DEPARTMENT OF CHEMISTRY , VEER NARMAD SOUTH GUJARAT UNIVERSITY, SURAT. Dr. RAKESH SHARMA Spectroscopy Infrared Spectra

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HISTORY OF IR Sir William Herschel

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Spectroscopy is the branch of science dealing with the study of interaction of electromagnetic radiation with matter The most important consequence of such interaction is that energy is absorbed or emitted by the matter in discrete amount called quanta

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Electromagnetic Radiation • Electromagnetic radiation : light and other forms of radiant energy • Wavelength (λ): the distance between consecutive peaks on a wave. • Frequency (ν): the number of full cycles of a wave that passing through a fixed point in a second. • Wave number: The number of waves per centimeter.

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low high Frequency ( n ) Energy X-RAY ULTRAVIOLET INFRARED MICRO- WAVE RADIO FREQUENCY Ultraviolet Visible Vibrational infrared Nuclear magnetic resonance 200 nm 400 nm 800 nm 2.5 m m 15 m m 1 m 5 m short long Wavelength ( l ) high low THE ELECTROMAGNETIC SPECTRUM BLUE RED

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Infra means beyond and infrared i.e. beyond red region

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REGION WAVE LENGTH λ (μm) WAVE NUMBER υ (cm -1 ) FREQUENCY RANGE Hz NEAR 0.78 - 2.5 12800 - 4000 3.8x10 14 -1.2x10 14 MIDDLE 2.5 - 50 4000 - 200 1.2x10 14 - 6x1 12 FAR 50 - 1000 200 -10 6x10 12 - 30x10 11 MOST USED 2.5 - 15 4000 - 670 1.2x10 14 -2x10 13 IR-REGION: 12,800 - 10 cm -1

Use of Wave number than wavelength offer several advantages Wave number are directly proportional to frequency and are expressed in much more convenient numbers (in this region of the spectrum), 5000-500 cm -1 Because the wave number is directly proportional to frequency and energy, the use of wave numbers allows spectra to be displayed linear in energy , which is a distinct aid in sorting out related vibrational transitions

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n = 1 2 p c K m n : Frequency in cm -1 c : Velocity of light => 3 * 10 10 cm/s K : Force constant => dynes /cm m : masses of atoms in grams m = m 1 m 2 m 1 + m 2 = M 1 M 2 M 1 + M 2 (6.02 * 10 23 ) If we wish to express the radiation in wave number

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From above eq. IR measurements permits the evaluation of force constant for various types of chemical bond But generally above eq. can be used to estimate the wave number of the fundamental absorption peak, or the absorption peak due to the transition from the ground state to first excited state for a variety of bond types Now we calculate the approximate wave number and WL of the fundamental absorption peak due to the stretching vibration

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= 1 2 p c n K m larger K, higher frequency larger atom masses, lower frequency constants 2150 1650 1200 C=C > C=C > C-C = C-H > C-C > C-O > C-Cl > C-Br 3000 1200 1100 750 650 increasing K increasing m

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Calculating stretching frequencies Hooke’s law : n = 1 2 p c K m n : Frequency in cm -1 c : Velocity of light => 3 * 10 10 cm/s K : Force constant => dynes /cm m : masses of atoms in grams m = m 1 m 2 m 1 + m 2 = M 1 M 2 M 1 + M 2 (6.02 * 10 23 ) n = 4.12 K m C —C K = 5* 10 5 dynes/cm C =C K = 10* 10 5 dynes/cm C C K = 15* 10 5 dynes/cm

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Calculating stretching frequencies C =C K = 10* 10 5 dynes/cm n = 4.12 K m m = M 1 M 2 M 1 + M 2 = (12)(12) 12 + 12 = 6 n = 4.12 10* 10 5 6 = 1682 cm -1 n Experimental 1650 cm -1 C —H K = 5* 10 5 dynes/cm n = 4.12 5* 10 5 .923 = 3032 cm -1 m = M 1 M 2 M 1 + M 2 = (12)(1) 12 + 1 = 0.923 n Experimental 3000 cm -1 C —D K = 5* 10 5 dynes/cm n = 4.12 5* 10 5 .923 = 2228 cm -1 m = M 1 M 2 M 1 + M 2 = (12)(2) 12 + 2 = 1.71 n Experimental 2206 cm -1

SELECTION RULE:

