344 Organic Chemistry Laboratory Fall 2013 Lecture 3 Gas - PowerPoint PPT Presentation

344 Organic Chemistry Laboratory Fall 2013 Lecture 3 Gas Chromatography and Mass Spectrometry June 19 2013

Chromatography

Chromatography – separation of a mixture into individual components Paper, Column, Thin Layer (TLC), High-Pressure Liquid (HPLC) All feature a stationary phase and a mobile phase Focus on Gas Chromatography (GC) (coupled with mass spec. – see later) Stationary phase is a packed column (non-volatile liquid on a solid support) Mobile phase is a gas Sample needs to be volatile (You will use TLC in the lab to monitor the progress of reactions)

GC instrument schematic He or N 2 mobile gas phase chromatogram stationary phase TCD or FID

A typical GC-MS instrument Sample vial carousel Mass Spectrometer Gas Chromatograph

GC trace – mixture of aromatic hydrocarbons 2.4 min 0.7 min + 1.2 min 140 o C 80 o C 2.3 min 2.7 min 110 o C Detector output 144 o C 136 o C Retention Time (min)

Features of the GC trace Number of signals Corresponds to number of resolvable components in mixture Order of elution Components are eluted according to their retention time Retention time Governed by extent of interaction with stationary phase For our purpose, this follows the relative order of boiling points Other considerations Peak area approximates amount of each compound Compound polarity (polar compounds move slower than non/less-polar cpds) Column polarity (compounds move slower on polar columns) Column temperature (raising column temperature speeds up elution of all cpds) Column length (longer columns lead to better separation/resolution of signals).

EI-MS Electron Impact or Electron Ionization Mass Spec (both OK to use) Uses high energy electron beam (70 eV), sample in gas phase Ionization energy for most organic molecules 8-15 eV Gives info on molecular mass and formula of compound (m/z, isotopes) Gives info on connectivity of molecule (fragmentation pattern)

C 60 Buckminsterfullerene (C 60 ) was discovered as an anomalous signal in a mass spec experiment! H.W. Kroto, R. E. Smalley and R. F. Curl won the 1996 Nobel Prize in Chemistry for the discovery of C 60 . Nature , 1985 , 318 , 162-163

Molecular Ion [M] .+ [M] .+ gives the mass (m) of the molecule Molecule Molecular Fragments Fragments give info on connectivity of the molecule You’ve all seen Star Wars, right?

Mass Spectrum of Methanol CH 3 OH 31 Base peak m/z = mass/charge ratio [M] +. 32 Only consider singly charged species (z = 1) The most intense peak is referred to as the base peak 29 All other peaks scaled relative to base peak 15

From where on the molecule is the electron lost? Alkanes – sigma bond e Alkenes – pi bond e Heteroatom compounds (O, N, S, etc.) – non-bonding lone pairs e O O

✗ ✔ Cation Radical [B-C] [A] + [A-B-C] Molecular Ion [B-C] ✔ [A] Radical cation + ✗ ✔ Radical Cation Even electron fragments [EE]+

[C] [A-B-C] [A-B] + ✗ ✔ Neutral Radical cation Molecular Ion ✔ Radical cation Only CATIONS and RADICAL CATIONS detected by Mass Spec Radicals and other neutrals (CO, H 2 O) NOT detected by Mass Spec Odd electron fragments [OE] . +

Mass Spectrum of Octane 43 57 29 85 71 [M] .+ 114

3 o > 2 o > 1 o > Me R 3 C R 2 CH RCH 2 CH 3 Increasing stability of cation or radical

Electron lost from this C-C bond Both fragmentations involve formation of a Me radical or a Me cation

Electron lost from this C-C bond ✔ ✔ Stability of cation and radical is important

Electron lost from this C-C bond ✔ ✔

Electron lost from this C-C bond ✔ Stability of cation and radical is important Fragmentations involving formation of a Me species are disfavored

Mass Spectrum of Octane CH 3 CH 2 CH 2 43 Why is m/z = 43 the base peak? Is there any special stability associated with CH 3 CH 2 CH 2 + ? CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 CH 2 CH 2 CH 2 CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 CH 2 57 29 85 71 + Not much CH 3 [M] .+ but it is there! No m/z = 99 114 CH 3 15

Mass Spectrum of Octane CH 3 CH 2 CH 2 43 It can rearrange to the isopropyl cation (a 2 o cation) H H 3 C C CH 3 CH 3 CH 2 CH 2 1 o 2 o [M] .+ 114

Mass Spectrum of 2-methylpentane 43 71 CH 3 CH 2 [M] . 29 + 86

Branched alkanes fragment either side of the branch point(s) The m/z = 43 fragment is the base peak – why? H 3 C CH 3 CH 3 CH 2 CH 2 C H mass = 43 m/z = 43 ✔ As in octane, CH 3 CH 2 CH 2 + will rearrange to Me 2 CH + H 3 CH 2 CH 2 C CH 3 [CH 3 ] C H m/z = 71 ✔ mass = 15

