+ 2 e- radical cation

Transcription

1 Spectroscopy Beauchamp 1 Basics of Mass Spectroscopy The roots of mass spectroscopy (MS) trace back to the early part of the 20th century. In 1911 J.J. Thomson used a primitive form of MS to prove the existence of isotopes with neon-20 and neon- 22. urrent, easy-to-use, table-top instruments of today are a very recent luxury. In less than a day, you could be running samples on a mass spectrometer. owever, it would take you longer to learn the many intricacies of MS, something we cannot pursue in a book such as this. We will mainly look at electron impact mass spectrometry (EI) and briefly mention chemical ionization (I) as they pertain to determining an organic structure. The technique of MS only requires very small amounts of sample ( g-ng) for high quality data. For that reason, it is the preferred method to evaluate product structures in combinatorial chemistry, forensic laboratories and with complicated biological samples. Generally, in these situations, you have some indication of the structure(s) possible. MS can be coupled to separation techniques such as gas chromatography (G) and high pressure liquid chromatography (PL) to make a combination technique (G-MS and L-MS). G can separate components in relatively volatile mixtures and PL can separate components in relatively less volatile mixtures. There are also options for direct inlet of solid samples and sampling methods for high molecular weight biomolecules and polymers. But, these are beyond the scope of this book. MS is different from the other spectroscopies (UV-Vis, I, NM) in that absorption or emission of electromagnetic radiation is not used. ather, the sample (molecule) is ionized by some method (often a high energy electron beam = electron impact = EI). An electron is knocked out of a bonding molecular orbital (M), forming a radical cation. Dications and anions can also be formed, but we will not consider these possibilities. high EI mass spec + energy e- + 2 e- radical cation The cations formed are accelerated in a high voltage field, focused and separated by mass to charge ratio (m/z or m/e) using a magnetic and/or electric fields. A detector indicates the intensity of each mass signal and the mass data (x axis) are plotted against this intensity (y axis) to produce a spectrum similar to that shown below. It is also possible that this same data can be printed in a tabulated, numerical form (shown in the side box). The most useful information from the MS is the molecular weight (the M+ peak), which can indicate what the formula is. The formula provides the degree of unsaturation, which gives important clues to the possible structures (rings and pi bonds). Fragment peaks that are detected provide hints as to the nature of the carbon skeleton, heteroatoms and functional groups present. The most abundant peak (largest) in the mass spectrum is called the base peak. It is assigned a value of 100% and all other detectable masses are indicated as a percent of the base peak. The molecular weight peak is called the mass peak or molecular ion peak or parent peak and symbolized with an M. Since this peak is a radical cation, it often also has a + or +. (plus sign and a dot) superscript as well. We will use M+. There is often ambiguity in the other fragment peaks because of high energy rearrangements that are possible. It is usually very difficult to assign a structure to a completely unknown molecule based solely on mass spectroscopy. But a mass spectrum can help provide a very important piece of the puzzle, the molecular weight. y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

2 Spectroscopy Beauchamp 2 base peak = largest peak in MS spectrum = 100% peak, other peaks are reported as a percent of this peak molecular ion = M = M+ = M+ = parent peak nly specific isotopic masses are found in the molecular formula. We do not see average masses that are listed in the periodic table. Also present will be M+1, M+2, etc. peaks due to other isotopes. n low resolution MS these peaks can help decide what the molecular formula is. In the MS example below, some of the peaks are very logical (57, 85 and 91 are logical) and some are less so (39, 41, 42, 51 and 55). It is also true that peaks that are logical are sometimes small or completely missing (119). Many of the other peaks will be explainable with certain assumptions about fragmentations discussed later in this chapter.. Tabulated Data Mass percent (base) = M percent relative intensity phenyl-2-hexanone 12 16, MW = mass charge 85 = base peak = 91 m e Many smaller peaks are not shown, but listed in data table to the left. M + peak 176

3 Spectroscopy Beauchamp 3 Typical MS Instrument Features. The moving charged cations (- + ) can be made to curve in their direction of flight in a magnetic or electric field. The amount of curvature is determined by the mass (m) of the ions as shown in the following equations (assuming the charge, e, is constant = +1). The magnetic field (B) and/or accelerator plate voltage (V) can be altered to cause each possible mass to impact the detector. The charged masses must survive about 10-6 to 10-5 seconds to make this journey to the detector. ften there is some rational feature to explain each peak s special stability that allows it to last long enough to reach the detector, where it becomes part of the data we examine. We will look at some of these features later in this discussion. We will not discuss other possibilities, such as metastable ions or +2 and negatively charged ions. ur main goal in this book is interpretation. m e r = = B 2 r 2 2V mv e y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc 1 B m = mass e = charge (usually +1) B = size of magnetic field r = radius of curvature V = voltage on accelerator plate Besides just seeing a positively charged mass at the detector, we must resolve it from nearby mass values. MS instruments can be either low resolution (LMS) or high resolution (MS). Low resolution MS instruments can generally resolve single amu values as high as about 2000 amu s (e.g. they can distinguish 300 amu from 301 amu). An atomic mass unit is defined as 1/12 the mass of a neutral carbon-12 atom ( 12 = , by definition). igh resolution MS instruments can resolve masses as close as the fourth decimal place (XXX.XXXX). With such accuracy, an exact molecular formula can be determined by a computer. A molecular formula can also be obtained from LMS,

