chapter 9

chapter 9 - Chapter 9 NMR (Nuclear Magnetic Resonance) and...

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Unformatted text preview: Chapter 9 NMR (Nuclear Magnetic Resonance) and Mass Spectrometry §9.2 NMR Spectrometers A sample is placed in a very strong, constant magnetic field produced by a superconducting magnet. The sample is irradiated with a short pulse of radio frequency energy that excites certain nuclei. The nuclei relax, emitting a signal that is converted to a spectrum by a Fourier transformation. §9.2 MRI MRI (magnetic resonance imaging) is an NMR where humans are the sample. The names (MRI vs NMR) are different because patients equate “nuclear” with radioactivity. Neither NMR nor MRI use radioactivity. §9.2 1H-NMR 1H-NMR 1. 2. 3. 4. spectra provide a lot of information: Chemical shift (gives functional group like IR) Integration (number of Hs per signal) Coupling (Hs on adjacent Cs) Number of signals (number of unique Hs) This is a lot of structural information from a single spectrum. This site offers an NMR game that will give you some experience solving NMR problems: http://www.spectralgame.com/ §9.2 A 1H-NMR Spectrum cyclobutene 1H-NMR spectra have a peak for each unique type of hydrogen in a molecule. x-axis units are parts per million (ppm) or δ. Left is “downfield”; right is “upfield”. ↤downfield upfield↦ §9.2A Chemical Shift Chemical shift – the x-axis position of an NMR signal. Nuclei of a given functional group usually appear in the same region. Chemical shift charts identify the most likely functional group for a given chemical shift. O Memorize Chemical shift – functional group R CH3 0.9 RR'N CH3 R O CH3 2.1 2.4 R CH2X 3.5 O O CH3 4.5 R O CH3 3.8 R2C CH2 5.3 R R H 9.7 §9.2B Integration Integration – the area under each signal. For 1H-NMR, integration is proportional to the number of H atoms producing each signal. Integration is calculated automatically by most NMRs. Not useful for 13C-NMR. 6:4 How many different types of Hs are there? What is the ratio of red to blue Hs? Which peak belongs to each H? §9.2C Coupling Coupling – NMR signals are split into 2+ smaller, closely spaced peaks by their neighbors. These Hs couple: HH CC H C H C X Y Z chir ality center vicinal Hs (3 bonds apart) CH2 next to a chirality center §9.2C Signals and Peaks There is one NMR signal for each unique nuclei. 13C-NMR: each nuclei signal has one peak. 1H-NMR: each nuclei signal can have 1-7 peaks. 13C-NMR 1H-NMR Number of Peaks 1 2 3 4 5 6 7 Signal Name singlet doublet triplet quartet pentet sextet septet §9.2C Signals vs Peaks How do I know if I’m looking at a triplet or 3 singlets? Doublet peaks have the same height and are close together. Other multiplets (non-singlets) are evenly spaced and pyramid shaped: sextet triplet §9.2C n + 1 Rule A H atom with n adjacent (vicinal, or neighboring) Hs has a signal of n + 1 peaks. HH H H H H H H adjacent H b a c d d not adjacent H Hydrogen a: ___ adjacent H's --> Hydrogen b: ___ adjacent H's --> Hydrogen c: ___ adjacent H's --> Hydrogen d: ___ adjacent H's --> ___ peaks ___ peaks ___ peaks ___ peaks Singlet No adjacent H Doublet 1 adjacent H Triplet Quartet 3 adjacent H 2 adjacent H H H C H O C O H C H H C H H s quartet q t singlet triplet O R O CH3 4.5 O R CH3 2.1 R CH3 0.9 §9.2C Coupling Equivalent Hs do not couple to each other. SH §9.9 Integration and Splitting What is the integration and splitting of each of these Hs? O O H H H H N 2H, triplet 1H, doublet 6H, doublet 3H, triplet One H is at the same carbon and the other H is at a vicinal carbon 1H, doublet ----> of doublets §9.3 Interpreting an 1H-NMR 1. Count the signals (= unique protons). 2. Correlate chemical shift with structure. 3. Determine the relative area of each signal (integration) that is proportional to the relative number of protons. 4. Interpret the splitting pattern for each signal to give neighboring hydrogen atoms; sketch possible molecular fragments. 