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Infrared Spectroscopy Part 1

Infrared Spectroscopy Part 1. Lecture Supplement: Take one handout from the stage. Midterm Exam 1. Date: Monday April 30 Time: 5:00-6:50 PM Topics: Molecular Structure: Introduction and Review  Carbohydrates Location: last name A-H in Franz 1178

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Infrared Spectroscopy Part 1

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  1. Infrared Spectroscopy Part 1 Lecture Supplement: Take one handout from the stage

  2. Midterm Exam 1 • Date: Monday April 30 • Time: 5:00-6:50 PM • Topics: Molecular Structure: Introduction and Review  Carbohydrates • Location: last name A-H in Franz 1178 • last name I-Sh in Haines 39 • last name Si-Z in PAB 1425 • Question and Answer Session • Lecture time, Monday April 30 • Submit questions to harding@chem.ucla.edu • Label as “Question for Q&A” • Deadline for possible inclusion: noon Sunday April 29 • Extra Office Hours (Steve J): Sunday 2-4 PM, Young Hall 3077F

  3. Fundamental principle Absorption of photons causes changes in molecular vibrations • Bending (H-O-H) • Motion not along bond axis • Less important than stretching Infrared Spectroscopy (IR)Molecular Vibrations • Molecular Vibrations • Bonded atoms move around in space • Very fast: one vibration cycle ~10-15 seconds • Stretching (H-Cl) • Atoms move along bond axis

  4. add energy Molecular Vibrations • Vibration energy •  vibration energy  average bond length Excited state higher energy Ground state lower energy

  5. Excited vibrational state DE = hn Vibrational state energy Ground vibrational state Molecular Vibrations • Vibration energy • Vibrational energy is quantized (only certain energy values are possible) n = stretching frequency For bond vibrations: DE = dependent on bond = ~5 kcal mol-1 = lower energy than red light photons = infrared photons

  6. Few photons absorbed Many photons absorbed The Infrared Spectrum Spectrum = plot of photon energy versus photon quantity Typical infrared spectrum: Number of photons absorbed Stretching frequency Proportional to photon energy

  7. Molecular Structure from IR Spectrum • How does spectrum give information about molecular structure? • Structure controls number of photons absorbed • Structure controls stretching frequency

  8. Vector sum of bond dipoles Intensity of IR peak d+ X Y d- Structure versus Photon Quantity From quantum mechanics: Chance of photon absorption controlled by change in dipole moment (m) Useful approximation Consider only one bond

  9. d+ X Y d- Absorption Intensity versus Bond Dipoles • Bond dipole ~ (magnitude of electronegativity difference) x (bond length) •  DEN  dipole •  bond length  dipole •  bond dipole  absorption In practical terms: • Highly polar bond  strong peak • Symmetrical (nonpolar) or nearly symmetrical bond  peak weak or absent

  10. Absorption Intensity versus Bond Dipoles Examples: C=O peak strong C=C peak absent (or maybe weak) C=C peak present but weak • Caution! • Weak peaks not always discernable • Be careful when excluding symmetrical functional groups base on absence of peak

  11. spring stiffness increasing atoms bond stretching frequency atom masses Functional groups determine IR stretching frequencies Structure versus Stretching Frequency • Hooke’s Law (1660) • Stretching frequency of two masses on a spring  bond order C-C C=C CC

  12. 1 2 3 4 5 Fingerprint region Characteristic Stretching FrequenciesThe Five Zones IR spectrum divided into five zones (groups) of important absorptions

  13. Characteristic Stretching FrequenciesThe Five Zones * attached to benzene ring **attached to alkene

  14. Characteristic Stretching FrequenciesThe Five Zones C=O frequencies 20-40 cm-1 lower when conjugated to a pi bond

  15. Characteristic Stretching FrequenciesThe Five Zones • Complete table: Thinkbook, inside front cover • What do I have to know? • Functional groups in each zone  Learn by working lots of problems • Do not memorize stretching frequencies; table given on exam