4 Spectroscopy Beauchamp 4 through a slightly more involved procedure. MS instruments tend to be more expensive and less common. Exact Masses We need to be precise in our calculation of possible masses for each collection of atoms because the atoms in any cation hitting the detector are specific isotopes. The atomic weights listed in the periodic table are average weights based on the abundance and mass of all of the naturally occurring isotopes of each element. For example, the atomic weight of bromine in the periodic table is 79.9, even though there is no bromine isotope with a mass of 80. The 79.9 atomic weight is a result of an approximate 50/50 mixture of two stable isotopes of mass 78.9 and Because of this complication, we will require data on the exact masses and the relative abundance of the common isotopes that we expect to encounter. Those most useful to us in organic chemistry and biochemistry are listed below. Average Element Atomic Weight Nuclides Exact Mass elative Abundance* hydrogen (D) carbon nitrogen oxygen fluorine silicon phosphorous sulfur chlorine 35,453 bromine N N F Si Si Si P S S S l l Br Br iodine I *The most abundant nuclide is assigned 100% and the others assigned a fractional percent of that value. oincidently, in the examples listed in the table above with more than one isotope, the lowest mass isotope is the 100% isotope.

5 Spectroscopy Beauchamp 5 btaining a molecular formula from a MS is relatively straight forward Each possible molecular mass is unique when calculated to 3-4 decimal places and computers can do the calculations for us. Try the problems below. Unfortunately, here you have to do the calculations yourself. Problem 1 - A low-resolution mass spectrum of 1,10-phenanthroline showed the molecular weight to be 180. This molecular weight is correct for the molecular formulas 14 12, 13 8 and 12 8 N 2. A high-resolution mass spectrum provided a molecular weight of Which of the possible molecular formulas is the correct one? What is the degree of unsaturation in 1,10-phenanthroline? Problem 2 Isopalhinine A, a natural product was found by low-resolution mass spectrometry to have a molecular weight of 291. Possible molecular formulas include N 5, NN 4, and N 3. igh-resolution mass spectrometry indicated that the precise molecular weight was What is the correct molecular formula of isopalhinine? What is the degree of unsaturation? To obtain a molecular formula from a LMS requires more sophistication. Various possible formulas can be generated using the molecular ion peak and the rule of 13. The first possible formula assumes that only carbon and hydrogen are present. The molecular mass (M+) is divided by 13 generating an integer (n) and a remainder (r). The number 13 represents the mass of one carbon atom and one hydrogen atom. The formula becomes n n+r. All molecular hydrocarbons have even mass molecular weights. Each of these masses = 13 amu = + (We assume there are "n" of them if the unknown was a hydrocarbon. This is our starting point formula.) These are left over hydrogen atoms = r M 13 = n + r M = molecular weight n = number of units = quotient r = left over hydrogens = remainder Possible hydrocarbon molecular formula = n n+r (as a hydrocarbon always an even mass) The degree of unsaturation can be calculated for this formula and possible rings and/or pi bonds can be considered (discussed in the introduction, p 10). If oxygen and/or nitrogen (and other elements) are present, the / numbers in the molecular formula must be changed by an amount equal to the new element s isotopic mass. It is assumed, when substituting atoms, that the major isotope is used in all cases (always the lowest mass isotope, for us), =1, =12, N=14, =16, S=32, l=35, Br=79. Since oxygen weighs 16, we can subtract 4 (= 16) from the formula and substitute in the oxygen atom. If two oxygen atoms were present, we would subtract 2x( 4 ) = 2 8 and so forth. Nitrogen-14 would substitute for 2 and n nitrogen atoms would substitute for ( 2 )x(n). If we did not have enough hydrogen atoms for some reason (it happens), we could take away one carbon atom and add in 12 hydrogen atoms, or if there were too many hydrogens, you could do it the other way around and add one carbon and take away 12 hydrogen atoms. Information concerning the possible number of nitrogen atoms in the molecular formula is also available in the molecular mass. If the molecular mass is an even number, then the number of y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

