Mass Spectroscopy PDF

MASS SPECTROSCOPY Theory In mass spectrometry, a substance is bombarded with high electron beam having enough energy to fragment the molecule. The positive fragments which are produced (cations and radical cations) are accelerated in a vacuum through a magnetic field and are sorted on the basis of m...

MASS SPECTROSCOPY Theory In mass spectrometry, a substance is bombarded with high electron beam having enough energy to fragment the molecule. The positive fragments which are produced (cations and radical cations) are accelerated in a vacuum through a magnetic field and are sorted on the basis of mass to charge ratio (m/e or m/z). Since the bulk of the ions produced in the mass spectrometer carry a unit of positive charge, the value m/e or m/z is equivalent to the molecular weight of the fragment. The analysis of mass spectroscopy information involves the re- assembling of fragments, working backwards to generate the original molecule. Ion Source: converts sample to gas phase and ionizes sample Mass Analyzer: “sorts” ions according to mass-to-charge ratio Ion Detector: converts ions to electrical signal proportional to # of ions. 1-A molecule is bombarded with a high energy electron beam having sufficient energy to fragment the molecule 2-The molecule is ionized into a Molecular ion (radical cation/ odd number of electrons). 3-The radical cation is then fragmented into cations or radical cations and neutral fragments (only positive fragments are detected in MS  4-These cations are accelerated in vacuum through a magnetic field and are sorted on the basis of mass-to-charge ratio (m/e or m/z). 5-The value of m/e or m/z is equivalent to the molecular weight of the fragment. Mass spectrum 1- A plot of relative intensity (indicates relative abundance of the ion) (y- axis) vs the mass to charge ratio (m/z) (x-axis). 2- Peaks are sharp vertical lines. 3- Analysis of mass spectrum involves reassembly of peaks. Most important peaks 1-Base Peak - most intense peak (100%intensity( other peaks are reported relative to its intensity. It represents the most stable fragment of a molecule 2- Molecular Ion Peak - represents the highest mass of molecule (molecule-radical cation), it - has odd number of electrons and it can form important fragment ions in the spectrum. Also it can be base peak of aromatics, conjugated alkenes, cycloalkanes. 3- Isotope Peak:- peak that represents presence of isotope higher than 13 37 81 molecular ion peak by 1 (e.g. C) or 2 (e.g. Cl, Br). It has much lower intensity. Fragmentation Patterns: 1-Fragmentation occurs at weak bonds C-C weaker than C-O *Molecular ions from aromatic rings, conjugated alkenes, and cycloalkanes are the most stable. *Alcohols, ethers and highly branched alkanes generally show the greatest tendency towards fragmentation. 2-Gives the most stable fragments (neutral + radical-cation). The behavior of an unbranched alkane. Since there are no heteroatoms in this molecule, there are no non-bonding valence shell electrons. Consequently, the radical cation character of the molecular ion (m/z = 170) is delocalized over all the covalent bonds. Fragmentation of C-C bonds occurs because they are usually weaker than C-H bonds, and this produces a mixture of alkyl radicals and alkyl carbocations. The positive charge commonly resides on the smaller fragment, so we see a homologous series of hexyl (m/z = 85), pentyl (m/z = 71), butyl (m/z = 57), propyl (m/z = 43), ethyl (m/z = 29) and methyl (m/z = 15) cations. These are accompanied by a set of corresponding alkenyl carbocations (e.g. m/z = 55, 41 &27) formed by loss of 2 H. All of the significant fragment ions in this spectrum are even-electron ions. In most alkane spectra the propyl and butyl ions are the most abundant. ***These three examples are hydrocarbons having no functional groups. Hexane shows the same fragmentation pattern as other unbranched alkanes. Thus, alkyl carbocations at m/z=15, 29, 43 and 57 Da provide the dominant peaks in the spectrum. The m/z=57 butyl cation (M-29) is the base peak, and the m/z=43 and 29 ions are also abundant. 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. Nitrogen rule: if molecular weight of compound is even, it must contain even number of nitrogen or not contain nitrogen at all. If molecular weight of compound is odd, it must contain odd number of hydrogen atoms.  If S is present, M+2 will be 4% of M+. Two-bond cleavages include Retro Diels-Alder McLafferty rearrangement