12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy

Transcription

1 12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy

2 Determining the Structure of an Organic Compound The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants In the 19 th and early 20 th centuries, structures were determined by synthesis and chemical degradation that related compounds to each other Physical methods now permit structures to be determined directly. We will examine: mass spectrometry (MS) infrared (IR) spectroscopy nuclear magnetic resonance spectroscopy (NMR) ultraviolet-visible spectroscopy (VIS) 2

3 Why this Chapter? Finding structures of new molecules synthesized is critical To get a good idea of the range of structural techniques available and how they should be used 3

4 12.1 Mass Spectrometry of Small Molecules:Magnetic-Sector Instruments Bonds in cation radicals begin to break (fragment) Sample vaporized and subjected to bombardment by electrons that remove an electron creating a cation radical Mass to charge (m/z) ratio is measured Molecular ion (M + ) shows molecular weight 4

5 The Mass Spectrum Plot mass of ions (m/z) (x-axis) versus the intensity of the signal (roughly corresponding to the number of ions) (y-axis) Tallest peak is base peak (100%) Other peaks listed as the % of that peak Peak that corresponds to the unfragmented radical cation is parent peak or molecular ion (M + ) Propane MW = M + = 44 H H C H H C H H C H H 5

6 12.2 Interpreting Mass Spectra Molecular weight from the mass of the molecular ion Double-focusing instruments provide high-resolution exact mass atomic mass units distinguishing specific atoms Example MW 72 is ambiguous: C 5 H 12 and C 4 H 8 O but: C 5 H amu exact mass C 4 H 8 O amu exact mass Result from fractional mass differences of atoms 16 O = , 12 C = , 1 H = Instruments include computation of formulas for each peak If parent ion not present due to electron bombardment causing breakdown, softer methods such as chemical ionization are used 6

7 M + peak Example: Possible formulas can be determined from the mass. (M + ) Predict possible molecular formulas containing C, H, and maybe O if : M + = 86 M + = 156 7

8 M + peak Example: Possible formulas can be predicted from the mass. (M + ) Predict possible molecular formulas containing C, H, and maybe O if : M + = 86 M + = 156 C 6 H 14 C 5 H 10 O C 4 H 6 O 2 C 3 H 2 O 3 C 11 H 24 C 12 H 12 C 11 H 8 O C 10 H 20 O C 10 H 4 O 2 C 9 H 16 O 2 C 8 H 12 O 3 C 7 H 8 O 4 C 6 H 4 O 5 8

9 M + peak : Peaks above the molecular weight appear as a result of naturally occurring heavier isotopes in the sample (M+1) from 1.1% 13 C in nature (M and M+2) in 75.8%/24.2% ratio = 35 Cl and 37 Cl (M and M+2) in 50.7%/49.3% ratio = 79 Br and 81 Br Propane MW = M + = 44 H H C H H C H H C H H M+1 = 45 From 1.1% 13 C 9

10 Click on image to enlarge M + peak: Halides M + and M +2 in 75.8%:24.2% (~ 3:1) ratio = 35 Cl and 37 Cl CH 3 Cl M + and M +2 in 50.7%:49.3% (~ 1:1) ratio = 79 Br and 81 Br Br 10

11 Mass-Spec Fragmentation Patterns Molecular ions break into characteristic fragments that can be identifed Serves as a fingerprint for comparison with known materials in analysis (used in forensics) Positive charge goes to fragments that best can stabilize it 11

12 Mass-Spec Fragmentation Patterns M -15 = 57 CH 3 =15 MW=72; M + peak not seen 12

13 Mass Spec: Fragmentation of Hexane Hexane (m/z = 86 for parent) has peaks at m/z = 71, 57, 43, 29 71=Loss of CH 3 (15) 57=Loss of CH 3 CH 2 (29) 13

14 Learning Check: CH 3 MW=98 MW=98 Identify the following MS as that of methylcyclohexane or ethylcyclopentane. 14

15 Solution: CH 3 MW=98 MW=98 Identify the following MS as that of methylcyclohexane or ethylcyclopentane =29 loss of (CH 3 CH 2 ) 98-83=15 (loss of CH 3 ) M + =98 15

16 Learning Check: CH 3 MW=98 MW=98 Identify the following MS as that of methylcyclohexane or ethylcyclopentane. 16

17 Solution: CH 3 MW=98 MW=98 Identify the following MS as that of methylcyclohexane or ethylcyclopentane =15 (loss of CH 3 ) M + =98 17

