1. Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray) Winter 2012 Problems: 4–7, 15, 22. Winter 2011 Problems: 1–11, 23, 28, 29. Spring 2011, Lecture 1 Problems: 1–5, 12, 20. Spring 2011, Lecture 2 Problems: 1, 2, 9–13, 15, 16, 20–22. Fall 2011 Problems: 8, 20–24. Suggested problems to attempt before attending the review:

2. Infrared Spectroscopy Part 1 Lecture Supplement page 112

3. Midterm 1 1 hour exam (in class on Friday, May 4) Will cover: –Intro & Review up through Carbohydrates (Mass Spectrometry and IR will not be on exam) Last name A-K A-P in CS50 Last name L-Z Q-Z in Franz 1260 Tools –Pen and/or pencil –Eraser –Model kit –No calculators or cell phone

4. How should I study? Review past “Exam 1”s on Hardinger’s website http://www.chem.ucla.edu/harding/index.html (on left frame, click “Ch14C” then in middle frame click “Current and Past Exam and Keys”) http://www.chem.ucla.edu/harding/index.html

5. Exam 1 Review: Tuesday, 5/1 YH3069 from 7-9pm (Ray) Winter 2012 Problems: 4–7, 15, 22. Winter 2011 Problems: 1–11, 23, 28, 29. Spring 2011, Lecture 1 Problems: 1–5, 12, 20. Spring 2011, Lecture 2 Problems: 1, 2, 9–13, 15, 16, 20–22. Fall 2011 Problems: 8, 20–24. Suggested problems to attempt before attending the review:

6. Infrared Spectroscopy (IR) Molecular Vibrations Fundamental principle Absorption of photons causes changes in molecular vibrations Molecular vibrations Bonded atoms move around in space Very fast: One vibration cycle = ~10 -15 seconds Bending (H-O-H) Motion not along bond axis Stretching (H-Cl) Atoms move along bond axis Movie files: “HCl stretch.mov” (left) and “water bend.mov” (right)

7. Molecular Vibrations Vibration energy Ground state lower energy add energy Excited state higher energy  vibration energy causes  average bond length water_groundstate.mov water_excitedstate.mov

8. Molecular Vibrations Vibration energy Excited vibrational state  E = h For bond vibrations:  E = dependent on atoms and bond order = ~5 kcal mol -1 = lower energy than red light photons = infrared photons = stretching frequency Unit = wavenumber = cm -1 Ground vibrational state Vibrational state energy Vibrational energy is quantized (only certain energy values are possible)

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

10. Molecular Structure from IR Spectrum Structure controls number of photons absorbed (IR spectrum y-axis) Structure controls stretching frequency (IR spectrum x-axis) How does spectrum give information about molecular structure?

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

12. Bond Dipoles Control Absorption Intensity Bond dipole ~ (magnitude of electronegativity difference) x (bond length)   EN causes  bond dipole  bond length causes  bond dipole  + X Y  -  bond dipole causes  absorption In practical terms: Highly polar bond  strong peak Symmetrical (nonpolar) or nearly symmetrical bond  peak weak or absent }

13. Absorption Intensity versus Bond Dipoles C=O peak strongC=C peak absent or weakC=C peak present Examples  

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

15. 100 Transmittance (%) 0 4000 3000 1500 Stretching frequency (cm -1 )10002000 Characteristic Stretching Frequencies The Five Zones IR spectrum divided into five zones (groups) of important absorptions 12 3 4 5 Fingerprint region

16. Characteristic Stretching Frequencies The Five Zones BondStretching FrequencyIntensity and Shape Zone 1: 3700-3200 cm -1 Alcohol O-H3650-3200 cm -1 usually strong and broad Alkyne  C-H 3340-3250 cm -1 usually strong and sharp Amine or amide N-H3500-3200 cm -1 medium; often broad Zone 2: 3200-2700 cm -1 Aryl * or vinyl ** sp 2 C-H3100-3000 cm -1 variable Alkyl sp 3 C-H2960-2850 cm -1 variable Aldehyde C-H~2900, ~2700 cm -1 medium; two peaks Carboxylic acid O-H3000-2500 cm -1 usually strong; very broad * attached to benzene ring **attached to alkene

