2 GeneralThe mass spectrum is a plot of ion abundance versus m/e ratio (mass/charge ratio).The most abundant ion formed in the ionization chamber gives to rise the tallest peak in the mass spectrum, called the base peak.By using one of the many ionization methods, the simple removal of an electron from a molecule yields a positively charged radical cation, known as the molecular ion and symbolized as [M]+.After formation of molecular Ion in the ionization chamber, excess energy causes further fragmentation of the molecular ion. Various positively charged masses (and/or positively charged radical cations) show up in the spectrum.The mass of the molecular ion can be determined to an accuracy of ± of a mass unit, which yields a high resolution (hi-res) mass spectrum.Ionization MethodsThe most common method of ionization involves Electron impact (EI) in which molecule is bombarded with high-energy electrons. This is the strongest ionization method which usually causes further fragmentation. Since the molecular ion gets fragmented, the molecular ion usually produces a small peak with this method.
3 Mass Spectrum of Isobutyrophenone 105(base peak)C6H5CO+molecularweight = 14877C6H5+molecular ion, M(148)M+1
4 Facts Concerning the Molecular Ion Peak The peak must correspond to the ion of highest mass excluding theusually much smaller isotopic peaks that occur at M+1, M+2, etc.To be a molecular ion, the ion must contain an odd number of electrons. One electron is lost, forming a radical-cation.To determine this, calculate the IHD. It must be a whole number.Consider an ion at m/z = A possible molecular formula is C6H8O2.The IHD = 3 (a whole number), so this could be the molecular ion.However, an ion at 105 could correspond to a molecular formula ofC7H5O. The IHD is 5.5 (not a whole number), so this can’t be themolecular ion.The ion must be capable of producing smaller, fragment ions by lossof neutral fragments of predictable structure.
5 The Lifetimes of Various Molecular Ions Aromatics > conjugated alkenes > alicyclic compounds> organic sulfides >unbranched hydrocarbons > mercaptans > ketones > amines > esters >ethers > carboxylic acids > branched hydrocarbons > alcoholsTherefore, alcohols produce small or non-existent molecular ions becausetheir lifetimes are too short. They fragment before they can be detected.CH3CH2CH2CH2OHMW = 74No Mvisible
6 The Nitrogen RuleThe nitrogen rule states that an odd number of nitrogen atoms will forma molecular ion with an odd mass number. An even number of nitrogenatoms (or none at all) will produce a molecular ion with an even massnumber. This occurs because nitrogen has an odd-numbered valence.Examples: C6H5CH2NH2 MW = 107H2NCH2CH2NH2 MW = 60
7 Determining Possible Molecular Formulas from the Molecular Ion: Rule of 13 Rule of Thirteen: Based upon the assumption that CnHn and its mass of 13 is present in most organic compounds.Divide the molecular ion by 13. This gives a value for n and any remainder (R) = additional H’s.For a M+ = 106, n = 8(106/13) with a R of 2. A possible molecular formula for this ion is C8H8+2 = C8H10For each CH there are heteroatom equivalents.
8 Rule of 13 Heteroatom Equivalents ElementCH Equivalent1H12C31PC2H716OCH432SC2H814NCH216O32SC416O14NC2H635ClC2H1119FCH779BrC6H728SiC2H4127IC10H7
9 Candidate molecular formulas for M+. = 108 108/13 = 8, R = 4 C8H12O = CH4; therefore, C8H12 – CH4 + O = C7H8Oor C8H12 – 2(CH4) + 2O = C6H4O2or C8H12 – C6H7 + Br = C2H5BrCalculate three candidate molecular formulas for C10H18.
10 When an odd amu M+. is seen, suspect one nitrogen or an odd multiple. Candidate molecular formulas for a M+. = 121 are:121/13 = 9, R = 4 C9H13; IHD = 3.5 so it can’t be M+.O = CH4; N = CH2or C9H13 – CH2 + N = C8H11N; IHD = 3, so it may be M+.or C9H13 – 2(CH2) + 2N = C7H9N2; IHD = 4.5 so it can’t be M+.or C9H13 – 3(CH2) + 3N = C6H7N3; IHD = 5, so it may be M+.or C9H13 – (CH2) – (CH4) + N + O = C7H7NO IHD = 5,so it may be M+.
