Infrared Spectroscopy and Mass Spectrometry Organic Chemistry Second Edition David Klein Chapter 15 Infrared Spectroscopy and Mass Spectrometry Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Please note: If your clicker system can only hold 5 multiple choice answers, we have provided ‘alternate answers’ for those questions in which the author originally had more than 5 choices. These answers appear on the slides with a green background. If your clicker system can hold more than 5 multiple choice answers, please delete the slides with the green backgrounds, and use the original answers the author has listed. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Section 15.1 1. Arrange the following from LARGEST to SMALLEST frequency.   Infrared Microwaves X rays Cosmic rays Visible Ultraviolet Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: D, C, F, E, A, B. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Infrared Microwaves X rays Cosmic rays Visible Ultraviolet Alternate Answer: Section 15.1 1. Arrange the following from LARGEST to SMALLEST frequency.   Infrared Microwaves X rays Cosmic rays Visible Ultraviolet A, B, D, C, F, E B, A, E, F, C, D B, A, F, E, C, D D, C, E, F, A, B D, C, F, E, A, B Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: E.   Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

releasing heat and getting colder remaining warm. 15.1 2. Medically speaking, IR is used in cancer detection. A patient is exposed to cold air, the nervous system reduces blood flow to the surface, conserving heat. The cancerous cells respond by releasing heat and getting colder remaining warm. recording a spike in temperature. Increasing their number. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: B. Remaining warm. While the non-cancerous cells reduce their temperature, the cancerous cells remain warm, thereby allowing easy detection via IR of the location and size of the tumor. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

 Section 15.2 3. IR radiation causes _______excitation in covalent bonds. The excitation can either be a _______ or a _______ (which can either be in-plane and called _______ or out-of-plane and called _______). Bend Scissoring Stretch Twisting Vibrational Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: E, C, A, B, D. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Alternate Answer: Section 15. 2 3 Alternate Answer:  Section 15.2 3. IR radiation causes _______excitation in covalent bonds. The excitation can either be a _______ or a _______ (which can either be in-plane and called _______ or out-of-plane and called _______). Bend Scissoring Stretch Twisting Vibrational D, B, A, C, E D, B, C, A, E E, C, A, B, D E, C, B, A, D E, C, D, B, A Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: C. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

15.2 4. Infrared radiation of molecules results in vibrational excitation of bonds excitation of metals excitation of water relaxation of bonds. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: A. vibrational excitation of bonds. This vibrational excitation will cause bonds to stretch and compress. The bonds will also bend in- and out-of-plane. All these changes can be monitored by infrared. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

15.2 5. Wavenumber is Proportional to frequency so large wavenumbers represent higher energy Frequency of light, divided by a constant (the speed of light, c) Has units of inverse centimeter (cm-1) All of the above. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: D. All of the above. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

I > II > III > IV IV > I > III > II 15.2 6. According to Hooke’s Law, the wavenumber of absorption is dependent upon bond strength and atom mass. If the bond’s strength is similar (for example all single bonds), rank the following in order of decreasing wavenumber. C—H C—Br C—O C—D I II III IV I > II > III > IV IV > I > III > II I > IV > III > II II > III > IV > I Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: C. I > IV > III > II Hooke’s Law uses the reduced mass (mred) and it is in the denominator so the greater the mass the smaller the wavenumber. Hydrogen with the smallest mass will have the larger wavenumber. C—H has n ~3000 cm-1 and C—Br n ~700 cm-1 See Conceptual Checkpoint 15.1-15.4 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

 Section 15.3, 15.4, 15.5, 15.6 7. What IR signal (dilute solution) would we expect to see for the indicated functional group on this molecule?  a. intense stretch at 3620  b. intense stretch at 1580  c. intense stretch at 1720 d. intense stretch at 1650  e. intense stretch at 1780  f. broad peak at 2550  g. broad peak at 3250 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: d.   IR spectroscopy reads vibrations caused on organic bonds by IR light. For organic molecules, different functional groups will vibrate at different frequencies based upon their structure. The double bond will stretch back and forth like a spring between frequencies of 1600 and 1700. For more examples of this type of problem, see Skillbuilder 15.1. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

a. intense stretch at 3620 b. intense stretch at 1580 Alternate Answer: Section 15.3, 15.4, 15.5, 15.6 7. What IR signal (dilute solution) would we expect to see for the indicated functional group on this molecule?  a. intense stretch at 3620  b. intense stretch at 1580  c. intense stretch at 1700 d. intense stretch at 1650  e. intense stretch at 1780  f. broad peak at 2550  g. broad peak at 3250 A B C D E Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: D. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