6 Spectroscopy Beauchamp 6 nitrogen atoms has to be zero or an even number (= 0, 2, 4...). If the molecular mass is an odd number, then the number of nitrogen atoms has to be odd (= 1, 3, 5...). emember, each nitrogen atom in the formula adds an extra bonding position. N N N n 2n+2 x n 2n+3 N 1 n 2n+4 N 2 (N is odd) (N is even) = even mass = even mass = even mass = even mass = odd mass = even mass = even mass = even mass = even mass MW = even mass MW = odd mass MW = even mass Problem 3 - An unknown compound produces a molecular weight of 108. What are all possible formulas having only carbon and hydrogen or having carbon, hydrogen and an oxygen atom ( two oxygen atoms) or having carbon hydrogen and nitrogen (what is the minimum of nitrogen atoms that would have to be present)? What is the degree of unsaturation for each of these possibilities? Is it possible that the formula has only a single nitrogen? If so what would the formula be? If not, why not? What if the molecular weight was 107? (Same questions.) To choose among the various formulas generated from the rule of 13, we can consider the other possible isotopes present and their relative abundances to calculate the size of the peaks just one mass unit (M+1) and two mass units (M+2) larger than the molecular ion peak (M+). For each possible formula, percents of the M+1 and M+2 peaks versus the M+ peak are calculated. In this calculation the M+ peak is assumed to be 100% for comparisons with M+1 and M+2, regardless of the base peak. These calculated values are compared to the experimental values to determine the most likely formula. The reason for this is that the relative sizes of the M+1 and M+2 peaks are determined by the number and isotopic abundance of the elements present. The presence of either chlorine, bromine or sulfur significantly changes the M+2 peak. If there are multiple halogens (l and Br), the M+2, M+4, M+6 and beyond can be calculated and compared to the experimental mass spectrum. This approach only works if the M+ peak is large enough so that M+1 and M+2 are significant. If the M+ peak is too small, we can t tell what the relative fractions of M+1 and M+2 are. Let s take a look at how one could calculate the relative size of these peaks (M+1 and M+2). Sample calculation using M+, M+1, M+2 peaks to identify the molecular formula by LMS We will assume an actual formula that is owever, we will pretend we don t know this. ow could the M+1 and M+2 lead us to the correct formula? The molecular mass of 4 10 is 74 and that would produce our molecular ion peak, M+. We would have an extra amu in the mass if we had a different isotope one amu higher. We could do this 4 ways with carbon (because there are four 13 atoms) 10 ways with hydrogen ( 2 = D) and 1 way with oxygen ( 17 ). The probabilities for these possibilities are shown below for the M+1 peak. If we add all of these together we can see the total probability for getting an M+1 peak relative to for getting the M+ peak. Using a similar strategy we can estimate the probability for getting an M+2 peak, which will be considerably lower since we have to get two 13 or two 2 or one 13 and one 2. The main contribution to the M+2 peak is the 18 isotope. Taken together, these three peaks would predict the indicated distribution for M+, M+1 and M+2 for this collection of atoms ( 4 10 ).

7 Spectroscopy Beauchamp 7 molecular ion peak = M+ = 4x( 12 ) + 10x( 1 ) + 1x( 16 ) = 74 amu as a fraction = as a percent = 100% M+1 peak - arises from different possibilities of one additional amu = 75 amu Whatever the size of this peak, it is assumed to be 100% for comparison with the M+1 and M+2 peaks. one 13 = (4 ways) = one 2 = (10 ways) = D D one 16 = (1 ways) = sum of possibilities = (0.0439) + (0.0015) + (0.0004) = M+1 peak as a percent of M+ peak = (0.0458)x(100%) = 4.58% M+2 peak - arises from different possibilities of two additional amu = 76 amu two 13 = two 2 = one 18 = x 3 2 x 1 (1 ways) = = (0.0439) 2 (6 ways) = x 9 2 x 1 = (2.25x10-8)(45 ways) = 1 x 10-6 = = too small to consider one 13 and one 2 = = 1 x 10-6 = = (4 ways) x sum of possibilities = (0.0007) + (0.0020) + (0.0001) = (10 ways) M+2 peak as a percent of M+ peak = (0.0028)x(100%) = 0.28% 100% M+ = molecular ion peak Exact Mass (M+1) (M+2) (formulas) N N N N etc N ere is our compound. "mini" probability theory There are 4 ways of picking the first carbon and 3 ways of picking the second carbon (=4x3) and since all carbon is the same, we can't tell what carbon was picked first and second, so we divide by two facorial (2x1). 4.58% 0.28% M+ M+1 M+2 To find a possible molecular formula using the M+1 and M+2 peaks, we first find the correct molecular weight for our molecule (in this case mass = 74). Then we look through the M+1 and M+2 values for two values that match our mass spec data. In this case we see that 4 10 is a very close match and it becomes our best guess. Since the molecular weight is even, the number of nitrogens atoms must be even (0,2,4...). Any formulas with an odd number of nitrogen atoms must be part of a fragment. M+ = molecular ion peak Exact Mass (M+1) (M+2) (formulas) N N Data tables exist with many values already calculated for comparisons. y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

8 Spectroscopy Beauchamp 8 Problem 4 a. alculate the relative intensities (as a percent) of M +, M+1 and M+2 for propene ( 3 -= 2 ) and diazomethane ( 2 =N=N). an these two formulas ( 3 6 vs 2 N 2 ) be distinguished on the basis of their M+1 and M+2 peaks? alculate the exact mass (four decimal places) for both of these formulas. an they be distinguished on the basis of exact mass? elpful data are on page 4. b. Both + and have fragment masses of approximately 29, yet + has a M+1 peak of 1.13% and M+2 peak of 0.20%, whereas has a M+1 peak of 2.24% and M+2 peak of 0.01%. igh resolution mass spec shows + to have a different fragment mass than Explain these observations and show all of your work. elpful data are on page 4. hlorine, bromine and sulfur, when present, have very characteristic M+2 peaks (32.6% for l, 96.9% for Br and 4.4% for S). If multiple l s and/or Br s are present M+2, M+4 and beyond are indicative of the number and type of halogen(s) present. The various patterns are available in many references. owever, you can calculate these values yourself, as was done above for the M+1 and M+2 peaks above. one l comparison of M+ peak ( 35 l) to M+2 peak ( 37 l) M+ peak relative size probability of 35 l = 100 (1 way) = (assigned a referenced value of 100%) M+2 peak relative size probability of 37 l = percent of M+ peak = (1 way) = % 32% (100%) = 32% M+ M+1 M+2 one Br comparison of M+ peak ( 79 Br) to M+2 peak ( 81 Br) M+ peak relative size probability of 79 Br = 100 (1 way) = (assigned a referenced value of 100%) M+2 peak relative size probability of 81 Br = percent of M+ peak = (1 way) = % 97% (100%) = 97% M+ M+1 M+2

