22_MassSpec - Mass Spectroscopy This is powerfu tool for...

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Unformatted text preview: Mass Spectroscopy This is powerfu tool for the idntification of ’ materials. o It relies of the production of ions from a parent " compound and the subsequent characterization of the patterns that are produced. 0 While it is a qualitative tool, it can be coupled " with other methods. 2}: o We'll look at some of the basic equipment and ‘4 types of data obtained. oderor h roeto wrk itmut I I " conducted under vacuum conditions - 10"4 - 10-5 torr. 5. Major reason - increase the mean-free path. "The average distance that our ions and molecules will travel before colliding with another ion or molecule." A high mean-free path insures predictable and reproducible fragmentation, high sensitivity and reliable mass analysis. Turbomolecular pump A turbomolecular pump relies on a series of blades or airfoils j that spin at 30,000 - 90,000 RPM. This tends to deflect gas molecules down and out the outlet. rotating blade fixed blade to rough Dump lubrication access MS block diagram The vacuum system ‘ vcuu H rocd sn ciatin o ' pumps - two stage vacuum system. EA rotary pump serves as our rough or fore pump. It can produce a vacuum of 10'2 - 10‘4 torr. :59 Either a turbomolecular or diffusion pump is A used achieve vacuums in the 10'5 torr range. fiThese pumps actually acts like compressors. Oil diffusion pump A diffusion pump is another commonly used type of high vacuum pump. to _ roughing : ammpactmss ‘--r_-..;.-;-~.-..-.-. . .A‘ a. . ..-.‘.__..,-.. - ‘. ,3... "y. .e. ... _ .. .w-;.-._;.-.;-$._-...-.-. A number of ionization techniques exist. . A“ 9'30"“ (70 9") SW93 3 - . 3 neutral molecule. This knocks out a second 70 W e_. A d B _ c . - I' electron, producing a EIBCtron 'mpaCt (El) . j; molecular ion - with some _ I _ - fi additional energy. ‘7” EV e "w energy B Chemical ionization (CI) 3 - This additional energy is A3533 internalized via a series of .7 \\.A é + .c vibrations, rotations and N + B --c FaSt atom bombardment (FAB) molecular rearrangements A B t g _ . o + . A_ 1...: . . . , “ This results in fragmentation of f \ I \ Field Ionization the molecule. 3+1, 0 B + c+ Mg Am, B Increasmg energy and fragmentation Plasma desorption Electron impact process I Electron impact source 70 eV electrons re typiclly sd. __ At lower energies, you only get a small number 3 Filament Lens 543°" of ions being produced. - ' .// At higher energies, the fragmentation is too great and little information is obtained. low ionization high fragmentation - Repeller information obtained Target - anode Electron impact source .....$.. t r._.._.,. ... ‘..._ . ..-.‘.r.. w . ...r. - ‘. H... ..y. .e. ..__ u -\‘ ;.-.._,‘.-.;-$. Filament - Typically made of Re- OUI' SOUFCB 0f . When a sample molecule enters the source, it 70 eV electrons. '. 3. passer through the electron beam and is ionized. Target - anode used in association with the ‘ :1 filament to produce electrons. - ' The repellerinsures : _ U : thatthe ion is rapidly _jj_ Repeller - posrtlvely Charged electrode used to pushed “tonne ‘push' positive ions out of the ionization source. towards the source ‘3 lens stack. _ . . . - Any negative ions are Lens stack series of Increasmgly more negative pulled "Home electrodes used to accelerate our ions to i repairer, constant kinetic energy. : 5" 3 . simply passes directly ,' Electron impact source 3. ' As an ion accelerates towards the first lens, it comes under the influence of the next, more negative lens. lt passes the first lens and accelerates . towards the next. . By the final lens, it is traveling so fast that it I Mass analyzers .. 'w :1.\e-;.-...-.w..--.-._~ r -- .—.-" 4'::;u;v¢-_-1\':.i'.‘¥_} mean—q 5-{kc'a‘ - :th- _g_:.--:—c -za: ; 5-1-3 - ' A mass analyzer or filter is the portion of a mass ' spectrometer that is responsible for resolving different mass fragments. ' Typically, all ions are accelerated to the same kinetic energy (1/2 mvz). I' Some aspect of these accelerated ions is then into the analyzer- exploited as the basis for resolving them. ' - constant KE I Quadrupole _ pi:wr-w.~ ' as --.‘ \-.":'4.'.r:‘_‘.'=::'«:'¢: .