SELECTION RULE 1) ABSORPTION OF CORRECT WAVELENGTH OF RADIATION (MATCHING OF FREQUENCY A molecule will absorb suitable radiation when its natural frequency of vibration matches with the frequency of incident radiation (i.e. a net transfer of energy takes place and it results in a change in the amplitude of molecular vibration ) Natural frequency of HCl molecule is 8.7 X 10 13 Hz (vib/sec)or(2890 cm -1 ) When IR radiation will allow to pass on HCl sample and transmitted radiation is analyzed . It is observed that part of radiation which has same frequency 8.7X10 13 HZ is absorb. Thus remaining has been transmitted so gives characteristic value of HCl

2) DIPOLE MOMENT :

2) DIPOLE MOMENT “A molecule will absorb IR radiation if the change in the vibrational state is associated with the change in the dipole moment of the molecule” μ = q x r A dipole moment arises from a separation of charges in a molecule : μ =dipole moment (Coulomb ·meters) q =magnitude of charges r = Distance between charges

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H Cl δ + δ - r A hetronuclear diatomic molecule is composed of 2 diff. atoms. if this atoms exhibit diff. electron withdrawing powers , the e - density will be shifted towards more electronegative atom such molecule are said to be polar and possesses an electric dipole moment OR “We can say, dipole moment arise as a consequence of asymmetrical partial charge distribution”

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μ = q x r q = magnitude of charge r = Distance between charges Here charge is measured in coulomb & distance in meters So SI unit of μ = Cm (i.e. coulomb meter) but for convenience μ is often given in unit “DEBYE” 1D = 3.336X10 -30 Cm but atomic charge is q x e μ = q x e x r q + q - δ + δ - r e = 1.602 X 10-19 C

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If HCl has dipole moment 1.83D bond length And r is 92 pm from eq. q = 0.41 Then this value indicates that charge in HCl molecule is distributed (asymmetrically ) such that Cl atom has effectively gained 0.41 of e - and H atom has lost 0.41of an electron. In polyatomic molecule, the total dipole moment is the vector sum of the dipole moment of the individual bond. In symmetrical molecule such as CCl 4 there is no overall dipole moment ,althoug C Cl bond is polar from μ = q x e x r

3)SELECTION RULE OF VIBRATIONAL QUANTUM NUMBER:

3)SELECTION RULE OF VIBRATIONAL QUANTUM NUMBER Based on the harmonic oscillator model (Δv = ±1) Here + absorption & - emission or Vibrational quantum number change by unity

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Harmonic oscillator Anharmonic oscillator Overtones and fundamental band

TYPES OF MOLECULAR VIBRATION:

TYPES OF MOLECULAR VIBRATION Stretching vibration Symmetric Stretching Asymmetric Stretching Symbol nu Symbol nu Isolated vibration Without change bond axis Without change bond angle

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Bending Vibration In plane Scissoring Rocking Symbol s Symbol rho ρ

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Out of plane Twisting Wagging Symbol ω Symbol tau,τ Bending Vibration

Number of possible vibrational modes:

Number of possible vibrational modes CAN WE KNOW THE POSSIBLE VIBRATION? YES, but how? 3N-5 for linear molecules 3N-6 nonlinear molecules N: number of atoms in a molecules BUT WHAT IS 3N, 5 & 6. HOW’S IT COME?

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Atoms are never fixed in the space but move about continuously Each atom may be said to posses three degrees of freedom of movement , and thus in N-atom (polyatomic molecule) Molecule there will be 3N degrees of freedom (That means 3 coordinates are needed to locate a point in space. And each coordinate corresponds to one degree of freedom for one of the atom ,so a molecule containing N-atom is said to have 3N degree of freedom )

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From this 3 types of motion 1) TRANSLATION MOTION Definition of translation motion require 3 coordinates and thus this common motion requires 3 of the 3N degree of freedom

2)ROTATIONAL MOTION:

Another 3 degree of freedom are needed to describe the rotational motion of the molecule. (i.e. poly atomic molecule has generally 3 degree of rotation freedom) But in the special case of linear molecule all the atom lie on a straight line and only 2 rotation can be define (because rotation around the bond axis is not possible) 2)ROTATIONAL MOTION

3)VIBRATIONAL MOTION:

Now the rest i.e. substrate transition & rotation motion from 3N degree of freedom i.e. 3N-6 ( for non linear molecule) 3 of translation motion + 3 of rotational motion=6 & 3N-5(for linear molecule) 3 of translation motion + 2 of rotational motion=5 3)VIBRATIONAL MOTION

