0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 C h a p t e r S i x t e e n: Nuclear Magnetic Resonance Spectroscopy An 1 NMR FID of ethanol
Note: Problems with italicized numbers are more challenging. You may want to try them last. Copyright 2012 by Martin ulce. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior permission of the copyright holder
CM 323: Summary of Important Concepts YConcepts for Chapter 16: Nuclear Magnetic Resonance Spectroscopy I. Electromagnetic radiation A. Can be considered to be described as oscillating photons moving at the velocity of light. Their energy to proportional to their amplitude of oscillation: E = hν, where E = energy h = Planck's constant (6.6.2 x 10-27 erg @ sec -1 ) ν = frequency of oscillation in ertz B. Molecules can interact with photons. A molecule in its lowest energy state (called the ground state) will absorb the energy of photons when photons collide with it. This generates an excited state of the molecule, providing it with excess energy which it releases by doing something, such as vibrating or rotating. C. Typical molecule-photon interactions: 1. Radio waves: Nuclear Magnetic Resonance Spectroscopy (NMR) a. Atomic nuclei in molecules absorb Rf energy, causing the magnetic moments of the nuclei to wobble or precess 2. eat waves: Infrared Spectroscopy (IR) a. Chemical bonds in molecules absorb heat, causing them to vibrate 3. Visible and ultraviolet light: UV/VIS Spectroscopy a. Electrons in molecular orbitals of conjugated molecules absorb light, causing them to be promoted to low-lying antibonding (usually π*) molecular orbitals II. NMR A. Functions as if nuclei in molecules ring like bells when placed in a strong magnetic field and exposed to Rf energy B. Tells you three important things useful in structure determination: 1. The number of chemically nonequivalent types of atoms 2. The relative ratios of the chemically nonequivalent types of atoms 3. "Nearest neighbor" atoms C. 1 NMR 1. Chemically nonequivalent s are s with different electronic environments. They appear in different places in the NMR spectrum: a. Shielded, upfield, at relatively low frequency, closer to δ = 0 i. electron rich environment b. Deshielded, downfield, at relatively high frequency, closer to δ = 10 ii. election sparse environment c. For example: CCl 3 C 2 Cl 2 C 3 Cl C 4 δ 7.3 5.3 3.1 0.9 89
2. Integration of the different signals in the NMR spectrum indicates how many of each chemically nonequivalent there are in the molecule. a. Provides the empirical formula ratio; can be a multiple thereof b. For example: C 3 OC 2 CN δ 4.2, integrates for 2 δ 3.5, integrates for 3 3. s can "hear" chemically nonequivalent, neighboring s 2 or 3 σ bonds away. This is called coupling or splitting; the resonances of s that "hear one another in this way are split into multiple resonances in the NMR spectrum according to the nearest neighbor rule: a. Multiplicity = n + 1, where n = number of nearest neighbor s b. Common multiplets: i. singlets (no nearest neighbors) ii. doublets (1 nearest neighbor) iii. triplets (2 nearest neighbors) iv. quartets (3 nearest neighbors) c. Exceptions to the nearest neighbor rule: i. nearest neighbors 3 σ bonds away are most common ii. nearest neighbors that are connected through a C=O or an atom with lone pairs on it usually do not count as nearest neighbors; the coupling is usually not seen in the NMR spectrum iii. it is rare to see nearest neighbor effects on aromatic rings unless: I. there is an strongly EWG or EDG on the ring II. there is more than one substituent on the ring d. For example: δ 3.7, 2, t δ 1.2, 6, d O O O δ 4.4, 1, s δ 4.1, 1, m δ 2.2, 2, t 90
