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04/15/1504/15/15 INSTERUMENTAL ANALYSISINSTERUMENTAL ANALYSIS 11
Infra Red spectroscopyInfra Red spectroscopy
OutlineOutline
1.1. Infra Red, ReviewInfra Red, Review
2.2. Fundamental I.R.Fundamental I.R.
3.3. ExamplesExamples
4.4. Functional GroupsFunctional Groups
5.5. Differences between I.R. and U.V.Differences between I.R. and U.V.

04/15/15 INSTERUMENTAL ANALYSIS2
Infra Red, ReviewInfra Red, Review
 Definition –Definition – Infra Red:Infra Red: (I.R): extends(I.R): extends
from the red end of the visible region tofrom the red end of the visible region to
the micro wave region.the micro wave region.
 14000 ― 20 cm14000 ― 20 cm-1-1
νν
 0.7 ― 500 µ λ0.7 ― 500 µ λ

Infra Red, ReviewInfra Red, Review
 I.R. is rather qualitative than quantitative.I.R. is rather qualitative than quantitative.


I.RI.R


Near I.R Fundamental I.R Far I.RNear I.R Fundamental I.R Far I.R

Fundamental I.R.Fundamental I.R.
It is the range of greatest use.It is the range of greatest use.
(Middle I.R. region(Middle I.R. region((
υ =4000 -200 cmυ =4000 -200 cm-1-1
λ = 2.5 – 50 µλ = 2.5 – 50 µ

 Absorption of radiation in the u.v.visAbsorption of radiation in the u.v.vis
results in electronic transition, whileresults in electronic transition, while
in the I.R. region it results inin the I.R. region it results in
molecular transition that occur asmolecular transition that occur as
vibration energy,vibration energy,

which could be of stretching or bendingwhich could be of stretching or bending
vibration (e.g CH3-CO-Hvibration (e.g CH3-CO-H((
Therefore the I.R. spectrum is a result ofTherefore the I.R. spectrum is a result of
transitions between two differenttransitions between two different
vibrational energy levels. The vibrationalvibrational energy levels. The vibrational
motion of molecule resembles themotion of molecule resembles the
motion of a ball attached to a springmotion of a ball attached to a spring

If a spring with a ball attached to it is setIf a spring with a ball attached to it is set
into vibration, the frequency of vibrationinto vibration, the frequency of vibration
(No# of oscillations/sec( of the system(No# of oscillations/sec( of the system
can be calculated, which is a function ofcan be calculated, which is a function of
the mass of the ball (m( and the forcethe mass of the ball (m( and the force
constant of the spring (k(.constant of the spring (k(.

V α 1 ½ , V α kV α 1 ½ , V α k½½
mm
m ↓v , ↑K ↑Vm ↓v , ↑K ↑V↑↑
K=force constantK=force constant
Hooke's lawHooke's law
V= 1 k/mV= 1 k/m
2Π2Π

ExamplesExamples
ExampleExample(1(1(:-(:-
E.g. if a spring attached to a ball of a massE.g. if a spring attached to a ball of a mass
= 1gm, has a force constant= 1gm, has a force constant
K=4 x 10K=4 x 10 -5-5
N/cmN/cm
What is the frequency (ν) of the systemWhat is the frequency (ν) of the system
when it is set into vibration?when it is set into vibration?

 Note:-Note:-
1x101x10-5-5
N = 1 gm.cm / SN = 1 gm.cm / S 22
∴∴ 4x104x10-5-5
N=4 gm.cm/SN=4 gm.cm/S22

ExamplesExamples
NoteNote::
1x101x10-5-5
N = 1 gm.cm / SN = 1 gm.cm / S22
∴∴ 4x104x10-5-5
N=4 gm.cm/SN=4 gm.cm/S22

V =  1      4 x 10V =  1      4 x 10-5-5
N/cmN/cm
            2Π         1 gm2Π         1 gm


    =        =    11      44gm.cm/cmgm.cm/cm
                22Π       gm.SΠ       gm.S22

   =     =  11        44  x Sx S-2-2
          22ΠΠ

  =    =  11    SS-1-1
= 0.033 cycle/sec = 0.033 cycle/sec
        ΠΠ

= = 11  cycle/3seccycle/3sec..
= = 11  cycle/3seccycle/3sec..

