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1 Typical Spectroscopy Problem Given: C 10 H 12 O 3 and M + = - PDF document

Dec 28, 2022 •128 likes •172 views

Lecture 7 Spectroscopy; Organometallic Compounds Mass Spectrometry Summary of techniques Typical spectroscopy problems Types of organometallic compounds, nomenclature Electronegativity and carbon-metal bonds Preparation

  1. Lecture 7 – Spectroscopy; Organometallic Compounds • Mass Spectrometry • Summary of techniques • Typical spectroscopy problems • Types of organometallic compounds, nomenclature • Electronegativity and carbon-metal bonds • Preparation of organometallic compounds • Basicity of organometallics The mass spectrum of benzene The mass spectrum of chlorobenzene Figure 13.35 Figure 13.36 Cl • Cl 34.9689 • C 6 H 5 77.0391 The mass spectrum of propylbenzene Figure 13.39 Why 112 and 114? – Isotopic distribution (Cl 35 and Cl 37) • C 2 H 5 29.0391 C 7 H 7 • 91.0548 Summary of spectroscopy/spectrometry techniques Use on tests : you’ll be given the “M + ” (molecular ion), which is the same as the formula (molecular weight) of the molecule. 1 H NMR – tells you the number, type and environment of protons 13 C NMR – same for C but no coupling or integration e.g. Given formula = C 3 H 5 O 2 (total = 73 a.m.u.) IR – types of bonds and functional groups in a molecule but M + = 146 a.m.u. UV-Vis – conjugation in pi systems Therefore actual formula of unknown must be C 6 H 10 O 4 Mass Spec – formula weight of molecules and fragmentation patterns e.g. Given formula = C 5 H 10 O 3 (total = 118 a.m.u.) Can use combinations of techniques to work out and M + = 118 a.m.u. molecular structure Therefore the formula given is the actual formula 1

  2. Typical Spectroscopy Problem Given: C 10 H 12 O 3 and M + = 180.08 3H 1 H NMR (ppm) 3.36 (s, 3H), 3.62 (t, 2H, J = 6.9 Hz), 4.36 (t, 2H, J = 2H 6.9 Hz), 7.48 (m, 3H), 8.00 (m, 2H) 2H 13 C NMR (ppm) 58.7, 64.6, 70.2, 128.6, 128.8, 129.1, 132.9, 166.0 1H 2H 2H IR (cm -1 ) 1740 8 7 6 5 4 3 2 1 0 PPM Structure? 1 H NMR (ppm) 3.24 (s, 3H), 3.83 (t, 2H, J = 6.9 Hz), 4.42 (t, 2H, J = 6.9 Hz), 7.37 (m, 2H), 7.47 (m, 1H), 7.97 (m, 2H) O O CH 3 O Starting points: MS – M+ matches given formula, therefore C 10 H 12 O 3 is correct 1 H NMR – protons at 7 ppm – benzene ring ; integration = 5 H, monosubstituted 1 H NMR – singlet for 3H at 3.24 ppm, CH 3 must be on the substituent, close to O IR and 13 C NMR – C=O present, 166 ppm says its an ester 1 H NMR – 2 x 2H signals that couple (same J value) – XCH 2 CH 2 X 180 160 140 120 100 80 60 40 20 0 PPM 13 C NMR (ppm) 59.0, 64.0, 71.6, 128.7, 129.9, 130.2, 133.1, 166.0 Chapter 14: Organometallics, structure and nomenclature C more electronegative than M δ - δ + δ + δ - M CHR 3 H CR 3 Cl CR 3 M CR 3 similar E.N. different E.N. different E.N. Li no dipoles dipoles dipoles CH 2 =CHNa (CH 3 CH 2 ) 2 Mg H Cyclopropyl Vinyl Diethyl lithium sodium magnesium radical carbon carbon CH 3 MgI (CH 3 CH 2 ) 2 AlCl chemistry electrophiles nucleophiles Methyl magnesium iodide Diethylaluminum chloride 2

  3. Electrostatic potential maps of (a) methyl fluoride and (b) methyllithium Electronegativity Differences in Organometallics Figure 14.1 Preparation of Organometallic Compounds Preparation of Organometallic Compounds Organolithiums Organomagnesium compounds – Grignard reagents M + X - + R X + 2 M RM R X + Mg RMgX 2 Li Mg Cl Li + LiCl ether Br MgBr ether -30 o C RT 2 Li Li + LiBr Br Mg ether Br MgBr 35 o C ether 35 o C 3

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