e. Coupling or splitting is constant for a given set of nearest neighbors. It is reported by the couping constant J and is reported in hertz. f. For example: 1.0 0.5 1.5 4.0 3.8 3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 Integrated 1 NMR spectrum of ethanol, C 3 C 2 O, showing coupling constant of 6.8 z between the methylene quartet and the methyl triplet D. 13 C NMR 1. Works just like 1 NMR; spectrum displays resonances of all chemically nonequivalent Cs in molecule. Normally, all resonances are singlets. The chemical shift axis is larger: about 0-200 ppm. 2. Tells you three important things for structure determination: a. The number of chemically nonequivalent Cs (i.e., acts as a "carbon counter") b. Although not normally integrated, the relative intensities are an indication of how many of each type of C as well as how many s are attached to each C. Generally, relative intensies are: i. C 3 > C 2 > C > C c. The number of s attached to each carbon in the 13 C NMR spectrum can be determined using a DEPT experiment, which sorts the 91
signals in the spectrum according to their identity as being: i. Methyl (C 3 ) carbons ii. Methylene (C 2 ) carbons iii. Methine (C) carbons iv. Quarternary carbons with no s attached do not appear in DEPT spectra. They are identifed by comparison with the normal 13 C NMR spectrum D. Refer to tables 13.5 and 13.10 in the textbook, appendices 4 and 5 in the textbook, or to the following two tables for chemical shift data which can aid in determining structures from NMR spectra 92
BASIC 1 NMR CEMICAL SIFTS (R = or Alkyl) Type Approximate δ, ppm 1E (methyl) 0.9 2E (methylene) 1.3 3E (methine) 1.5 R 2 C=CR (vinylic) 4.6-5.9 RC/C (terminal akynyl) 2-3 R 2 C=CRC 2 R (allylic) 1.7 Ar (aryl) 6.5-8.0 Ar CR R (benzylic) 2.2-3 Cl CR 2 3-4 Br CR 2 2.5-4 I CR 2 2-4 RO CR 2 3.3-4 RCO 2 CR 2 3.7-4.1 O 5 2-2.6 R 3 C C CR 2 O 5 9-10 R C O 5 10.5-13 R C O R O R 2 N anywhere anywhere 93
BASIC 13 C NMR CEMICAL SIFTS C Type Approximate δ, ppm Alkanes Methyl 0-30 Methylene 15-55 Methine 25-55 Quaternary 30-40 Alkenes C=C 80-145 Alkynes C/C 70-90 Arenes 110-170 Benzene 128.7 Alcohols and Ethers C O - 90 Amines C N 40-60 alogens C F 70-80 C Cl 25 - C Br 10-40 C I -20-10 Carbonyls R 2 C=O 190-220 RXC=O (X = O or N) 1-180 94
NOTE: The NMR spectra in this problem set are high-resolution graphics. Details such as mutiplicities and coupling constants can be examined by magnifying page images with your pdf viewer. 1. Using the 1 NMR spectra below, give structures for compounds A, B, and C: a. C 8 10 O 2 4 6 1.0 1.8 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 95
b. C 9 9 ClO 2 2 2 3 2 1.1 1.1 1.7 1.1 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 96
c. C 6 12 O 2 2 3 2 2 3 1.1 1.7 1.1 1.1 1.7 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 97
2. Provide short answers for each of the following questions: a. Label each chemically nonequivalent type of hydrogen and assign the multiplicies of the 1 NMR resonances for each type of hydrogen in the following molecule: C O C C C O Br C C C C C b. At 1.41 T magnetic field strength, the hydrogens in benzene resonate at a frequency of 377 z relative to an internal standard. At a magnetic field strength of 11.75 T, will these hydrogens resonate at that same frequency, a higher frequency, or a lower frequency? c. A 1:1 mixture of water, 2 O, and hydrogen sulfide, 2 S, propvide two signals when the 1 NMR spectrum of the mixture is recorded. Which signal is upfield relative to the other? 98