Example (2):Example (2):
In case of a two ball (systemIn case of a two ball (system)?)?
What is the frequency of the systemWhat is the frequency of the system??
K = 5 x 10K = 5 x 10-5-5
N/cmN/cm
mm11 = 1gm , m= 1gm , m22 = 1 gm= 1 gm
mm11 m m22
Reduced mass =Reduced mass = m m11+m+m22 = M= M

V  =   1       kV  =   1       k
              2Π    m2Π    m11mm22         M         M
                                                            mm11+m+m22

      V =  1    5 x 10V =  1    5 x 10 -5 -5
N/cmN/cm
            2Π2Π          m          m11mm22
                                        mm11+m+m22

 M= reduced mass = 1/ (1+1( = ½=0.5 gmM= reduced mass = 1/ (1+1( = ½=0.5 gm
      V =  1      k/mV =  1      k/m
                  2Π2Π

      V =  1  5gm.cm/cmV =  1  5gm.cm/cm
            2Π     0.5gm.S2Π     0.5gm.S22

 V =  1  5V =  1  5gm.cm/cmgm.cm/cm
                  2Π   0.52Π   0.5gm.Sgm.S22

= = 0.50.5  cycle per secondcycle per second
Or 1cycle in two secondOr 1cycle in two second

 In a molecular system the vibrationalIn a molecular system the vibrational
frequency is a function of the mass offrequency is a function of the mass of
the atoms and the bond strength.the atoms and the bond strength.
∴∴ The same equation will be usedThe same equation will be used
V = 1 KV = 1 K
2Π reduced mass2Π reduced mass

Example:Example:
Calculate the stretching vibrational frequency ofCalculate the stretching vibrational frequency of
the following chemical bondthe following chemical bond
C-H, O-H and C=C if the masses areC-H, O-H and C=C if the masses are
C = 20 x 10C = 20 x 10-24-24
gmgm
H = 1.6 x 10H = 1.6 x 10 -24-24
gmgm
O= 25.6 x 10O= 25.6 x 10 -24-24
gmgm

AndAnd
K of C-H=5 N/cmK of C-H=5 N/cm
K of O-H=5 N/cmK of O-H=5 N/cm
K of C=C=1 x 10 N/cmK of C=C=1 x 10 N/cm
V = 1 KV = 1 K
2Π reduced mass (M2Π reduced mass (M((

 ∴∴For C-H, v=9.3 x 10For C-H, v=9.3 x 101313
ss-1-1
(cps)(cps)
 For O-H, v=9.2 x 10For O-H, v=9.2 x 101313
ss-1-1
(cps(cps
 For C=C, v=5.1 x 10For C=C, v=5.1 x 101313
ss-1-1
(cps)(cps)

 EE vibrationvibration = (v= (v--
+ ½) hv+ ½) hv
 vv--
= Quantum #.0,1,2,3,…,etc= Quantum #.0,1,2,3,…,etc
 h= plank's constanth= plank's constant
 v= frequencyv= frequency


The chemical bond differs from the 2The chemical bond differs from the 2
ball system in that only certainball system in that only certain
vibrational energy is considered, e.gvibrational energy is considered, e.g
the vibrational energy is quantized.the vibrational energy is quantized.

