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chapter13 - 13. Structure Determination: Nuclear Magnetic...

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Unformatted text preview: 13. Structure Determination: Nuclear Magnetic Resonance Spectroscopy Based on McMurry's Organic Chemistry, 7th edition The Use of NMR Spectroscopy Used to determine relative location of atoms within a molecule Most helpful spectroscopic technique in organic chemistry Maps carbon-hydrogen framework of molecules Depends on very strong magnetic fields 2 Why This Chapter? NMR is the most valuable spectroscopic technique used for structure determination More advanced NMR techniques are used in biological chemistry to study protein structure and folding 3 13.1 Nuclear Magnetic Resonance Spectroscopy 1H or C nucleus spins and the internal magnetic field aligns parallel to or against an aligned external magnetic field (See Figure 13.1) Parallel orientation is lower in energy making this spin state more populated Radio energy of exactly correct frequency (resonance) causes nuclei to flip into anti-parallel state Energy needed is related to molecular environment (proportional to field strength, B) see Figure 13.2 13 4 The NMR Measurement The sample is dissolved in a solvent that does not have a signal itself and placed in a long thin tube The tube is placed within the gap of a magnet and spun Radiofrequency energy is transmitted and absorption is detected Species that interconvert give an averaged signal that can be analyzed to find the rate of conversion Can be used to measure rates and activation energies of very fast processes 5 NMR Instrument and Sample Tubes Chemagnetics 400 MHz ~$500,000 Superconducting electromagnet, cooled with liquid He 6 Fig. 13-4, p. 444 7 ABSENCE OF MAGNETIC FIELD APPLIED MAGNETIC FIELD (BO) Explain Magnetic Dipole on Board Fig. 13-1, p. 441 8 Photons in Microwave Region Cause Magnetic Dipoles to Flip Over to Oppose the Applied Field Frequency of oscillating magnetic field in photon must equal the oscillating magnetic field of the precessing hydrogen magnetic dipole. Photon energy (microwave) is needed to flip (excite) a spinning hydrogen nucleous (proton) from having its magnetic dipole aligned with the applied (external) magnetic field (Bo) to being aligned opposite to the applied field. The stronger the applied magnetic field, the higher photon energy photon required to flip the hydrogen dipole. Fig. 13-2, p. 441 9 Table 13-1, p. 442 10 13.2 The Nature of NMR Absorptions Electrons in bonds shield nuclei from magnetic field Different signals appear for nuclei in different environments Hydrogen (1H) Spectrum Carbon (13C) Spectrum 11 13.3 Chemical Shifts The relative energy of resonance of a particular nucleus resulting from its local environment is called chemical shift NMR spectra show applied field strength increasing from left to right Left part is downfield is upfield Nuclei that absorb on upfield side are strongly shielded. Chart calibrated versus a reference point, set as 0, tetramethylsilane [TMS] 12 Hydrogen Chemical Shift Shielding aromatic Ar-H bare proton H+ deshielded protons -CH2-Cl -COCH3 shielded protons -CH2Si(CH3)4 chemical shift ( ) Bo Bo = BP (bare proton), nucleus sees the full applied magnetic field Ho Bo = BP + BE1 (shielded nucleous), nucleus is shielded from applied field by electron cloud around nucleous. Electrons generate small magnetic field in opposition to the applied filed. A higher Bo values is needed to cause the nuclear dipole to flip. Bo = BP + BE2 (deshielded nucleous), nucleus is less shielded from applied field by electron cloud around nucleous because of electron withdrawing of Cl or O atoms. Fig. 13-5, p. 445 13 Equivalent H's Two H's that are in identical environments (homotopic) have the same NMR signal Test by replacing each with X if they give the identical result, they are equivalent Protons are considered homotopic Same chemical shift 14 Nonequivalent H's Replacement of each H with "X" gives a different constitutional isomer Then the H's are in constitutionally heterotopic environments and will have different chemical shifts they are nonequivalent under all circumstances 15 Table 13-3, p. 458 16 13.9 