Chapter 12: Mass Spectrometry: molecular weight of the sample

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1 Structure Determination: hapter 12: Mass Spectrometry- molecular weight of the sample; formula hapter 12: Infrared Spectroscopy- indicated which functional groups are present hapter 13: Nuclear Magnetic Resonance- map of the - framework hapter 14: Ultraviolet-Visible Spectroscopy- p-bonds hapter 12: Mass Spectrometry: molecular weight of the sample formula The mass spectrometer gives the mass to charge ratio, therefore the sample (analyte) must be an ion. Mass spectrometry is a gas phase technique- the sample must be vaporized. Electron-impact ionization ionization chamber e Sample Inlet torr R- R- (M ) mass analyzer electron beam 70 ev (6700 KJ/mol) proton neutron electron u u u 1

2 mass charge m = = z B 2 r 2 2V B= magnetic field strength r= radius of the analyzer tube V= voltage (accelerator plate) Ions of non-selected mass/charge ratio are not detected Ions of selected mass/charge ratio are detected Magnetic Field, B o Ionization chamber The Mass Spectrometer Molecular Ion (parent ion, M)= molecular mass of the analyte; sample minus an electron Nitrogen rule: organic molecules with an odd mass must have an odd number of nitrogens. rganic molecules with an even mass have an even number of nitrogens (0 is an even number) Base peak- largest (most abundant) peak in a mass spectra; arbitarily assigned a relative abundance of 100%. If the mass can be determined accurately enough, then the molecular formula can be determined (high resolution mass spectrometry) - double focusing mass spectrometer 2

3 Exact Masses of ommon Natural Isotopes Isotope mass natural abundance Isotope mass natural abundance (1.11%) 14 N N (0.38%) (0.04%) (0.20%) 19 F l l (32.5%) 79 Br Br (98%) 127 I igh resolution mass spectrometry can give the formula of the sample mass= 58 N N N N N N N N For mass = 200, there are at least 51 empirical formulas with,,n, 3

4 The radical cation (M ) is unstable and will fragment into smaller ions - e m/z = 15 charge neutral not detected m/z = 14 charge neutral not detected m/z=15 m/z=14 (M) m/z=17 (M1) - e - e m/z = 44 m/z = 43 charge neutral not detected -or- m/z = 29 charge neutral charge neutral m/z = 15 not detected not detected m/z=29 m/z=43 m/z=15 (M) m/z=45 (M1) 4

5 3 - e m/z = 72 m/z = 57 charge neutral not detected m/z=57 No molecular ion m/z=72 Mass spectra can be quite complicated and interpretation difficult. Some functional groups have characteristic fragmentation (sect. 12.4) It is difficult to assign an entire structure based only on the mass spectra. owever, the mass spectra gives the mass and formula of the sample which is very important information. To obtain the formula, the molecular ion must be observed. soft ionization techniques (chemical ionization, fast atom bombardment) Methods have been developed to get large molecules such as polymers and biological macromolecules (proteins, peptides, nucleic acids) into the vapor phase (matrix-assisted laser desorption ionization and electrospray ionization) 5

6 Spectroscopy and the Electromagnetic Spectrum Electromagnetic (EM) radiation = light matter photon (quanta) energy Spectroscopy: the interaction of EM radiation with matter (organic molecule) light (hn) organic molecule (ground state) absorption organic molecule (higher energy state) * relax organic emission molecule (ground state) quantized energy levels: the energy states are at discrete levels; there is no continuum between these levels hn E = h n n = c/l E = h c l E a n n a l -1 E a l -1 c = speed of light (3 x cm sec -1 ) l = wavelength (distance of one wave) n = frequency: number of waves per unit time (sec -1, z) h = Planck s constant = 6.62 x J sec -1 6

7 The Electromagnetic Spectrum Increasing Energy Longer wavelength (l) Infrared: molecular vibrations; stretching and bending of bonds - identify functional groups (h. 12) Radiowaves: nuclear magnetic resonance spectroscopy; nuclear magnetic moments versus an external magnetic field - gives a and map of the molecule (h. 13) UV-vis: valance electron transitions; types of p-bonds (h. 14) 7

8 Infrared Spectroscopy: Vis Near IR Infrared (IR) Far IR microwave x 10-4 cm 2.4 x 10-3 cm mm 25 mm 4000 n 400 n E a 1 l l is expressed as n (wavenumber), reciprocal cm (cm -1 ) n = 1 therefore E a n l Absorption of infrared radiation causes bonds within a molecule to vibrate, stretch and bend. Bonds behave likes springs. Stretch- deforms bond length Symmetric stretch Antisymmetric stretch Bend- deform bond angle In-plane bend ut-of-plane bend 8

9 Bond Stretch: ooke s Law X Y n = 1 2 p c f m x m y m x m y 1 2 n = vibrational frequency c = speed of light m x = mass of X m y = mass of Y m x m y = reduced mass (m) m x m y E a n a f f = spring constant; type of bond between X and Y (single, double or triple) ooke s law simulation: Interpretation of an Infrared Spectra: organic molecules contain many atoms. As a result, there are many stretching and bending modes- IR spectra have many absorption bands Four distinct regions of an IR spectra X- single bond region triple bond region double bond region fingerprint region 4000 cm cm cm cm cm -1 9

10 Fingerprint region ( cm -1 )- low energy single bond stretching and bending. The fingerprint region is unique for a given organic compound and is thus used to identify them. owever there are few diagnostic absorptions. Double-bond regions ( cm -1 ) = cm -1 = cm -1 Triple-bond region: ( cm -1 ) cm -1 (weak, often not observed) N cm -1 X- Single-bond region ( cm -1 ) cm -1 N cm cm -1 sp cm -1 sp2 = cm -1 sp - ~3300 cm -1 arbonyl groups: aliphatic aldehyde aliphatic ketone aliphatic ester 1730 cm cm cm conjugated aldehyde conjugated ketone conjugated ester 1705 cm cm cm aromatic aldehyde 1705 cm -1 aromatic ketone 1690 cm -1 aromatic ester 1715 cm -1 onjugation moves the = stretch to lower energy (right) 10

11 yclic Ketones: aliphatic ketone conjugated ketone 1715 cm cm -1 aromatic ketone 1690 cm cm cm cm cm -1 Ring strain moves the = absorption to higher energy (left) Table 12.1 (page 458) Alkenes = cm -1 medium - strong = cm -1 medium Aromatic = cm -1 strong = cm -1 weak cm -1 strong Alkynes cm -1 strong weak - medium Alcohols cm -1 strong cm -1 strong and broad Amines -N cm -1 medium N cm -1 medium arbonyl = cm -1 strong arboxylic acids cm -1 strong and very broad Nitrile N medium 11

12 Problem = 12

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