitational field

59. The frequency of precession is called Larmour frequency. When this
frequency becomes identical to the radiofrequency resonance occurs
and flipping of proton takes place, with the absorption of radiation.

61. •The energy difference between the two states is given
by ΔE = γhBo/2π = hν
•ν = γBo/2π, ν is the frequency of the radiation which
comes into resonance with the proton ν
•γ is the gyro magnetic ratio which is a fundamental
nuclear constant which is unique for a particular
nucleus

62. CHEMICAL SHIFT
The energy of resonance (The field strength required to
attain Larmour frequency) is dependent upon the
electronic environment about the nucleus.
When a molecule is placed in an external magnetic field
its electrons are caused to circulate and thus they
produce a secondary magnetic field.
The induced magnetic field may oppose or reinforce the
applied field

64. Shielding
If the induced magnetic field opposes the applied field, the net field
felt by the proton in a molecule will be less than the applied field.
The proton is now said to be shielded. The more shielded a proton
the greater must be the strength of the applied field in order to
achieve resonance with the radiofrequency. The more shielded
proton absorbs at higher field strength (upfield)

65. Deshielding
Whenthesecondarymagneticfieldproducedbytheelectrons
reinforcestheappliedfield,thenetfieldbytheprotoninthe
moleculewillbegreaterthantheappliedfieldandtheprotonis
saidtobedeshielded.(protonsinanaromaticring).Theseprotons
absorb at a a lower field strength (downfield)

67. Differentprotonsinamoleculegivesignalsatdifferentfield
strengths.Thisshiftintheresonancepositionofanucleusthat
results from its molecular environment is called its chemical shift.
Eg. CH
3-OCH
2CN
CH
3 and CH
2 protons give signals at different field strengths
Chemicalshiftmeasurementsarebasedontheresonancepositionof
astandard.ThesubstanceselectedforhydrogenNMRistetramethyl
silane (TMS); Si(CH
3
)
4

68. Siismoreelectropositivethancarbonandhenceit
pusheselectrondensitytowardsthemethylprotonsand
hencetheseprotonsaremoreshieldedthantheprotons
of most of the organic compounds
TheresonanceofTMSaresharpandintenseasallthe12
hydrogennucleiareequivalentandhenceabsorbat
exactly the same position. It is a low boiling substance
The chemical shift is measured in parts per million
Chemical shift = ν
sample-ν
TMS/operating frequency in MHz

69. The NMR spectra are displayed with the field increasing from left to
right which places TMS resonance to the extreme right
Twomeasurementunitsareusedinchemicalshift(δscaleandτscale).δ
scalesetsTMSresonanceat0ppmforboth
1
Hand
13
CNMR.TheTMS
resonanceisattheextremeright.Theδscalenumbersincreasesfrom
right to left in the direction of weaker shielding.
InτscaletheTMSresonanceisat10ppmandscalenumbersdecreases
to the left
When
1
HNMRspectrumistakenusinga100MHzinstrumentthesignalfor
theprotoninchloroform(CHCl
3)appearsat728HzdownfieldfromTMS
signal. Calculate the chemical shift
Chemical shift δ = 728-0/100 = 7.28 ppm

70. Factors influencing chemical shift
1As the electron density around the proton decreases shielding effect
decreases and delta value increases
The chemical shift values of CH
3 protons
Eg. CH
3F - 4.3 , CH
3Cl -3.1 , CH
3Br –, CH
3I – 2.2
CH
3F -4.3, CH
3-O-CH
3 – 3.2, (CH
3)
3N – 2.2, CH
3-CH
3 – 0.9
2Cumulative effect of electronegative substituents
CHCl
3 – 7.3, CH
2Cl
2 – 5.3 , CH
3Cl – 3.1
3Distance from the electronegative atom
CH
3-CH
2-CH
2Cl
1.0 1.4 3.4
4 Deshielding effect increases the delta value (Eg aromatic protons)

71. Magnetic Anisotropy of π system
Magnetic Anisotropy (Non-uniform magnetic field).
Electrons in the π system of the aromatic compounds ,
alkenes, alkynes etc interact with the applied magnetic
field that causes anisotropy. It causes both shielding
and deshielding. Eg benzene

