Alkanes undergo extensive fragmentation CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH3 Relative intensity 43 57 100 80 60 40 20 Decane 71 85 99 142 20 40 60 80 100 120 m/z 13

More stable carbocations will be more abundant.

CH3 | CH3 - CH2 -- C -- CH3 m/e = 71 + +CH3 m/e = 15 CH3 | +C - CH3 m/e = 57 CH3 - CH2 + m/e = 29 CH3 | +C - H m/e = 43

MS of 2,2,4-trimethylpentane Branched alkanes have small or absent M+

CH2—CH2CH3 Propylbenzene fragments mostlyat the benzylic position 100 Relative intensity 100 80 60 40 20 91 CH2—CH2CH3 120 20 40 60 80 100 120 m/z + 91 14

Cycloalkanes Loss of side chain Loss of ethylene fragments Ionization followed by fragmentation Ejection of electron fragmentation Radical/cation Splitting out of ethylene.

MS of methylcyclopentane:

Alkene Fragmentation Fairly prominent M+ Fragment ions of CnH2n+ and CnH2n-1+ Terminal alkenes lose allyl cation if possible to form resonance-stabilized allylic cations

Cycloalkenes + Prominent molecular ion Retro Diels-Alder cleavage • Observed! Typically you see both. More stable cation will predominate Also works for hetero-substituted (e.g. make enol) Both EI (shown) and in CI. (protonated molecular ion, cleave, then reprotonation)

Cyclohexenes give a 1,3-diene and an alkene, a process that is the reverse of a Diels-Alder reaction.

Mass spectrum of 4-terpineol as a good example for Retro Diels Alder fragmentation EI Mass Spectrum + + + 4-terpineol(MW 154) mz 86 mz 68

Alkyne Fragmentation Molecular ion readily visible Terminal alkynes readily lose hydrogen atom Terminal alkynes lose propargyl cation if possible (resonance-stabilized propargyl cation or a substituted propargyl cation).

Aromatic Hydrocarbon Fragmentation Molecular ion usually strong Alkylbenzenes cleave at benzylic carbon tropylium ion formation

Some molecules undergo very little fragmentation Benzene is an example. The major peak corresponds to the molecular ion. Relative intensity 100 80 60 40 20 m/z = 78 20 40 60 80 100 120 m/z 9

Benzene Compounds - - - - 7 memberd ring Benzonium ion O H C H + O H · 2 O H · - CO - H · + m/z 107 m/z 108 7 memberd ring + H H H H - H 2 + H H Benzonium ion H Phenyl ion m/z 79 m/z 77 [C H ] + 6 7 [C H ] + 6 5 CH3 - CH3 · + · H + · + H CH3 CH3 Tropylium ion m/z 91

Mass Spectrum of n-Octylbenzene Tropylium ion

Mass Spectrum of benzyl bromide Tropylium ion Bromine pattern

Alcohol Fragmentation One of the most common fragmentation patterns of alcohols is loss of H2O to give a peak which corresponds to M-18 and no M peak. Molecular ion strength depends on substitution primary alcohol weak M+ secondary alcohol VERY weak M+ tertiary alcohol M+ usually absent Another common pattern is loss of an alkyl group from the carbon bearing the OH (-Cleavage) to give a resonance-stabilized oxonium ion and an alkyl radical (largest R group lost as radical).

MS of 1-butanol is an example demonstrating both processes. Figure 14.10 Mass spectrum of 1-butanol. Elimination of water. 74 – 56 = 18 (water). Elimination of propyl radical. 74 – 31 = 43 (C3H7)

Mass Spectrum of 3-methyl-1-butanol

Mass Spectrum of 3-methyl-2-butanol

Carbonyl Compounds Fragmentation Dominant fragmentation pathways: a-cleavage b-cleavage McLafferty rearrangement

Characteristic fragmentation patterns are: Aldehydes and Ketones Characteristic fragmentation patterns are: Cleavage of a bond to the carbonyl group (-cleavage). Note that an a cleavage of an aldehyde could produce a peak at M – 1 by eliminating H atom. This is useful in distinguishing between aldehydes and ketones.

McLafferty rearrangement. Splitting out an alkene (neutral molecule) and producing a new radical/cation.