<.w:v-—-'-_-5--je;u' «hp—.3ng -za: ".1549!" Types of mass analyzers .3 Magnetic :3 Electrostatic .3.) Time of Flight :3 Quadrupole Mass Filter :3 Quadrupole ion storage (ion trap) quadrupole analyzer x-ray /lens \ / ; ionization '\ ‘ source detector We‘ll look at the Quadrupole Mass Filter — it is commonly used in GC/MS systems. Ions are produced and accelerated. They enter a region I with four poles. The potential of the poles varies resulting in ions . _ of a specific mass ' ' This type of analyzer is the most commonly used for reaching the '. GC/MS work. A large number of spectra are available '. detector. ' - ‘- for comparison to your unknowns. ' : Quadrupole : Quadrupole Scanning :- At any set of conditions, only ions of a specific MIZ . When scanning over a mass range, the DC potential can successfully travel through the entire filter. is varied linearly with the RFIDC ratio held constant. Others are drawn into the rods. This is how entire mass spectra are obtained. Mass range: 10 - 2000 MlZ 10 - 800 M12 is typical An entire mass spectra (1 0-800 MIZ) can be acquired in about one second. i: A electron multiplier is the most - z; common type of detector used. The data system is the ' “ heart of our MS system. f_ The inner surface of the detector is an -‘_ electroemissive material. When struck by an ion, electrons are ejected. MOdern Ms Sysmms rely D t m b _ t t, I th on computers for rapid Lie 0 e Increasing po en la, e . electrons are accelerated and when :\ _ data couecuon‘ 3' they strike another surface, even more " ~ - ':j' processmg and electrons are ejected. j; ' identification of mass spectra. EM voltage This significantly amplifies our signal. -3000 Relative Abundance Abundance Mass i' Charge I Mass l‘ Charge I A simple mass spectrum : A simple mass spectrum Relative Abundance 01 N Mass i Cha 3 : Mass 1 Charge ' A simple mass spectrum 7 A simple mass spectrum ‘ The mases in thisexample should look familiar. Well we know what most of the lines are due to but 3. This is a spectrum for air - a mixture. 3'. What 90”” 16 and 1 7 be fmm? 3 '5. 1m K- r .‘I a "2 Nitrogen - N2 15N14N - an isotope peak. Its small because 15M is not very common in nature Relative Abundance Relative Abundance E M355 1 Charge Mass I Che : B Atomic masses and isotopes 'i- The atomic weights that e find on the periodic ' table represent an average for all naturally occurring isotopes. Example -' Chlorine - atomic weight = 35.453 A mixture of 35CI - 75.77%, 370i - 24.23% We must not only account for these isotopes, we can exploit them for determining the number and types of elements that are Mass 1' Cha 3 present. Relatlve Abundance : Atomic masses and isotope To help with some of our calculations, we commonly work with relative abundance of the isotopes. The most abundant isotope is set to 100% and the the other isotopes are normalized to it. For most elements in organic molecules, the most abundant element is also the lightest. Lets look at some of the more common elements. Isotope classes Te A, +, A+2 class syte i used to indicat the types of isotopes present. A - Only a single isotope. ‘7 This may mean that the element only exists as a single isotope (like fluorine) The abundance of all but one isotope is too small to use. (Example. deuterium - 0,0015%) .,..-.... ‘.-. .... -... . a: .7...” . .. .. -\‘ .. .. . .. -.v w... The remaining parts of the presentation are for you to look at all on your own. You will find the information useful as it provides a very brief summary of how to evaluate mass spectra. Due to time constraints, we’ll not be covering it and as such, you will not be tested on it. : Atomic masses and isotope Mass FIB Mass FIB Mass RB 1.0078 100 12.0000 100 13.0034 1.1 14.0031 100 15.0001 0.37 15.9949 100 18.9984 100 319720 100 32.9715 0.8 339679 4.4 34.9989 100 36.9659 32.5 78.9183 100 90.9163 98 17.9992 0.2 RB = relative abundance Isotope classes A+1 ' Two isotopes. Examples. C and N A+2 "’ At least two isotopes with the highest mass isotope being +2 from the lowest mass isotope. Examples. 