Isotopes Atoms exist as isotopes (different # neutrons, same # protons) 12 C is most abundant isotope of carbon ~1.08 % of C-atoms in a sample are 13 C isotope (NMR active, useful) ~0.016% of H-atoms in a sample are 2 H isotope (D) ~0.38% of N-atoms in a sample are 15 N isotope Atomic weight Cl = 35.48 g/mol 35 Cl 75.8 % 37 Cl 24.2 % ~3:1 ratio of isotopes Atomic weight Br = 79.90 g/mol 79 Br 50.7 % 81 Br 49.3 % ~1:1 ratio of isotopes

Mass Spectrum of 1-Bromobutane 57 Br = 79.90 g/mol 79 Br 50.7 % Think about structure of m/z = 29, 41, 57 81 Br 49.3 % MW = 136 and 138 41 81 Br] + . ( 138 ) – CH 3 CH 2 . [C 4 H 9 ( 29 ) =[C 2 H 4 81 Br] + ( 109 ) 79 Br] + . ( 136 ) – CH 3 CH 2 . [C 4 H 9 ( 29 ) =[C 2 H 4 79 Br] + ( 107 ) C 4 H 9 81 Br 29 138 C 4 H 9 79 Br 136 C 2 H 4 79 Br C 2 H 4 81 Br 79 Br 81 Br 107 109 79 81

Mass Spectrum of Benzyl chloride [C 7 H 7 ] + Cl = 35.48 g/mol 91 35 Cl 75.8 % 37 Cl 24.2 % [C 7 H 7 37 Cl] + . ( 128 ) – 37 Cl . = [C 7 H 7 ] + ( 91 ) MW = 126 and 128 [C 7 H 7 35 Cl] + . ( 126 ) – 35 Cl . = [C 7 H 7 ] + ( 91 ) C 7 H 7 37 Cl m/z = 91 is an aromatic cation! 128 C 7 H 7 35 Cl Cycloheptatrienyl cation 126 Tropylium cation [C 7 H 7 ] +

Mass Spectrum of Methanol CH 3 OH 31 Base peak identity of base peak? 32 [M] +. CH 3 OH – H = [CH 2 OH] + or [CH 3 O] + . H a O H a -H a 29 H b C O H b H a H a H a m/z = 31 m/z = 32 oxonium cation 15

Mass Spectrum of 2-octanone 58 43 Me C O via a - cleavage via a - cleavage 128 R C O [M] .+ 113

a -cleavage at a C=O group Me C O + ✔ m/z = 43 γ α α β O CH 3 C O Molecular ion + ✔ m/z = m/z = 128 113 Practice drawing the fragmentation pattern for α - cleavage

GC trace – mixture of aromatic hydrocarbons 2.4 min 0.7 min + 1.2 min 140 o C 80 o C 2.3 min 2.7 min 110 o C 144 o C 136 o C Retention Time (min) GC-MS - a mass spectrum is obtained for each compound as it elutes

Synthesis of 1-phenylcyclohexene Use GC-MS to gauge success of reaction/purification Br MgBr Mg C 12 H 10 m/z: 154 C 6 H 5 Br Biphenyl by-product C 6 H 10 O Grignard m/z: 156, 158 m/z: 98 reaction O Why not just use 1 H-NMR? 1 H-NMR could get difficult to interpret because signals from starting H 2 SO 4 materials, by-product and final OH product would overlap. E1 reaction C 12 H 16 O C 12 H 14 But would certainly use 1 H-NMR to m/z: 176 m/z: 158 characterize the purified product. You will perform both an E1 and a Grignard reaction in CHEM 344 lab!

SAMPLE INFORMATION 1-Phenylcyclohexene (target product) GC DATA GAS CHROMATOGRAM 2 resolvable components in reaction mixture Biphenyl (byproduct) MASS SPECTRUM OF COMPONENT 1 Product, starting material, or something else? 154 Biphenyl (byproduct) MASS SPECTRUM OF COMPONENT 2 Product, starting material or something else? 1-Phenylcyclohexene (target product) 158

Synthesis of 1-phenylcyclohexene Did the reaction work? YES …..but we need to purify the product a little more. Br MgBr Mg C 12 H 10 m/z: 154 C 6 H 5 Br C 6 H 10 O Biphenyl m/z: 156, 158 Component 1 is indentified as m/z: 98 O biphenyl based on the molecular ion m/z = 154 H 2 SO 4 OH The absence of a 79 Br/ 81 Br isotope pattern tells us that bromobenzene is not present C 12 H 16 O C 12 H 14 m/z: 176 m/z: 158 Component 2 is identified as 1-phenylcyclohexene based upon the molecular ion m/z = 158. GC shows that the reaction mixture is ~95% 1-phenylcyclohexene