  16. C-C Guided Tour of Functional GroupsTerminal Alkyne

  17. Guided Tour of Functional GroupsTerminal Alkene

  18. broad C-O Guided Tour of Functional GroupsAlcohol

  19. very strong 1718 cm-1 Guided Tour of Functional GroupsKetone

  20. Infrared Spectroscopy Part 2 Lecture Supplement: Take one handout from the stage

  21. } Functional groups Infrared Spectroscopy Part 1 Summary • Infrared photons cause excitation of molecular vibrations • Photon absorption probability higher with more polar bonds • Energy of photons absorbed depends on: Bond order Masses of atoms bonded • IR spectrum divided into five zones • Each zone analyzed for absence or presence of functional groups • Stretching frequency, peak shape both important Alcohol O-H usually gives broad peak C=O stretch gives strong peak

  22. very strong 1718 cm-1 Guided Tour of Functional GroupsKetone (again)

  23. very strong ~2900 cm-1 usually obscured 1718 cm-1 Guided Tour of Functional GroupsAldehyde

  24. 1720 cm-1 Guided Tour of Functional GroupsKetone with Alkene Conjugation Conjugation with pi bond lowers C=O stretch by 20-40 cm-1

  25. 1743 cm-1 Guided Tour of Functional GroupsEster

  26. very broad 1711 cm-1 Guided Tour of Functional GroupsCarboxylic Acid

  27. May be two peaks Guided Tour of Functional GroupsBenzene Ring

  28. Five Zone IR Spectrum AnalysisExample #1: C6H12O2 1700 cm-1 Step 1: Calculate DBE DBE = C - (H/2) + (N/2) + 1 = 6 - (12/2) + (0/2) +1 = 1 One ring or one pi bond

  29. Present Five Zone IR Spectrum AnalysisExample #1: C6H12O2 1700 cm-1 • Step 2: Analyze IR Spectrum • Zone 1 (3700-3200 cm-1) Alcohol O-H: N-H: C-H: Absent - no N in formula Absent - not enough DBE

  30. Present Five Zone IR Spectrum AnalysisExample #1: C6H12O2 1700 cm-1 • Zone 2 (3200-2700 cm-1) Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Probably not (not enough DBE) Absent - no 2700 cm-1 Absent - not broad enough

  31. Five Zone IR Spectrum AnalysisExample #1: C6H12O2 1700 cm-1 • Zone 3 (2300-2000 cm-1) Alkyne CC: Nitrile CN: Absent - no peaks; not enough DBE Absent - no peaks; not enough DBE

  32. Present @ 1700 cm-1 Five Zone IR Spectrum AnalysisExample #1: C6H12O2 1700 cm-1 • Zone 4 (1850-1650 cm-1) C=O: Possibilities: ketone ester - not enough oxygens aldehyde - no 2700 cm-1 peak carboxylic acid - zone 2 not broad amide - no nitrogen

  33. Actual structure: Five Zone IR Spectrum AnalysisExample #1: C6H12O2 1700 cm-1 • Zone 5 (1680-1450 cm-1) Benzene ring: Alkene C=C: Absent - no peak ~1600 cm-1; not enough DBE Absent - no peak ~1600 cm-1; not enough DBE

  34. Five Zone IR Spectrum AnalysisExample #2: C8H7N Step 1: Calculate DBE DBE = C - (H/2) + (N/2) + 1 = 8 - (7/2) + (1/2) +1 = 6 Six rings and/or pi bonds Possible benzene ring

  35. Five Zone IR Spectrum AnalysisExample #2: C8H7N • Step 2: Analyze IR Spectrum • Zone 1 (3700-3200 cm-1) Alcohol O-H: N-H: C-H: Absent - no oxygen in formula Absent - peaks too small Absent - peaks too small

  36. Five Zone IR Spectrum AnalysisExample #2: C8H7N • Zone 2 (3200-2700 cm-1) Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Present - peaks > 3000 cm-1 Present - peaks < 3000 cm-1 Absent - no 2700 cm-1; no C=O in zone 4 Absent - not broad enough; C=O in zone 4

  37. } Five Zone IR Spectrum AnalysisExample #2: C8H7N • Zone 3 (2300-2000 cm-1) Alkyne CC: Nitrile CN: Possible Possible

  38. Five Zone IR Spectrum AnalysisExample #2: C8H7N • Zone 4 (1850-1650 cm-1) C=O: Absent - no peak; no oxygen in formula

  39. Absent - not enough DBE for alkene plus benzene plus triple bond Actual structure: Five Zone IR Spectrum AnalysisExample #2: C8H7N • Zone 5 (1680-1450 cm-1) Benzene ring: Alkene C=C: Present - peaks ~1600 cm-1 and ~1500 cm-1

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