15. 3 8. There are two regions of the IR spectrum 15.3 8. There are two regions of the IR spectrum. One is the diagnostic region and the other is the fingerprint. Which statement best describes the differences? The diagnostic region has signals for the different functional groups present while the finger print region will tell how these are connected. The diagnostic region is for single bonds only; and the fingerprint region has signals from multiple bonds. The diagnostic region has signals for notable functional groups while the fingerprint region can be used to pinpoint specific compounds. The diagnostic region shows connections between groups while the fingerprint region shows how many groups are present. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: C. The diagnostic region has signals for notable functional groups while the fingerprint region can be used to pinpoint specific compounds. The diagnostic region is that above 1000 wavenumbers. This region contains mostly double, triple and H-X bonds which have distinctive vibrational frequencies when exposed to IR light. The fingerprint region (wavenumber less than 1000cm-1) is where many of the C-C and C-X bonds will be found. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

 Section 15.3, 15.4, 15.5, 15.6 9. What IR signal (dilute solution) would we expect to see for the indicated functional group on this molecule?  a. intense stretch at 3620  b. intense stretch at 1580  c. intense stretch at 1700  d. intense stretch at 1650  e. intense stretch at 1780  f. broad peak at 2550  g. broad peak at 3250 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: c. IR spectroscopy reads vibrations caused on organic bonds by IR light. For organic molecules, different functional groups will vibrate at different frequencies based upon their structure. The double bond of a ketone will stretch back and forth like a spring between frequencies of 1680 and 1750. For more examples of this type of problem, see Skillbuilder 15.1.   Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

 Section 15.3, 15.4, 15.5, 15.6 9. What IR signal (dilute solution) would we expect to see for the indicated functional group on this molecule? Alternate Answers:  a. intense stretch at 3620  b. intense stretch at 1580  c. intense stretch at 1700  d. intense stretch at 1650  e. intense stretch at 1780  f. broad peak at 2550  g. broad peak at 3250 A B C D E Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: C Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Section 15.3, 15.4, 15.5, 15.6 10. What IR signal (dilute solution) would we expect to see for the indicated functional group on this molecule?  a. intense stretch at 3620  b. intense stretch at 1580  c. intense stretch at 1700  d. intense stretch at 1650  e. intense stretch at 1780  f. broad peak at 2550  g. broad peak at 3250 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: a. IR spectroscopy reads vibrations caused on organic bonds by IR light. For organic molecules, different functional groups will vibrate at different frequencies based upon their structure. The single bond between an oxygen and a hydrogen of an alcohol (in dilute solution, non-hydrogen bonded) will stretch back and forth like a spring between frequencies of 3590 and 3650. For more examples of this type of problem, see Skillbuilder 15.1. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Alternate Answer: Section 15. 3, 15. 4, 15. 5, 15. 6 10 Alternate Answer: Section 15.3, 15.4, 15.5, 15.6 10. What IR signal (dilute solution) would we expect to see for the indicated functional group on this molecule?  a. intense stretch at 3620  b. intense stretch at 1580  c. intense stretch at 1700  d. intense stretch at 1650  e. intense stretch at 1780  f. broad peak at 2550  g. broad peak at 3250 A B C D E Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: A. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

C=C at 1620 cm-1 C=O at 1690 cm-1 O-H at 3220 cm-1 N-H at 3400 cm-1 15.3 and 15.6 11. Which representative IR peak(s) would you expect for the structure below? C=C at 1620 cm-1 C=O at 1690 cm-1 O-H at 3220 cm-1 N-H at 3400 cm-1 I and IV II and III I and III I only Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: C. I and III. The important structural features present are hydroxyl group with distinctive O-H signal between 3200 and 3600 cm-1 and the C=C bond with a signal at 1600-1700 cm-1. See Skillbuilder 15.1 for more examples. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Presence of O-H for an alcohol in structure I 15.6 and 15.7 12. What IR characteristics indicated below would be used to differentiate between structure I and structure II? I. II. Presence of O-H for an alcohol in structure I The C-H of the aldehyde in structure II Presence of C=O for aldehyde in structure II Presence of C=O for ketone in structure I II only II and III I and IV III and IV Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: A. II only The presence of the C-H bond with its IR stretch at 2750-2850 cm-1 would show up in IR spectrum for structure II but not I. Both the aldehyde and ketone C=O have signals in the region of 1680-1750 cm-1. See Skillbuilder 15.1 and 15.2. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

