Infrared Spectroscopy
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Infrared Spectroscopy

Course Number: CHEM 30B Chem 30B U, Summer 2009

College/University: UCLA

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CFQ & PP: Infrared Spectroscopy Reading Brown and Foote: Chapter 12, Section 18.3A Lecture Supplement Infrared (IR) Spectroscopy (page 18 of this Thinkbook) Suggested Text Exercises Brown and Foote Chapter 12: 3 12 Optional Interactive Organic Chemistry CD and Workbook Supporting Concepts: IR Spectroscopy Tutorial (p. 76) Spectroscopy: IR (p. 58) Optional Web Site Exercises Webspectra...

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& CFQ PP: Infrared Spectroscopy Reading Brown and Foote: Chapter 12, Section 18.3A Lecture Supplement Infrared (IR) Spectroscopy (page 18 of this Thinkbook) Suggested Text Exercises Brown and Foote Chapter 12: 3 12 Optional Interactive Organic Chemistry CD and Workbook Supporting Concepts: IR Spectroscopy Tutorial (p. 76) Spectroscopy: IR (p. 58) Optional Web Site Exercises Webspectra (http://www.chem.ucla.edu/~webspectra) Practice Five Zone analysis. Optional Software IR Tutor (Available on the computers in the Science Learning Center, Young Hall, 4th floor) You do not need to memorize the typical stretching frequencies for each functional group, as given in the lecture handout. This data will be given on an exam if needed. However you are expected to be intimately familiar with the distribution of functional groups in the Five Zone Analysis. Concept Focus Questions 1. Briefly explain the molecular events that result in an infrared spectrum. 2. What molecular structure features control the intensity of an infrared absorption? 3. Explain why similar functional groups absorb infrared photons of similar energies. 4. Briefly explain the Five Zone approach to analysis of an infrared spectrum. 5. The infrared absorption bands of most common functional groups have characteristic features (e.g., two absorptions, a broad peak, etc.) in addition to the photon energies. Briefly describe the important examples of this in the Five Zone analysis. 6. What is the effect of conjugation on the energy of an infrared absorption band? 7. What is the fingerprint region of the infrared spectrum? Why is it usually ignored? CFQ & PP: Infrared Spectroscopy 97 Concept Focus Questions Solutions 1. Absorption of an infrared photon results in the excitation of the molecule to a higher vibrational quantum state. 2. For a stretching vibration, a photon will be absorbed and the molecule excited to a higher vibrational quantum state only if that vibration results in a change in bond dipole. The bond dipole is a product of bond length and charge difference of the bonded atoms. As a bond vibrates, the bond length changes, so this criterion is met. The charge difference is determined by the electronegativity of the bonded atoms. If these atoms are not identical then they have a difference in electronegativity and thus a difference in charge. If these atoms are identical, they will have equal electronegativity and thus no charge difference. In this case, the product of bond length change and charge difference (zero) is zero, so no photon is absorbed. If the atoms are even slightly different, then a small change in bond dipole will occur, along with the corresponding absorption of an infrared photon. Thus, all of the bond stretches of H2C=O will show up on the IR spectrum because all of the bonds consist of unequal atoms. For H2C=CH2, the C-H bond stretches will give IR bands, but the C=C bond (which is made up of two identical carbon atoms) will not. The C=C bonds of H2C=CH-CH=CH2 will show up in the IR because the carbons are not equal. One carbon is a CH2 and the other is a CH-CH. In summary, asymmetrical bonds give IR absorptions whereas symmetrical bonds do not. 3. The energy of the photon necessary to excite the molecule to a higher vibrational quantum level is controlled by the masses of the attached atoms and the bond strength. Similar functional groups are made up of similar atoms. Thus molecules with similar functional groups will absorb photons of similar energies. This is why we can use IR spectroscopy to determine what functional groups are present or absent in a compound by simply looking at the positions of the IR peaks. 