5. Join all fragments to make a molecule consistent with the data. A compound with formula C5H10O2 has the following 1HNMR. Propose a reasonable structure. O O IHD = 1 H H C H O H C H O C H C H H H C H 3H 2H 2H 3H What molecule has this 1H-NMR? Y X Y X X Y C9H10O2 IHD = 5 -OCH3 O C H3 3H Y 3H X 2H O O 2H §9.4 Nuclear Spin Some nuclei have a spin quantum number of ½ and can have two spin states, +½ and -½. When these nuclei spin, they create a tiny magnetic field: §9.4 Nuclear Spin Outside a magnetic field, the magnetic moments of nuclei with I = ½ are oriented in all directions. Within a magnetic field, the nuclei orient themselves with or against the field. NMRs observe energy absorptions as spins flip from α to β. Spin states aligned with the magnetic field (α) are lower in energy than those against it (β). §9.6 Induced Field In a strong magnetic field (the applied field), electrons circulate in various patterns. This produces a minor magnetic field (induced field) opposed or aligned with the applied magnetic field. Shielding – the induced field opposes the applied field at the nuclei. Deshielding – the induced field is aligned with the applied field at the nuclei. §9.6 Shielding and Deshielding Least shielded Most shielded sp2 (aromatic) < sp2 (alkene) < sp < sp3 Cl Cl H H H I H H C H H Least shielded Downfield Most shielded upfield §9.6 Shielding and Deshielding Least shielded Most shielded sp2 (aromatic) < sp2 (alkene) < sp < sp3 Cl Cl H H H I H H C H H Least shielded Downfield Most shielded upfield §9.6 Shielding and Deshielding Least shielded Most shielded sp2 (aromatic) < sp2 (alkene) < sp < sp3 Cl Cl H H H I H H C H H Least shielded Downfield Most shielded upfield §9.6 Shielding and Deshielding Least shielded Most shielded sp2 (aromatic) < sp2 (alkene) < sp < sp3 Cl Cl H H H I H H C H H Least shielded Downfield Most shielded upfield Shielding by ! "#lectrons The highly polarizable !-electrons circulate in the presence of a magnetic field which produces an induced magnetic field, b. Aromatics b a "ring current" is produced from the circulating electrons benzene lines of force of b are perpendicular to plane of ring Bo Because of the orientation of the induced magnetic field relative to the planar benzene ring, there are defined "shielding" (Bo - b) and "deshielding" (Bo + b) sectors. Shielding means the resonance is shifted to higher field and deshielding means the resonance is shifted to lower external magnetic field. 28 §9.6 Alkyne Shielding A strong magnetic field is applied to an alkyne. The 4 π electrons of the alkyne begin to circulate around the C≡C axis. Movement of the π electrons creates an induced magnetic field opposed to the applied field. §9.6 Shielding and Deshielding Electronegative groups deshield a C─H bond by withdrawing electron density. H H C H H Cl C H H Cl C H Cl Cl C H Cl Cl H H §9.6 Shielding and Deshielding Electronegative groups deshield a C─H bond by withdrawing electron density. H H C H H Cl C H H Cl C H Cl Cl C H Cl Cl H H §9.6 Shielding and Deshielding Electronegative groups deshield a C─H bond by withdrawing electron density. H H C H H Cl C H H Cl C H Cl Cl C H Cl Cl H H §9.6 Shielding and Deshielding Electronegative groups deshield a C─H bond by withdrawing electron density. H H C H H Cl C H H Cl C H Cl Cl C H Cl Cl H H §9.7 Tetramethylsilane The standard used most often in 13C and 1H-NMR is tetramethylsilane (TMS): C H3 H3C Si C H3 C H3 It has a chemical shift of zero and is used to calibrate NMR spectra. §9.7 ppm The units of the x-axis are ppm (or δ): Units of ppm are dimensionless. ppm (= parts per million) refers to the 106 multiplier. §9.8A Homotopic Protons Homotopic or chemically equivalent nuclei have the same chemical shift. An easy test to determine if protons are homotopic: - Replace the H's in question with another atom. - If the new compounds are the same, the H's are