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

Transcription

1 344 Organic Chemistry Laboratory Fall 2013 Lecture 3 Gas Chromatography and Mass Spectrometry June

2 Chromatography

3 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)

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

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

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

7 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).

8 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)

9 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,

10 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?

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

12 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

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

14 [A-B-C] Molecular Ion Radical cation [C] + Neutral [A-B] 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]. +

15 Mass Spectrum of Octane [M].+ 114

16 3 o > 2 o > 1 o > Me R 3 C R 2 CH RCH 2 CH 3 Increasing stability of cation or radical

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

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

19 Electron lost from this C-C bond

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

21 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 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 CH CH 3 CH 2 CH 2 CH 2 CH 2 Not much CH but it is there! [M] No m/z = CH

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

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

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

25 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 = g/mol 35 Cl 75.8 % 37 Cl 24.2 % ~3:1 ratio of isotopes Atomic weight Br = g/mol 79 Br 50.7 % 81 Br 49.3 % ~1:1 ratio of isotopes

26 Mass Spectrum of 1-Bromobutane 57 Br = g/mol Think about structure of m/z = 29, 41, Br 50.7 % 81 Br 49.3 % MW = 136 and 138 [C 4 H 9 81 Br] +. (138) CH 3 CH 2. (29) =[C2 H 4 81 Br] + (109) [C 4 H 9 79 Br] +. (136) CH 3 CH 2. (29) =[C2 H 4 79 Br] + (107) 29 C 4 H 9 81 Br Br Br 81 C 2 H 4 79 Br C 2 H 4 81 Br C 4 H 9 79 Br 136

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

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

29 Mass Spectrum of 2-octanone Me C O via a-cleavage via a-cleavage R C O [M].+

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

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

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

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

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