72. Interpretation of Chemical Shift
1The number of signals tells how many different kinds of
protons are present
2The position of signal (δ value) tells about the nature of
protonic environment
3The intensity of signals measured by the area under each
peak tells us the relative ratio of different kinds of proton
CH
3-OCH
2CN

73. Spin-Spin coupling
Eachhydrogeninamoleculespinsandgeneratestheirown
magneticfield.Theprotonsonneighboringcarbonscan
generatemagneticfieldswhosemagneticmomentwill
interactwithmagneticmomentoftheexternalmagnetic
field. This results in the splitting of the NMR signal.
ThefinestructureintheCH
3andCH
2peaksarisesfromthe
phenomenon of spin- spin splitting.
Splittingarisesduetothecouplinginteractionsbetweenthe
neighboringprotonsandisdecidedbythenumberof
possiblespinorientationswhichtheseneighboringprotons
can adopt.

74. •The splitting of signals will give information about the
electronic environment of each of the proton

75. NMR spectrum of ethyl bromide
The CH
2 signal (a) is more deshielded than CH
3 signal (b) as it is attached to
the electronegative oxygen atom The CH
3 signal is split into a triplet under
the influence of CH
2 protons and the intensity ratio is 1:2:1. The CH
2 signal
is spit into a quartet under the influence of CH
3 protons and the intensity
ratio is 1:3:3:1

76. The distance between the peaks in a multiplet is called the coupling
constant and is represented by the letter j. It is a measure of the
effectiveness of spin-spin coupling

77. The CH
3 signal splits into a triplet under the influence of CH
2 protons in the intensity ratio
1:2:1. Similarly CH
2
signal splits into a quartet under the influence of CH
3
protons and the
intensity ratio is 1:3:3:1
Spin orientations of protons
CH
2 group which causes splitting of CH
3 signal
Spin orientations of protons CH
3 group which causes splitting of CH
2 signal

78. Predict the NMR spectrum 1,1dichloroethane
The NMR spectrum is characterized by a doublet centered at 2.1 ppm for the methyl
protons (CH
3) and a quartet centered at 5.9 ppm for CH proton. The CH
3 signal splits
into a doublet under the influence of C-H proton and the intensity ratio is 1:1. The spin
of the C-H proton can orient in two ways aligned with the magnetic field or aligned
against the magnetic field.
Similarly the C-H signal spits into a quartet under the influence of CH
3 protons in the
intensity ratio 1:3:3:1 . The spin of the CH
3 protons can align in four different ways

79. • If the signal is not split is called singlet
•If the signal is split into two peaks it is called a doublet
•If it is split into two three it is called triplet
•Four-quartet
•Five-pentet
•Six-sextet
•Seven-septet
•Eight-octet

80. NMR spectrum of CH
3-CHCl
2

82. NMR spectrum of CH
3-CHCl
2

83. The number of peaks into which the signal for a particular proton is split is
called the multiplicity
•The simple rule to predict the multiplicity of a signal is n+1 where n is
equal to the number of equivalent protons that are vicinal.
The peak intensity ratio is given by the Pascal’s triangle in general

84. Predict the NMR spectra of

86. The CH
3 signal is split into a triplet under the influence of CH
2 protons and
the intensity ratio is 1:2:1 & CH
2 signal is split into a quartet under the
influence of CH
3 protons and the intensity ratio is 1:3:3:1

87. CH
2 (c) signal is most deshielded due to the presence of electron withdrawing carbonyl group,
which is followed by CH
3
(d). CH
2
(c) is split int to a triplet under the influence of CH
2
(b) and the
intensity ratio is 1:2:1. CH
3
(d) signal is a singlet as it is not having a neighbouring proton. CH
3
(a) is
least deshielded and it is split into a triplet under the influence of CH
2
(b) and the peak intensity
ratio is 1:2:1. The CH
2
(b) is split into a sextet (6) as there are five neighbouring protons (CH3 (a)
and CH2 ©)and the intensity ratio is 1:5:10:10:5:1

88. There are two types of protons. Two CH
3 groups which are
identical (a) and CH
2 (b). The CH
2 signal will be more
deshielded than CH
3 protons and hence it has a higher delta
value. The CH
3 signal is a 6 proton triplet and the intensity
ratio is 1:2:1
The CH
2 signal is appearing as a 2 proton septet and the
intensity ratio is 1:6:15:20:15:6:1.

