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04/15/2304/15/23 INSTERUMENTAL ANALYSISINSTERUMENTAL ANALYSIS 11

Infra Red spectroscopy Infra 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.


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 to from the red end of the visible region to the micro wave region. the micro wave region. 14000 ― 20 cm14000 ― 20 cm-1-1νν 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.vis Absorption of radiation in the u.v.vis results in electronic transition, while results in electronic transition, while in the I.R. region it results in in the I.R. region it results in molecular transition that occur as molecular transition that occur as vibration energy, vibration energy,

Fundamental I.R.Fundamental I.R.


which could be of stretching or bending which could be of stretching or bending vibration (e.g CH3-CO-H)vibration (e.g CH3-CO-H)

Therefore the I.R. spectrum is a result of Therefore the I.R. spectrum is a result of transitions between two different transitions between two different vibrational energy levels. The vibrational vibrational energy levels. The vibrational motion of molecule resembles the motion 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 set If a spring with a ball attached to it is set into vibration, the frequency of vibration into 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 of can be calculated, which is a function of the mass of the ball (m) and the force the mass of the ball (m) and the force constant of the spring (k).constant of the spring (k).

Fundamental I.R.Fundamental I.R.


V α 1 ½ , V α kV α 1 ½ , V α k ½½

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


Examples Examples

ExampleExample(1)(1)-:-:

E.g. if a spring attached to a ball of a mass E.g. if a spring attached to a ball of a mass = 1gm, has a force constant= 1gm, has a force constant

K=4 x 10 K=4 x 10 -5-5 N/cm N/cm

What is the frequency (ν) of the system What is the frequency (ν) of the system when it is set into vibration? when it is set into vibration?


Note:- Note:-

1x101x10-5-5N = 1 gm.cm / S N = 1 gm.cm / S 22

4x104x10-5-5N=4 gm.cm/SN=4 gm.cm/S22


ExamplesExamplesNoteNote : :

1x101x10-5-5N = 1 gm.cm / SN = 1 gm.cm / S 22

4x104x10-5-5N=4 gm.cm/SN=4 gm.cm/S22

V = 1 4 x 10V = 1 4 x 10-5-5N/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 (system)In 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/cm N/cm

mm1 1 = 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 -5N/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 vibrational In a molecular system the vibrational frequency is a function of the mass of frequency 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 of Calculate the stretching vibrational frequency of the following chemical bond the 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 gm gm

H = 1.6 x 10 H = 1.6 x 10 -24-24 gm gm

O= 25.6 x 10 O= 25.6 x 10 -24-24 gm gm


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 (M)2Π reduced mass (M)


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


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


The difference ∆ Ε between two The difference ∆ Ε between two successive vibration energy levels can successive vibration energy levels can be calculated by the following equation:be calculated by the following equation:

∆ ∆ΕΕ = = ΕΕvib1vib1 – Ε – Ε vib0 vib0

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 hv hv


∆ ∆ 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 the Absorption of radiation is equal to the difference between 2 successive difference between 2 successive vibrational energy levels. Light of vibrational energy levels. Light of this kind of energy is found in the this kind of energy is found in the I.R. regionI.R. region..



Transition of the Transition of the zerozero ground level state ground level state to the next Lto the next L11 state, requires high state, requires high

energy intensity of the fundamental energy intensity of the fundamental bandband..

Transition from LTransition from Loo→L→L22 or →L or →L33

absorption is weak giving rise to absorption is weak giving rise to overtone.overtone.



V= 1 kV= 1 k

22 Π mΠ m

differencedifference ∆ Ε ∆ Ε = = = (v + = (v + ½)½) h v h v

differencedifference ∆ Ε ∆ Εvibvib= (v + = (v + ½)½) h k h k

2Π m2Π m


Functional GroupsFunctional Groups

a) Intensity of OH peak is ↑than the intensity of a) 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 the The 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 intensity c) 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 in The 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 decrease The 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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