Chapter 20 Mass Spectroscopy
Mass Spectrometry (MS) Mass spectrometry is a technique used for measuring the molecular weight and determining the molecular formula of an organic compound.
Mass Spectrometry (MS) In a mass spectrometer, a molecule is vaporized and ionized by bombardment with a beam of high-energy electrons. The energy of the electrons is ~ 1600 kcal (or 70 ev). Since it takes ~100 kcal of energy to cleave a typical s bond, 1600 kcal is an enormous amount of energy to come into contact with a molecule. The electron beam ionizes the molecule by causing it to eject an electron. The ions formed may remain intact (as molecular ions, M+), or they may fragment The resulting mixture of ions is sorted by mass/charge (m/z) ratio, and detected Molecular weight and chemical formula may be derived from the M+ and M+1 ions Molecular structure may be deduced from the distribution of fragment ions
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Introduction When the electron beam ionizes the molecule, the species that is formed is called a radical cation, and symbolized as M +. The radical cation M + is called the molecular ion or parent ion. The mass of M + represents the molecular weight of M. Because M is unstable, it decomposes to form fragments of radicals and cations that have a lower molecular weight than M +. The mass spectrometer analyzes the masses of cations. A mass spectrum is a plot of the amount of each cation (its relative abundance) versus its mass to charge ratio (m/z, where m is mass, and z is charge). Since z is almost always +1, m/z actually measures the mass (m) of the individual ions. 5
Mass Spectrometry Consider the mass spectrum of CH 4 below: The tallest peak in the mass spectrum is called the base peak. The base peak is also the M peak, although this may not always be the case. Though most C atoms have an atomic mass of 12, 1.1% have a mass of 13. Thus, 13 CH 4 is responsible for the peak at m/z = 17. This is called the M + 1 peak. 7
Mass Spectrometry Since the molecular ion is unstable, it fragments into other cations and radical cations containing one, two, three, or four fewer hydrogen atoms than methane itself. Thus, the peaks at m/z 15, 14, 13 and 12 are due to these lower molecular weight fragments. 8
The mass spectrum of ethanol base peak M + SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/1/09)
Mass Spectrometry (MS) A mass spectrometer produces a spectrum of masses based on the structure of a molecule. A mass spectrum is a plot of the distribution of ion masses corresponding to the formula weight of a molecule and/or fragments derived from it The x-axis of a mass spectrum represents the masses of ions produced The y-axis represents the relative abundance of each ion produced The pattern of ions obtained and their abundance is characteristic of the structure of a particular molecule 12
Components of a mass spectrometer
Mass Spectrometer The mass spectrometer records a mass spectrum Schematic of a mass spectrometer 14
16 ANALYZERS
The Mass Spectrum Data from a mass spectrometer can be represented as a graph or table The most abundant (intense) peak in the spectrum is called the base peak and is assigned a normalized intensity of 100% The masses are based on rounding of atom masses to the nearest whole number (in low resolution mass spectroscopy) The data and fragmentation patterns for ammonia are as follows The base peak for ammonia is the molecular ion, but this is often not the case 17
18 Isotopic Abundance
Easily Recognized Elements in MS Nitrogen: Odd number of N = odd MW CH 3 CN M + = 41 SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/2/09)
The Nitrogen Rule Mass Spectrometry Hydrocarbons like methane (CH 4 ) and hexane (C 6 H 14 ), as well as compounds that contain only C, H, and O atoms, always have a molecular ion with an even mass. An odd molecular ion indicates that a compound has an odd number of nitrogen atoms. The effect of N atoms on the mass of the molecular ion in a mass spectrum is called the nitrogen rule: A compound that contains an odd number of N atoms gives an odd molecular ion. A compound that contains an even number of N atoms (including zero) gives an even molecular ion. Two street drugs that mimic the effect of heroin illustrate this principle. 20
Easily Recognized Elements in MS Bromine: M + ~ M+2 (50.5% 79 Br/49.5% 81 Br) 2-bromopropane M + ~ M+2 SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/1/09)
Easily Recognized Elements in MS Chlorine: M+2 is ~ 1/3 as large as M + Cl M + M+2 SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/2/09)
23 Mass Spectrometry Alkyl Halides and the M + 2 Peak Most elements have one major isotope. Chlorine has two common isotopes, 35 Cl and 37 Cl, which occur naturally in a 3:1 ratio. Thus, there are two peaks in a 3:1 ratio for the molecular ion of an alkyl chloride. The larger peak, the M peak, corresponds to the compound containing the 35 Cl. The smaller peak, the M+2 peak, corresponds to the compound containing 37 Cl. Thus, when the molecular ion consists of two peaks (M and M + 2) in a 3:1 ratio, a Cl atom is present. Br has two isotopes 79 Br and 81 Br,inaratioof~1:1. Thus, when the molecular ion consists of two peaks (M and M + 2) in a 1:1 ratio, a Br atom is present.