90. There three types of protons, belonging to CH
3 (a), CH
2(b)
and CH
3 © is most deshielded as it is attached to the
electronegative oxygen. It is a singlet. CH
2 signal has the
next higher value of δ and it is appearing as a quartet. (
intensity ratio 1:3:3:1). The CH
3
(a) is least deshielded
and it giving rise to a triplet (1:2:1)

94. There are three kinds of protons corresponding to CH
3, CH
2
and OH protons. OH proton is most deshielded due to the
presence of highly electronegative oxygen atom, followed by
CH
2 group and CH
3 is the least deshielded and hence has the
lowest value. The area ratio is (CH
3:CH
2:OH is 3:2:1)
Spin-spin splitting
CH
3 peak is split into a triplet under the influence of CH
2
protons and the peak intensity ratio is 1:2:1
The CH
2 signal is split into quartet under the influence of CH
3
and the intensity ratio is 1:3:3:1
The OH proton gives a singlet.

96. High resolution spectrum of ethanol
This is observed when ethanol is very pure. If traces of water are present in
ethanol , first exchange of proton takes place between water and ethanol and
splitting due to OH are not observed in CH
2
and vice versa
There are three kinds of protons corresponding to CH
3, CH
2 and OH protons. OH
proton is most deshielded due to the presence of highly electronegative oxygen
atom, followed by CH
2
group and CH
3
is the least deshielded and hence has the
lowest δ value. The area ratio is (CH
3
:CH
2
:OH is 3:2:1)
Spin-spin splitting
CH
3 peak is split into a triplet under the influence of CH
2 protons and the
peak intensity ratio is 1:2:1
The CH
2 signal is split into eight peaks (double quartet) under the influence
of CH
3 and OH protons. The coupling constant of OH proton is different
from CH
3 protons as the spin is transferred though oxygen in the former
case.
The OH proton gets spit into a triplet under the influence of CH
2 protons.

97. How many signals are observed in the
1
H NMR spectrum of Cl-CH
2-CH
2-Cl?
Substantiate your answer
•There is only one peak as the two CH
2 groups are equivalent. The
spectrum consists of a singlet due to the 4 hydrogen atoms as there is
no splitting of the CH
2 signals

98. Applications of NMR spectroscopy
•Magnetic Resonance Imaging (MRI)
MRI is a diagnostic procedure to visualize the detailed
internal structure of the body.
It is based on the fact that body tissue contains a large
amount of water and hence protons. When a strong
magnetic field is applied the protons get aligned. The
MRI machine creates a strong magnetic field around
the person.

99. •A radiofrequency current is briefly turned on
producing radio waves of the right frequency known
as the resonance frequency
•Radio waves are absorbed and flipping of the proton
spin takes place in the magnetic field
•When the radio frequency is turned off spin of the
proton returns to a thermodynamic equilibrium called
relaxation
•During relaxation a radio frequency signal is generated
from the body. This is called free induction decay
response signal (FID response signal)

100. •This signal can be measured with conductive coils
placed around the object being imaged
•It can recorded and mapped
•Relaxation times for molecules to regain their natural
alignment varies depending on the type of tissue being
imaged. Computer will record the amount of time
taken by the molecule to realign themselves

101. 2NMR is used for the structural elucidation of many
inorganic and organic compounds
3Quantitative analysis: The areas under the peaks are
directly proportional to the number of protons causing
the respective peaks. This is the basis of quantitative
analysis
4Detection of hydrogen bonding: Both inter and intra
molecular hydrogen bonding shift the absorption
position of H bonded proton to downfield.
Intermolecular H-bonding is concentration dependent
whereas intramolecular H-bonding is not concentration
dependent

102. 1Explain spin-spin splitting in the
1
HNMR spectrum of ethanol
2Which molecule will absorb at a longer wavelength in UV. Explain
3Compare the strengths of C-H bond and C=O bond if the absorption
frequencies are 3000 cm
-1
and 1700 cm
-1
respectively.

