CHEM 241 UNIT 5: PART A DETERMINATION OF ORGANIC STRUCTURES BY SPECTROSCOPIC METHODS [MASS SPECTROMETRY]

Transcription

1 CHEM 241 UNIT 5: PART A DETERMINATION OF ORGANIC STRUCTURES BY SPECTROSCOPIC METHODS [MASS SPECTROMETRY] 1

2 Introduction Outline Mass spectrometry (MS) 2

3 INTRODUCTION 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 (NMR) spectroscopy ultraviolet-visible (UV) spectroscopy We shall focus on MS and IR in this Part. 3

4 Introduction Cont d What is spectroscopy? Interaction of radiant energy with matter 4

5 Structure Determination and Spectroscopy Mass spectrometry Infrared Spectroscopy Ultraviolet Spectroscopy Nuclear Magnetic Resonance Spectroscopy What size and formula? What functional groups are present? Is a conjugated pi electron system present? What carbon hydrogen framework is present? 5

6 Mass Spectrometry (MS) Measures molecular weight Sample vaporized and subjected to bombardment by electrons that remove an electron Creates a cation-radical Bonds in cation radicals begin to break (fragment) Charge to mass ratio is measured (see next page) 6

7 A Mass Spectrometer 7

8 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 + ) 8

9 Molecular Ion Molecular ion (M + ): the species formed by removal of a single electron from a molecule For our purposes, it does not matter which electron is lost; radical cation character is delocalized throughout the molecule. Therefore, we write the molecular formula of the parent molecule in brackets with a plus sign to show that it is a cation a dot to show that it has an odd number of electrons 9

10 MS Examples: Methane and Propane Methane produces a parent peak (m/z = 16) and fragments of 15 and 14 (See Figure 12-2 a) The MS of propane is more complex (Figure 12-2 b) since the molecule can break down in several ways 10

11 Methane Methane m/z=16 M + Base peak M

12 Propane m/z=44 Base peak M + M

13 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 13

14 Other Mass Spectral Features If parent ion not present due to electron bombardment causing breakdown, softer methods such as chemical ionization are used Peaks above the molecular weight appear as a result of naturally occurring heavier isotopes in the sample (M+1) from 13 C that is randomly present 14

15 Interpreting Mass-Spectral Fragmentation Patterns The way molecular ions break down can produce characteristic fragments that help in identification Serves as a fingerprint for comparison with known materials in analysis (used in forensics) Positive charge goes to fragments that best can stabilize it 15

16 Mass Spectral Fragmentation of Hexane Hexane (m/z = 86 for parent) has peaks at m/z = 71, 57, 43, 29 16

17 Mass-Spectral Behavior of Some Common Functional Groups Functional groups cause common patterns of cleavage in their vicinity 17

18 Mass Spectral Cleavage Reactions of Alcohols Alcohols undergo α-cleavage (at the bond next to the C-OH) as well as loss of H-OH to give C=C 18

19 Mass Spectral Cleavage of Amines Amines undergo α-cleavage, generating radicals 19

20 Fragmentation of Ketones and Aldehydes A C-H that is three atoms away leads to an internal transfer of a proton to the C=O, called the McLafferty rearrangement Carbonyl compounds can also undergo α cleavage 20

21 Mass Spectrum of Toluene Base peak 21

22 Fragmentation of Toluene m + CH 3 H H CH 2 m-1 Tropylium ion m/e=65 22

23 Common Fragments 23

24 CH 3 CH 2 CH 2 + CH 3 CH 2 + CH 3 CH 2 CH 2 CH 2 + M + = 86 CH 3 CH 2 CH 2 CH 2 CH

25 Interpreting Mass Spectral Fragmentation Patterns The mass spectra provides a fingerprint of the compound. There is little likelihood that two spectra will have the same spectra. 25

26 Interpreting Mass Spectra Molecular Weight Knowing the molecular weight allows one to determine possible formulas. Degree of unsaturation # C - (# H + X)/2 +(# N)/ (10)/2 + 1 = 3 m/z = 110 C 8 H 14 C 7 H 10 O C 6 H 6 O 2 C 6 H 10 N 2 26