4. We divide the IR spectrum into five zones, based upon the energy of photons absorbed by twelve common functional groups. The zones and functional groups therein are summarized below. Zone 1 (3700 - 3200 cm-1): OH, NH, terminal alkyne C-H Zone 2 (3200 - 2700 cm-1): sp3 C-H (alkane), sp2 C-H (alkene and benzene ring), aldehyde C-H, carboxylic acid O-H Zone 3 (2300 - 2100 cm-1): alkyne and nitrile triple bonds Zone 4 (1950 - 1650 cm-1): carbonyl functional groups, aromatic overtones Zone 5 (1680 - 1450 cm-1): C=C functional groups (alkene and benzene ring) For sample Five Zone analyses, see the practice problems below. 98 CFQ & PP: Infrared Spectroscopy 5. Alcohol, Amine: The O-H and N-H stretches are usually broadened due to hydrogen bonding. Carboxylic acid: Must have two bands: broad O-H stretch in Zone 2 and C=O stretch in Zone 4. Aldehyde: Must have two sets of bands: 2900 and 2700 cm-1 in Zone 2 (2900 cm-1 often hidden by sp3 C-H stretches), and C=O stretch in Zone 4. Alkyne: The triple bond stretch will not be observed if the alkyne is symmetric or nearly so. Carbonyl: Often the most intense peak in the entire spectrum. Benzene ring: Must have bands at about 1600 cm-1 (may be two bands close together) and another at about 1500 cm-1. Since most benzene ring compounds have at least one hydrogen atom attached to the benzene ring, an sp2 C-H stretch will also be seen in Zone 2. Aromatic overtones sometimes may be observed as a set of 4 bands around 1800 cm-1. 6. Conjugation shifts a band to lower energy. For example, a ketone typically absorbs in the region of 1750 - 1705 cm-1, whereas an enone ((O=C-C=C; also called an ,unsaturated ketone) absorbs at 1685 - 1665 cm-1, and an aryl ketone (ketone conjugated with a benzene ring) absorbs at 1700 - 1680 cm-1. This shift to a lower energy is seen when any functional group is conjugated. 7. The region from about 1450 to 400 cm-1 is called the fingerprint region. This region is often complex because it contains many absorptions. It is usually ignored in routine IR analysis because it can be difficult to make precise assignments. For example, a peak at 1360 cm-1 could be a CH2 bend or a C-N stretch. Without precise assignments, it can be difficult to determine the absence or presence of functional groups from the fingerprint region. However, the fingerprint regions of two IR spectra of the same compound will be identical, in the same way that two fingerprints left by the same person in different places will be identical. Thus the fingerprint region of an unknown compound can be compared with the fingerprint region of a known sample, and if they match exactly, the unknown and known are identical. Practice Problems 1. Several functional groups have absorptions in more than one of the Five Zones. To conclude that the functional group is present, both absorptions must be seen. Prepare a table listing these functional groups and the corresponding absorptions. Hint: There are at least six such functional groups. Some of them require three distinct absorptions. CFQ & PP: Infrared Spectroscopy 99 2. Perform Five Zone analyses for the following IR spectra. (a) C7H8O2 (b) C6H12O2 (c) C6H6BrN 100 CFQ & PP: Infrared Spectroscopy (d) C8H10 (e) C6H10O 3. Decide which structure is the best fit for the IR spectrum, and briefly explain your reasoning. OH O O OH C CH 2-phenylethanol ethyl phenyl ether 4,4-dimethyl-2,5cyclohexadienone 1-ethynylcyclohex-2-en-1-ol CFQ & PP: Infrared Spectroscopy 101 4. Describe the essential features expected in the infrared spectrum for this compound. O HO NH O 5. Rank the following bond stretches in order of expected absorption intensity and stretching energy: C-F, C-Cl, C-Br, and C-I. Briefly explain your reasoning. 6. Rank the absorption intensity of the C=O stretches of formaldehyde, acetone, and hexafluoroacetone. Briefly explain your reasoning. 7. Briefly state what can be concluded about the structure of unknown compounds from this data: Formula C12H12O; no IR absorption 1650 cm-1 - 1750 cm-1. 