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Chapter 14

Chapter 14. Mass Spectroscopy. Mass Spectrometer. Positive ions are detected. Neutral species are undetected. p.545. Parent peak due to molecular radical cation . Figure 14.2, p.546. Detecting nitrogen, N. Consider some simple molecules and their nominal mass.

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Chapter 14

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  1. Chapter 14 Mass Spectroscopy

  2. Mass Spectrometer Positive ions are detected. Neutral species are undetected.

  3. p.545

  4. Parent peak due to molecular radical cation. Figure 14.2, p.546

  5. Detecting nitrogen, N Consider some simple molecules and their nominal mass. What is unusual about the N compound? The parent peak should have an odd mass for an odd number of nitrogens.

  6. p.546

  7. One way to distinguish between molecules having the same about the same mass is to measure their mass very precisely. About the same mass for these species formed from the most common isotopes. In these cases we can actually determine the molecular formula from high resolution mass spectroscopy. p.547

  8. 35.453 = (100 * 34.9689 + 31.98 * 36.9659) / 131.98 Recall that the atomic weight is the average mass for all isotopes found in nature. For example chlorine…. Table 14.1, p.548

  9. Low resolution mass spec does not involve itself with precise mass measurements. Low resolution examines the various peaks produced. First consider the parent radical cation: if an element has two naturally occurring isotopes then two peaks will be produced. Bromine has two isotopes 79Br and 81Br in about equal amounts. Obtain two peaks at 122 and 124. Figure 14.4, p.550

  10. Further comments on presence of chlorine and bromine. Both Cl and Br have two common isotopes separated by two mass units. Given the natural abundances we may calculate the ratio of the M and M+2 peaks for various combinations of Cl and Br being present. The presence of peaks at X, X+2… for the molecular ion or fragment hopefully with close to the expected ratio is taken as indication of Cl or Br. Ratio of peaks calculated as Ratio of peaks calculated as

  11. Molecular Peaks, M+1 Have seen that for Cl and Br, having two common isotopes, two radical cation peaks produced. What about other elements having more than one isotope? We know what the isotopes are and their natural occurrence. For the M+1 peak, one atom must be using an isotope heavier by one.

  12. Here is the data. We will use isotopic occurrence data for H, C, O for the M + 1 peak.

  13. The M+2 peak Recap: The M+1 peak has contributions from one atom being a heavier isotope by 1. • The M+2 peak can have contributions from • One atom being a heavier isotope by 2. • Two atoms being heavier by 1 each.

  14. M+2 peak, contributions from one atom and two atoms. Recap: The M+1 peak has contributions from one atom being a heavier isotope by 1. (M+1)/M = ca. 1.1% * no. of C atoms + 0.36% * no. of N atoms • The M+2 peak can have contributions from two sources • One atom being a heavier isotope by 2. Mainly O (excluding S, Cl and Br) • Two atoms being heavier by 1 each. Mainly C atoms. • (M+2)/M = ca. (0.20% * no. of O atoms) + (1.1 * no. of C atoms)2/200% Example 1: C5H5N [(A + 1)+]/[A+] = 5 x 1.1% + 1 x 0.36% = 5.9% [(A + 2)+]/[A+] = 5.52/200 % = 0.15% Example 2: C7H5O [(A + 1)+]/[A+] = 7 x 1.1% = 7.7% [(A + 2)+]/[A+] = 7.72/200 % + 0.20% = 0.50%

  15. Technique to obtain molecular formula using intensities of M, M+1, M+2 peaks. Example. Given the data. Looking at M+2 there is no Br, Cl or S. There could be oxygen. Even mass for M means there could only be even number of Nitrogen Consider the M+1 peak, nominal mass + 1. If we know the formula we should be able to calculate the relative intensity of that peak due to the contributions from each of the atoms present. Here are the major contributors to M+1. Here are major contributors to M+2.

  16. Technique to obtain molecular formula using intensities of M, M+1, M+2 peaks. Example. Given the data. Equations M+1: (1.11% x # of C) + (0.38 x # of N+ small contributions from O M+2: (0.20 x # of O) + (1.1 x # of C)2/200 Examine reasonable formulae. Calculate M+1, M+2 peaks We can have 0 or 2 nitrogens. Even number. We can have 0,1,2,3,4 oxygens. 0.88/0.2 < 5 Can have 0,1,2,3,4,5,6,7,8,9 carbons. 10.2/1.11 <10 Find molecular formulas having reasonable M+1 peaks

  17. Return to Fragmentation of Molecular Radical Cation, M+ The highly energetic radical/cation can undergo fragmentation. • First distinguish between some species • Radical/cation produced by ejection of electron from structure. It contains all the atoms of the original molecule just minus one electron. Example C4H10+. • Carbocation which is not a radical, is electron deficient and is a fragment of a stable molecule. Example C4H9+ • Radical is not charged and is a fragment of a stable molecule.Example C4H9.