The student is correct in her interpretation. 15.6 13. A student indicated that the spectrum with very broad peak centered at 2986 cm-1, a strong stretch at 1715 cm-1, an in-plane bend at 1415 cm-1, and medium stretch at 1240 cm-1 represents a carboxylic acid. What statement best represents this student’s interpretation? The student is correct in her interpretation. The student is wrong because this represents an alcohol with the broad peak for O-H at 2986 The student is wrong because this would be an aldehyde due to the C=O stretch at 1715 The student is wrong because there is way to know without seeing the actual spectrum Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: A. The student is correct in her interpretation. She cannot tell exactly which carboxylic acid is present because not enough information is given for the fingerprint region. Neither B or C are possible because more than just those peaks are present. D is not correct because one can make some predictions from representative peaks but it is always better to see the actual spectrum and the shape of peaks. See Skillbuilder 15.1 and 15.2 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Section 15.7 14. Indicate the peak(s) from infrared spectroscopy that will be lost and gained during the following transformations.    A. Loss of broad peak between 3200-3600 cm-1 B. Gain of peak between 2100-2200 cm-1 C. Loss of peak between 500-600 cm-1 D. Gain of peak between 3200-3600 cm-1 E. Gain of peak between 1600-1700 cm-1 F. Gain of peak between 1680-1750 cm-1 G. Gain of peak between 500-600 cm-1 H. Gain of peak at 3300 cm-1 I. Loss of peak between 1680-1750 cm-1 J. Gain of peak between 3000-3100 cm-1 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

For more examples of this type of problem, see Skillbuilder 15.2. Answer: I: I, D II: A, E III: A, E, J IV: C, B, H For more examples of this type of problem, see Skillbuilder 15.2.    Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Alternate Answers: Section 15.7 14. Indicate the peak(s) from infrared spectroscopy that will be lost and gained during the following transformations.    A. Loss of broad peak between 3200-3600 cm-1 B. Gain of peak between 2100-2200 cm-1 C. Loss of peak between 500-600 cm-1 D. Gain of peak between 3200-3600 cm-1 E. Gain of peak between 1600-1700 cm-1 F. Gain of peak between 1680-1750 cm-1 G. Gain of peak between 500-600 cm-1 H. Gain of peak at 3300 cm-1 I. Loss of peak between 1680-1750 cm-1 J. Gain of peak between 3000-3100 cm-1 I = A, F; II = D, I; III = F, A ,J; IV = C, B, H I = A, F; II = A, E; III = A, E, J; IV = B, C, H I = I, D; II = F, A; III = F, A, J; IV = B, C, H I = I, D; II = A, E; III = A, E, J; IV = C, B, H I = I, D; II = A, F; III = A, F, J; IV = C, B, H Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: D Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Section 15. 8 15. In mass spectrometry, a compound is Section 15.8 15. In mass spectrometry, a compound is _______and converted into _______, which are then separated and _________.   The most common technique to do this involves bombarding the compound with high-energy _______. This technique will generate a _______ that is called a _______, which will often undergo _______ to form more stable species. Detected Electrons Fragmentation Ions Molecular Ion Radical Ion Vaporized Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: G, D, A, B, F, E, C.    Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Alternate Answer: Section 15.8 15. In mass spectrometry, a compound is _______and converted into _______, which are then separated and _________.   The most common technique to do this involves bombarding the compound with high-energy _______. This technique will generate a _______ that is called a _______, which will often undergo _______ to form more stable species. Detected Electrons Fragmentation Ions Molecular Ion Radical Ion Vaporized A, B, F, E, C, D, G B, F, E, C, G, A, D C, E, F, B, A, D, G D, G, C, F, E, B, A G, D, A, B, F, E, C Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: E. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

M+. = m/z = 132, relative abundance 38.3% Section 15.8 16. What is the molecular formula of a compound with the following data: M+. = m/z = 132, relative abundance 38.3% (M+1)+. = m/z = 133, relative abundance 2.95% C6H14 C6H14O C6H14O2 C7H16 C7H16O C7H16O2 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

For more examples of this type of problem, see Skillbuilder 15.3. Answer: F For more examples of this type of problem, see Skillbuilder 15.3. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