8. Decide which structure that best fits the spectrum. Briefly explain why the other two structures were excluded. O H3CC CCH2OCH3 O A B C 102 CFQ & PP: Infrared Spectroscopy 9. When recording an IR spectrum in the lab, we take no special precautions to account for the presence of oxygen, nitrogen or argon in the air. What are the consequences of this? Practice Problems Solutions 1. Functional Group Absorptions terminal alkyne 3300 cm-1 (C-H stretch) 2260 - 2100 cm-1 (CC stretch) 3100 - 3000 cm-1 (sp2 C-H stretch) 1950 - 1750 cm-1 (aromatic overtones) ~1600 cm-1 and 1500 - 1450 cm-1 (benzene ring stretch) ~2900 and ~2700 cm-1 (C-H stretch) 1740 - 1720 cm-1 (C=O stretch) 3000 - 2500 cm-1 (O-H stretch) 1725 - 1700 cm-1 (C=O stretch) 1700 - 1680 cm-1 (C=O stretch) 1950 - 1750 cm-1 (aromatic overtones) ~1600 cm-1 and 1500 - 1450 cm-1 (benzene ring stretch) 1685 - 1665 cm-1 (C=O stretch) 1680 - 1620 cm-1 (C=C stretch) aromatic ring aldehyde carboxylic ketone enone There acid aryl are a few uncommon exceptions to this table. For example, acetylene is a terminal alkyne that displays a Zone 1 C-H stretch, but not a Zone 3 CC stretch, due to the symmetry of the triple bond. A benzene ring in which all six ring hydrogens have been substituted (e.g., hexamethylbenzene) has Zone 4 and 5 peaks but not Zone 2 sp2 C-H stretches. Alkenes bearing four substituents, such as (CH3)2C=C(CH3)2 are sufficiently common as to be excluded from this table. 2. When given only a formula and an IR spectrum it may not be possible to determine the exact structure of a molecule. The actual compound answers given here are meant to help you verify how the structure corresponds to the particular spectrum. (a) DBE = 4 Possible benzene ring Zone 1: 3388 cm-1: alcohol O-H No N-H: no nitrogen in formula No terminal alkyne: no Zone 3 absorption Zone 2: >3000 cm-1: sp2 C-H (vinyl or aryl) <3000 cm-1: sp3 C-H CFQ & PP: Infrared Spectroscopy 103 No carboxylic acid or aldehyde: no Zone 4 carbonyl Zone 3: No triple bonds: no absorptions Zone 4: No C=O or aromatic ring: no absorptions Zone 5: 1599 cm-1 and 1494 cm-1: benzene ring No alkene: not enough DBE for alkene plus benzene ring Note that the IR analysis includes the functional groups that are present as well as those functional groups that are absent. This will be useful later on when we use IR spectra to deduce the structure of unknown compounds. OH Actual compound: OCH3 The use of information from more than one source, such as formula (from mass spectrum) and IR spectral features, is a common and useful practice when solving spectroscopy problems. Think "outside the spectrum"! (b) DBE = 1 One double bond or one ring Zone 1: No alcohol O-H, N-H, or terminal alkyne: no absorptions Zone 2: No vinyl or aryl C-H <3000 cm-1: sp3 C-H No carboxylic acid O-H: not broad No aldehyde C-H: no 2700 cm-1 Zone 3: No triple bonds: no absorptions Zone 4: No aromatic ring: no overtones 1745 cm-1: not aldehyde: no 2700 cm -1 probably ester or ketone Zone 5: No C=C or benzene ring: no absorptions O Actual compound: O (c) DBE = 4 Possible benzene ring Zone 1: No alcohol O-H: No oxygen in formula 104 CFQ & PP: Infrared Spectroscopy 3458 cm-1 and 3363 cm-1: N-H.* No terminal alkyne: No Zone 3 triple bond absorption Zone 2: >3000 cm-1: vinyl or aryl C-H No sp3 C-H (no significant Zone 2 peak below 3000 cm-1) No CO2H or aldehyde: no oxygen in formula, no C=O in Zone 4 Zone 3: No triple bonds: no absorption Zone 4: No C=O: no absorption; no oxygen in formula Aromatic overtones present Zone 5: ~1600 cm-1 and ~1500 cm-1: benzene ring No C=C: not enough DBE for benzene ring and alkene * A primary amine (RNH2) gives two N-H bands, arising from symmetric and asymmetric stretching modes. A secondary amine (R2NH) gives one N-H band. A tertiary amine (R3N) yields no N-H stretching absorptions, as it has no N-H. Br Actual compound: NH2 (d) DBE = 4 Possible benzene ring Zone 1: No alcohol O-H, N-H, or terminal alkyne: no absorptions; no nitrogen or oxygen in formula Zone 2: >3000 cm-1: vinyl or aryl C-H <3000 cm-1: sp3 C-H No CO2H or aldehyde: no oxygen in formula; no C=O in Zone 4 Zone 3: No triple bonds: no absorption; no nitrogen in formula Zone 4: No C=O: no absorption; no oxygen in formula Aromatic overtones present Zone 5: 1600 cm-1 and 1500 cm-1: benzene ring No C=C: not enough DBE for benzene ring plus alkene CH3 Actual compound: CH3 CFQ & PP: Infrared Spectroscopy 105 (e) DBE = 2 Two double bonds or one triple