Mass Spec of 2-octanone displays both a cleavage and McLafferty CH3CH2CH2CH2CH2CH2CO+ resulting from a cleavage. CH3CO+ resulting from a cleavage. Figure 14.11 Mass spectrum of 2-octanone. Ions of m/z 43 and 113 result from -cleavage. The ion at m/z 58 results from McLafferty rearrangement.

Carboxylic Acids Characteristic fragmentation patterns are: -cleavage to give the ion [CO2H]+ of m/z 45. McLafferty rearrangement if g H present. Loss of water, especially in CI Loss of 44 is the loss of CO2

MS of butanoic acid Figure 14.12 Mass spectrum of butanoic acid. Common fragmentation patterns of carboxylic acids are -cleavage to give the ion [COOH]_ of m/z 45 and McLafferty rearrangement.

Fatty Acids

Esters -cleavage and McLafferty rearrangement:

MS of methyl butanoate Figure 14.13 Mass spectrum of methyl butanoate. Characteristic fragmentation patterns of esters are -cleavage and McLafferty rearrangement.

Ether Fragmentation a-cleavage C-O cleavage (requires stable cation to lose) Ion rearrangement Ms of 2-Chloroethylphenyl ether 94 100 C l H 107 80 O 94 60 Intensity 40 156 107 20 20 40 60 80 100 120 140 160

Amines The most characteristic fragmentation pattern of 1°, 2°, and 3° aliphatic amines is -cleavage

Cyclic amines will lose adjacent H•, form iminium ion In CI, NH+ can eliminate adjacent alkene, reprotonate

Halide Fragmentation Loss of halogen atom Elimination of HX a-cleavage

•+ Chlorobenzene •+ + 77 M+2 = 114 M+ = 112

Sulfur Compounds Fortunately there is an [M+2]+ of 4% for the natural abundance of 34S. This is diagnostic for S vs 2x16O Aliphatic thiols can split out H2S, [M-34] a-cleavage at carbon bearing the sulfur in thiols, thioethers, similar to ethers, etc.

Nitroaromatics m/z = 93 Loss of •N=O Loss of CO Aromatic! m/z=65 (this can form from lots of different origins) Loss of •N=O N + O• O Loss of CO Aromatic! m/z=65 Good test for aryloxy

Clusters of Ions m/z Spaced by unit mass Each peak is for the same molecular formula Different peaks because there are some molecules with 13C, 2H etc. Especially significant for Cl, Br The Nominal mass is m/z of the lowest member of the cluster. This is the isotopomer that has all the C’s as 12C, all protons as 1H, all N’s as 14N, etc. m/z

Analyzing Ion Clusters: a way to rule candidate structures Mass spectrometry “sees” all the isotopomers as distinct ions An ion with all 12C is one mass unit different from an ion with one 13C and the rest 12C Since the isotope distribution in nature is known for all the elements (13C is 1.1%), the anticipated range and ratios of ions for a given formula can be predicted and calculated

H H H 79 79 78 93.4% 6.5% 0.1% all H are 1H and all C are 12C Isotopic Clusters H H H 79 79 78 93.4% 6.5% 0.1% all H are 1H and all C are 12C one C is 13C one H is 2H 10

visible in peaks for molecular ion Isotopic Clusters in Chlorobenzene 35Cl 37Cl visible in peaks for molecular ion Relative intensity 100 80 60 40 20 112 114 m/z 20 40 60 80 100 120 11

+ H Isotopic Clusters in Chlorobenzene no m/z 77, 79 pair; therefore ion responsible for m/z 77 peak does not contain Cl Relative intensity 100 80 60 40 20 77 m/z 20 40 60 80 100 120 11

Isotopic Abundance 81Br

Molecules with Heteroatoms Isotopes: present in their usual abundance. Hydrocarbons contain 1.1% C-13, so there will be a small M+1 peak. If Br is present, M+2 is equal to M+. If Cl is present, M+2 is one-third of M+. If iodine is present, peak at 127, large gap. If N is present, M+ will be an odd number. If S is present, M+2 will be 4% of M+.

