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INDEX OF HYDROGEN DEFICIENCY THE BASIC THEORY OF THE BASIC THEORY OF INFRARED SPECTROSCOPY and.

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Presentation on theme: "INDEX OF HYDROGEN DEFICIENCY THE BASIC THEORY OF THE BASIC THEORY OF INFRARED SPECTROSCOPY and."— Presentation transcript:

1 INDEX OF HYDROGEN DEFICIENCY THE BASIC THEORY OF THE BASIC THEORY OF INFRARED SPECTROSCOPY and

2 WHAT CAN YOU LEARN FROM A MOLECULAR FORMULA ? MOLECULAR FORMULA ? YOU CAN DETERMINE THE NUMBER OF RINGS AND / OR DOUBLE BONDS.

3 Saturated Hydrocarbons CH 4 C 5 H 12 C3H8C3H8 C 4 H 10 C2H6C2H6 C 9 H 20 C n H 2n+2 GENERAL FORMULA branched compounds also follow the formula

4 -2H -4H -2H FORMATION OF RINGS AND DOUBLE BONDS Formation of each ring or double bond causes the loss of 2H.

5 Determine the expected formula for a noncyclic, saturated compound ( C n H 2n+2 ) with the same number of carbon atoms as your compound. Correct the formula for heteroatoms Subtract the actual formula of your compound The difference in H’s divided by 2 is the Index of Hydrogen Deficiency CALCULATION METHOD (explained later) Index of Hydrogen-deficiency

6 C5H8C5H8 C 5 H 12 C5H8C5H8 H4H4 Index = 4/2 = 2 double bond and ring in this example = ( C n H 2n+2 ) Two Unsaturations

7 O or S -- doesn’t change H in calculated formula N or P -- add one H to the calculated formula F, Cl, Br, I -- subtract one H from calculated formula Index of Hydrogen Deficiency C-HC-X C-HC-O-H C-HC-NH 2 +N,+H -H,+X +O CORRECTIONS FOR ATOMS OTHER THAN HYDROGEN

8 C4H5NC4H5N C 4 H 10 C 4 H 11 N H6H6 Index = 6/2 = 3 two double bonds and ring in this example = ( C n H 2n+2 ) add one H for N C 4 H 5 N

9 The index gives the number of one ring and the equivalent of three double bonds gives an index of 4 If index = 4, or more, expect a benzene ring double bonds or triple bonds or rings in a molecule Benzene

10 PROBLEM A hydrocarbon has a molecular formula of C 6 H 8. It will react with hydrogen and a palladium catalyst to give a compound of formula C 6 H 12. Give a possible structure. INDEX C 6 H 14 -C 6 H 8 H6H6 Index = 6/2 = 3 C6H8C6H8 +2 H 2 C 6 H 12 HYDROGENATION Pd Hydrogenation shows only two double bonds. Therefore, there must also be a ring.

11 A FEW POSSIBLE ANSWERS..... there is still work required to fully solve the problem

12 INFRARED SPECTROSCOPY

13 lowhigh Frequency ( ) Energy X-RAYULTRAVIOLETINFRARED MICRO- WAVE RADIOFREQUENCY Ultraviolet Visible Vibrational infrared Nuclear magnetic resonance 200 nm 400 nm800 nm 2.5  m15  m 1 m5 m short long Wavelength ( ) highlow THE ELECTROMAGNETIC SPECTRUM BLUERED

14 X-ray UV/Visible Infrared Microwave Radio Frequency Bond-breaking Electronic Vibrational Rotational Nuclear and Electronic Spin REGIONENERGY TRANSITIONS Types of Energy Transitions in Each Region of the Electromagnetic Spectrum of the Electromagnetic Spectrum (NMR)

15 Detection Electronics and Computer Infrared Source Determines Frequencies of Infrared Absorbed and plots them on a chart Sample Simplified Infrared Spectrophotometer NaCl plates Absorption “peaks” Infrared Spectrum frequency intensity of absorption (decreasing) focusing mirror

16 4-Methyl-2-pentanone C-H < 3000, 1715 cm -1 KETONE WAVELENGTH (cm-1) %TRANSMITTANCE%TRANSMITTANCE AN INFRARED SPECTRUM

17 ( ) = wavenumbers (cm -1 ) = 1 (cm) = wavelength (cm) THE UNIT USED ON AN IR SPECTRUM IS = frequency = c c = speed of light WAVENUMBERS (  ) c = 3 x cm/sec wavenumbers are directly proportional to frequency = = or c 1 cm/sec cm = sec 1 c

18 Two major types : STRETCHING BENDING C C C C Molecular vibrations both of these types are “infrared active” ( excited by infrared radiation )

19 BONDING CURVES AND VIBRATIONS AND VIBRATIONS MORSE CURVES STRETCHING

20 r avg decreasing distance energyenergy r min r max zero point energy  (average bond length) BOND VIBRATIONAL ENERGY LEVELS MORSE CURVE

21 bond dissociation energy r avg distance energyenergy r min r max zero point energy vibrational energy levels (average bond length) BOND VIBRATIONAL ENERGY LEVELS Bonds do not have a fixed distance. They vibrate continually even at 0 o K (absolute). The frequency for a given bond is a constant. Vibrations are quantized as levels. The lowest level is called the zero point energy.

22 FREQUENCY (cm -1 ) WAVELENGTH (  m) O-HC-H N-H C=OC=N Very few bands C=C C-Cl C-O C-N C-C X=C=Y (C,O,N,S) C N C Typical Infrared Absorption Regions (stretching vibrations) (stretching vibrations) N=O * * nitro has two bands

23 MATHEMATICAL DESCRIPTION OF THE OF THE VIBRATION IN A BOND VIBRATION IN A BOND …. assumes a bond is like a spring HARMONIC OSCILLATOR

24 HOOKE’S LAW x0x0 x1x1 xx K -F = K(  x) m1m1 m2m2 K Molecule as a Hooke’s Law device restoring force = stretch compress force constant

25 Harmonic Oscillator Morse Curve (anharmonic) THE MORSE CURVE APPROXIMATES AN HARMONIC OSCILLATOR HOOKE’S LAW ACTUAL MOLECULE Using Hooke’s Law and the Simple Harmonic Oscillator approximation, the following equation can be derived to describe the motion of a bond…..

26 = 1 2c2c K   = m 1 m 2 m 1 + m 2 = frequency in cm -1 c = velocity of light K = force constant in dynes/cm m = atomic masses SIMPLE HARMONIC OSCILLATOR >> multiple bonds have higher K’s   =  reduced mass ( 3 x cm/sec ) THE EQUATION OF A This equation describes the vibrations of a bond. where

27 = 1 2c2c K  larger K, higher frequency larger atom masses, lower frequency constants C=C > C=C > C-C = C-H > C-C > C-O > C-Cl > C-Br increasing K increasing 

28 DIPOLE MOMENTS

29 Only bonds which have significant dipole moments will absorb infrared radiation. Bonds which do not absorb infrared include: Symmetrically substituted alkenes and alkynes Many types of C-C Bonds Symmetric diatomic molecules H-H Cl-Cl

30 The carbonyl group is one of the strongest absorbers Also O-H and C-O bonds ++ -- STRONG ABSORBERS ++ -- oscillating dipoles couple and energy is transferred infrared beam

31 RAMAN SPECTROSCOPY Another kind of vibrational spectroscopy that can detect symmetric bonds. Infrared spectroscopy and Raman spectroscopy complement each other.

32 RAMAN SPECTROSCOPY In this technique the molecule is irradiated with strong ultraviolet light at the same time that the infrared spectrum is determined. Ultraviolet light promotes electrons from bonding orbitals into antibonding orbitals. This causes formation of a dipole in groups that were formerly IR inactive and they will absorb infrared radiation... ++ -- h UV   * **.. transition absorbs IR induced dipole no dipole symmetric ….. we will not talk further about this technique

33 SUGGESTED SOFTWARE

34 Select ChemApps folder Select Spectroscopy icon Select IR Tutor icon IR TUTOR IR TUTOR ACTUALLY ILLUSTRATES INFRARED VIBRATIONS AND THEORY WITH ANIMATIONS


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