106. Instrumental methods
Thermal Analysis
Thermal analysis comprise a group of techniques in which a physical
property of a substance is measured as a function of temperature while the
substance is subjected to a controlled temperature program.
Thermogravimetric analysis (TGA)
This is a technique in which mass of a substance is monitored as function of
temperature, as the sample is heated from room temperature to a
temperature as high as 1200
o
C, in a controlled atmosphere.
As the temperature increases the sample may undergo physical or chemical
changes which will be accompanied by mass loss.

107. Thermogravimetric analysis (TGA)
•Thermal stability of the material and fraction of
volutile components can be studied
•The measurement is carried out in air or in inert
atmosphere (N
2, He, Ar etc)
•The weight is recorded as a function of
temperature
•The graph obtained is called TG thermogram (a
plot of mass vs temperature)

108. A Typical thermogram

109. •The horizontal portion AB indicates region where
there is no mass change
•This gives information about the temperature upto
which the material is stable.
•The slanting down portion BC indicates the region
which represent the weight loss due to dehydration,
dissociation etc.

110. •The components of the TGA apparatus are
1.Sample holder 2.furnace with temperature
programming facility 3. thermobalance 4.
temperature sensor 5. environment control
equipment 5. detector and recorder
The sample to be analyzed is taken in a sample holder
(~3 mg). The sample holder is surrounded by a furnace
with temperature programming facility. The heating
rate can be suitably adjusted according to the
requirement of the experiment.

111. •The environment control equipment provides suitable
atmosphere for analysis such as N
2, He etc.
•The sample holder is attached to a thermobalance
which automatically measures the mass of the sample
whenever the temperature changes. It is highly
temperature sensitive.
•The temperature sensor records the sample
temperature
The signals are amplified and recorded. The graph
obtained is a plot of mass vs temperature

112. Block Diagram of TGA Instrument

113. TGA of Calcium oxalate (CaC
2O
4.H
2O)
•Thermogravimetric analysis can be used for the
identification of inorganic compounds, polymers etc.

115. •Removal of water starts at 100
o
C and get completed
at 226
o
C.
•The horizontal portion between 226
o
C to 346
o
C
shows that anhydrous CaC
2O
4 is stable in this range
•Slightly above 346
o
C anhydrous CaC
2O
4 decomposes
to give CaCO
3. This process is completed at 420
o
C
•CaCO
3 is stable upto 660
o
C. Above 660
o
C it
decomposes to CaO and CO
2. This process is
completed at 840
o
C.
•The horizontal portion above this represents stable
CaO

116. •Removal of water starts at 100
o
C and get completed
at 226
o
C.
•The horizontal portion between 226
o
C to 346
o
C
shows that anhydrous CaC
2O
4 is stable in this range
•Slightly above 346
o
C anhydrous CaC
2O
4 decomposes
to give CaCO
3. This process is completed at 420
o
C
•CaCO
3 is stable upto 660
o
C. Above 660
o
C it
decomposes to CaO and CO
2. This process is
completed at 840
o
C.
•The horizontal portion above this represents stable
CaO

117. Study of the thermal stability of polymers
•TG thermogram of some polymers are given in the
figure. This gives valuable information regarding the
decomposition mechanism of polymers which can be
used for their identification
•PVC starts decomposing at low temperature
compared to LDPE. This is due to the elimination of
HCl from PVC
•PTFE is having high thermal stability owing to the
presence of strong C-F bond

119. Applications
Qualitative analysis
•Identification of substances and purity determination
•Stability of the substance at elevated temperatures
•Decomposition mechanism of polymers, inorganic salts etc
Quantitative analysis
•Amount of particular constituent in a sample of substance
can be estimated
•Amount of filler in a polymer can be estimated

120. Limitations of TGA
•TGA can be used only if there is a weight change
accompanying a physical change
•Eg. oxidation, decomposition, vaporization etc
•Phase changes like fusion, crystalline transition
where there is no weight change cannot be studied