M+. = m/z = 132, relative abundance 38.3% Alternate Answer: Section 15.8 16. What is the molecular formula of a compound with the following data: M+. = m/z = 132, relative abundance 38.3% (M+1)+. = m/z = 133, relative abundance 2.95% C6H14 C6H14O C6H14O2 C7H16 C7H16O C7H16O2 B C D E F Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: E.   Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Section: 15.10, 15.11 17. In a mass spectrum of a compound, the following data (intensity) is given. M+ = 80, M+ +1 = 8.8, M+ +2 = 26. This compound will have  ______  carbons and will contain a  ______ .  a. 8, sulfur  b. 9, silicon  c. 10, chlorine  d. 11, bromine Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: c. For carbons, 8.8/80 × 100 = 0.11 × 100 = 11 %/1.1 = 10 carbons. For other atoms, 26/80 × 100 = 0.325 × 100 = 32.5 %. For M+ +2, Si = 3.35%, S = 4.4 %, Cl = 32.5% and Br = 98 %. Therefore, this contains chlorine. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

15.12 18. In mass spectroscopy, signals at M-15 and M-29 indicate the loss of _______ and _______ groups respectively. vinyl, methyl methyl, propyl Ethyl, allyl methyl, ethyl Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: D. methyl, ethyl. A peak at M-15 indicates the loss of a methyl radical .CH3 with 1 C and 3 H for total mass of 15. A peak at M-29 would represent the loss of an ethyl radical .CH2CH3 with 2 C and 5 H for total mass of 29. See Skillbuilder 15.3 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

4 degrees of unsaturation 3 degrees of unsaturation 15.16 19. Calculate the degrees of unsaturation for the formula C8H12O. 4 degrees of unsaturation 3 degrees of unsaturation 2 degrees of unsaturation 1 degree of unsaturation Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: B. 3 degrees of unsaturation in C8H12O. If the compound were fully saturated, the 8 carbon atoms would need 2n+2 hydrogen atoms or 18 H. There are only 12 H so there is a deficiency of 6 (18-12), this corresponds to 3 degrees of unsaturation. Remember the oxygen atoms do not affect the 2n+2 rule. See Skillbuilder 15.4. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

4 degrees of unsaturation 3 degrees of unsaturation 15.16 20. Calculate the degrees of unsaturation for the formula C5H9NO. 4 degrees of unsaturation 3 degrees of unsaturation 2 degrees of unsaturation 1 degree of unsaturation Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: C. 2 degrees of unsaturation in C5H9NO. If the compound were fully saturated, the 5 carbon atoms would need 2n+2 hydrogen atoms or 12 H. Nine H atoms are present, the N atom causes us to subtract 1 H and the oxygen can be ignored. This leaves 8 H to be accounted (9-1). The hydrogen deficiency is then 4 (12-8) and this means 2 degrees of unsaturation. See Skillbuilder 15.4. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

One triple bond, one double bond One triple bond, two rings 15.16 21. If a compound has 4 degrees of unsaturation, which set of conditions is NOT possible for the structure? One triple bond, one double bond One triple bond, two rings Two rings and two double bonds One triple bond, one double bond, one ring Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: A. One triple bond and one double bond Answer: A. One triple bond and one double bond. A triple bond is equivalent to 2 HDI, double bond and rings are both equivalent to 1 HDI. The presence of one triple bond (2) and one double bond (1) would be 3 degrees of unsaturation not 4. See Skillbuilder 15.4. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Section: 15.12 22. Calculate the HDI (hydrogen deficiency index) for each of the following. A. C20H40   B. C16H24O   C. C8H10Br2Cl4   D. C9H14Br3O2N Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

For more examples of this type of problem, see Skillbuilder 15.4. Answer: A: 1 B: 5 C: 1 D: 2 For more examples of this type of problem, see Skillbuilder 15.4. Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Alternate Answer: Section: 15. 12 23 Alternate Answer: Section: 15.12 23. Calculate the HDI (hydrogen deficiency index) for each of the following.   A. C20H40   B. C16H24O   C. C8H10Br2Cl4   D. C9H14Br3O2N A = 1; B = 4; C = 2; D = 2 A = 1; B = 5; C = 1; D = 2 A = 2; B = 4; C = 1; D = 2 A = 2; B = 5; C = 2; D = 1 A = 2; B = 5; C = 1; D = 2 Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.

Answer: B.   Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.