9 Spectroscopy Beauchamp 9 one S comparison of M+ peak to M+1 to M+2 peak M+ peak relative size probability of 32 S = (1 way) = (assigned a referenced value of 100%) M+1 peak relative size probability of 33 S = percent of M+ peak = M+2 peak relative size probability of 34 S = percent of M+ peak = (100%) = 0.8% (100%) = 4.4% (1 way) = (1 way) = % 0.8% 4.4% M+ M+1 M+2 one Br and one l comparison of M+ peak to M+2 and M+4 peaks M+ peak relative size probability of 79 Br = (from above) probability of 35 l = (from above) (probability of 79 Br)(probability of 35 l) = (0.508) (0.758)(1 way) = (assigned a referenced value of 100%) M+2 peak relative size probability of 81 Br = (from above) probability of 37 l = (from above) (probability of 79 Br)(probability of 37 l)(1 way) = (0.508) (0.242)(1) = (probability of 81 Br)(probability of 35 l)(1 way) = (0.492) (0.758)(1) = total = percent of M+ peak = (0.496/0.373)x100% = 129% M+4 peak relative size (probability of 81 Br)(probability of 37 l)(1 way) = (0.492) (0.242)(1) = percent of M+ peak = (0.119/0.373)x100% = 31% 129% 100% 31% M+ M+2 M+4 two l comparison of M+ peak to M+2 peak to M+4 peaks M+ peak relative size probability of two 35 l = (0.758) 2 (1 way) = (assigned a referenced value of 100%) M+2 peak relative size probability of 37 l = (from above) (probability of 35 l)(probability of 37 l)(2 ways) = (0.758) (0.242)(2) = percent of M+ peak = (0.367/0.602)x100% = 61% M+4 peak relative size (probability of 37 l)(probability of 37 l)(1 way) = (0.242) 2 (1) = percent of M+ peak = (0.059/0.602)x100% = 10% 100% 61% 10% M+ M+2 M+4 y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

10 Spectroscopy Beauchamp 10 Problem 5 - alculate the relative intensities (as a percent) of M+, M+2 and M+4 for Br 2. Use the probabilities from above. Problem 6 - alculate the relative intensities (as a percent) of M +, M+2, M+4 and M+6 for Brl 2 and Br 2 l. int: All of the data you need to perform these calculations are in the examples above. Use the probabilities from above. Energetics of Fragmentation of simple hydrocarbon patterns Bonds are broken in fragmentations, forming radicals and/or cations. The energy costs for radicals and cations of common hydrocarbon patterns are worked out in the tables that follow. We first assume a - bond is homolytically broken (each atom gets one electron, no charge is formed). Next, we take away the cost of making the hydrogen atom (the same for every - bond) to find out what the cost is for forming only the carbon free radical. Lower energy possibilities are favored over higher energy possibilities. A few problems are provided just below the following tables to illustrate these points. A similar diagram is constructed to estimate the energy costs of forming carbocations. We start out the same, but in this diagram we include the ionization potential of the carbon free radical, a value that can be measured experimentally. We again take away the energy to make the hydrogen free radical and also take away the energy change when the hydrogen atom attracts the extra electron (electron affinity) to become a hydride. What remains is an estimate of the energy to make only the carbocation. This is a considerably larger amount of energy than to make the carbon free radical (because we are stealing away an electron).

11 Spectroscopy Beauchamp 11 General Energy ycle for arbocations - relative energy to form carbocations (all energy values in kcal/mole) e- heterolytic bond energy f o ( ) = -52 heat of formatio of hydrogen atom, common to all cycles ionization potential of f o ( electron affinity) = homolytic bond energy f o ( ) = + value (see table) Energy to form carbocation ompound ( 3 ) 2 - ( 3 ) 3-2 = adical (BE) (hydrogen carbocation) (methyl carbocation) (primary carbocationl) 98 ( 3 ) 2 (secondary carbocation) 95 ( 3 ) 3 (tertiary carbocation) 92 2 = 2 (allylcarbocation) (benzyl carbocation) 88 I.P. E.A.() f o ( ) o f ( ) = [BE+IP-EA- o f ( )] =energytomake (104) + (313) - (17) - (52) = +348 (105) + (227) - (17) - (52) = +263 (98) + (193) - (17) - (52) = +222 (95) + (169) - (17) - (52) = +195 (92) + (154) - (17) - (52) = +177 (86) + (186) - (17) - (52) = +203 (88) + (165) - (17) - (52) = +184 ommon arguments for relative stabilities of free radicals and carbocations are inductive effects/hyperconjugation and resonance. Inductive effects and hyperconjugation argue that switching out a hydrogen for a carbon group allows greater electron donation to the electron deficient carbon atom (free radical or carbocation) because of increased pairs of electrons polarized towards the electron deficient centers. arbocations are much more electron deficient than free radicals and benefit much more from this effect. The resonance argument states that an adjacent pi bond or lone pair can spread electron density through parallel p orbitals, thus reducing the energy to form a cation or free radical. y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