Chemical Shifts in 1H NMR Spectroscopy Proton signals range from 0 to 10 Lower field signals are H's attached to sp2 C Higher field signals are H's attached to sp3 C Electronegative atoms attached to adjacent C cause downfield shift Aldehydes ( ~ 10) Acids ( ~ 11-12) Upfield (higher Bo) Downfield (lower Bo) 17 13.10 Integration of 1H NMR Absorptions: Proton Counting The relative intensity of a signal (integrated area) is proportional to the number of protons causing the signal This information is used to deduce the structure For example in ethanol (CH3CH2OH), the signals have the integrated ratio 3:2:1 For narrow peaks, the heights are the same as the areas and can be measured with a ruler Integration line. Height of line displacement is proportional to the area of the peak, which is proportional to the number of hydrogens giving that signal. 18 13.10 Integration of 1H NMR Absorptions: Proton Counting The relative intensity of a signal (integrated area) is proportional to the number of protons causing the signal This information is used to deduce the structure For example in ethanol (CH3CH2OH), the signals have the integrated ratio 3:2:1 For narrow peaks, the heights are the same as the areas and can be measured with a ruler Measure length of red arrows in arbitrary units. The ratio of arrow is proportional to ratio of hydrogens in the molecule. Relative height = 9 Ratio of hydrogens 9:3 C6H12O 2 Relative height = 3 C6H12O2 3 + 9 = 12 19 13.11 Spin-Spin Splitting in 1H NMR Spectra Peaks are often split into multiple peaks due to interactions between nonequivalent protons on adjacent carbons, called spin-spin splitting The splitting is into one more peak than the number of H's on the adjacent carbon ("n+1 rule") The relative intensities are in proportion of a binomial distribution and are due to interactions between nuclear spins that can have two possible alignments with respect to the magnetic field The set of peaks is a multiplet (2 = doublet, 3 = triplet, 4 = quartet) 20 Simple Spin-Spin Splitting: Magnetic Field of Neighbors Effects Proton Signal An adjacent CH3 group can have four different spin alignments as 1:3:3:1 ratio of the adjacent H signal gives a ratio of 1:2:1 This gives peaks in An adjacent CH2 The separation of peaks in a multiplet is measured is a constant, in Hz J (coupling constant) 21 Table 13-4, p. 462 22 Example of Spin-Spin Splitting PEAK INTEGRATION CH3CH2Br 3 peaks (triplet) h=3 n + 1 rule CH3CH2Br 4 peaks (quartet) h=2 These peaks are expanded views of the peaks in the spectrum, making it easier to identify the number of peaks. 23 Rules for Spin-Spin Splitting Equivalent protons do not split each other The signal of a proton with n equivalent neighboring H's is split into n + 1 peaks Protons that are farther than two carbon atoms apart do not split each other 24 Predict the NMR Spectrum for Each of the Following Compounds CH3CH3 (discussed in lecture) CH3CHBrCH3 (2-bromopropane) CH3CH(OH)CH3 (isopropyl alcohol or 2-propanol) CH3COCH3 (acetone or 2-propanone) CH3CO2CH2CH3 (ethyl acetate) (CH3)3CCH2NH2 (neopentyl amine or 1- amino-2,2-dimethyl- propane) SPECTRAL DATABASE: http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng 25 Predict the NMR Spectrum for Each of the Following Compounds CH3CH3 (discussed in lecture) CH3CHBrCH3 (2-bromopropane) CH3CH(OH)CH3 (isopropyl alcohol or 2-propanol) CH3COCH3 (acetone or 2-propanone) CH3CO2CH2CH3 (ethyl acetate) (CH3)3CCH2NH2 (neopentyl amine or 1- amino-2,2-dimethyl- propane) SPECTRAL DATABASE: http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng 26 Example of Spin-Spin Splitting 7 peaks (heptet) 2 peaks (doublet) 27 Predict the NMR Spectrum for Each of the Following Compounds CH3CH3 (discussed in lecture) CH3CHBrCH3 (2-bromopropane) CH3CH(OH)CH3 (isopropyl alcohol or 2-propanol) CH3COCH3 (acetone or 2-propanone) CH3CO2CH2CH3 (ethyl acetate) (CH3)3CCH2NH2 (neopentyl amine or 1- amino-2,2-dimethyl- propane) SPECTRAL DATABASE: http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng 28 Isopropyl alcohol IR Spectrum Fingerprint region O-H stretch aliphatic C-H stretch C-O stretch H1 NMR Spectrum Can you identify which peaks correspond to protons A, B, and C? 2 peaks (doublet) C3 H A