homotopic. H F H F 3-fluoro-3,4-methylhexane The new compounds are the same. ∴ These H's are homotopic. §9.8B Enantiotopic and Diasterotopic Hs Enantiotopic Hs – two Hs in a molecule which, if replaced, would generate enantiomers. Enantiotopic Hs have the same chemical shift. §9.8B Enantiotopic and Diasterotopic Hs Diastereotopic Hs – two Hs in a molecule which, if replaced, would generate diastereomers. Diastereotopic Hs have different chemical shifts. §9.8B Diasterotopic Protons H O OH H H H and H shifts are very close, but still distinct. H2N 8 7 6 5 4 3 §9.8 Equivalent Protons Are these Hs homotopic, enantiotopic or diastereotopic? H H H X X H diastereotopic Cl Cl H H X Cl X H H enantiotopic homotopic diastereotopic H H X X H C H C F Br X C H C F Br H C X C F Br §9.9 Signal Splitting A proton’s signal is split because its neighbors increase or decrease the induced magnetic field around it. Pascal’s triangle gives the relative height of multiplet peaks. 1 1 1 1 1 1 1 6 5 4 10 15 20 3 6 2 3 4 10 15 5 6 1 1 1 1 1 1 §9.9C Coupling Constants The peaks of a multiplet are evenly spaced apart. Coupling constants – the difference in Hz between peaks of a multiplet. §9.9C Coupling Constants Hs that split each other have signals with identical coupling constants. These coupling constants are the same. §9.9C Coupling Constants How many neighbors do these Hs have? Pretend they have the same coupling constant. H H C H H C ? H C H H septet 6 neighboring Hs doublet 1 neighboring H Isopropyl groups have septet-doublet patterns. If these signals have the same coupling constant, what alkyl group do they represent? §9.9E Complicated Spectra How many peaks would you expect for cyclopentanone’s 1H-NMR? O There are two, but they are overlapping so it’s hard to tell. §9.9E Complicated Spectra Aromatic compounds are notoriously difficult to analyze due to complex splitting and similar chemical shift. CF3 §9.9F Complex Splitting Consider the Hc J cb J cd d splitting of Hc in this compound: Cl Cl c split by Hb split by Hd J cd CH3CHCHCH CH3 Cl a b This pattern is a doublet of doublets, and all peaks (should) have equal intensity. §9.9F Complex Splitting Hb What splitting pattern would you predict for Hb of 2-chlorobutane? Jab > Jbc. split by Ha a bc d CH3CHCH2CH3 Cl split by Hc Start with the larger coupling first. quartet of triplets §9.10 Hydrogen-Bonding Hs on O and N rapidly exchange. Sometimes these do not appear in the NMR. This exchange can often be “seen” by NMR. O─H and N─H peaks are often broad because the Hs are constantly being pulled on and off. O─H and N─H don’t usually split vicinal protons. The range of O─H and N─H shifts is extremely broad: ROH 0.5 – 6.0 δ RNH2 1.0 – 5.0 δ §9.10 Deuterium Exchange ROH D2O ROD O OH D2O R C OD In D2O, O─H and N─H protons are not seen. Hs on O and N instantly exchange with D. D is not observed on 1HNMR. O R C RNH2 O R C NH2 D2O RND2 O R C ND2 D2O §9.10 Deuterium Exchange O H C OD H O C OH in D2O in CDCl3 §9.11 13C-NMR 13C-NMR spectra have a peak for each unique carbon atom in a molecule. The x-axis units are parts per million (ppm) or δ. Left is “downfield”; right is “upfield”. ↤downfield upfield↦ §9.11 13C NMR and Symmetry If a molecule is symmetrical, there will be fewer signals than C atoms in the molecular formula. 1 unique C; 1 signal benzene How many signals in 13C NMR? §9.11 13C NMR and Symmetry How many signals does this compound have in 13C NMR? O O 4 signals Is there a plane of symmetry? Yes. Atoms on either side of the plane are equivalent. §9.11 13C NMR Cs attached to electronegative atoms appear more downfield. 