Molecular Formula as a Clue to Structure 1

Molecular Weights One of the first pieces of information we try to obtain when determining a molecular structure is the molecular formula. We can gain some information about molecular formula from the molecular weight. Mass spectrometry makes it relatively easy to determine molecular weights. 6

Mass spectrometry can measure exact masses. Exact Molecular Weights CH3CO O CH3(CH2)5CH3 Heptane Cyclopropyl acetate Molecular formula C7H16 C5H8O2 Molecular weight 100 100 Exact mass 100.1253 100.0524 Mass spectrometry can measure exact masses. Therefore, mass spectrometry can be used to distinguish between molecular formulas. 6

The “Nitrogen Rule” Molecules containing atoms limited to C,H,O,N,S,X,P of even-numbered molecular weight contain either NO nitrogen or an even number of N. This is true as well for radicals as well. Not true for pre-charged, e.g. quats, (rule inverts) or radical cations. In the case of Chemical Ionization, where [M+H]+ is observed, need to subtract 1, then apply nitrogen rule. Example, if we know a compound is free of nitrogen and gives an ion at m/z=201, then that peak cannot be the molecular ion.

The Nitrogen Rule NH2 93 138 NH2 O2N 183 NH2 O2N NO2 A molecule with an odd number of nitrogens has an odd molecular weight. A molecule that contains only C, H, and O or which has an even number of nitrogens has an even molecular weight. NH2 93 138 NH2 O2N 183 NH2 O2N NO2 6

Index of Hydrogen Deficiency Degree of Unsaturation Relates molecular formulas to multiple bonds and rings For a molecular formula, CcHhNnOoXx, the degree of unsaturation can be calculated by: index of hydrogen deficiency = ½ (2c + 2 - h - x + n) 6

index of hydrogen deficiency = Molecular Formulas Knowing that the molecular formula of a substance is C7H16 tells us immediately that is an alkane because it corresponds to CnH2n+2 C7H14 lacks two hydrogens of an alkane, therefore contains either a ring or a double bond Index of Hydrogen Deficiency index of hydrogen deficiency = 1 (molecular formula of alkane – molecular formula of compound) 2 6

Index of hydrogen deficiency Example 1 C7H14 Index of hydrogen deficiency 1 2 = (molecular formula of alkane – molecular formula of compound) 1 2 = (C7H16 – C7H14) 1 2 = (2) = 1 Therefore, one ring or one double bond. 6

Example 2 C7H12 1 2 = (C7H16 – C7H12) 1 2 = (4) = 2 Therefore, two rings, one triple bond, two double bonds, or one double bond + one ring. 6

Index of hydrogen deficiency = Oxygen has no effect CH3(CH2)5CH2OH (1-heptanol, C7H16O) has same number of H atoms as heptane Index of hydrogen deficiency = 1 (C7H16 – C7H16O) = 0 2 No rings or double bonds 6

Index of hydrogen deficiency = CH3CO O Cyclopropyl acetate Index of hydrogen deficiency = 1 (C5H12 – C5H8O2) = 2 2 One ring plus one double bond 6

Treat a halogen as if it were hydrogen. If halogen is present Treat a halogen as if it were hydrogen. H Cl C3H5Cl C same index of hydrogen deficiency as for C3H6 H CH3 6

Rings versus Multiple Bonds Index of hydrogen deficiency tells us the sum of rings plus multiple bonds. Using catalytic hydrogenation, the number of multiple bonds can be determined. 6

Interpretation of Mass Spectra Select a candidate peak for the molecular ion (M+) Examine spectrum for peak clusters of characteristic isotopic patterns Test (M+) peak candidate by searching for other peaks correspond to reasonable losses Look for characteristic low-mass fragment ions Compare spectrum to reference spectra

Isotopic Abundances

Formula Matching Basics Atomic weights are not integers (except 12C) 14N = 14.0031 amu; 11B = 11.0093 amu; 1H = 1.0078 amu 16O = 15.9949 amu; 19F = 18.9984 amu; 56Fe = 55.9349 amu Sum of the mass defects depends on composition Hydrogen increases mass defect, oxygen decreases it Accurate mass measurements narrow down the possible formulae for a particular molecular weight 301 entries (150 formulae) in NIST’02 with nominal MW = 321 4 compounds within 0.0016 Da (5 ppm) of 321.1000. Mass spectrum and user info complete the picture Isotope distributions indicate/eliminate elements (e.g. Cl, Br, Cu) User-supplied info eliminates others (e.g. no F, Co, I in reaction) Isomers are not distinguished in this analysis