O C H C C3 H H C A 7 peaks (heptet) 1 peak (singlet) 29 Predict the NMR Spectrum for Each of the Following Compounds CH3CH3 (discussed in lecture) CH3CHBrCH3 (2-bromopropane) CH3CH(OH)CH3 (isopropyl alcohol or 2-propanol) CH3COCH3 (acetone or 2-propanone) CH3CO2CH2CH3 (ethyl acetate) (CH3)3CCH2NH2 (neopentyl amine or 1- amino-2,2-dimethyl- propane) SPECTRAL DATABASE: http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng 30 Fingerprint region impurity C=O stretch C-H stretch 3000 2000 1500 1000 C3 H O C C3 H 1 peak (singlet) acetone 31 Predict the NMR Spectrum for Each of the Following Compounds CH3CH3 (discussed in lecture) CH3CHBrCH3 (2-bromopropane) CH3CH(OH)CH3 (isopropyl alcohol or 2-propanol) CH3COCH3 (acetone or 2-propanone) Assignment: Use database below to verify your prediction for the last two compounds. CH3CO2CH2CH3 (ethyl acetate) propane) (CH3)3CCH2NH2 (neopentyl amine or 1- amino-2,2-dimethyl- SPECTRAL DATABASE: http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng 32 Work Handout #20 What information can you obtain from the Mass Spectrum? What information can you obtain from the Infrared Spectrum? What information can you obtain from the NMR spectrum? What is the compound structure? 33 Propose a structure for C3H6Br2 that fits the structure below. h=4 h=8 34 Propose a structure for C3H6Br2 that fits the structure below. BrCH2CH2CH2B r BrCH2CH2CH2B r BrCH2CH2CH2Br 35 The spectrum below corresponds to a compound with the formula C4H7O2Cl. The compound also has an infrared peak at 1740 cm-1. There is no broad peak at 3500 cm-1. Suggest a structure for the compound. quartet singlet h=3 h=2 h=2 triplet 36 C4H7O2Cl Triplet at 1.4 ppm Singlet at 4.1 ppm quartet at 4.3 ppm Table 13-3, p. 458 37 The spectrum below corresponds to a compound with the formula C4H7O2Cl. The compound also has an infrared peak at 1740 cm-1. There is no broad peak at 3500 cm-1. Suggest a structure for the compound. Integration 2:2:3 = 7 this fits formula C4H7O2Cl quartet singlet triplet deshielded 38 The spectrum below corresponds to a compound with the formula C4H7O2Cl. The compound also has an infrared peak at 1740 cm-1. There is no strong broad peak at 2500 - 3100 cm-1, nor at ~3500 cm1. Suggest a structure for the compound. Possible Candidates. Which is correct? Explain! C=O ~1740 cm-1 No strong broad peak at ~3100-2500 cm-1, means not a carboxylic acid No strong broad peak at ~3400 3600 cm-1, means not an alcohol Compound must be an ester. Triplet + quartet suggests an ethyl group. 39 The spectrum below corresponds to a compound with the formula C4H7O2Cl. The compound also has an infrared peak at 1740 cm-1. There is no strong broad peak at 2500 - 3100 cm-1, nor at ~3500 cm1. Suggest a structure for the compound. Possible Candidates. Which is correct? Explain! A C3C2 HH B CH lC 2 O C OHC C2 l O C=O ~1740 cm-1 C OHC 3 C2 H No strong broad peak at ~3100-2500 cm-1, means not a carboxylic acid No strong broad peak at ~3400 3600 cm-1, means not an alcohol Compound must be an ester. Triplet + quartet suggests an ethyl group. 40 The spectrum below corresponds to a compound with the formula C4H7O2Cl. The compound also has an infrared peak at 1740 cm-1. Suggest a structure for the compound. Compound B is the correct answer. A C3C2 HH B CH lC 2 O C OHC C2 l O C OHC 3 C2 H O-CH2-CH3 Cl-CH2-CO CH2-CH3 This singlet would be shifted much further from TMS, ~ 5-6 ppm 41 Propose a structure for C4H9Br, which has the NMR spectrum below. 2 peaks area = 6 2 peaks area = 2 9 peaks area = 1 area = 6 area = 1 area = 2 deshielded slightly deshielded shielded 42 Propose a structure for C4H9Br, which has the NMR spectrum below. C3 H C 3C C 2B H HH r CH3-CH-CH2-Br CH3-CH-CH2-Br CH3-CH 43 Propose a structure for C4H8Cl2 , which has the NMR spectrum below. 3 peaks area = 4 6 peaks area = 2 4 peaks area = 4 2 peaks area = 6 44 Propose a structure for C4H8Cl2 , which has the NMR spectrum below. 3 peaks area = 2 4 peaks area = 2 6 peaks area = 1 2 peaks area = 3 45 Propose a structure for C4H8Cl2 , which has the NMR spectrum below. 3 peaks area = 2 6 peaks area = 1 Cl -CH2-CH2 -Cl 4 peaks area = 2 CH-CH2-CH22 peaks area = 3 Cl C l C 3C C 2C 2C H HH H l CH3-CH-CH2- CH3-CH- 46 13C NMR Spectrum of 1-pentene Can you assign the peaks? 