0, 8, Cl, Br. In organic M8, the most abundant isotopes are also the lowest mass. ; Interpretation of mass spectral Evaluate the general appearance I Find the molecular ion if possible I Determine the elemental composition I Identify structural features I Propose possible structures I Orcurrn st f initoic ses I " is based on 120 = 1200000000. As a result of this and the mass lost to binding energy, the mass of every other element is NOT a whole number. With high resolution MS, we can use this mass ' defect to specifically identify the elemental composition of each line. With unit resolution MS, this can present a problem. Our unit resolution MS may miss assign the ‘ mass if a peak if it is g 0.5. We can predict how many hydrogen must be present for this to cause a problem. =0.5/(1.0078- 1) # Hydrogen = 64 hydrogen 128 hydrogen would give us a +1 shift. Working with the molecular ion n of he oeul eas o Sis h I ability to determine the molecular formula from the molecular ion. The molecular ion must: Be the highest mass ion - exclusive of isotope related ions which also must be measurable. Be an ‘odd electron ion' Agree with the rest of the spectrum. In orgaic S, w are delin pimrily with H, C, N, and O. The most common isotopes of H and N both weigh more than their assigned ‘whole number' atomic mass. H - 1.0078 N - 14.0031 Hydrogen presents the greatest problem “ because our compounds contain many. A kan wul reen h wot ases let ' determine how large it must be to hit the 0.5 limit. alkane = CNH(2N+2, 2N + 2 = 64 N = 31 MW=31*12+64=434 So its only a real problem at high mass and most " systems can correct for it. Working with the molecular ion Lts asuethat you hae otained the I following mass spectral data: Which line would be the molecular ion? Working with the molecular ion A good uss would be 4. Why? We 44 is too big compared to 43 mile 45 is too small compared to 44 Also, remember that most of our elements consist of a mixture of isotopes. Masses greater that the molecular ion are to be expected (hoped for). Normalization 0 Select a potential molecular ion along with any lines of higher mass. 0 Normalize these lines so that the potential molecular ion is 100. o The error of the isotope lines is estimated as being 10.2 absolute or 110% relative. 0 Using these values, you can begin constructing a normalization table. Normalization table 3 Sta b puttig the msss nd abundance in the first two columns. Then calculate your normalization factor (NF) where NF = 100 f rel. abund ofA rel. Norm fl unmmm-II HI...- Atomic masses and isotope Mass FiB Mass FiB Mass RB Class H 1.0078 100 A C 12.0000 100 13.0034 1.1 A+1 N 14.0031 100 15.0001 0.37 A+1 0 15.9949 100 17.9992 0.2 A+2 F 18.9984 100 A S 31.9720 100 32.9715 0.8 33.9679 4.4 A+2 CI 34.9989 100 36.9659 32.5 A+2 Br 78.9183 100 90.9163 98 A+2 RB = relative abundance Normalization Select a potential molecular ion ' mle abund 44 31440 (potential molecular ion) 45 1026 The values can now be plugged into your normalization table. Normalization table - a simple way to track your calculations. Normalization table Next, calculate the normalized abundances for your isotope lines - A+1 in this example, + 10% relative error We rel. Norm C abund abund "*1 flflmmm-I- 3.25 m...- Normalization table _ Bd pa, r pr ob " three carbons. We’ll look a 3i1 carbons. In the next columns, calculate the mass due to C ' l" and the A+1 abundances. Mass of parent ion Mass of 3 carbon Remaining mass A mass of 8 is pretty small. Most likely it is due ‘ to hydrogen. Potential formula - 03H8 (propane) Carbon - an A+1 element When we normalize to the molecular ion, the A +1 element can help us determine the number of carbons in our compound. Roughly speaking, for every 100 carbons we measure, one of them will be a 13C. When normalized to the molecular ion, this equates to an increase in the A+1 peak of 1.1 for every carbon. 