17. Characteristic Stretching Frequencies The Five Zones BondStretching FrequencyIntensity and Shape Zone 3: 2300-2000 cm -1 Alkyne C  C 2260-2000 cm -1 variable and sharp Nitrile C  N 2260-2220 cm -1 variable and sharp Zone 4: 1850-1650 cm -1 Ketone C=O1750-1705 cm -1 strong Ester C=O1750-1735 cm -1 strong Aldehyde C=O1740-1720 cm -1 strong Carboxylic acid C=O1725-1700 cm -1 strong Amide C=O1690-1650 cm -1 strong C=O frequencies 20-40 cm -1 lower when conjugated to a pi bond

18. Characteristic Stretching Frequencies The Five Zones BondStretching FrequencyIntensity and Shape Zone 5: 1680-1450 cm -1 Alkene C=C1680-1620 cm -1 variable Benzene C=C ~1600 cm -1 and ~1500-1450 cm -1 variable; 1600 cm -1 often two peaks Fingerprint region (below 1450 cm -1 ): Not useful for Chem 14C What do I need to know from this table? Functional groups in each zone  Learn by working lots of problems Do not memorize stretching frequencies; table given on exam Complete table: Lecture Supplement and Thinkbook, inside front cover

19. Guided Tour of Functional Groups Terminal Alkyne 100 Transmittance (%) 0 4000 3000 1500 Stretching frequency (cm -1 ) 1000 2000 12 3 4 5 Fingerprint region

20. Guided Tour of Functional Groups Terminal Alkene 100 Transmittance (%) 0 4000 3000 20001000 Stretching frequency (cm -1 ) 1500 12 3 4 5 Fingerprint region

21. Guided Tour of Functional Groups Alcohol 4000 3000 2000 1500 1000 Stretching frequency (cm -1 ) 0 100 Transmittance (%) broad C-O 12 3 4 5 Fingerprint region

22. Guided Tour of Functional Groups Ketone 40003000 2000 15001000Stretching frequency (cm -1 ) 0 100 Transmittance (%) sp 3 C-H 1709 cm -1 Very strong 12 3 4 5 Fingerprint region

23. Infrared Spectroscopy Part 2 Lecture Supplement page 122 100 Transmittance (%) 0 4000 3000 1500 Stretching frequency (cm -1 )10002000 12 3 4 5 Fingerprint region

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

25. Guided Tour of Functional Groups Ketone 40003000 2000 15001000Stretching frequency (cm -1 ) 0 100 Transmittance (%) sp 3 C-H 1709 cm -1 Very strong

26. Guided Tour of Functional Groups Aldehyde ~2900 cm -1 usually obscured very strong 1718 cm -1 100 Transmittance (%) 0 4000 3000200015001000 Stretching frequency (cm -1 ) ~2700 cm -1

27. Guided Tour of Functional Groups In general: Conjugation with C-C pi bond lowers C=O stretching frequency by 20-40 cm -1 1709 cm -1 Effect of pi bond conjugation? 1667 cm -1 1686 cm -1

28. Guided Tour of Functional Groups Ester 1743 cm -1 100 Transmittance (%) 0 40003000 2000 1500 1000 Stretching frequency (cm -1 )

29. Guided Tour of Functional Groups Carboxylic Acid very broad 1711 cm -1 100 Transmittance (%) 0 4000 3000 2000 1500 1000Stretching frequency (cm -1 )

30. Guided Tour of Functional Groups Benzene Ring Sometimes two peaks 100 Transmittance (%) 0 4000 3000 2000 15001000 Stretching frequency (cm -1 )

31. Five Zone IR Spectrum Analysis Example #1: C 6 H 12 O 2 DBE= C - (H/2) + (N/2) + 1 = 6 - (12/2) + (0/2) + 1 = 1 100 Transmittance (%) 0 4000 3000 2000 15001000 Stretching frequency (cm -1 ) 1700 cm -1 One ring or one pi bond Step 1: Calculate DBE

