MASS SPECTROMETRY: BASIC EXPERIMENT

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1 relative abundance Pavia MASS SPECTROMETRY: BASIC EXPERIMENT scienceaid.co.uk -e Molecule Molecule e base peak [Fragments] +. fragment peaks molecular ion molecular ion cluster m/z Electron Ionization Effect of ionizing energy on spectrum 1-propanol (C 3 H 8 O, M = 60) 70 ev 20 ev M 14 ev 11 ev e - + 2e -.

2 Chemical Ionization Effect of ionization on spectrum EI MW=142 CH + 5 CH 4 M CH 4 CH + 5 CH 4 (M+H) + CH 4 CH 5 + CH 4 (CH 3 ) 2 CHCH 3 CH 4 + e CH e CH 4+ + CH 4 CH 5+ + CH 3 CH + 5 (M+H) + + CH 4 FRAGMENTATION Pavia 8.8 Radical cations formed upon electron impact are unstable! One or the other of these process will predominate Large peaks result from fragmentations that are energetically most favorable Remember: You will only see peaks for positively charged species (cations and radical cations), X + and Y + not neutral molecules or radicals.

3 Fragmentation of Radical Cations Radical cations are very reactive species Radical cations are very unstable. Fragmentations are unimolecular chemical reactions. The relative stability of cations and radical cations can often be understood in terms of substitution patterns and resonance. Remember: Ease of formation of carbocations less stable more stable c.f. ease of S N 1 reactions of R-Hal, or E1 dehydration of ROH Alkyl groups are inductive electron donating substituents Allylic and benzylic cations are resonance stabilized

4 Spectral Database for Organic Compounds (SDBS) Alkanes Pavia 8.8 I hexane (C 6 H 14 ; M = 86) M M M 71 M base peak small M 15 Molecular ions fragment by cleavage of C-C bonds to give one cation and one radical. The cations are detected. 29 M M M M 15 86

5 Peaks appear in clusters separated by 14 mass units. M 15, M 29, M, M, M 71, M 85, M 99,. with other fragments corresponding to 1 and 2 mass units fewer by lose hydrogen atoms (radicals): 2-methylpentane M 15, M 29, M, M, M 71, M 85, M 99,. 29 M M M M base peak larger M 15 smaller

6 3-methylpentane M 15, M 29, M, M, M 71, M 85, M 99,. M M base peak even smaller 2,2-dimethylbutane M 15, M 29, M, M, M 71, M 85, M 99,. M M M M-?!? Prominent M-15 No!!

7 Branched alkanes undergo more fragmentation - small peak, more predominant M-15 Carbocations rearrange Which is which? (all C 6 H 14 ) hexane 2-methylpentane 2,2-dimethylbutane 86 71

8 Which is which? (all C 10 H 22 ; M = 142) 3,4-diethylhexane 3,3-dimethyloctane 5-methylnonane Summary: Alkanes For straight chain alkanes - Molecular ion peak is usually present but weak (esp. for branched alkane) - Clusters of fragments appear spaced by m/z of 14 - The largest peak in each cluster corresponds to an alkyl radical cation, C n H 2n+1 - A peak for M CH 3 is often weak or absent - More fragmentation at highly branched positions

9 Cycloalkanes Pavia 8.8 J cyclohexane (C 6 H 12 ; M = 84) 56 M 28 E 69 M methylcyclopentane (C 6 H 12 ; M = 84) base peak even m/z, M 28 from rearrangement strong!! Alkenes Pavia 8.8 K 1-butene (C 4 H 8 ; M = 56) 41 M 15 Fragmentation to give allylic cation 55 M E-2-hexene (C 6 H 12 ; M = 84) 55 M 29 strong peak 69 M 15 84

10 Pavia 8.8 K 1-pentene (C 5 H 10 ; M = 70) 42 M M Hexenes (C 7 H 12 ; M = 84) Pavia 8.8 K 41 M M 28 M M M M M 29 Not possible to determine location or E/Z stereochemistry of C=C

11 39 M M M-1 67 M-15 82

12 Summary: Alkenes and Alkynes 1. Relatively strong ion (removal of p electron does not result in bond breaking 2. Strong peak arises from cleavage of alkyl radical to form a resonance stabilized allylic cation (M-15, M-29, M-, ) terminal alkene CH 2 CH=CH 2 (m/z = 41) 3. McLafferty rearrangement loss of alkene M 28, M 42, M 56, M 70, M 84, M 98,. 3. Difficult to identify position of p bond since the double bond migrates easily. No information about E/Z isomers. cyclohexene (C 6 H 10 ; M = 82) 54 M Retro-Diels-Alder Reaction

13 limonene (C 10 H 16 ; M = 136) 68 M Which is which? (all C 8 H 16 ; M = 112) propylcyclopentane 2-ethyl-1-hexene 2,5-dimethyl-1-hexene

14 Aromatic Hydrocarbons toluene (C 7 H 8 ; M = 92) 91 M 1 92 Pavia 8.8 M ethylbenzene (C 8 H 10 ; M = 106) 91 M butylbenzene (C 10 H 14 ; M = 134) 91 M 134 -bromotoluene (C 7 H 7 Br, M = 170[ 79 Br]) 91 M M+2

15 What are all those m/z = 91 peaks for PhCH 2 X? or C 7 H 7 (m/z = 91) 1,4-diethylbenzene, M =134

16 1-butylbenzene (C 10 H 14 ; M = 134) 91 M 92 M Alcohols, Ethers and Amines Pavia 8.8 N,O,T 1-hexanol (C 6 H 14 O, M = 102) 31 M 71 M M M M 18 MW 102 m/z = 31 M 18 M 18: Dehydration m/z = 31 (M-71): -cleavage of an alkyl radical

17 M 46: Dehydration with loss of alkene 3-methyl-3-heptanol (C 8 H 18 O, M = 130) 73 M 101 M M 15 MW = heptanol (C 7 H 16 O, M = 116) 59 M 69 M M 29 MW = 116

18 Spectral Database for Organic Compounds (SDBS) 2,2-dimethyl-3-pentanol (C 7 H 16 O, M = 116) M M 87 M 29 MW=116 Which C 6 H 12 O alcohol? 45 M 69 M M M

19 56 70 Which is which? (all C 7 H 16 O; M = 116) heptanol 2,2-dimethyl-3- pentanol 3-ethyl-3-pentanol Methyl 1-propyl ether (C 4 H 10 O, M = 74) 45 M Diethyl ether (C 4 H 10 O, M = 74) 31 M 59 M 15 74

20 ethyl 1-butyl ether (C 6 H 14 O, M = 102) 31 M M M = ethoxypentane (C 7 H 16 O, M = 116) 59 M 87 M 29 MW =116 What are the formulas of each labeled fragment? Explain how these fragments are formed. 45 M M M M M M 42

21 Spectral Database for Organic Compounds (SDBS) Which C 6 H 12 O ether? Bear in mind that methyl ethers do not fragment by cleavage of the CH 3 -O bond. 45 M 87 M N,N-dibutylamine (C 8 H 19 N, M = 129) 86 M 129

22 2-octanamine (C 8 H 19 N, M = 129) 44 M 85 MW = 129 2,4,4-trimethylpentan-2-amine (C 8 H 19 N, M = 129) 58 M 71 MW = Identify each compound There is one alcohol, one ether and one amine dipropylether 3,3-dimethyl-2-butanol triethylamine

23 science.widener.edu/svb/massspec/ci.html E.I and C.I of tri-n-butylamine C 12 H 27 N (MW = 185) C.I. (CH 4 ) Summary: Alcohols, Ethers and Amines -cleavage to give resonance-stabilized cation Dehydration (M-H 2 O) McLafferty (-H 2 O, -CH 2 =CH 2 ) -cleavage (-Rˑ) followed by elimination (-C n H 2n ) from another substitutent (ethers, 2 /3 amines)

24 Spectral Database for Organic Compounds (SDBS) Spectral Database for Organic Compounds (SDBS) Aldehydes and ketones Pavia 8.8 P-Q propanal (C 3 H 6 O, MW = 58) 29 M M 1 2-butanone (C 4 H 8 O, MW = 72) M 29 M 15 72

25 hexanal (C 6 H 12 O, MW = 100) 44 M M 44 2-hexanone (C 6 H 12 O, MW = 100) M 58 M M M

26 Spectral Database for Organic Compounds (SDBS) benzaldehyde (C 7 H 6 O, MW = 106) 77 M M acetophenone (C 8 H 8 O, M = 120) 77 M 105 M

27 Spectral Database for Organic Compounds (SDBS) Which is which? Which ketone? 105 M 77 M

28 Spectral Database for Organic Compounds (SDBS) Summary: Aldehydes and Ketones 1. -cleavage: C C(=O) Loss of larger R predominates M-1, M-15, M-29, M-,. 2. McLafferty rearrangement: transfer of g-h to C=O Becomes more important as R M-28, M-42, M-56,. Carboxylic acids propanoic acid (C 3 H 6 O 2, M = 74) 29 M Pavia 8.8 R,S 45 M 29 M M 1

29 Spectral Database for Organic Compounds (SDBS) Spectral Database for Organic Compounds (SDBS) butanoic acid (C 4 H 8 O 2, M = 88) 60 M 28 M M 71 M benzoic acid (C 7 H 6 O 2, M = 122) 77 M M

30 Spectral Database for Organic Compounds (SDBS) Summary: Carboxylic acids 1. M-1 2. Cleavage of C(O)-OH 3. Cleavage of R-C(O) [M-R] + and [CO 2 H] +ˑ m/z = McLafferty rearrangement for CH-C-C-C(=O)OH Esters methyl acetate (C 3 H 6 O 2, M = 74) M M 15 74

31 Modes of fragmentation of esters Cleavage of alkoxy radical (R O ) from carbonyl Cleavage of alkyl (R) radical or cation from carbonyl [M OR ] + m/z = M 31, M 45, M 59,... [M R] + m/z = M 15, M 29, M, R + m/z = (15,29),,, 71,... McLafferty rearrangement - Elimination of an alkene (C n H 2n ) from the alkyl substituent to give M C n H 2n + [M C n H 2n ] +. m/z = 28, 42, 56, 70, 84,... Alkyl ( R ) versus alkoxy cleavage ( OR ) and McLafferty rearrangement ( C n H 2n ) of alkyl substituent 29 M-59 M M-59 M M M-42 M=88 M= M-56 M= M=130 α-cleavage of R increases as R ; but McLafferty begins to dominate α-cleavage of CH 3 O becomes less important

32 More modes of fragmentation of esters McLafferty rearrangement on alkoxy side of the ester Elimination of a carboxylic acid from the alkoxy substituent to give C n H 2n+ [C n H 2n ] +. m/z = 28, 42, 56, 70,... "McLafferty plus one" rearrangement Transfer of two hydrogen atoms from the alkoxy group. Loss of C n H 2n-1 to give RCO 2 H 2 + [M C n H 2n-1 ] + m/z = M 41, M 55, M 69, M 83,... and finally, fragmentation from OR by loss of alkyl from -position Alkoxy cleavage, McLafferty rearrangement, McLafferty+1 rearrangement and a-cleavage of R O-C(=O)R 28 M-60 M-45 M M-60 M-73 M M M M M M M M M- 70 M M- M = 88 M = α-cleavage of R O predominates. McLafferty rearrangement with elimination of RO 2 H becomes more important as R M = 116 M = 130

33 Predict the peaks that appear in the MS of hexyl butanoate, C 10 H 20 O 2 (MW=172) Cleavage of R + and R from C=O Loss of C n H 2n+1 O (M 31,45,59, ) M = 129 Loss of RO from C=O Loss of C n H 2n+1 O (M 31,45,59, ) M 101 = 71 McLafferty rearrangement (alkyl) Loss of C n H 2n (M 28,42,56,70, ) M 28 = 144 McLafferty rearrangement (alkoxy) Loss of C n H 2n O 2 (M 48,62,76, ) 84 McLafferty + 1 Loss of C n H 2n-1 (M 41,55,69, ) M 83 = 89 -cleavage of -b C-C bond of alkoxy Loss of C n H 2n+1 (M 15,29,,, ) M 71 =

34 Which ester? 29 M M-107 M M M-55 M-55: McL+1 C 4 H 9 O- 103 = C 4 H 9 CO 2 H + H + M-73: loss RO C 4 H 9 O- 85 = C 6 H 13+ or C 4 H 9 CO + M-101: loss of C 6 H 13 O = C 4 H 9+ or C 2 H 5 CO + MW = 158 Alkyl groups Acyl groups C n H 2n+1 C n H 2n+1 CO n M M : C n H 2n-1 lost: McLafferty+1-1: alkene C n H 2n lost: McLaffery observed: McLafferty (alkoxy substituent of ester) 0: alkyl C n H 2n+1 observed: alkyl cation, R + or C n H 2n+1 CO + +2: lost: alkoxy C n H 2n+1 O

35 Identify each ester C 6 H 12 O 2 esters, MW = tert-butyl acetate CH 3 C(=O)OC(CH 3 ) Methyl pentanoate CH 3 CH 2 CH 2 CH 2 COOCH 3 Isopropyl propionate CH 3 CH 2 COOCH(CH 3 ) Halogen Containing Compounds 42 M Cl Pavia 8.8 V 1-chloropropane (C 3 H 7 Cl, M = 78 [ 35 Cl]) M Cl M 35 2-chloropropane (C 3 H 7 Cl, M = 78 [ 35 Cl]) 63 M