  18. Example. Identify this molecule Ejection of an H Molecular radical ion Due to heavier isotopes H2O

  19. Example 2 carbon Oxygen H ejection parent Heavier isotopes CH2O

  20. Fragmentation of the radical/cation can lead to 1. A radical and carbocation as separate species. Usually a bond is split. Choice of bond to split is frequently controlled by the carbocation stability. Carbocation, CH3CH2+ Radical/ cation, C4H10+. Radical, CH3CH2. Remember only the positive species are detected. Neutral species are invisible. 2. Rearrangements can occur including elimination of neutral molecules to produce a different radical/cation. Radical/ cation, C5H12O+. Radical/ cation, C5H10+. p.550

  21. How to think about it… For this fragmentation Think of it this way… This C becomes + One electron in this bond. When it splits we get a carbocation (observable in MS) and a radical (not observable in MS). The fragmentation is written this way. This C bears the .

  22. In fragmentation to produce a carbocation stability of the carbocation is an important factor in determining where the fragmention will occur. p.551

  23. For simple linear alkanes fragmentation will occur towards the middle of the chain. Figure 14.5, p.552

  24. Caution. Sometimes a peak will occur which cannot be explained such as the ethyl peak below. Figure 14.6, p.552

  25. Technique: recall that the neutral molecules split out do not produce a peak. The mass of the neutral particle (invisible) may sometimes be obtained by subtracting the mass of the newly formed positive ion (detected) from the mass of the original radical carbocation. In this example the parent molecule has mass of 84. The mass of a positive fragment is 56. The peak at 56 is subtracted from the mass of the original ion, 84, yielding 28, the mass of ethylene which is taken as the invisible neutral molecule. Figure 14.7, p.553

  26. Ionization followed by fragmentation Ejection of electron fragmentation Radical/cation Splitting out of ethylene. p.553

  27. Alkenes can yield allylic stabilized carbocations by fragmentation, splitting out a radical. Difference is 15, the methyl radical

  28. Alcohols have several characteristic fragmentation patterns. • 1. An alcohol radical/cation can undergo a fragmentation to produce a radical and a resonance stabilized carbocation. The one “electron bond” p.555

  29. 2. Alcohol radical/cations can split out water to produce a new alkene radical/cation which may be detected. Here is an example demonstrating both processes. Elimination of water. 74 – 56 = 18 (water). Elimination of propyl radical. 74 – 31 = 43 (C3H7) Figure 14.10, p.555

  30. Aldehydes and Ketones Several characteristics reactions the radical/cations 1. a cleavage: break bond to carbonyl group Note that an a cleavage of an aldehyde could produce a peak at M – 1 by eliminating H atom. This is useful in distinguishing between aldehydes and ketones.

  31. 2. McLafferty Rearrangement: splitting out an alkene (neutral molecule) and producing a new radical/cation. Note that the process involves a six membered ring for a transition state. p.557

  32. Mass Spec of 2-octanone displays both a cleavage and McLafferty The “invisible” radical C6H13 CH3 radical “Invisible” pent-1-ene CH3CH2CH2CH2CH2CH2CO+ resulting from a cleavage here. CH3CO+ resulting from a cleavage at this bond.

  33. Carboxylic Acids can also undergo a cleavage and McLafferty rearrangement. The peak at 60 is usually prominent for a carboxylic acid.

  34. Likewise for esters, 2 a cleavages (around C=O) and McLafferty C

  35. An unexpected observation. p.559

  36. Figure 14.14, p.559

  37. Some general principles The relative height of the parent radical cation peak is greatest for straight chain compounds and decreases for branched structures. Easier cleavage. Cleavage is favored at branched carbons due to increased stability of substituted carbocations. Double bonds or aromatic rings stabilize the parent radical cation increasing the size of its peak. Double bonds (aromatic rings) favor cleavage yielding allylic (benzylic or tropylium) carbocations. Bonds beta to a hetero atom having lone pairs are frequently cleaved. Elimination of small, stable molecules (water, alkenes, etc.) can occur to yield a new radical cation.

  38. Example Given spectra for an unknown compound: IR, MS, and NMR. Identify the compound. First, let’s do the MS

  39. Mass Spectrum Now the IR…. Base peak, the largest, others are relative to it. Want M+1 and M+2 relative to M not base peak. • Observations: • Parent mass is even. Even number of nitrogens: 0, 2, 4 • No Br or Cl since M+2 is much too small. • At most about 9 carbons since 9.9/1.1 =9. • What molecular formulas could we have with good values for M+1 between 9 and 11? Molecular peak, parent peak. Best fit for both M+1 and M+2. Maybe aromatic!

  40. Infra Red. Assumed formula C9H10O2 Carbonyl, no OH C-O stretch Looks as if it may be an ester with an aromatic group. Next the NMR….

  41. NMR . Assumed formula C9H10O2 First the hydrogen counts.

  42. Chemical Arithmetic. Assumed formula C9H10O2 Consistent with the lack of splitting could be either. C9H10O2 • - C6H5 • CH2 • CH3 Tentatively identified parts, Subtract to get CO2

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