12 Spectroscopy Beauchamp 12 The differences in relative carbocation stabilities parallel the trend seen in free radicals, but are greatly enhanced versus the free radical stabilities. ne could also make a steric argument for tertiary being the most stable free radical or carbocation. The geometry changes from 109 o (sp 3 ) bond angles to 120 o bond angles (sp 2 ). The ground state of a tertiary - bond would start at higher potential energy from crowding, which would be relieved somewhat when the fourth group is removed, providing, perhaps, part of the advantage in the tertiary reaction over secondary over primary over methyl when forming tertiary free radicals and carbocations. Breaking a bond is a large uphill energy transformation, but less so with a sterically crowded starting point, so E a is a little smaller than expected. more crowded as sp 3 center = higher potential energy starting point with 3-4 larger groups around tetrahedral carbon less crowded as sp 2 with 3 groups around trigonal planar carbon is slightly more stable than it would be if groups were smaller

13 Spectroscopy Beauchamp 13 Problem 7 onsider the possible fragmentation of 2-methylbutane (isopentane). There are 3 types of - bonds that could break (b,d,f) and 4 types of - bonds that could break (a,c,e,g). nly consider breaking the - bonds (b,d,f) and the tertiary - bond (c). Each bond could break in two ways: either atom could be a cation and either atom could be a free radical. alculate the energy cost for each possibility (each bonded atom as a radical and each atom as a cation). For each possibility what are the masses that would be observed at the detector (we only see cations)? This problem will require eight calculations for the four bonds considered. 2 a 3 b d f 2 c e g 2-methylbutane (isopentane) high energy electron beam 2 a 3 b d f 2 c e g radical cation Possible fragmentations? Energy to rupture bonds (eight calculations). b b c c d d f f Actual Mass Spectrum tabulated and graphical. mass percent = base = M+ Peaks 15, 29, 43, 57 and 72 are logical. In our discussions of fragmentation we will see how many of the other peaks are explainable. percent relative intensity ev isopentane = base peak 41,42 57 Many smaller peaks not shown M+ peak 72 MW = mass m = charge e y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

14 Spectroscopy Beauchamp 14 Problem 8 onsider the possible fragmentation of 2,2,4-trimethylpentane. There are four types of - bonds that could break (a, b, d, f) and 4 types of - bonds that could break (a, c, e, g). nly consider breaking the - bonds (a, b, c, d). Each bond could break in two ways: either atom could be a cation and either atom could be a free radical. alculate the energy cost for each possibility (each bonded atom as a radical and each atom as a cation). For each possibility what are the masses that would be observed at the detector? This problem will require eight calculations for the four bonds considered (we only see cations) a b c d 3 high energy electron beam a b c d 3 Possible fragmentations? 3 3 2,2,4-trimethylpentane 3 3 radical cation Energy to rupture bonds (eight calculations). a a b b c c d d Actual Mass Spectrum tabulated and graphical. mass percent = base = M+ (missing)

15 Spectroscopy Beauchamp 15 Problem 9 - Predict reasonable fragmentation patterns for n-octane and where the major ion peaks should appear. ationalize your predictions on the basis of energetics. The mass spectrum is provided for comparison. Some of the less logical peaks will become explainable after our discussions on fragmentation. Is there a logical peak that is missing? Actual Mass Spectrum tabulated and graphical. mass percent = base = M+ percent relative intensity ev octane base peak mass m charge = Z Many smaller peaks not shown M + peak y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

16 Spectroscopy Beauchamp 16 Special patterns of fragmentation from organic functional groups Alkanes - Key Points (see examples above) 1. Lower mass alkyl branch fragments (2-6 s, masses = 29, 43, 57, 71, 85) are more intense than higher mass fragments ( 6). The loss of the smaller branch as the cation more commonly reaches the detector. 2. The major carbocations that form follow carbocation stabilities ( + = 3 o > 2 o > 1 o > Me). It is also quite possible that less stable carbocations rearrange to more stable carbocations before they reach the detector. We can t tell by only observing the mass since they have the same number. less stable primary carbocation more stable tertiary carbocation proposed fragmentation probable rearrangement mass = 57 can't tell which mass = Linear alkanes more often have observable molecular ion peaks, while increased branching weakens the molecular ion peak. Fragmentation is more common at branch points. Loss of a methyl from a straight chain is considerably weaker than loss of a methyl at a branch point. M+ = 114 (6%) base peak = 43 (M - 15) = 99 peak (0%) M+ = 114 (3%) base peak = 43 (M - 15) = 99 peak (1%) M+ = 114 (0%) base peak = 57 (M - 15) = 99 peak99 peak (6%) 4. Linear fragments often differ by 14 amu (different size branches split off between carbons in different molecules, 2 = 14). Take another look at problem 9, just above. 5. There are often clusters of peaks around main peaks. Very large fragment peaks will have a trailing M+1 peak due to 13 isotopes (about 1% for every carbon present). A rough guide for any large peak is that it will have M+1 peak that is about 1% its size for every carbon in the fragment due to 1% 13 isotopes at each carbon. For example, if a fragment mass had an 80% value in a five carbon fragment, the next mass peak would be expected to be 0.05x80% 4% size based on 13 isotopes. If there were 10 carbons, the next mass peak would be expected to have 0.10x80% 8% size just based on the 13 isotopes (in addition to any real fragments that might come at that value.