The difference ∆ Ε between twoThe difference ∆ Ε between two
successive vibration energy levels cansuccessive vibration energy levels can
be calculated by the following equation:be calculated by the following equation:
∆∆ΕΕ == ΕΕvib1vib1 – Ε– Εvib0vib0
But ΕBut Εvib1vib1 = (v= (v--
++ ½)½)hvhv
For ΕFor Εvib0vib0= (0 + ½(hv = ½hv= (0 + ½(hv = ½hv
For ΕFor Εvib1vib1= (1 + ½( hv = ³/= (1 + ½( hv = ³/22 hvhv

∴∴∆∆ Ε ˉv1 –v0 =³/2 hv - ½ hv = 1 hvΕ ˉv1 –v0 =³/2 hv - ½ hv = 1 hv
Absorption of radiation is equal to theAbsorption of radiation is equal to the
difference between 2 successivedifference between 2 successive
vibrational energy levels. Light ofvibrational energy levels. Light of
this kind of energy is found in thethis kind of energy is found in the
I.R. regionI.R. region..

Transition of theTransition of the zerozero ground level stateground level state
to the next Lto the next L11 state, requires highstate, requires high
energy intensity of the fundamentalenergy intensity of the fundamental
bandband..
Transition from LTransition from Loo→L→L22 or →Lor →L33
absorption is weak giving rise toabsorption is weak giving rise to
overtone.overtone.

V= 1 kV= 1 k
22Π mΠ m
differencedifference ∆ Ε∆ Ε == = (v += (v + ½)½) h vh v
differencedifference ∆ Ε∆ Εvibvib= (v += (v + ½)½) h kh k
2Π m2Π m

Functional GroupsFunctional Groups
 a) Intensity of OH peak is ↑than the intensity ofa) Intensity of OH peak is ↑than the intensity of
NH peak.NH peak.
b) free OH occurrs at 3750-3500 cm-1.b) free OH occurrs at 3750-3500 cm-1.
 Linked –OH bond at 3450-3200 cm-1.Linked –OH bond at 3450-3200 cm-1.
 The intensity of the free OH is less than theThe intensity of the free OH is less than the
intensity of the linked OH bond.intensity of the linked OH bond.

 Free NH occurs at 3400-3200 cm-1.Free NH occurs at 3400-3200 cm-1.
 Linked NH occurs at 3500-3000 cm-1.Linked NH occurs at 3500-3000 cm-1.
c) NH peaks are weaker than OH peaks in intensityc) NH peaks are weaker than OH peaks in intensity
and so appear sharper.and so appear sharper.

 -C-H aromatic 3200 -3000 cm-1-C-H aromatic 3200 -3000 cm-1
 -C-H aliphatic 3000-2850 cm-1.-C-H aliphatic 3000-2850 cm-1.
 -C≡N 2250 cm-1.-C≡N 2250 cm-1.
 -C-O-C-O
O 1670 cm-1O 1670 cm-1
 -C-OH-C-OH

 *ethylene CH-bending region 660-960*ethylene CH-bending region 660-960
 *substitution on Benzene ring*substitution on Benzene ring
 mono subs. 750-700 cm-1.mono subs. 750-700 cm-1.
 Di – subs. 750 (ortho)Di – subs. 750 (ortho)
 Meta-subs. 810-780 cm-1.Meta-subs. 810-780 cm-1.
 Para – subs. 850-800 cm-1.Para – subs. 850-800 cm-1.

Differences between I.R. and U.V.Differences between I.R. and U.V.
 * Intensity of absorption in U.V. is 105μ but I.R.* Intensity of absorption in U.V. is 105μ but I.R.
it is 103μ.it is 103μ.
 *I.R. shows the peaks of many functional groups,*I.R. shows the peaks of many functional groups,
but U.V. shows only one peak.but U.V. shows only one peak.
 *I.R. is rather qualitative where as U.V. is*I.R. is rather qualitative where as U.V. is
quantitative.quantitative.

Differences between I.R. and U.V.Differences between I.R. and U.V.
 The thickness of the cell in U.V≈1cm, where inThe thickness of the cell in U.V≈1cm, where in
I.R. the cell is less in thickness.I.R. the cell is less in thickness.
 The width of the cell in I.R. is also decreaseThe width of the cell in I.R. is also decrease
 in order to increase I.R. intensity.in order to increase I.R. intensity.

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