1. 13.21 Mass Spectrometry

2. Atom or molecule is hit by high-energy electron Principles of Electron-Impact Mass Spectrometry e–e–e–e–

3. Atom or molecule is hit by high-energy electron electron is deflected but transfers much of its energy to the molecule e–e–e–e– Principles of Electron-Impact Mass Spectrometry

4. Atom or molecule is hit by high-energy electron electron is deflected but transfers much of its energy to the molecule e–e–e–e– Principles of Electron-Impact Mass Spectrometry

5. This energy-rich species ejects an electron. Principles of Electron-Impact Mass Spectrometry

6. This energy-rich species ejects an electron. Principles of Electron-Impact Mass Spectrometry forming a positively charged, odd-electron species called the molecular ion e–e–e–e– +

7. Molecular ion passes between poles of a magnet and is deflected by magnetic field amount of deflection depends on mass-to-charge ratio highest m/z deflected least lowest m/z deflected most Principles of Electron-Impact Mass Spectrometry +

8. If the only ion that is present is the molecular ion, mass spectrometry provides a way to measure the molecular weight of a compound and is often used for this purpose. However, the molecular ion often fragments to a mixture of species of lower m/z.

9. The molecular ion dissociates to a cation and a radical. Principles of Electron-Impact Mass Spectrometry +

10. The molecular ion dissociates to a cation and a radical. Principles of Electron-Impact Mass Spectrometry + Usually several fragmentation pathways are available and a mixture of ions is produced.

11. mixture of ions of different mass gives separate peak for each m/z intensity of peak proportional to percentage of each ion of different mass in mixture separation of peaks depends on relative mass Principles of Electron-Impact Mass Spectrometry + + + + + +

12. mixture of ions of different mass gives separate peak for each m/z intensity of peak proportional to percentage of each atom of different mass in mixture separation of peaks depends on relative mass ++++ ++ Principles of Electron-Impact Mass Spectrometry

13. 20406080100 120 m/z m/z = 78 10080 60 40 20 0 Relative intensity Some molecules undergo very little fragmentation Benzene is an example. The major peak corresponds to the molecular ion.

14. HHH HH H HHH HH H HHH HH H all H are 1 H and all C are 12 C one C is 13 C one H is 2 H Isotopic Clusters 78 7979 93.4%6.5%0.1%

15. 20406080100 120 m/z10080 60 40 20 0 Relative intensity 112 114 Isotopic Clusters in Chlorobenzene visible in peaks for molecular ion 35 Cl 37 Cl

16. 20406080100 120 m/z Relative intensity 77 Isotopic Clusters in Chlorobenzene no m/z 77, 79 pair; therefore ion responsible for m/z 77 peak does not contain Cl H H H H H+ 10080 60 40 20 0

17. Alkanes undergo extensive fragmentation m/z Decane 142 43 57 71 85 99 CH 3 —CH 2 —CH 2 —CH 2 —CH 2 —CH 2 —CH 2 —CH 2 —CH 2 —CH 3 Relative intensity 10080 60 40 20 0 20406080100 120

18. Propylbenzene fragments mostly at the benzylic position 20406080100 120 m/z Relative intensity 120 91 CH 2 —CH 2 CH 3 10080 60 40 20 0

19. 13.22 Molecular Formula as a Clue to Structure

20. Molecular Weights CH 3 (CH 2 ) 5 CH 3 Heptane CH 3 CO O Cyclopropyl acetate Molecular formula Molecular weight C 7 H 16 C5H8O2C5H8O2C5H8O2C5H8O2 100100 Exact mass 100.1253100.0524 Mass spectrometry can measure exact masses. Therefore, mass spectrometry can give molecular formulas.

21. Molecular Formulas Knowing that the molecular formula of a substance is C 7 H 16 tells us immediately that is an alkane because it corresponds to C n H 2n+2 C 7 H 14 lacks two hydrogens of an alkane, therefore contains either a ring or a double bond

22. Index of Hydrogen Deficiency relates molecular formulas to multiple bonds and rings index of hydrogen deficiency = 1 2 (molecular formula of alkane – molecular formula of compound)

23. Example 1 index of hydrogen deficiency 12 (molecular formula of alkane – molecular formula of compound) C 7 H 14 12 (C 7 H 16 – C 7 H 14 ) = =12 (2) = 1 = Therefore, one ring or one double bond.

24. Example 2 C 7 H 12 12 (C 7 H 16 – C 7 H 12 ) =12 (4) = 2 = Therefore, two rings, one triple bond, two double bonds, or one double bond + one ring.

25. Oxygen has no effect CH 3 (CH 2 ) 5 CH 2 OH (1-heptanol, C 7 H 16 O) has same number of H atoms as heptane index of hydrogen deficiency = 1 2 (C 7 H 16 – C 7 H 16 O) = 0 = 0 no rings or double bonds

26. Oxygen has no effect index of hydrogen deficiency = 1 2 (C 5 H 12 – C 5 H 8 O 2 ) = 2 = 2 one ring plus one double bond CH 3 CO O Cyclopropyl acetate

27. If halogen is present Treat a halogen as if it were hydrogen. C C CH 3 Cl H H C 3 H 5 Cl same index of hydrogen deficiency as for C 3 H 6

28. Rings versus Multiple Bonds Index of hydrogen deficiency tells us the sum of rings plus multiple bonds. Catalytic hydrogenation tells us how many multiple bonds there are.