47 1H NMR Spectrum of 1-pentene SOLUTION! F B,C A D E 48 13.12 More Complex Spin-Spin Splitting Patterns Example: trans-cinnamaldehyde 49 Fig. 13-20, p. 466 50 Table 13-3, p. 458 51 13.4 13C NMR Spectroscopy: Signal Averaging and FT-NMR Carbon-13: only carbon isotope with a nuclear spin Natural abundance 1.1% of C's in molecules Sample is thus very dilute in this isotope Sample is measured using repeated accumulation of data and averaging of signals, incorporating pulse and the operation of Fourier transform (FT-NMR) All signals are obtained simultaneously using a broad pulse of energy and resonance recorded Frequent repeated pulses give many sets of data that are averaged to eliminate noise Fourier-transform of averaged pulsed data gives spectrum (see Figure 13-6) 52 13.5 Characteristics of 13C NMR Spectroscopy Provides a count of the different types of environments of carbon atoms in a molecule 13 C resonances are 0 to 220 ppm downfield from TMS (Figure 13-7) Chemical shift affected by electronegativity of nearby atoms O, N, halogen decrease electron density and shielding ("deshield"), moving signal downfield. sp3 C signal is at 0 to 9; sp2 C: 110 to 220 C(=O) at low field, 160 to 220 53 13C NMR spectrum of CH3CH2CH2CH2OH Single spectrum Average of 200 spectra Fig. 13-6a, p. 447 54 Chemical shift for 13C nuclei. 55 Spectrum of 2-butanone is illustrative- signal for C=O carbons on left edge NMR signal comes beyond normal scan. Short scan beyond 200 ppm is made and added to the spectrum as an insert. 56 Spectrum of 4-bromoacetophenone is illustrative- signal for C=O carbons on left edge and aromatics at ~130 ppm. 57 p. 450 58 Can You Assign the Carbons to the Peaks in the NMR Spectrum? Fig. 13-9, p. 450 59 Can You Assign the Carbons to the Peaks in the NMR Spectrum? 1 2 3 4 Fig. 13-9, p. 450 60 13.6 DEPT 13C NMR Spectroscopy Improved pulsing and computational methods give additional information DEPT-NMR (distortionless enhancement by polarization transfer) Normal spectrum shows all C's then: Obtain spectrum of all C's except quaternary (broad band decoupled) Change pulses to obtain separate information for CH2, CH Subtraction reveals each type (See Figure 13-10) 61 Normal 13C Spectrum All carbons viewed. 13C DEPT-90 Only CH Viewed CH Only 13C DEPT-135, CH, CH3 are viewed as positive signals & CH2 is negative signal. CH & CH3 Only CH2 Only Fig. 13-10, p. 451 62 p. 452 63 13.7 Uses of13C NMR Spectroscopy Provides details of structure Example: product orientation in elimination from 1chloro-methyl cyclohexane Difference in symmetry of products is directly observed in the spectrum 1-chloromethylcyclohexane has five sp3 resonances ( 20-50) and two sp2 resonances 100-150 64 13.8 1H NMR Spectroscopy and Proton Equivalence Proton NMR is much more sensitive than C and the active nucleus (1H) is nearly 100 % of the natural abundance Shows how many kinds of nonequivalent hydrogens are in a compound Theoretical equivalence can be predicted by seeing if replacing each H with "X" gives the same or different outcome Equivalent H's have the same signal while nonequivalent are different 13 There are degrees of nonequivalence 65 Enantiotopic Distinctions If H's are in environments that are mirror images of each other, they are enantiotopic Replacement of each H with X produces a set of enantiomers The H's have the same NMR signal (in the absence of chiral materials) 66 Diastereotopic Distinctions In a chiral molecule, paired hydrogens can have different environments and different shifts Replacement of a pro-R hydrogen with X gives a different diastereomer than replacement of the pro-S hydrogen Diastereotopic hydrogens are distinct chemically and spectrocopically 67 13.13 Uses of 1H NMR Spectroscopy The technique is used to identify likely products in the laboratory quickly and easily Example: regiochemistry of hydroboration/oxidation of methylenecyclohexane Only that for cyclohexylmethanol is observed 68 69 ...
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This note was uploaded on 04/07/2008 for the course CHGN 222 taught by Professor Cowley during the Spring '08 term at Mines.

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