17 Spectroscopy Beauchamp ycloalkanes tend to have stronger molecular ion peaks (two bonds have to break) and their fragment patterns are more complicated to interpret (and we won t try to interpret every possibility). Alkene fragmentation peaks are often subfeatures of the fragmentation pattern. Loss of 2 2 (= 28) is common, if present. M+ = 112 (59%) M-28 = 84 (39%) M+ = 114 (6%) M-28 = 86 (2%) 7. Two masses that seem to show up in nearly every mass spectrum are 39 and 41. These may arise from resonance stabilized carbocations formed by rearrangements in the high energy electron beam. Look for peaks that extend those patterns by units of 14 (insertion of a 2 ),. which are also commonly observed masses. 8. Even masses of 30, 44, 58, 72, etc. on occasion can be due to radical-cation alkanes that form from high energy rearrangements. Some of these masses form from other fragmentations too. But if there is no other logical reason to see one of these masses, this could be a possible explanation. ommon alkane fragmentations occur at branch points; more branches lead to more stable carbocations. owever, skeletons can rearrange in almost any conceivable way possible to form more stable carbocations (e.g. 3 o + > 2 o + > 1 o + > 3 +). Also, alkanes can lose 2 or - to form alkenes, so we have to consider possible alkene rearrangements for alkanes too (see our next functional group). Smaller masses tend to be more prominent than larger masses in the mass spectrum. Perhaps they don t have as many options for falling apart as the larger fragments do. Also, when larger fragments fall apart, they make smaller fragments. y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

18 Spectroscopy Beauchamp M+ = octane - all alkane fragments are observed, except = base M+ = not observed in octane nly cations reach detector, so only the part with positive charge is observed at the detector. A positive charge is written on all fragments to indicate that either part could retain the positive charge (in a rearranged stable form). ften you can see the mass of both cations of a possible fragmentation. It is useful to look for both fragment masses in the mass spectrum. Peaks related to alkene fragmentations are discussed in the next functional group. The typical appearance of a mass spectrum is shown below. Data is also often presented as shown to the right. The intensity of the peaks tends to decrease as the fragment masses get larger. Larger fragments are less likely to survive the 10-5 second trip to the detector. Mostly peaks greater than 4% of the base peak are shown. M+ = MW = 114 actual peaks in octane mass % none alkane peaks base M+

19 Spectroscopy Beauchamp 19 3,4-dimethylhexane - has branches possible alkyl fragments 15 / 99 MW = 114 Loss of hydrogen (-) or an alkane (-) fragment generates alkenes so alkene fragmentation patterns are also observed from alkane structures (see on next page). (- 2 ) (-) alkenes (see the next functional group) / / / = base (4%) elimination reaction similar to - 2 in alcohols to form alkene 56 (100%) The base peak (56) is likely from an alkene, 4 8. emarkably, it is the major peak in the spectrum! Mostly peaks greater than 4% of the base peak are shown. 85 MW = 114 actual peaks in 3,4-dimethylhexane mass % none alkane peaks the base peak is not expected M M+ = It is very common to see alkene fragments in the mass spectra of alkanes, though it is very surprising to see one as the base peak, as is the case here. In the next functional group, we will compare fragmentations of alkenes and alkanes. Alkenes - Key Points 1. A pi electron is likely to be ionized first from the M of the alkene as the least tightly held electrons. Alkenes often produce stronger molecular ion peaks than alkanes because of this. + e- emaining sigma bond holds skeleton together. octane, MW =114 (M+ = 6%) oct-1-ene, MW =112 (M+ = 20%) 2. The double bond can migrate through the skeleton (this makes it difficult to distinguish among positional isomers sharing a common skeleton). These alkenes all look similar. y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

20 Spectroscopy Beauchamp Allylic cleavage is common due to resonance stabilization of cation fragment. The mass can vary depending on the groups attached to the allylic part. Look for peaks that extend this pattern by units of 14 (insertion of 2 x1, x2, ). ' ionization ' fragmentation resonance stabilized carbocation mass = 41 ( = ) 55 ( = 3 ) 69 (= 2 3 ) 83 ( = 3 7 ) etc. ' free radical is sucked away 4. McLafferty-like rearrangements are possible (similar to carbonyl pi bonds). Again, bond migration is possible. Also look for some of these fragment peaks in alkane mass spectra that have lost 2. 2 McLafferty-like rearrangement 2 even mass fragmentation It is possible to see the cation charge on either fragment. Both fragments will be even unless an odd number of nitrogen atoms is present. 2 2 mass = 42 ( = ) 56 ( = 3 ) 70 (= 2 3 ) 84 ( = 3 7 ) 28 ( = ) 42 (1 extra ) 56 (2 extra ) 70 (3 extra ) 5. yclohexenes often undergo retro Diels-Alder reactions. 1 2 fragmentation is a retro-diels-alder reaction 1 diene dienophile nly cations reach the detector. Either fragment could be positive, but usually the diene would be the more stable cation. Both fragments will be even unless an odd number of nitrogen atoms is present. 2