18 12.3 Mass Spec: Common Functional Groups Alcohols: Alcohols undergo α-cleavage (at bond next to the C-OH) as well as loss of H-OH (18) to give C=C Loss of 18 18

19 Mass Spec: Alcohols H 3 C CH 3 CH 19

20 Mass Spec: Amines Nitrogen Rule: Amines with odd # of N s have Odd M + Amines undergo α-cleavage, generating radicals 20

21 Mass Spec: Carbonyl Compounds A C-H that is three atoms away leads to an internal transfer of a proton to the C=O, called the McLafferty rearrangement Loss of 28 Carbonyl compounds can also undergo α cleavage 21

22 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments Most biochemical analyses by MS use: - electrospray ionization (ESI) - Matrix-assisted laser desorption ionization (MALDI) Linked to a time-of-flight mass analyzer MALDI-TOF MS of chicken eg-white lysozyme 22

23 12.5 Spectroscopy and the Electromagnetic Spectrum Radiant energy is proportional to its frequency (cycles/s = Hz) as a wave (Amplitude is its height) Different types are classified by frequency or wavelength ranges 23

24 Electromagnetic Spectrum High frequency (υ) Short wavelength (λ) = High energy Low frequency (υ) Long wavelength (λ) = Low energy 24

25 Absorption Spectra Organic compound exposed to electromagnetic radiation, can absorb energy of certain wavelengths. Changing wavelengths to determine which are absorbed and which are transmitted produces an absorption spectrum Energy absorbed is shown as dips in spectrum High frequency (υ) Short wavelength (λ) = High energy Infrared Absorption of Ethyl Alcohol CH 3 CH 2 OH Low frequency (υ) Long wavelength (λ) = Low energy 25

26 12.6 Infrared Spectroscopy High frequency = High E Low frequency = Low E IR region lower energy than visible light (< red produces heating as with a heat lamp) m to m region used by organic chemists for structural analysis IR energy in a spectrum is usually measured as wavenumber (cm -1 ), the inverse of wavelength and proportional to frequency Specific IR absorbed by organic molecule related to its structure 26

27 Infrared Spectroscopy IR energy absorption corresponds to atomic movements, such as vibrations and rotations from bending and stretching of bonds between groups of atoms Energy is characteristic of the bonding of atoms in a functional group Bond stretching dominates higher energy modes Light objects connected to heavy objects vibrate fastest: C-H, N-H, O-H For two heavy atoms, stronger bond requires more energy: C C, C N > C=C, C=O, C=N > C-C, C-O, C-N, C-halogen 27

28 12.7 Interpreting Infrared Spectra Most functional groups absorb at a characteristic energy cm -1 N-H, C-H, O-H (stretching) N-H, O-H 3000 C-H cm -1 double bonds C=C, C=O, C=N (stretching) C=O C=C cm -1 Below 1500 cm -1 fingerprint Single bonds C-C, C-O, C-N, C-X (vibrations) cm -1 C C and C N (stretching) 28

29 Infrared Spectra: Functional Grps Characteristic higher energy IR absorptions used to confirm the existence of the presence of functional groups 29

30 12.8 IR Spectra: Functional Grps Alkane -C-H C-C Alkene Alkyne 30

31 IR: Aromatic Compounds (Subsituted benzene teeth ) C C 31

32 IR: Alcohols and Amines O-H broadens with Hydrogen bonding CH 3 CH 2 OH C-O Amines similar to OH N-H broadens with Hydrogen bonding 32

33 IR: Alcohols: O-H stretch Gas phase (no H-bonding) CCl 4 sln (0.25M) (some H-bonding) Liquid Film (Lots of H-bonding) 33

34 IR: Alcohols C-O 34

35 IR: Amines Primary and secondary amines exhibit a characteristic broad IR N-H stretching absorption between 3250 and 3500 cm -1. Primary amines show two strong peaks in this range, whereas secondary amines show only one. Primary amines also show a band near 1600 cm -1 due to a scissoring motion of the NH 2 group. Tertiary amines do not show any of these signals since they do not have a hydrogen bound to nitrogen. Amines similar to OH N-H broadens with Hydrogen bonding 35

36 IR: Amines Examples 36

37 IR: Carbonyls: C=O Aldehydes Carbonyls in general: Strong, sharp C=O peak 1670 to 1780 cm 1 Conjugation lowers stretching frequency 37

