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

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

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

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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

Figure 13.35

The mass spectrum of propylbenzene

Figure 13.39

C2H5
  • 29.0391
C7H7
  • 91.0548

The mass spectrum of chlorobenzene

Figure 13.36

Cl C6H5
  • 77.0391
Cl• 34.9689

Why 112 and 114? – Isotopic distribution (Cl 35 and Cl 37) Use on tests: you’ll be given the “M+” (molecular ion), which is the same as the formula (molecular weight) of the molecule. e.g. Given formula = C3H5O2 (total = 73 a.m.u.) but M+ = 146 a.m.u. Therefore actual formula of unknown must be C6H10O4 e.g. Given formula = C5H10O3 (total = 118 a.m.u.) and M+ = 118 a.m.u. Therefore the formula given is the actual formula

Summary of spectroscopy/spectrometry techniques

1H NMR – tells you the number, type and environment of protons 13C NMR – same for C but no coupling or integration

IR – types of bonds and functional groups in a molecule UV-Vis – conjugation in pi systems Mass Spec – formula weight of molecules and fragmentation patterns Can use combinations of techniques to work out molecular structure

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Typical Spectroscopy Problem

Given: C10H12O3 and M+ = 180.08

1H NMR (ppm) 3.36 (s, 3H), 3.62 (t, 2H, J = 6.9 Hz), 4.36 (t, 2H, J =

6.9 Hz), 7.48 (m, 3H), 8.00 (m, 2H)

13C NMR (ppm) 58.7, 64.6, 70.2, 128.6, 128.8, 129.1, 132.9, 166.0

IR (cm-1) 1740 Structure?

1 2 3 4 5 6 7 8 PPM

1H 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)

2H 2H 2H 2H 1H 3H 20 40 60 80 100 120 140 160 180 PPM 13C NMR (ppm) 59.0, 64.0, 71.6, 128.7, 129.9, 130.2, 133.1, 166.0

O O O CH3

Starting points:

MS – M+ matches given formula, therefore C10H12O3 is correct

1H NMR – protons at 7 ppm – benzene ring; integration = 5 H, monosubstituted 1H NMR – singlet for 3H at 3.24 ppm, CH3 must be on the substituent, close to O

IR and 13C NMR – C=O present, 166 ppm says its an ester

1H NMR – 2 x 2H signals that couple (same J value) – XCH2CH2X

Chapter 14: Organometallics, structure and nomenclature

H CR3 Cl CR3 δ- δ+ M CR3 δ- δ+ similar E.N. no dipoles different E.N. dipoles different E.N. dipoles radical chemistry carbon electrophiles carbon nucleophiles

M CHR3

C more electronegative than M

Li H CH2=CHNa (CH3CH2)2Mg CH3MgI (CH3CH2)2AlCl

Cyclopropyl lithium Vinyl sodium Diethyl magnesium Methyl magnesium iodide Diethylaluminum chloride

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Electronegativity Differences in Organometallics

Electrostatic potential maps of (a) methyl fluoride and (b) methyllithium Figure 14.1

Preparation of Organometallic Compounds

Organolithiums

X R 2 M RM M+X- + + Cl 2 Li Li ether

  • 30oC

+ LiCl Br 2 Li ether 35oC Li + LiBr Preparation of Organometallic Compounds

Organomagnesium compounds – Grignard reagents

X R Mg RMgX +

Br Br Mg Mg MgBr MgBr ether RT ether 35oC