21 Spectroscopy Beauchamp 21 Alkenes Fragmentation Patterns (Many of those below can also be found in octane, an alkane.) nly cations reach detector, so only the part with positive charge is seen at the detector. A positive charge is written on both fragments to indicate that either could retain the positive charge (in a rearranged stable form). ften you can see both as cations from different fragmentations. The following peaks are explained by common alkene fragmentations (data on the right). Many of them are found in fragment peaks of octane, an alkane (see data on the following page). A pi bond can migrate through the skeleton to almost any conceivable position, leading to almost any variation conceivable. McLafferty rearrangements allylic fragmentations 112 (0%) 112 (0%) 112 (0%) 42 (15%) 70 (12%) 56 (18%) 56 (18%) 70 (12%) 42 (15%) 112 (0%) 112 (0%) 112 (0%) 112 (0%) 71 (20%) 41 (44%) 57 (34%) 55 (11%) 69 (2%) 43 (100%) 3 15 (0%) 97 (0%) actual peaks from octane mass % (0%) 84 (7%) 28 (4%) 112 (0%) 83 (0%) 29 (27%) y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

22 Spectroscopy Beauchamp 22 Similar fragmentation patterns for 8 16 alkenes. Notice that octane (an alkane) has many of these same fragments. A E G B D F A B D E F G octane 1-octene trans-2-octene cis-2-octene trans-3-octene cis-3-octene cis-4-octene trans-4-octene not available alkyl branch fragments = 15, 29, 43, 57, 71, 85, 99 allylic fragments = 27, 41, 55, 69, 83, 97 McLafferty fragments = 28, 42, 56, 70, 84,

23 Spectroscopy Beauchamp 23 Another Alkene Example (7 alkene) alkenes (1-heptene, 2-heptene, 3-heptene, all of them look similar because the pi bond can migrate through the skeleton) This example starts with hept-1-ene McLafferty rearrangements A pi bond can migrate through the skeleton to almost any conceivable position. allylic fragmentations 7 14 = 98 (14%) 42 (55%) 56 (100%) 7 14 = 98 (14%) 41 (97%) 57 (31%) 43 (16%) 7 14 = 98 (14%) 56 (100%) 42 (55%) 7 14 = 98 (14%) 55 (68%) 7 14 = 98 (14%) 70 (44%) 28 (5%) 7 14 = 98 (14%) 69 (31%) 29 (56%) 3 = 15 (1%) 7 14 = 98 (14%) 83 (31%) all peaks > 1% alkyl branches 15 (1%) 29 (56%) 43 (16%) 57 (31%) 71 (3%) 85 (0%) allylic fragments 27 (26%) 41 (97%) 55 (68%) 69 (31%) 83 (4%) McLafferty fragments 28 (5%) 42 (55%) 56 (100%) 70 (44%) y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

24 Spectroscopy Beauchamp 24 Alkynes - Key Points 1. Terminal alkynes have weak or missing M+ peaks (they often lose radical hydrogen), though M-1 can be very strong. M+ (M-1)+ 2. The triple bond can migrate through the skeleton (this makes it difficult to distinguish among positional isomers sharing a common skeleton). These alkynes all look similar. 3. All alkynes give a reasonably strong m/e = 39 peak from propargylic cleavage (resonance is K, but more electronegative sp carbocation resonance form reduces contribution). This mass can also be explained by rearrangement to from a very stable aromatic cyclypropenyl carbocation. If you look at a lot of mass spectra, this mass always shows up, even if no alkyne is present. Look for peaks that extend this pattern by units of 14 (insertion of 2 x1, x2, ). ' fragmentation radical cation ' nly cations reach the detector. Mass 39 is in every EI mass spectrum. This could be because the cation is really an aromatic carbocation. resonance also works for mass = 39 ( = ) 53 ( = 3 ) 67 (= 2 3 ) 4. Small peaks at M=26 are probably ethyne (acetylene). M = McLafferty-like rearrangements are possible (similar to the alkene above and a carbonyl pi bond) radical cation even mass fragmentation on one or the other. 40 ( = ) 28 ( = ) 54 ( = 3 ) 42 ( = 3 ) 68 (= 2 3 ) 56 (= 2 3 ) 82 (= 3 7 ) 70 (= 3 7 ) Either fragment can be observed and both show an even mass.

25 Spectroscopy Beauchamp 25 Example peaks from hept-1-yne: alkynes (1-heptyne, 2-heptyne, 3-heptyne, all of them look similar because the pi bonds can migrate through the skeleton) McLafferty rearrangements A pi bond can migrate through the skeleton to almost any conceivable position. allylic fragmentations 7 12 = 96 (1%) 40 (12%) 56 (26%) 7 12 = 96 (1%) 39 (30%) 57 (28%) 7 12 = 96 (1%) 54 (35%) 42 (8%) 7 12 = 96 (1%) 53 (18%) 43 (4%) 15 (0.5%) 29 (46%) 43 (4%) 57 (28%) 71 (0.2%) 85 (0%) 7 12 = 96 (1%) 68 (30%) 28 (4%) 7 12 = 96 (1%) 67 (44%) 29 (46%) 3 15 (0.5%) 7 12 M+ = 96 1-heptyne 7 12 M+ = 96 2-heptyne 7 12 M+ = 96 3-heptyne 7 12 = 96 (1%) 81 (100%) mass % mass % mass % mass % mass % y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