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Now out of this normal vibration no. of stretching and bending vibration can be calculated For stretching vibration =N -1 For bending vibration =2N -5(non linear) i.e. [(3N - 6)-(N -1)]=2N -5 =2N-4(linear) i.e. [(3N - 5)-(N -1)] =2N – 4 Examples: O 2 , N 2 , Cl 2 : 3N-5 =3x2-5 = 1, only one vibration mode CO 2 3N-5 =3x3-5 = 4(linear molecule) H 2 O 3N-6 =3x3-6 = 3

HYDROGEN BONDING :

HYDROGEN BONDING INTERMOLECULAR H - BONDING INTRAMOLECULAR H - BONDING o - Nitro phenol Formic acid dimer

How intermolecular and intramolecular h-bonding can be differentiate with the use of IR spectra?:

How intermolecular and intramolecular h-bonding can be differentiate with the use of IR spectra? The inter and intramolecular h-bonding can be distinguished by dilution . The intramolecular H-bonding is independent of the conc. Because it is an internal effects and hence intramolecular H-bonds remains unaffected on dilution and so absorption band also remains unaffected giving bonded O-H absorption But intermolecular H-bonding is dependent on dilution the H-bonds are broken on dilution (or at low conc.) and hence there is decrease in the bonded O-H absorption on successive dilution the intensities of the bands due to intermolecular H-bonding gradually decrease and finally disappear

INSTRUMENTATION:

INSTRUMENTATION

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SOURCE SAMPLE REFRENCE TRANSDUCER MC Infrared Spectrum

INTERPRETATON:

INTERPRETATON The IR region ( 4000-670 cm-1) is of great importance in studying of organic molecule It is divided in to 2 region 1)Function group region: 4000-1300 cm-1 2)Finger print region : 1300-670 cm-1 The higher frequency group region is called the function group region It is useful to identify group.

FINGER PRINT REGION:

FINGER PRINT REGION 1300-670 cm -1 is called fingerprint region because the pattern of absorption in this region is unique to any particular comp. Each org. comp. has its own unique absorption in this region Although similar type of comp. show similar spectra in the function group region but there is always clear difference in the fingerprint region. (analogue to fact that finger print of a person is always different than another person) The pattern of the IR spectra in the finger print region are very sensitive and change even with minor chemical or stereo chemical alteration in molecule

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There are no rigid rule for interpreting an IR spectrum. But some strategies for interpreting IR spectra. 1) The spectrum must be adequately resolved. 2) instrument should be calibrated 3) a good rule to remember that always place more reliance upon the negative evidence . The absence of band in a particular region is a sure indication of the absence of group absorbing in that region. 4) To distinguish between inter and intra H bonding. spectra of the sample are scanned at two different concentration. IMP

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Some characteristic infrared absorption frequencies BOND COMPOUND TYPE FREQUENCY RANGE, cm-1 C-H alkanes 2850-2960 and 1350-1470 alkenes 3020-3080 (m) and RCH=CH2 910-920 and 990-1000 R2C=CH2 880-900 cis -RCH=CHR 675-730 (v) trans -RCH=CHR 965-975 aromatic rings 3000-3100 (m) and monosubst. 690-710 and 730-770 ortho -disubst. 735-770 meta -disubst. 690-710 and 750-810 (m) para -disubst. 810-840 (m) alkynes 3300

C 9 H 12 C-H unsat’d & sat’d 1500 & 1600 benzene mono C 9 H 12 – C 6 H 5 = -C 3 H 7 isopropylbenzene n -propylbenzene?

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n -propylbenzene

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isopropyl split 1370 + 1385 isopropylbenzene

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C 8 H 6 C-H unsat’d 1500, 1600 benzene mono C 8 H 6 – C 6 H 5 = C 2 H phenylacetylene 3300 C-H

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C 4 H 8 1640-1680 C=C 880-900 R 2 C=CH 2 isobutylene CH 3 CH 3 C=CH 2 Unst’d

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Which compound is this? 2-pentanone 1-pentanol 1-bromopentane 2-methylpentane 1-pentanol

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What is the compound? 1-bromopentane 1-pentanol 2-pentanone 2-methylpentane 2-pentanone

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In a “matching” problem, do not try to fully analyze each spectrum. Look for differences in the possible compounds that will show up in an infrared spectrum.

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Research Group

THANK YOU:

THANK YOU

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