Presentation on theme: "Organic Spectroscopy Mass Spectrometry. General The mass spectrum is a plot of ion abundance versus m/e ratio (mass/charge ratio). The most abundant ion."— Presentation transcript:
Organic Spectroscopy Mass Spectrometry
General The 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 Methods The 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.
105 (base peak) molecular ion, M (148) Mass Spectrum of Isobutyrophenone molecular weight = 148 C 6 H 5 CO + 77 C 6 H 5 + M+1
Facts Concerning the Molecular Ion Peak 1.The peak must correspond to the ion of highest mass excluding the usually much smaller isotopic peaks that occur at M+1, M+2, etc. 2.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 = 112. A possible molecular formula is C 6 H 8 O 2. The IHD = 3 (a whole number), so this could be the molecular ion. However, an ion at 105 could correspond to a molecular formula of C 7 H 5 O. The IHD is 5.5 (not a whole number), so this can’t be the molecular ion. 3.The ion must be capable of producing smaller, fragment ions by loss of neutral fragments of predictable structure.
Aromatics > conjugated alkenes > alicyclic compounds> organic sulfides > unbranched hydrocarbons > mercaptans > ketones > amines > esters > ethers > carboxylic acids > branched hydrocarbons > alcohols Therefore, alcohols produce small or non-existent molecular ions because their lifetimes are too short. They fragment before they can be detected. The Lifetimes of Various Molecular Ions CH 3 CH 2 CH 2 CH 2 OH MW = 74 No M visible
The Nitrogen Rule The nitrogen rule states that an odd number of nitrogen atoms will form a molecular ion with an odd mass number. An even number of nitrogen atoms (or none at all) will produce a molecular ion with an even mass number. This occurs because nitrogen has an odd-numbered valence. Examples: C 6 H 5 CH 2 NH 2 MW = 107 H 2 NCH 2 CH 2 NH 2 MW = 60
Determining Possible Molecular Formulas from the Molecular Ion: Rule of 13 Rule of Thirteen: Based upon the assumption that C n H n 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 C 8 H 8+2 = C 8 H 10 For each CH there are heteroatom equivalents.
Rule of 13 Heteroatom Equivalents ElementCH Equivalent ElementCH Equivalent 1 H 12 C 31 PC2H7C2H7 16 OCH 4 32 SC2H8C2H8 14 NCH 2 16 O 32 SC4C4 16 O 14 NC2H6C2H6 35 ClC 2 H FCH 7 79 BrC6H7C6H7 28 SiC2H4C2H4 127 IC 10 H 7
Candidate molecular formulas for M +. = /13 = 8, R = 4 C 8 H 12 O = CH 4 ; therefore, C 8 H 12 – CH 4 + O = C 7 H 8 O or C 8 H 12 – 2(CH 4 ) + 2O = C 6 H 4 O 2 or C 8 H 12 – C 6 H 7 + Br = C 2 H 5 Br Calculate three candidate molecular formulas for C 10 H 18.
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 C 9 H 13 ; IHD = 3.5 so it can’t be M +. O = CH 4 ; N = CH 2 or C 9 H 13 – CH 2 + N = C 8 H 11 N; IHD = 3, so it may be M +. or C 9 H 13 – 2(CH 2 ) + 2N = C 7 H 9 N 2 ; IHD = 4.5 so it can’t be M +. or C 9 H 13 – 3(CH 2 ) + 3N = C 6 H 7 N 3 ; IHD = 5, so it may be M +. or C 9 H 13 – (CH 2 ) – (CH 4 ) + N + O = C 7 H 7 NO IHD = 5, so it may be M +.
Determining the Molecular Formula from the Molecular Ion: Isotope Ratio Data In this method, the relative intensities of the peaks due to the molecular ion and 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
ElementIsotope Relative Abun- dance Isotope Relative Abun- dance Isotope Relative Abun- dance Carbon 12 C C1.08 Hydrogen 1H1H100 2H2H0.016 Nitrogen 14 N N0.38 Oxygen 16 O O O0.20 Sulfur 32 S S S4.40 Chlorine 35 Cl Cl32.5 Bromine 79 Br Br98.0 Relative Abundances of Common Elements and Their isotopes
Example: 3-pentanone, C 5 H 10 O %(M + 1) = 100 (M + 1)/M = 1.08 x # C atoms x # H atoms x # O atoms = 1.08 x x x 1= 5.60 Actual spectrum: [1% (M +1)/17.4% (M)] x 100 = 5.75
HalogenMM + 2M + 4M + 6 Br Br Br Cl Cl Cl BrCl Br 2 Cl Cl 2 Br Relative Intensities of Isotope Peaks for Bromine and Chlorine
Determining the Molecular Formula from the Molecular Ion: High Resolution MS (HRMS) Using low resolution (LR) MS, you could not distinguish between the following molecular formulas, each of which has a mass of 60: C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 C 2 H 8 N 2 = (2 x 12) + (8 x 1) + (2 x 14) = 60 C 2 H 4 O 2 = (2 x 12) + (4 x 1) + (2 x 16) = 60 CH 4 N 2 O = 12 + (4 x 1) + (2 x 14) + 16 = 60 However, they can be distinguished using HRMS.
Precise Masses of Some Common Elements ElementAtomic WeightIsotopeMass Hydrogen H1H H2H Carbon C C Nitrogen N N Oxygen O O O Fluorine F Silicon Si Si Si Phosphorus P Sulfur S S S Chlorine Cl Cl Bromine Br Br Iodine I C 3 H 8 O = C 2 H 8 N 2 = C 2 H 4 O 2 = CH 4 N 2 O = Using precise masses:
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: CH 3 + < RCH 2 + < R 2 CH + < R 3 C + < CH 2 =CH-CH 2 + < C 6 H 5 -CH 2 + DifficultEasy Radical (not detected)
Fragmentation Patterns (cont.) Two-bond cleavage: The odd-electron molecular ion produces an odd-electron fragment ion and an even-electron neutral fragment (not detected). Not detected McLafferty Rearrangement
71 57 Cleavage at branch points MW = 86
87 MW = 102 -cleavage to hetero atoms cleavage to hetero atoms 43 CH 3 -CH=OH +
-cleavage to aromatic ring 91 MW = (from McLafferty Rearrangement)
43 Cleavage to carbonyl groups MW = 86
60 McLafferty rearrangement carboxylic acids
74 McLafferty rearrangement esters MW = 102
Hexane C 6 H 14 MW = Molecular ion peaks are present, possibly with low intensity. The fragmentation pattern contains clusters of peaks 14 mass units apart (which represent loss of (CH 2 )nCH 3 ).
3-Pentanol C 5 H 12 O MW = An alcohol's molecular ion is small or non-existent. Cleavage of the C-C bond next to the oxygen usually occurs. A loss of H 2 O may occur as in the spectrum below.
3-Phenyl-2-propenal C 9 H 8 O MW = Cleavage of bonds next to the carboxyl group results in the loss of hydrogen (molecular ion less 1) or the loss of CHO (molecular ion less 29).
3-Methylbutyramide C 5 H 11 NO MW = Primary amides show a base peak due to the McLafferty rearrangement.
n-Butylamine C 4 H 11 N MW = Molecular ion peak is an odd number. Alpha-cleavage dominates aliphatic amines.
n-Methylbenzylamine C 8 H 11 N MW = 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.
Naphthalene C 10 H 8 MW = Molecular ion peaks are strong due to the stable structure.
2-Butenoic acid C 4 H 6 O 2 MW = In 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.
Fragments appear due to bond cleavage next to C=O (alkoxy group loss, -OR) and hydrogen rearrangements. Ethyl acetate C 4 H 8 O 2 MW = 88.11
Fragmentation tends to occur alpha to the oxygen atom (C-C bond next to the oxygen). Ethyl methyl ether C 3 H 8 O MW = 60.10
The presence of chlorine or bromine atoms is usually recognizable from isotopic peaks. 1-Bromopropane C 3 H 7 Br MW =
Major fragmentation peaks result from cleavage of the C-C bonds adjacent to the carbonyl. 4-Heptanone C 7 H 14 O MW =
Which structure supports the following mass spectrum?
An unknown compound has the mass spectrum shown below. The IR spectrum shows peaks in the and the cm -1 ranges and a strong absorption at 1688 cm -1. Suggest a structure consistent with this data.
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.