32. Five Zone IR Spectrum Analysis Example #1 100 Transmittance (%) 0 4000 3000 2000 15001000 Stretching frequency (cm -1 ) 1700 cm -1 Step 2: Analyze IR Spectrum Present Absent - no N in formula Zone 1 (3700-3200 cm -1 ) C 6 H 12 O 2 DBE = 1 Alcohol O-H: Terminal alkyne  C-H: Amine or amide N-H: Absent - not enough DBE; no peak ~2200 cm -1

33. Five Zone IR Spectrum Analysis Example #1 100 Transmittance (%) 0 4000 3000 2000 15001000 Stretching frequency (cm -1 ) 1700 cm -1 Zone 2 (3200-2700 cm -1 ) C 6 H 12 O 2 DBE = 1 Aryl/vinyl sp 2 C-H: Alkyl sp 3 C-H: Aldehyde C-H: Carboxylic acid O-H: Absent - not enough DBE Absent - no 2700 cm -1 Absent - not broad enough Present

34. Five Zone IR Spectrum Analysis Example #1 100 Transmittance (%) 0 4000 3000 2000 15001000 Stretching frequency (cm -1 ) 1700 cm -1 Zone 3 (2300-2000 cm -1 ) C 6 H 12 O 2 DBE = 1 Alkyne C  C: Nitrile C  N: Absent - no peaks; not enough DBE

35. Five Zone IR Spectrum Analysis Example #1 100 Transmittance (%) 0 4000 3000 2000 15001000 Stretching frequency (cm -1 ) Zone 4 (1850-1650 cm -1 ) C 6 H 12 O 2 DBE = 1 1700 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 Present @ 1700 cm -1 Verify with 13 C-NMR

36. Five Zone IR Spectrum Analysis Example #1 100 Transmittance (%) 0 4000 3000 2000 15001000 Stretching frequency (cm -1 ) 1700 cm -1 Zone 5 (1680-1450 cm -1 ) C 6 H 12 O 2 DBE = 1 Benzene ring: Alkene C=C: Absent - no peak ~1600 cm -1 ; not enough DBE Absent - no peak ~1600 cm -1 ; not enough DBE (C=C plus C=O) Actual structure:

37. Five Zone IR Spectrum Analysis Example #2: C 8 H 7 N 100 Transmittance (%) 0 4000 3000 2000 1500 1000 Stretching frequency (cm -1 ) 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

38. Five Zone IR Spectrum Analysis Example #2 100 Transmittance (%) 0 4000 3000 2000 1500 1000 Stretching frequency (cm -1 ) Step 2: Analyze IR Spectrum Zone 1 (3700-3200 cm -1 ) C 8 H 7 N DBE = 6 Alcohol O-H: Amine or amide N-H: Terminal alkyne  C-H: Absent - no oxygen in formula Absent - peaks too small “No amine/amide” = false conclusion Example: (CH 3 ) 3 N has no N-H

39. Five Zone IR Spectrum Analysis Example #2 100 Transmittance (%) 0 4000 3000 2000 1500 1000 Stretching frequency (cm -1 ) Zone 2 (3200-2700 cm -1 ) C 8 H 7 N DBE = 6 Aryl/vinyl sp 2 C-H: Alkyl sp 3 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; no C=O in zone 4

40. Five Zone IR Spectrum Analysis Example #2 100 Transmittance (%) 0 4000 3000 2000 1500 1000 Stretching frequency (cm -1 ) Zone 3 (2300-2000 cm -1 ) C 8 H 7 N DBE = 6 Alkyne C  C: Nitrile C  N: Possible }

41. Five Zone IR Spectrum Analysis Example #2 100 Transmittance (%) 0 4000 3000 2000 1500 1000 Stretching frequency (cm -1 ) Zone 4 (1850-1650 cm -1 ) C 8 H 7 N DBE = 6 C=O:Absent - no peak; no oxygen in formula

42. Five Zone IR Spectrum Analysis Example #2 100 Transmittance (%) 0 4000 3000 2000 1500 1000 Stretching frequency (cm -1 ) Zone 5 (1680-1450 cm -1 ) C 8 H 7 N DBE = 6 Benzene ring: Alkene C=C: Present - peaks ~1600 cm -1 and ~1500 cm -1 Absent - not enough DBE for alkene plus benzene plus triple bond Actual structure: