PowerPoint Presentation:
DEPARTMENT OF CHEMISTRY , VEER NARMAD SOUTH GUJARAT UNIVERSITY, SURAT. Dr. RAKESH SHARMA Spectroscopy Infrared Spectra PowerPoint Presentation:
??????????????????? PowerPoint Presentation:
HISTORY OF IR Sir William Herschel PowerPoint Presentation:
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 PowerPoint Presentation:
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. PowerPoint Presentation:
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 PowerPoint Presentation:
Infra means beyond and infrared i.e. beyond red region PowerPoint Presentation:
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 PowerPoint Presentation:
Francis A. Carey, Organic Chemistry, Fourth Edition. Copyright © 2000 The McGraw-Hill Companies, Inc. All rights reserved. 2000 3500 3000 2500 1000 1500 500 Wave number, cm -1 Infrared Spectrum of Hexane CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 C—H stretching bending bending bending %T 100 PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
= 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 PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
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” PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
Bending Vibration In plane Scissoring Rocking Symbol s Symbol rho ρ PowerPoint Presentation:
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? PowerPoint Presentation:
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 ) PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
O-H alcohols or phenols 3200-3640 (b) C=C alkenes 1640-1680 (v) aromatic rings 1500 and 1600 (v) C≡C alkynes 2100-2260 (v) C-O primary alcohols 1050 (b) secondary alcohols 1100 (b) tertiary alcohols 1150 (b) phenols 1230 (b) alkyl ethers 1060-1150 aryl ethers 1200-1275(b) and 1020-1075 (m) all abs. strong unless marked: m, moderate; v, variable; b, broad PowerPoint Presentation:
IR spectra of ALKANES C—H bond “saturated” (sp3) 2850-2960 cm-1 + 1350-1470 cm-1 -CH2- + 1430-1470 -CH3 + “ and 1375 -CH(CH3)2 + “ and 1370, 1385 -C(CH3)3 + “ and 1370(s), 1395(m) PowerPoint Presentation:
n -pentane CH 3 CH 2 CH 2 CH 2 CH 3 3000 cm -1 1470 &1375 cm -1 2850-2960 cm -1 sat’d C-H PowerPoint Presentation:
CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 n -hexane PowerPoint Presentation:
2-methylbutane (isopentane) PowerPoint Presentation:
2,3- dimethylbutane PowerPoint Presentation:
cyclohexane no 1375 cm -1 no –CH 3 PowerPoint Presentation:
IR of ALKENES =C—H bond, “unsaturated” vinyl (sp 2 ) 3020-3080 cm -1 + 675-1000 RCH=CH 2 + 910-920 & 990-1000 R 2 C=CH 2 + 880-900 cis -RCH=CHR + 675-730 (v) trans -RCH=CHR + 965-975 C=C bond 1640-1680 cm -1 (v) PowerPoint Presentation:
1-decene 910-920 & 990-1000 RCH=CH 2 C=C 1640-1680 unsat’d C-H 3020-3080 cm -1 PowerPoint Presentation:
4-methyl-1-pentene 910-920 & 990-1000 RCH=CH 2 PowerPoint Presentation:
2-methyl-1-butene 880-900 R 2 C=CH 2 PowerPoint Presentation:
2,3-dimethyl-1-butene 880-900 R 2 C=CH 2 PowerPoint Presentation:
IR spectra BENZENE s =C—H bond, “unsaturated” “aryl” (sp 2 ) 3000-3100 cm -1 + 690-840 mono-substituted + 690-710, 730-770 ortho -disubstituted + 735-770 meta -disubstituted + 690-710, 750-810(m) para -disubstituted + 810-840(m) C=C bond 1500, 1600 cm -1 PowerPoint Presentation:
ethylbenzene 690-710, 730-770 mono- 1500 & 1600 Benzene ring 3000-3100 cm -1 Unsat’d C-H PowerPoint Presentation:
o -xylene 735-770 ortho PowerPoint Presentation:
p -xylene 810-840(m) para PowerPoint Presentation:
m -xylene meta 690-710, 750-810(m) PowerPoint Presentation:
styrene no sat’d C-H 910-920 & 990-1000 RCH=CH 2 mono 1640 C=C PowerPoint Presentation:
2-phenylpropene mono 880-900 R 2 C=CH 2 Sat’d C-H PowerPoint Presentation:
p -methylstyrene para PowerPoint Presentation:
IR spectra ALCOHOLS & ETHERS C—O bond 1050-1275 (b) cm -1 1 o ROH 1050 2 o ROH 1100 3 o ROH 1150 ethers 1060-1150 O—H bond 3200-3640 (b) PowerPoint Presentation:
1-butanol CH 3 CH 2 CH 2 CH 2 -OH C-O 1 o 3200-3640 (b) O-H PowerPoint Presentation:
2-butanol C-O 2 o O-H PowerPoint Presentation:
tert -butyl alcohol C-O 3 o O-H PowerPoint Presentation:
methyl n -propyl ether no O--H C-O ether PowerPoint Presentation:
2-butanone C=O ~1700 (s) PowerPoint Presentation:
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? PowerPoint Presentation:
n -propylbenzene PowerPoint Presentation:
isopropyl split 1370 + 1385 isopropylbenzene PowerPoint Presentation:
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 PowerPoint Presentation:
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 PowerPoint Presentation:
Which compound is this? 2-pentanone 1-pentanol 1-bromopentane 2-methylpentane 1-pentanol PowerPoint Presentation:
What is the compound? 1-bromopentane 1-pentanol 2-pentanone 2-methylpentane 2-pentanone PowerPoint Presentation:
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. PowerPoint Presentation:
1 PowerPoint Presentation:
2 PowerPoint Presentation:
3 PowerPoint Presentation:
4 PowerPoint Presentation:
5 PowerPoint Presentation:
Research Group THANK YOU:
THANK YOU