26 Spectroscopy Beauchamp 26 Benzenoid Structures - Key Points 1. Generally, aromatics compounds show a strong M+ peak. 2. A side chain alkyl branch ( 2 -) can fragment at the benzylic position, which is proposed to rearrange to the tropylium ion showing a m/e = 91 peak. Analogous rearrangements are possible in more substituted benzenoid compounds producing different, but predictable, masses. ' 2 fragmentation ' 2 lots of resonance rearrangement ' radical cation nly cations reach the detector. This mass is 91 (if = ) and even though it is a very stable cation, it rearranges to a more stable 'tropylium' carbocation. Any branches or heteroatoms would change the '91' mass. tropylium ion, an aromatic cation (lots of resonance) ' = mass N Isomeric benzenes are difficult to distinguish among, as a group. Even though the structures are different, the mass spectra of the compounds are pretty much alike due to high energy rearrangements. These isomers have similar looking mass spectra. 4. McLafferty-like rearrangements are possible, if a simple alkyl chain of three more carbons is present (oxygen can also be in the branch) and a hydrogen atom is on the gama atom. This fragmentation produces an even mass of m/e = 92 for an unsubstituted carbon chain. Substituted rings will have different masses depending on the additional atoms. emember that part of the 92 peak is -13 isotopes in the 91 peak (about 7x0.01 = 0.07). Even mass, if there is not an odd number of nitrogen atoms. = mass N 107 or can be on either fragment Both have even masses, if there is not an odd number of nitrogen atoms. = mass

27 Spectroscopy Beauchamp 27 or can be on either fragment M+ = 122 (35%) 94.0 = 100% 28.0 = 1% Examples: McLafferty rearrangements benzylic fragmentations = 148 (27%) 92 (74%) 56 (0.4%) 91 (100%) 57 (4%) 105 (11%) bridging phenyl group 43 (1%) 15 (0.2%) 29 (6%) 43 (1%) 57 (4%) 71 (0%) 85 (0.2%) 105 (11%) 65 (9%) 77 (4%) M+ = 162 hexylbenzene M+ = t-butyl-3-ethylbenzene M+ = 135 p-propyltoluene * * nly about 7% is due to 13 isotopes. mass % mass % mass % mass % mass % Notice that "91" is not logical, but it shows up y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc

28 Spectroscopy Beauchamp 28 alogenated ompounds - Key Points 1. Fluorine (mass = 19) and iodine (mass = 127) have only one naturally occurring isotope, loss of either of these masses is informative (M-19, M-127). Fluorine compounds tend to show weak M+ peaks (or none at all). When iodine is lost, there can be a big hole (= 127) in the middle of the mass spectrum. 2. hlorine has two isotopes (35 and 37) which occur in a 3:1 ratio; this is easily observed when there is a molecular ion and in any fragments that retain the chlorine. An M-35 peak is informative, and M-36 corresponds to loss of l. 3. Bromine has two isotopes (79 and 81) which occur in a 1:1 ratio; this is easily observed when there is a molecular ion and in any fragments that retain the bromine. An M-79 peak is informative, and M-80 corresponds to loss of Br. 4. Loss of X is common (see above) and loss of X can occur with fluorine (M-20), chlorine (M-36), bromine (M-80). 5. Loss of an alkyl radical and formation of a five atom ring or three atom ring is possible with chains of 5 and longer with bridging chlorine, bromine or iodine (also true for sulfur). X fragmentation X ations reach the detector, will see this mass. Free radicals are sucked away by the vacuum pump. X = mass l 91 Br I X fragmentation X ations reach the detector, will see this mass. X = mass l 63 Br 107 I 155 Free radicals are sucked away by the vacuum pump.

29 Spectroscopy Beauchamp 29 Examples 1-chlorhexane 6 13 l = 120 l alkyl branches 15 (1%) 29 (32%) 43 (72%) 57 (15%) 71 (3%) 85 (0.7%) l 91 (100%) l 63 (5%) 84 (4%), minus l other alkene fragments McLafferty allylic 28 (5%) 27 (27%) 42 (45%) 41 (59%) 56 (56%) 63 (81%) 70 (3%) 84 (1%) l and 37 l mass % mass % bromorhexane 6 13 Br = 165 Br alkyl branches 15 (1%) 29 (21%) 43 (66%) 57 (100%) 71 (3%) 85 (18%) Br 135 (8%) Br 107 (1%) 84 (4%), minus Br other alkene fragments McLafferty 28 (3%) 42 (10%) 56 (5%) 70 (3%) 84 (1%) allylic 27 (16%) 41 (42%) 63 (0%) 35 l and 37 l 35 l and 37 l mass % mass % iodohexane 6 13 I = iodopropane 3 7 I = 170 I 2-iodopropane 3 7 I = 170 I I alkyl branches 15 (1%) 29 (15%) 43 (100%) 57 (11%) 71 (0%) 85 (50%) alkyl branches 15 (2%) (1%) 29 (0%) (0%) 43 (100%) (100%) 57 (0%) (0%) 71 (0%) (0%) 85 (0%) (0%) Almost identical mass spectra. I 183 (0%) I not possible I 155 (0%) I 107 (2%) 84 (4%), minus I other alkene fragments McLafferty allylic 28 (3%) 27 (14%) 42 (3%) 41 (25%) 56 (2%) 63 (0%) 70 (0%) 84 (0%) 42, minus I other alkene fragments McLafferty allylic 28 (3%) (2%) 27 (32%) (28%) 42 (3%) (4%) 41 (37%) (36%) 56 (0%) (0%) 63 (0%) (0%) 70 (0%) (0%) 84 (0%) (0%) I = (5%) (6%) I = (1%) (2%) mass % mass % iodopropane 2-iodopropane mass % mass % y:\files\classes\spectroscopy Book home\1 Spectroscopy Workbook, latest MS full chapter.doc