11 Determining the Molecular Formula from the Molecular Ion: Isotope Ratio DataIn this method, the relative intensities of the peaks due to the molecular ionand related isotopic peaks are examined.Advantage:Does not require an expensive high-res MS instrument.Disadvantages:Isotopic peaks may be difficult to locate.Useless when the molecular ion peak is very weak or does not appear.3-pentanone
12 Relative Abundances of Common Elements and Their isotopes Carbon12C10013C1.08Hydrogen1H2H0.016Nitrogen14N15N0.38Oxygen16O17O0.0418O0.20Sulfur32S33S0.7834S4.40Chlorine35Cl 37Cl32.5Bromine79Br81Br98.0
13 Example: 3-pentanone, C5H10O %(M + 1) = 100 (M + 1)/M = 1.08 x # C atoms x # H atoms x # O atoms= 1.08 x x x 1= 5.60Actual spectrum: [1% (M +1)/17.4% (M)] x 100 = 5.75
14 Halogen M M + 2 M + 4 M + 6 Br 100 97.7 Br2 195.0 95.4 Br3 293.0 286.0 Relative Intensities of Isotope Peaks for Bromine and ChlorineHalogenMM + 2M + 4M + 6Br10097.7Br2195.095.4Br3293.0286.093.4Cl32.6Cl265.310.6Cl397.831.93.47BrCl130.0Br2Cl228.0159.031.2Cl2Br163.074.410.4
19 Determining the Molecular Formula from the Molecular Ion: High Resolution MS (HRMS)Using low resolution (LR) MS, you could not distinguish between the followingmolecular formulas, each of which has a mass of 60:C3H8O = (3 x 12) + (8 x 1) + 16 = 60C2H8N2 = (2 x 12) + (8 x 1) + (2 x 14) = 60C2H4O2 = (2 x 12) + (4 x 1) + (2 x 16) = 60CH4N2O = 12 + (4 x 1) + (2 x 14) + 16 = 60However, they can be distinguished using HRMS.
20 Precise Masses of Some Common Elements Atomic WeightIsotopeMassHydrogen1.00971H2HCarbon12C13CNitrogen14N15NOxygen16O17O18OFluorine19FSilicon28.08628Si29Si30SiPhosphorus30.97431PSulfur32.06432S33S34SChlorine35.45335Cl37ClBromine79.90979Br81BrIodine127IUsing precise masses:C3H8O =C2H8N2 =C2H4O2 =CH4N2O =
21 Fragmentation Patterns Most common: one-bond cleavage to produce an odd-electron neutral fragment,(radical, which is not detected) and an even-electron carbocation. Ease of frag-mentation to form cations follows the scheme below:CH3+ < RCH2+ < R2CH+ < R3C+ < CH2=CH-CH2+ < C6H5-CH2+Difficult EasyRadical (not detected)
22 Fragmentation Patterns (cont.) Two-bond cleavage: The odd-electron molecular ion produces an odd-electronfragment ion and an even-electron neutral fragment (not detected).Not detectedMcLafferty Rearrangement
29 Hexane C6H14 MW = 86.18Molecular ion peaks are present, possibly with low intensity. The fragmentation pattern contains clusters of peaks 14 massunits apart (which represent loss of (CH2)nCH3).
30 3-Pentanol C5H12O MW = 88.15An alcohol's molecular ion is small or non-existent. Cleavage of the C-C bond next to the oxygen usuallyoccurs. A loss of H2O may occur as in the spectrum below.
31 3-Phenyl-2-propenal C9H8O MW = 132.16 Cleavage of bonds next to the carboxyl groupresults in the loss of hydrogen (molecular ion less 1)or the loss of CHO (molecular ion less 29).
32 3-Methylbutyramide C5H11NO MW = 101.15 Primary amides show a base peakdue to the McLafferty rearrangement.
33 n-Butylamine C4H11N MW = 73.13 Molecular ion peak is an odd number. Alpha-cleavage dominates aliphatic amines.
34 n-Methylbenzylamine C8H11N MW = 121.18 Another example is a secondary amine shown below. Again, the molecular ion peak is an odd number. The base peak is from the C-C cleavage adjacent to the C-N bond.
35 Molecular ion peaks are strong due to the stable structure. Naphthalene C10H8 MW =Molecular ion peaks are strong due to the stable structure.
36 2-Butenoic acid C4H6O2 MW = 86.09In short chain acids, peaks due to the loss of OH(molecular ion less 17) and COOH (molecular ion less 45)are prominent due to cleavage of bonds next to C=O.
37 Fragments appear due to bond cleavage next to C=O (alkoxy group loss, -OR) and hydrogen rearrangements. Ethyl acetate C4H8O2 MW = 88.11
38 Ethyl methyl ether C3H8O MW = 60.10 Fragmentation tends to occur alpha to theoxygen atom (C-C bond next to the oxygen).
39 1-Bromopropane C3H7Br MW = 123.00 The presence of chlorine or bromine atoms isusually recognizable from isotopic peaks.
40 4-Heptanone C7H14O MW =Major fragmentation peaks result from cleavageof the C-C bonds adjacent to the carbonyl.
41 Which structure supports the following mass spectrum?
42 Which structure supports the following mass spectrum?
43 Which structure supports the following mass spectrum?
44 An unknown compound has the mass spectrum shown below. The IR spectrum shows peaks in the and the cm-1 ranges and a strongabsorption at 1688 cm-1. Suggest a structure consistent with this data.
45 An unknown compound has the mass spectrum shown below An unknown compound has the mass spectrum shown below. The IR spectrum shows peaks in the cm-1 range and a strong absorption at 1718 cm-1. Suggest a structure consistent with this data.