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CHEM 222 – Review Session

CHEM 222 – Review Session. February 9, 2009. Topics to be Covered. Theory behind: Separations IR NMR MS Practice Problems. Separations. Chromatography: a powerful separation method used to separate a mixture of compounds

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CHEM 222 – Review Session

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  1. CHEM 222 – Review Session February 9, 2009

  2. Topics to be Covered Theory behind: • Separations • IR • NMR • MS • Practice Problems

  3. Separations • Chromatography: a powerful separation method used to separate a mixture of compounds • Involving passing through a mixture dissolved in a mobile phase through a stationary phase • Mobile phase: a gas (inert), a liquid • Stationary phase: e.g., silica gel/alumina

  4. Separations • Polarities of two phases determine how fast a compound travels through the column • More polar components are more retained by polar solvents (mobile phase) => elute from column faster

  5. Separations The components are separated accordingto the degree to which they are retained bythe stationary phase. AmobileAstationary Areas of peaks are proportional to their concentration in the mixture.

  6. Separations • Thin Layer Chromatography (TLC) • Identification/preparative method • The stationary phase is a powdered adsorbent fixed to glass or a plastic plate • The solvent moves up the plate causing separation of components

  7. Separations Finish Rf = distance travelled by compound distance travelled by solvent front Start

  8. Separations • Question: If the stationary phase is polar (e.g., silica gel) and a compound is applied to the TLC plate, is the compound that travels the farthest the most polar or the most non-polar component in the mixture? • Which is the most retained?

  9. Infrared Spectroscopy (IR) • A spectroscopic technique used to identify the presence of various functional groups, dealing with the infrared region of the electromagnetic spectrum

  10. Infrared Spectroscopy (IR) • Infrared radiation causes atoms or groups of atoms to vibrate • IR is not energetic enough to excite electrons • electronic > vibrational > rotational (E transitions) • This interaction of atoms/molecules with radiation (absorption) is given on an IR spectrum • It can be interpreted, knowing the characteristic bands (frequencies) of functional groups

  11. Infrared Spectroscopy (IR) • Theoretical background: • Energy is quantized (discrete) • E=hv = hc/λh = 6.626 x 10-34 J-S (Planck’s constant)c = 3.00 x 108 ms-1 (speed of light)λ (wavelength in m)

  12. Infrared Spectroscopy (IR) • Only certain discrete levels are allowed for a molecule • Radiation can be thought of as small packets of energy hv called photons • Therefore, molecules can only absorb/release energy of quantized amounts • The energy of the photon must match the energy difference between two energy levels in the molecule • This means that only certain λ of light are absorbed

  13. IR Basics • Requirement for IR absorption: Need a change in dipole momentO=C=O symmetric stretch (dipoles cancel each other out)

  14. IR Basics • The excitation of a molecule from one vibrational energy level to another occurs only when: • The compound absorbs a photon of energy: ΔE = hv (specific frequency)causing the molecule to vibrate

  15. IR Basics • The location of the IR absorption band is measured as a frequency:v = 1/ λ (cm-1) Reciprocal centimeters

  16. IR Basics • Several vibrational modes: Asymmetricalstretching Symmetricalstretching Scissoring Rocking Bending (change in bond angle relativeto original bond axis) Stretching (change in distance betweenatoms along bond axis)

  17. Infrared Spectroscopy (IR) • Basic Frequencies:-O-H 3600 cm-1 -C C 2150 cm-1 -C N 2150 cm-1 -C=C- 1640 cm-1 -C=O 1720 cm-1

  18. Note: conjugated pi systems including C=O functionality leads to a downward-shift from 1710 cm-1 to 1690 cm-1 • Why? • Resonance causes double-bonded C=O to move to C-O- • Double bonds are stronger than single bonds • Actual molecule is in between a single-bond and double-bond (weaker than if it just were a double-bond with no resonance) • Therefore, less energy is required to stretch the bond, appearing at a lower energy IR absorption band

  19. Sites of unsaturation= 2 x #C + 2 - #H - #Hal + #N 2 Double bond = 1 Ring = 1 Triple bond = 2

  20. Nuclear Magnetic Resonance (NMR) • A technique used to identify compounds by monitoring energy absorption by molecules placed in a strong magnetic field • NMR is based on the concept that certain nuclei behave as magnets spinning on their axis

  21. Nuclear Spin, I • Intrinsic quantum mechanical property • If # of both protons and neutrons are equal, I = 0 (no net spin) • However, other nuclei have I ≠ 0, making them useful in NMR studies • E.g., 1H, 13C, 19F, 31P

  22. Nuclear Spin, I • A non-zero spin is associated with a magnetic moment μ For a rotating positive charge (reverse direction of μ for negative charge)

  23. Spin behavior in a magnetic field • When placed in a strong external magnetic field, B0, its magnetic moment μ can be aligned with or against B0 β E ΔE = hv B0 α

  24. Spin behavior in a magnetic field • At thermal equilibrium, the energy of α and β states are degenerate so have an equal population • In a magnetic field: • Degeneracy is removed and alignment with the field is favored (higher population of α state) • Energy is required to flip the dipole to the less stable, higher energy β state • The correct frequency to match this energy difference in order for absorption to occur is the resonance condition

  25. NMR Basics • What to look for: 1. # of signals = # of types of protons All protons do not absorb energy at the same frequency due to difference in electronic environments

  26. NMR Basics 2. Position of signals = chemical environment of protons • Shielding • Electrons generate small magnetic fields called an induced field • The induced field opposes the external magnetic field B0 so that the actual magnetic field felt by the proton is slightly less than B0 • Smaller effective B = small energy diff. between α/β spin states = lower frequency energy absorbed

  27. NMR Basics 3. Intensity of signals = # of protonsIntegration of peaks gives relative amounts of each kind of proton.

  28. NMR Basics • Splitting of signals = environment of protons as a result of surrounding protons • Enantiotopic H-atoms • Have same chemical shift  only one 1H NMR signal • Disasterotopic H-atoms • Do not have the safe chemical shift different 1H NMR signal

  29. NMR Basics • General rule: n+1 splitting • For n neighbors of equivalent protons (same coupling constant), the hydrogen is split into ~ n+1 peaks

  30. Mass Spectroscopy The most intense (tallest) peak is the Base Peak. Its fixed at 100% Molecular Ion: last peak to the right But molecular ion is not always present

  31. Even Electron Rule If MW is even, most fragments will have odd mass If MW is odd, most fragments will have even mass ONLY compounds with an odd number of Nitrogenswill have an odd MW

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