38 IR: C=O: Aldehydes 38

39 IR: C=O: Ketones Conjugation with a double bond or benzene ring lowers the stretching frequency by 30 to 40 cm

40 IR: C=O: Ketones Ring strain increases frequency: Incorporation of the carbonyl group in a small ring (5, 4 or 3-membered), raises the stretching frequency. 40

41 IR: C=O: Esters 1735 cm 1 in saturated esters Electron donating O increased the frequency 1715 cm 1 in esters next to aromatic ring or a double bond Conjugation decreases the frequency 41

42 IR: Carboxylic Acids 42

43 Learning Check: 32. Which of the following represents cyclohexane and which cyclohexene? 43

44 Solution: 32. Which of the following represents cyclohexane and which cyclohexene? =C- H -C-H -C=C- 44

45 Learning Check: 41.Propose a structure for the following unknown hydrocarbon: 45

46 Solution: =15 loss of CH3 41.Propose a structure for the following unknown hydrocarbon: M-1=67 M=68 C 5 H 8 2 deg of unsat C- H -C-H -C C- 46

47 Learning Check: 42. Propose a structure for the following unknown hydrocarbon: 47

48 Solution: 42. Propose a structure for the following unknown hydrocarbon: =15 loss of CH3 C 5 H 10 M=70 =C-H -C-H -C=C- 48

49 In mass spectrometry, what term is used to describe the ion that results from the ejection of one electron from a molecule? 1. base peak 2. parent peak 3. fragment 4. analyte 5. none of these 20% 20% 20% 20% 20%

50 What quantity is detected by mass spectrometry? 1. the energy of a molecule 2. the number of electrons ejected from a molecule 3. the number of ions of a particular mass to charge ratio 4. the number of electrons needed to ionize a molecule 5. the number of hydrogen atoms in a molecule 20% 20% 20% 20% 20%

51 High-resolution mass spectrometry enables one to determine the molecular formula of a molecule. 1. True 2. False 50% 50% 1 2

52 High-resolution mass spectrometry would allow one to distinguish between the following molecules. Br Br 50% 50% 1. True 2. False 1 2

53 Which of the given C 6 H 12 O isomers would be expected to produce an m/z peak at M + -18? 25% 25% 25% 25% 1. O 2. O OH O

54 Which of the following reasons explains why some mass spectrum peaks are larger than others? 1. The larger peaks represent fragments that are more stable. 2. The larger peaks represent fragments that are less stable. 3. Cations tend to give larger peaks than radical cations. 4. Radical cations tend to give larger peaks than cations. 5. None of these 20% 20% 20% 20% 20%

55 Which of the following compounds is most likely to undergo a McLafferty rearrangement? 1. NH 2 2. O 20% 20% 20% 20% 20% 3. O 4. OH 5. OH

56 Frequency is commonly reported in units of: 1. Joules 2. nm 3. amu 4. Hz 5. m/z 20% 20% 20% 20% 20%

57 The higher the wavenumber of a molecular vibration, the lower the energy of the infrared radiation needed to stimulate it. 1. True 2. False 50% 50% 1 2

58 Which type of electromagnetic radiation possesses the highest energy? 1. IR light 2. UV light 3. microwaves 4. visible light 5. FM radio waves 20% 20% 20% 20% 20%

59 Infrared spectroscopy is based on excitation. 1. electronic 2. rotational 3. nuclear 4. vibrational 25% 25% 25% 25%

60 What functional groups are most likely present in a compound whose IR spectrum shows absorbances at 2217 and 1648 cm -1? 1. aldehyde and alkene 2. alkene and alcohol 3. alcohol and nitrile 4. alkyne and ketone 5. nitrile and alkene 20% 20% 20% 20% 20%

61 (R)-2-pentanol and (S)-2-pentanol give identical IR spectra. 50% 50% 1. True 2. False 1 2

62 Which of the following compounds will have its carbonyl absorb at the lowest frequency in IR spectroscopy? 1. O 3. O H 2. O O 20% 20% 20% 20% 20% 4. O O 5. O

63 What approximate frequency range is considered the fingerprint region in infrared spectroscopy? cm cm cm cm cm -1 20% 20% 20% 20% 20%

64 (E)-2-Butene and (Z)-2-butene give identical IR spectra. 1. True 2. False 50% 50% 1 2