27 Interpreting MS The only elements to give significant M + 2 peaks are Cl and Br. If no large M + 2 peak is present, these elements are absent Is the mass of the molecular ion odd or even? Nitrogen Rule: if a compound has zero or an even number of nitrogen atoms, its molecular ion will appear as a even m/z value an odd number of nitrogen atoms, its molecular ion will appear as an odd m/z value 27

28 CH 3 CH 2 CH 2 + CH 3 CH 2 + CH 3 CH 2 CH 2 CH 2 + M + = 86 CH 3 CH 2 CH 2 CH 2 CH

29 Isotopes Virtually all elements common to organic compounds are mixtures of isotopes Element Atomic weight Isotope Mass (amu) Relative Abundance hydrogen H 2 H carbon C 13 C nitrogen N 15 N

30 Isotopes oxygen O 17 O 18 O sulfur S 33 S 34 S chlorine Cl 37 Cl bromine Br 81 Br

31 Isotopes Carbon, for example, in nature is 98.90% 12 C and 1.10% 13 C. Thus, there are 1.11 atoms of carbon-13 in nature for every 100 atoms of carbon-12 ( M + 1/M) = maximum number of carbons (degree of unsaturation) 31

32 Isotopes What is the maximum number of carbons a unknown substance can have if the relative abundance of the parent peak is 40 and the M + 1 peak, 1.4? ( M + 1/M) 1.1 ( 1.4/40) 100 = maximum number of 100 = maximum number of carbons = 3 carbons 32

33 M+2 and M+1 Peaks The most common elements giving rise to M + 2 peaks are chlorine and bromine Chlorine in nature is 75.77% 35 Cl and 24.23% 37 Cl a ratio of M to M + 2 of approximately 3:1 indicates the presence of a single chlorine in a compound Bromine in nature is 50.7% 79 Br and 49.3% 81 Br a ratio of M to M + 2 of approximately 1:1 indicates the presence of a single bromine in a compound 33

34 Chloromethane 34

35 Bromomethane 35

36 M+2 and M+1 Peaks Sulfur is the only other element common to organic compounds that gives a significant M + 2 peak 32 S = 95.02% and 34 S = 4.21% Because M + 1 peaks are relatively low in intensity compared to the molecular ion and often difficult to measure with any precision, they are generally not useful for accurate determinations of molecular weight 36

37 Isotope Effects 1 chloride will contain a M + 2 peak approximately one third the intensity of the molecular ion peak because of the presence of 37 Cl. 2 chlorides will contain a M + 2 peak approximately one third the intensity of the molecular ion peak and a M + 4 peak about one third the intensity of the M + 2 peak. 37

38 Isotope Effects 1 bromide will contain a M + 2 peak approximately equal the intensity of the molecular ion peak because of the presence of 91 Br. 1 bromide will contain a M + 2 peak approximately twice the intensity of the molecular ion peak and a M + 4 peak about equal to the molecular ion peak. 38

39 Try this Assume you have two unlabeled samples, one of methylcyclohexane and the other of ethylcyclopentane. How could you use mass spectrometry to tell which is what? 39

40 Try this 40

41 Strategy Look at the two possible structures and decide how they differ. Then think about how any of these differences might give rise to differences in mass spectra. Methylcyclohexane has a CH 3 group, and ethylcyclopentane has a CH 2 CH 3 group, which should affect the fragmentation patterns. 41

42 Solution Both samples have a M + = 98, corresponding to C 7 H 14. Sample B has a base peak at m/z = 83, corresponding to the loss of a CH 3 group from M +. Sample A has a small peak at m/z = 83. Sample A has a base peak at m/z = 69, corresponding to the loss of a CH 2 CH 3 group. However, B has a rather small peak at m/z = 69. Therefore, we can assume that B is methyl cyclohexane and A, ethylcyclopentane. 3 42