36 M Br Pavia 8.8 V 1-bromopropane (C 3 H 7 Br, M =122 [ 79 Br]) M+1 M Br M+1 2-bromopropane (C 3 H 7 Br, M =122 [ 79 Br]) Halogen Containing Compounds Pavia 8.8 V 1-bromopropane (C 3 H 7 Br, M =122 [ 79 Br]) M Br M+1 2-bromopropane (C 3 H 7 Br, M =122 [ 79 Br]) M Br M+1

37 2-chloroheptane (C 7 H 15 Cl, M = 134 [ 35 Cl]) MW = 134 Summary: Alkyl halides -cleavage of R Dehydrohalogenation Dehydrohalogenation with loss of an alkene cleavage of X

38 SUMMARY: SOME COMMON FRAGMENT PEAKS Pavia Appendix 12 Peak Fragment Interpretation lost M 1 H aldehydes, 3 alcohols, cyclic amines M 2 Multiple H 2 alcohols M 3 Multiple H 1 alcohols M 15 CH3 methyl groups M 17 HO alcohols, phenols, carboxylic acids M 18 H 2 O alcohols M 26 HC CH M 27 HC=CH 2 M 28 CH 2 =CH 2 cyclic alkanes, alkenes CH 3 CH 2 CH 2 C(=O)X [McLafferty rearr.] M 19,35/37,79/81 halogen M 15,29,,,71 alkyl M 31,45,59,73,87 alkoxy esters, ethers Peak Fragment Interpretation observed Pavia Appendix H 2 C C CH from alkyne propargyl + cation 41 H 2 C CH=CH 2 from terminal alkene allyl cation + 42 H 2 C=C=NH 2 from 1 amine (R CH 2 NH 2 ) H 2 C=C=OH from 1 alcohol + 77 C 6 H 5 + from substituted phenyl 91 C 7 H from C-substituted phenyl tropylium cation

39 Spectral Database for Organic Compounds (SDBS) Compound X revisited, yet again Analysis of fragmentation STRUCTURAL information M Fragmentation: Structure contains: Reporting MS Spectra in Papers and Progress Reports A MS (EI, 70 kv): molecular ion m/z 91 (tropilium, 100%), 205 (M-29, 64%), 234(, 35%), B C base peak 235 (M+1, 4.6%). D Fragments that help in structure determination M+1 peak (or others) that indicate something about molecular structure based on isotope distribution A B C D Ionization technique/method (e.g., EI, CI, FAB, MALDI). Mass Assignment of the peak to a particular ion Height of the peak relative to the base peak (100%)

40 Reporting HRMS Spectra in Papers and Progress Reports A HRMS (EI, 70 kv): m/z , 39%; Calc. for C 4 H 8 N 2 O, = 2 ppm. B C D E A B C D E Ionization technique/method (e.g., EI, CI, FAB, MALDI). Mass Height of the peak relative to the base peak (100%) Suggested formula The difference between the experimental value of molecular weight and the value calculated for the putative molecular formula, in ppm: = M exp M calc 10 6 /M exp WORK AS MANY PROBLEMS AS POSSIBLE! Pavia, chap 8 questions: 7 (a)-(x), 8, 9, 10, (a)-(f), 11, (a)-(g), 12 (a)-(c), 13, 14, 15 (a)-(b)

41 Problem sets based on M.S. alone Intensities of peaks are provided, you need to download free spectral visualization software to view actual spectra Determine the structures of the compounds for which the mass spectra and IR spectra are provided on the following slides. In some cases you might be able to identify a single compound. For others, a 1 H NMR would really help!! 1. [10 points] Match spectra a-e to the following compounds (all C 11 H 16 ). a 133 b 91 c 91 d e 105

42 2. [80 points] Provide the structures of compounds F, G, H and I. Partial credit will be assigned for correct structural information that you provide (e.g., formula, functional groups, presence of specific alkyl groups, benzene substitution patterns, etc.) 105 H F

43 30 G H NMR spectrum integral ratios and peak multiplicities provided 2 13 t mult I C NMR spectrum 1 H NMR spectrum integral ratios and peak multiplicities provided t t m t Mass spectrum 71 molecular ion not observed

44 tert-butyl phenyl ether, Ph-O-C(CH 3 ) (a) Spectra a-d, four monosubstituted benzenes, C 11 H 16 (MW = 148) a 91 (M-) b (M-91) 92 (M-56) 94 c 119 (M-29) d 105 (M-)