bond or two rings or one double bond plus one ring. Zone 1: These absorptions are too small to be alcohol O-H, N-H, or terminal alkyne for a low molecular weight compound. These peaks are most likely due to impurities (water is a common impurity; the O-H shows up here), or an overtone of the strong peak at 1718 cm-1. Zone 2: <3000 cm-1: sp3 C-H No vinyl or aryl C-H: no >3000 cm-1 No aldehyde C-H: no 2700 cm-1 No carboxylic acid O-H: not broad; only one oxygen atom in formula Zone 3: No triple bonds: no absorptions; no nitrogen in formula Zone 4: 1718 cm-1: ketone not aldehyde: no 2700 cm-1 not CO2H: only one oxygen in formula; Zone 2 not broad No aromatic ring: Overtones appear to be very weak or absent Zone 5: No C=C or benzene ring: no absorptions; not enough DBE for benzene ring O Actual compound: Do not confuse "benzene ring" or "aromatic ring" with just any old ring! For example, cyclohexanone has a cyclohexane ring but not a benzene ring so it does not have benzene ring absorptions in Zone 5. 3. Ethyl phenyl ether is eliminated because the IR shows an alcohol O-H at 3350 cm-1. The dienone is eliminated because none of the Zone 4 peaks are strong enough to be a carbonyl absorption in a compound of this low molecular weight. The terminal alkyne is eliminated because there is no alkyne absorption in Zone 3. The remaining structure, 2-phenylethanol, is a reasonable fit. The IR spectrum shows a broad alcohol O-H stretch at 3350 cm-1, and the benzene ring peaks at 1498 cm-1 and 1456 cm-1. There are aromatic overtones in Zone 4. There are no functional groups indicated by the IR that are not present in 2-phenylethanol. 4. Positions of these predicted absorptions are approximate. 3400 cm-1 (broad): alcohol O-H 3400 cm-1: N-H (the N-H and O-H bands are likely to obscure each other) 106 CFQ & PP: Infrared Spectroscopy Complex pattern above and below 3000 cm-1: vinyl, aryl, and sp3 C-H 2100 - 1800 cm-1: aromatic overtones 1690 cm-1: aryl ketone carbonyl 1670 cm-1: amide carbonyl The C=C will probably be very weak, as it is very nearly symmetric 1600 cm-1 and 1500 cm-1: benzene ring 5. Intensity: The intensity of the absorption is proportional to the change in the bond dipole as the bond is stretched. A C-F bond has the greatest bond dipole on the list, so it will give the most intense absorption. A C-I bond is not very polar so we expect it to have the weakest absorption. The order is C-F > C-Cl > C-Br > C-I. Energy: Stronger bonds are harder to stretch, but the energy is also dependent on the reduced mass of the atoms involved (see the text). Thus, it is not always trivial to make this prediction. Actual observed values are: C-F: 1400 - 1000 cm-1, C-Cl: 830 600 cm-1 (or below), C-Br: < 700 cm-1 and C-I: < 600 cm-1. In this case, there is an obvious trend. 6. Because the intensity of an IR absorption is dependent on the bond dipole, we can compare the intensity of similar bonds by considering the bonds dipoles. Formaldhyde has two hydrogens attached to the C=O instead of the two methyl groups in acetone. Methyls are weak electron donating groups, whereas hydrogens are essentially electroneutral (neither donate or withdraw from carbon; a function of small electronegativity difference as well as small atomic radius). Thus, the carbonyl carbon of formaldehyde has a greater + than does the carbonyl carbon of acetone. This means the C=O of formaldehyde is more polar than the C=O of acetone, so we expect a stronger C=O absorption for formaldehyde than for acetone. Trifluoromethyl is an electron-withdrawing group, so it should make the C=O of hexafluoroacetone even more polar than formaldehyde. Thus, we predict the following intensity of C=O absorptions: (CF3)2C=O > H2C=O > (CH3)2C=O. 7. The compound probably does not contain a carbonyl. If it does contain a carbonyl, then it must be sufficiently conjugated to fall below the normal carbonyl range. Ring strain can shift a carbonyl band above this range. The C=O stretch in cyclobutanone, for example, falls at 1775 cm-1. 8. Structure A: This structure is the best fit. CFQ & PP: Infrared Spectroscopy 107 Structure B: The spectrum does not contain a nonconjugated carbonyl stretch at around 1700 cm-1. If the absorption at 1644 cm-1 is a carbonyl, the energy suggests it to be a conjugated carbonyl. Structure B is excluded. Structure C: The IR spectrum does not contain a carbon-carbon triple bond stretch around 2200 cm-1. Structure C is excluded. 