Mass Spectrometry Alkyl Halides and the M + 2 Peak Mass spectrum of 2-chloropropane [(CH 3 ) 2 CHCI] The compound contains a chlorine, because M + 2 peak is 1/3 the height of the molecular ion peak The base peak at m/z = 43 results from heterolytic cleavage of the C Cl bond The peaks at m/z = 63 and m/z = 65 have a 3:1 ratio, indicating the presence of a chlorine atom 24
cleavage results from the homolytic cleavage of a C C bond at the carbon 25
Mass Spectrometry Alkyl Halides and the M + 2 Peak Mass spectrum of 2-bromopropane [(CH 3 ) 2 CHBr] 26
The weakest bond is the C Br bond The base peak is at m/z = 43 [M 79, or (M + 2) 81] The propyl cation has the same fragmentation pattern it exhibited when it was formed in the cleavage of pentane 27
Easily Recognized Elements in MS Sulfur: M+2 larger than usual (4% of M + ) S M + Unusually large M+2 SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/1/09)
Easily Recognized Elements in MS Iodine I + at 127 Large gap Large gap M + ICH 2 CN I + SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/2/09)
Fragmentation Patterns Alkanes Fragmentation often splits off simple alkyl groups: Loss of methyl M + -15 Loss of ethyl M + -29 Loss of propyl M + -43 Loss of butyl M + -57 Branched alkanes tend to fragment forming the most stable carbocations. The impact of the stream of high energy electrons often breaks the molecule into fragments, commonly a cation and a radical. Bonds break to give the most stable cation. Stability of the radical is less important.
Fragmentation Patterns Mass spectrum of 2-methylpentane
Fragmentation Patterns Aromatics have a peak at m/z = 77 for the benzene ring. NO 2 77 77 M + = 123
Rule of Thirteen The Rule of Thirteen can be used to identify possible molecular formulas for an unknown hydrocarbon, C n H m. Step 1: n = M + /13 (integer only, use remainder in step 2) Step 2: m = n + remainder from step 1
Rule of Thirteen Example: The formula for a hydrocarbon with M + =106 can be found: C n H m Step 1: n = 106/13 = 8 (R = 2) Step 2: m = 8 + 2 = 10 Formula: C 8 H 10
Rule of Thirteen If a heteroatom is present, Subtract the mass of each heteroatom from the MW Calculate the formula for the corresponding hydrocarbon Add the heteroatoms to the formula
Rule of Thirteen Example: A compound with a molecular ion peak at m/z = 102 has a strong peak at 1739 cm -1 in its IR spectrum. Determine its molecular formula.
Problem 1
Mass 59 Odd-N IR-NH 2 59-16=43 C 3 H 9 N
Problem 2
Mass 72 IR-1700 cm-1 C=O 72-28=44 C 4 H 8 O
Problem 3
Mass- 120 Br present (M+2) 120-79=41 C 3 H 5 Br