3 - For our propane example Normalization table Three carbons looks like a good choice. Now we need to account for the remaining mass _ ‘ and come up with a formula. rel. Norm abund abund _ ml'e NF c.,_1 on c nummmuan 3.26 mums-mam A+2 Using isotopic abundances How did I know there were three carbons? No, | didn’tjust make it up. ' . The key to using the A+1 and A+2 lines is the known abundances of naturally occurring isotopes. In our simple example, all that was present was " H and C so lets consider them first. Carbon - an A+1 element _ As the number of carbons increase, the ‘ probability of there being a 13C also goes up. This is an additive process. For hydrocarbons. the intensity of the A+1 line H3C - CH2 - CH3 1.1 tells us the # carbon. H3C - CH2 - CH3 1.1 2H has too small of an H30 _ CH2 _ CH3 u intensity to be seen. % total 3 3 Other A+1 elements wiil ' also contribute to the intensity of this line. Carbon - an A+1 element There is also the chance that a molecule might contain more than one 13C. The probability of this happening is pretty rare - i- #C *(1.1)2 - 0.000364% for 3 C You instrument will not be able to see this. Its - less intense than the contribution from deuterium. fragments is most common. : 53 . . 30 In this example, you 1 can see that the _ j molecular ion is odd (73) and the other ' - 44 major Ions are even. 73 This is because the loss of odd mass ' 10 30 so 70 mass 1' charge The A+2 elements Chlorine and bromine are both common in organic compounds. They also produce very characteristic patterns due to the relatively high abundance of there isotopes. Lets look at some of their patterns. You need to be able to identify them if you ever plan of finding the molecular ion of a brominated or chlorinated compound. Nitrogen - the other A+1 element Nitrogen will also contribute to the A+1 line intensity (15N). A nitrogen contributes 0.37 to the A+1 line. ' . For organic MS work, we typically need to rely on another method to determine nitrogen (Nitrogen rule). 0.37/N is pretty hard to see. Compounds containing many N are not very volatile so are not often seen. The A+2 elements Several A+2 elements are seen in organic GClr MS work. 160 100 0.2 100 4.4 32.5 98 "50 is of too small of an abundance to be very useful in ' GCIMS work. Sulfur has an A+1 and A+2 isotope so you will need to ' correct the A+1 line when sulfur is present. By the time you have 4 chlorine, the A+2 line is significantly larger than the A line. Another example Dichloromethane - a common GCIMS solvent i 49 ' ; _ CHzcs 84 .2 Note the 0:2 _ _ pattern at 84 and the Cl . . . . pattern at 49 Bromine, which Is almost a 1:1 mixture of 79Br ; ;_j and 31Br also has a distinctive pattern. General appearance 7 General appearance Knowin te molcular fomua is nie but only I . Lets loo a a fe exampls so ou get an idea I part of the job. '- what we're talking about. There are a few basic rules and methods that . Lets look at some hydrocarbons can be used to help interpret simpler mass ' _ Straight chain SpeCtra' Branched Some factors to consider. _ f' Unsaturated Overall appearance Degree of fragmentation " Rings Presence Of Cluéteré Gefieral Shape ; 3: Each factor will have an effect on the overall Odd or even major lines In clusters. : appearance Saturated hydrocarbons : Saturated hydrocarbons 5-methylpentadecane 43 . This is a classic hydrocarbon pattern ' 35 141 Note the regular spacing of the clusters at intervals of 14 mass units. Also, the largest line in each cluster is an odd value. When a branch is present, _ . we see lines associate with _ ~ fragmentation on either side W/VW of the branch producing ; lines of higher intensity. n-hexadecane Saturated hydrocarbons : Unsaturated hydrocarbons Deviations from the normal distribution may indicat Manes the presence of a branch. As branching increases, '_ S t . .l t lk the number of lines goes up. _ -, pee ra are S'm' ar 0 a anes' . . Favored ionization process is ionization the the As a result, this approach is only - ;. - of limited use. : . J'lZ bond. Results in reduced fragmentation and normal hydrocarbon - increased parent peak intensity. diatributim‘ Fragments that contained the double bond are Presence of a single E offset by -2. branch. : 5 Unsaturated hydrocarbons : Unsaturated hydrocarbons Note the increased intensity f cycloamanes of the parent ion. ' L Compared to alkanes: The major ions are at - _ _ 41, 55, 59, 33, 97 . Reduced fragmentation is also observed. For an alkane they would be at The is an increase in the parent ion. 43, 57, 71, 85, 99, ._. Reason: Fragmentation requires that at W - ; least two bonds be broken. as _ 97 1-hexadecene ; O El ' +c-c—c-c-c-c Unsaturated