9. Molecular oxygen (O2) and nitrogen (N2) do not absorb infrared photons because their bond dipoles do not change when the bonds are stretched. Argon has no bond so it does not absorb photons in the IR region that we usually examine. Because these three materials that compose more than 99% of air do not absorb infrared photons in the region of interest, we generally do not take any special precautions about excluding air. However, water vapor and CO2, while present in small amounts, have significant IR absorptions, and can result in large peaks on the IR spectrum. 108 CFQ & PP: Infrared Spectroscopy

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2009H I G H E R S C H O O L C E R T I F I C AT E E X A M I N AT I O NInformation Processes and TechnologyTotal marks 100 S ection I Pages 28General Instructions Reading time 5 minutes Working time 3 hours Write using black or blue pen Draw diagrams us
École Normale Supérieure - GENERAL MA - General Ma
2009H I G H E R S C H O O L C E R T I F I C AT E E X A M I N AT I O N Legal StudiesTotal marks 100 S ection I Pages 29 25 marks This section has two parts, Part A and Part B Allow about 45 minutes for this section General Instructions Reading time 5
École Normale Supérieure - GENERAL MA - General Ma
2009H I G H E R S C H O O L C E R T I F I C AT E E X A M I N AT I O NSoftware Design and DevelopmentTotal marks 100 S ection I Pages 29General Instructions Reading time 5 minutes Working time 3 hours Write using black or blue pen Draw diagrams using
École Normale Supérieure - GENERAL MA - General Ma
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2002HI G HER S CHOO L CER T I FI CA T E EXA MINATIONArabic ContinuersTotal marks 80 Section I Pages 26 25 marks Attempt Questions 19 This section should take approximately 30 minutes Section II Pages 915 40 marks This secti
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2003HI G HER S CHOO L CER T I FI CA T E EXA MINATIONArabic ContinuersTotal marks 80 Section I Pages 25 25 marks Attempt Questions 18 This section should take approximately 30 minutes Section II Pages 915 40 marks This secti
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2004HI G HER S CHOO L CER T I FI CA T E EXA MINATIONArabic ContinuersTotal marks 80 Section I Pages 25 25 marks Attempt Questions 17 This section should take approximately 30 minutes Section II Pages 914 40 marks This secti
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2005HI G HER S CHOO L CER T I FI CA T E EXA MINATIONArabic ContinuersTotal marks 80 Section I Pages 25 25 marks Attempt Questions 17 This section should take approximately 30 minutes Section II Pages 915 40 marks This secti
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2006HIGHER SCHOOL CERTIFICATE EXAMINATIONArabic ContinuersTotal marks 80 S ection I Pages 26General Instructions Reading time 10 minutes Working time 2 hours and 50 minutes Write using black or blue pen Monolingual and/or
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2007HIGHER SCHOOL CERTIFICATE EXAMINATIONArabic ContinuersTotal marks 80 Section I Pages 25General Instructions Reading time 10 minutes Working time 2 hours and 50-minutes Write using black or blue pen Monolingual and/or b
École Normale Supérieure - GENERAL MA - General Ma
2007H I G H E R S C H O O L C E R T I F I C AT E E X A M I N AT I O NGeographyTotal marks 100 General Instructions Reading time 5 minutes Working time 3 hours Write using black or blue pen Board-approved calculators may be used A Stimulus Booklet is pr
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2008HIGHER SCHOOL CERTIFICATE EXAMINATIONArabic BeginnersListening SkillsGeneral Instructions You may NOT open the examination paper until instructed to do so on the recording Write using black or blue pen You may make no
École Normale Supérieure - GENERAL MA - General Ma
Centre NumberStudent Number2008HIGHER SCHOOL CERTIFICATE EXAMINATIONArabic ContinuersTotal marks 80 S ection I Pages 25General Instructions Reading time 10 minutes Working time 2 hours and 50 minutes Write using black or blue pen Monolingual and/or