1 / 34

INDEX OF HYDROGEN DEFICIENCY

INDEX OF HYDROGEN DEFICIENCY. and. THE BASIC THEORY OF INFRARED SPECTROSCOPY. WHAT CAN YOU LEARN FROM A MOLECULAR FORMULA ?. YOU CAN DETERMINE THE NUMBER OF RINGS AND / OR DOUBLE BONDS. Saturated Hydrocarbons. C n H 2n+2. GENERAL FORMULA. CH 4. C 2 H 6. C 3 H 8.

Ava
Télécharger la présentation

INDEX OF HYDROGEN DEFICIENCY

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

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

  3. Saturated Hydrocarbons CnH2n+2 GENERAL FORMULA CH4 C2H6 C3H8 C4H10 C5H12 C9H20 branched compounds also follow the formula

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

  5. Index of Hydrogen Deficiency CALCULATION METHOD • Determine the expected formula for a noncyclic, saturated compound ( CnH2n+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 (explained later) Index of Hydrogen-deficiency

  6. C5H8 C5H12 = ( CnH2n+2 ) C5H8 H4 Index = 4/2 = 2 Two Unsaturations double bond and ring in this example

  7. Index of Hydrogen Deficiency CORRECTIONS FOR ATOMS OTHER THAN HYDROGEN • 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 +0 C-H C-O-H +O +1 C-H C-NH2 +N,+H -1 C-H C-X -H,+X

  8. C4H5N C4H10 = ( CnH2n+2 ) C4H11N add one H for N C4H5 N H6 Index = 6/2 = 3 two double bonds and ring in this example

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

  10. INDEX C6H14 -C6H8 Index = 6/2 = 3 H6 HYDROGENATION Pd C6H8 + 2 H2 C6H12 Hydrogenation shows only two double bonds. Therefore, there must also be a ring. PROBLEM A hydrocarbon has a molecular formula of C6H8. It will react with hydrogen and a palladium catalyst to give a compound of formula C6H12. Give a possible structure.

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

  12. INFRARED SPECTROSCOPY

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

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

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

  16. 100 100 % T R A N S M I T T A N C E 80 80 60 60 40 40 20 20 0 0 3500 3000 2500 2000 1500 1000 500 WAVELENGTH (cm-1) KETONE 4-Methyl-2-pentanoneC-H < 3000, C=O @ 1715 cm-1 AN INFRARED SPECTRUM

  17. THE UNIT USED ON AN IR SPECTRUM IS WAVENUMBERS ( n ) n = wavenumbers (cm-1) 1 n l = wavelength (cm) = l (cm) c = speed of light n = frequency = nc c = 3 x 1010 cm/sec or ( ) 1 c cm/sec 1 n= c = = l l cm sec wavenumbers are directly proportional to frequency

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

  19. BONDING CURVES AND VIBRATIONS MORSE CURVES STRETCHING

  20. + + + + + + o o + + BOND VIBRATIONAL ENERGY LEVELS e n e r g y MORSE CURVE zero point energy rmin rmax decreasing distance ravg (average bond length)

  21. BOND VIBRATIONAL ENERGY LEVELS Bonds do not have a fixed distance. They vibrate continually even at 0oK (absolute). The frequency for a given bond is a constant. Vibrations are quantized as levels. The lowest level is called the zero point energy. e n e r g y bond dissociation energy vibrational energy levels zero point energy rmin rmax distance ravg (average bond length)

  22. 2.5 4 5 5.5 6.1 6.5 15.4 4000 2500 2000 1800 1650 1550 650 Typical Infrared Absorption Regions (stretching vibrations) WAVELENGTH (mm) C-Cl C=O C=N O-H C-H C N Very few bands C-O C=C N-H C C C-N X=C=Y C-C * N=O N=O (C,O,N,S) FREQUENCY (cm-1) * nitro has two bands

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

  24. HOOKE’S LAW force constant compress K stretch Dx x0 x1 restoring force -F = K(Dx) = Molecule as a Hooke’s Law device m1 m2 K

  25. THE MORSE CURVE APPROXIMATES AN HARMONIC OSCILLATOR HOOKE’S LAW Harmonic Oscillator ACTUAL MOLECULE Morse Curve (anharmonic) 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 K n = 2pc m m1 m2 m = m1 + m2 n = THE EQUATION OF A frequency in cm-1 SIMPLE HARMONIC OSCILLATOR c = velocity of light ( 3 x 1010 cm/sec ) K = force constant in dynes/cm where > > multiple bonds have higher K’s m= atomic masses This equation describes the vibrations of a bond. m=reduced mass

  27. 1 K n = 2pc m = C=C > C=C > C-C larger K, higher frequency larger atom masses, lower frequency increasing K constants 1650 2150 1200 increasing m C-H > C-C > C-O > C-Cl > C-Br 3000 1200 1100 750 650

  28. DIPOLE MOMENTS

  29. DIPOLE MOMENTS 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. + + d+ - - d- oscillating dipoles couple and energy is transferred STRONG ABSORBERS d- The carbonyl group is one of the strongest absorbers d+ Also O-H and C-O bonds 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. p* . induced dipole d+ d- . .. hn UV pp* absorbs IR transition no dipole symmetric ….. we will not talk further about this technique

  33. SUGGESTED SOFTWARE

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

More Related