hydrocarbons I Hydrocarbon examples , Aromatics Lets look at a series of simple examples. -" Aromatics are easy to ionize and produce _- i; _ stable molecular ions. . r All Will be 06 hydrocarbons This results in reduced fragmentation and ' The Spectra will only Show the major lines to abundant parent peaks. ' make it easier to view the changes. Unfortunately, it is difficult or impossible to _ _ _ i _ determine ring substitution patterns ~ F'rSt We “ IOOk at some "near 06 {3' hydrocarbons. Then we’ll evaluate some @ E' O. 1 cyclic ones. n-hexane ’ n-hexane vs. 1-hexene Note the molecular . ion at 86. ' ? n‘hexane We have a typical 7 Note how the molecular hydhl‘ocagbon Pzflern _ - 3" ion for 1-hexene (at 84) _ wit 5? eing t e -' - - - us more intense. most intense line. - j-. 0 2 60 so 0 4O 1-hex "9 Also, in each cluster, the CnH2n_1 and GHH2n 7; ions are of increased 55' intensity. 1'. o 20 40 an an relative abundance All three hexenes ’ cyclohexane 1 'hexene ‘ n-hexane For cycloalkanes, we see a large increase in the intensity of the Here you can see that the mo|ecu|ar ion, is not a large difference regardless of where the double bond is. This is because at least two bonds must be broken to form a fragment 0 20 4-D BD 30 The tendency for the bond 4 cyclohexane to migrate during 1‘ O ionization tends to make . g: the spectra look quite J El similar. O +CH2-CH2-CH2-CH2-CH2-CH2 o :a w an a Unsaturated rings Some other helpful tools °y°'°""’3’3 I I I I .‘ Some simle tool can b usd 0 help interpret ' your spectra. As the degree of . unsaturation increases, ; The odd electron rules. Used to confirm if a the relative intensity of ' spectral line could be due to a molecular ion. the molecular ion will 3 increase. The nitrogen rule. Helpful if an odd number of 9" N are present. Fragmentation is also a a; reduced. 3;. Logical losses. Do the other lines make sense based on your molecular ion? Odd-electron ions Ionization of our sample occurs by the loss of an " electron. The species we form is a ‘radical’ + M M ° Our original molecule had an even number of electrons. A molecular ion will have an odd number. This is a useful in identifying molecular ions. The nitrogen rule as A compound containing only C, H, O orX will have an even molecular weight e A compound with an odd number of nitrogen " will have an odd molecular weight. s A compound with an even number of nitrogen . “ will have an even molecular weight. So if the parent ion is odd, look for nitrogen. Logical losses _ We can also have neutral losses. It requires that " at least two bonds are broken so they are less common. 2 H2 27 HCN 36 HCI 17 NH3 00/0sz 44 co2 13 H20 30 CHZO 74 ch602 ' 20 HF 34 st so HBr ' We can use these ‘losses’ to help piece thing together. The nitrogen rule In organic compounds, there's a relationship " between the valance and the mass of the most common isotope for most elements. Even elements have an even valance. Odd elements have an odd valance. This leads to the ‘nitrogen rule.’ It assumes that we are limiting our elements to C, H, halogens, O and N. Logical losses Only a limited number of neutral fragments of ' low mass are commonly lost. 1 H 27 CZH3 15 CH3 29 CZH5 or CHO 16 NH2 31 OCH3 or CHZOH 17 OH 35 Cl 19 F 43 OCZH5 or COOH 26 CN 46 N02 These are for a single bond cleavage - radical losses. Logical losses example _ Lets see how logical losses can help. First, here is the information needed for the ' parent ion. M/e Rel. Abund 57 100.00 84 0.10 85 0-40 It's even numbered 15-51 so we either have no 87 1.00 N or an even number. 86 appears to be the parent ion. Logical losses example ' Logical losses example Now r-rliea etermeth mllar I I I I I I. I I. I. I. I I I. I formula. I -un-ance 86-72=14 CBH‘14 ' ‘9 o 4 a 1216202428323640444852566064637276303438 3' MIZ Another logical loss example - CHaclz You can also see how we have a Cl2 pattern at 84 and a Cl pattern at 49 So n-hexane appears to be the likely material. : This also confirms Mass, elements, lines and losses all make sense. " “‘3 '033 0f 3 0" ...
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22_MassSpec - Mass Spectroscopy This is powerfu tool for...

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