1 O 4 3 6 unique Cs; 6 signals 6 5 4 2 3 1 2 1 O 5 6 Isopentyl acetate How many signals in 13C NMR? §9.11 Cl Cl C Cl H Cl C Cl H Cl C H H Br C H Br Cl Cl Cl 13C NMR Cs attached to electronegative atoms appear more downfield. §9.11 Cl Cl C Cl H Cl C Cl H Cl C H H Br C H Br Cl Cl Cl 13C NMR Cs attached to electronegative atoms appear more downfield. §9.11 Cl Cl C Cl H Cl C Cl H Cl C H H Br C H Br Cl Cl Cl 13C NMR Cs attached to electronegative atoms appear more downfield. §9.11 Cl Cl C Cl H Cl C Cl H Cl C H H Br C H Br Cl Cl Cl 13C NMR Cs attached to electronegative atoms appear more downfield. §9.11 13C NMR Shifts You can predict functional group based on where a C atom appears on 13C NMR (chemical shift). carbonyl Note --> carbon aromatic alkene R-X alkane alkyne §9.13 Mass Spectrometry (MS) A technique that measures a sample’s mass. Except stereoisomers, every compound has a unique mass spectrum. It is possible to measure differences smaller than the mass of 1 electron! §9.13 A Typical Mass Spectrum The x-axis is mass versus charge. The y-axis is abundance. The highest peak (the base peak) sets the y-axis at 100. C H3 §9.14 Electron Impact Ionization Only ions are detected by MS, so molecules must be ionized. One common method is electron impact ionization (EI), where the sample is hit with high energy electrons. EI removes an electron from the sample to form a molecular ion: §9.15 Molecular Ion The molecular ion (M+·) is a cation radical. An electron was removed, not an atom. When possible, the electron is removed from a lone pair or a π bond: §9.15 Finding the Molecular Ion (M+) M+ is important since it H N O C H m/z = 95 molar mass = 95.10 gives a sample’s molar mass. M+ peaks may or may not be base peaks. Peaks in MS often appear in pyramidshaped groups. In the rightmost group, the tallest peak is the M+. §9.16A Fragmentation MS gives information about the M+· and its structure. +· Once it forms, the M breaks up into smaller pieces. Each break produces a cation (detected by MS) and a radical (not detected). Fragmentation normally occurs to give the most stable carbocation possible. C H3 C H 3C H CH3 C H 3C H CH 2C H 3 m / z 72 CH3 m / z 43 C H3 CH C H2 C H3 m / z 57 C H2 CH 3 43 57 72 §9.16B MS of Hexane m/z 86 §9.16C One-Bond Fragmentation Alkenes ionize to resonance-stabilized allyl cations. CH 2CH3 C─C bonds next to N, O or S readily ionize. O O CH 3 O §9.16C One-Bond Fragmentation C─CO bonds of aldehydes and ketones readily ionize. H 3C O H3 C C O H 3C CH 2 CH 3 C O Acylium ions are stable due to resonance. §9.16C One-Bond Fragmentation Benzyl groups rearrange to very stable tropylium ions. Y Y m/z 77 Monosubstituted benzenes give phenyl cations. H m/z 91 §9.16D Two-Bond Fragmentation Alcohols dehydrate in MS, giving strong M+· – 18 peaks. OH mo lar ma ss = 74 .12 mo lar m as s: 5 6. 11 §9.16D Two-Bond Fragmentation Carbonyl compounds with γ-Hs fragment to alkenes via the McLafferty rearrangement: §9.17 Iodomethane Consider the MS of iodomethane (142.0 g/mol): m/z 15.0 127.0 128.0 139.0 140.0 141.0 142.0 143.0 intensity 22.5 25.8 2.2 3.4 3.2 12.0 100.0 1.1 There is a small (1.1%) peak with m/z = 143.0. How is it possible to have an ion with an extra mass unit? C is not isotopically pure; it’s 1.1% 13C. This peak is the (M+1) +· peak. §9.17 Isotopes Cl and Br have these isotopic patterns: : 37Cl 1:1 79Br : 81Br 3:1 35Cl §9.17 Bromobenzene This is the MS of bromobenzene (157.0 g mol−1) m/z =156.0, 158.0. The intensity is ~1:1. What is the base peak at m/z = 77.0? PhBr+· → Ph+ + Br· §9.17 Molecular Formula from MS 1. 2. 3. 4. If M+· is even, the compound has an even number of N atoms. Number of C atoms ≈ (M+·+1 intensity) / 1.1 (M+·+ 2) peak intensities: S (4.4%), Cl (33%), Br (98%). Determine the number of O and H atoms. m/z % 100 6.8 31.8 peak M +· M+·+1 M+·+2 Cl 112.0 113.0 114.0 M+· is 112.0, ∴ an even # of Ns 6C + 1Cl = 6(12) + 35 = 107 112 – 107 = 5 ∴ C6H5Cl 6.8 / 1.1 ≈ 6 Cs 31.8% ≈ 33% ∴ Cl present Assigned problems Review problems. End of chapter problems: 23 - 27, 29 - 35, 37 - 47 ...
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This note was uploaded on 10/22/2010 for the course CHEM 322A at USC.

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