3. A bottle of a chemically pure compound labeled simply dimethylcyclohexene was subjected to catalytic hydrogenation. The product exhibited the following 13 C NMR spectrum: δ 36.2 32.1 24.0 18.3 Ozonolysis of the dimethylcyclohexene afforded a dialdehyde product, in which the aldehyde carbons were found to be chemically nonequivalent by 13 C NMR. Assuming the name on the label was correct, which constitutional isomer of dimethylcyclohexene did the bottle contain? 99
4. A molecule of molecular formula C 5 10 O 2 has the following 1 NMR spectrum. Provide a structural formula for the molecule. 3 1 6 1.0 0.3 2.1 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0
5. Match each following isomeric structure of methylcyclohexanone to its corresponding DEPT-tabulated 13 C NMR spectrum. DEPT ppm type 2133 C 45.5 C 41.9 C 2 36.3 C 2 28.1 C 2 25.3 C 2 14.8 C 3 O 200 A 175 1 125 75 25 DEPT ppm type 211.9 C 40.9 C 2 34.9 C 2 31.3 C 21.1 C 3 O B 200 DEPT 175 1 125 75 ppm type 211.6 C.1 C 2 41.2 C 2 C 34.3 C 33.5 C 2 25.3 C 2 22.1 C 3 25 O 200 175 1 125 75 25 101
6. A molecule of molecular formula C 7 5 NO 4 has the following 1 NMR and 13 C NMR spectra. Provide a structural formula for the molecule. 1.0 1.0 1.1 1.1 1.2 11 10 9 8 7 6 5 4 3 2 1 (solvent) (solvent) 200 175 1 125 75 25 102
7. A molecule of molecular formula C 9 10 O 2 has the following 1 NMR and 13 C NMR spectra. Provide a structural formula for the molecule. 1.0 2.0 2.0 2.1 2.8 10 9 8 7 6 5 4 3 2 1 (solvent) (solvent) 200 175 1 125 75 25 103
8. Provide structures for each molecule with the following 1 NMR spectra: a. C 6 12 O 1.65 1.70 1.75 1.80 1.85 1.90 (expansion of δ 2.4-0.8 ppm region) 1.0 2.0 3.0 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 b. C 6 10 O 1 1 2 3 3 10 9 8 7 6 5 4 3 2 1 104
c. C 6 12 O 3 (exchanges with D 2 O) 3 2 1 6 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 105
9. Using the IR and 1 NMR spectra below, provide structures for each of the following compounds: a. C 9 10 O 3 1 NMR 1 1 3 3 2 12 11 10 9 8 7 6 5 4 3 2 1 IR 105 95 90 85 80 %Transmittance 75 70 65 60 55 45 40 35 30 4000 30 3000 20 2000 Wavenumbers (cm-1) 10 0 0 106
b. C 4 8 O 1 NMR 1 2 2 3 10 9 8 7 6 5 4 3 2 1 IR 90 80 70 %Transmittance 60 40 30 20 10 0 4000 30 3000 20 2000 Wavenumbers (cm-1) 10 0 0 107
10. Vanillin gives vanilla extract its characteristic odor and flavor. It has mp 81-83 EC and molecular formula C 8 8 O 3. Its 1 and 13 C NMR and IR spectra are shown below. Provide a structural formula for this compound that is consistent with its spectra. INT: When the sample used to obtain the 1 NMR spectrum was shaken with D 2 O, the resonance integrating for 1 at δ 6.65 disappears. 1 NMR 1 2 1 1 3 10 9 8 7 6 5 4 3 2 1 13 C NMR 200 175 1 125 75 25 108
IR 90 80 70 %Transmittance 60 40 30 20 10 0 4000 30 3000 20 2000 Wavenumbers (cm-1) 10 0 0 109
11. Using the IR and NMR spectra below, provide structures for each of the following compounds: a. C 9 10 O 2 1 NMR 1 1 2 3 3 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 IR 90 80 70 %Transmittance 60 40 30 20 10 0 4000 30 3000 20 2000 Wavenumbers (cm-1) 10 0 0 110
b. C 6 12 O 2 1 NMR 1 2 3 6 8.5 8.0 IR 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 105 95 90 85 80 75 %Transmittance 70 65 60 55 45 40 35 30 25 20 4000 30 3000 20 2000 Wavenumbers (cm-1) 10 0 0 111
c. C 8 10 O 13 C NMR 200 175 1 125 75 25 IR 90 80 70 %Transmittance 60 40 30 20 10 0 4000 30 3000 20 2000 Wavenumbers (cm-1) 112 10 0 0
12. A chemical compound with molecular formula C 4 7 ClO was found to have the NMR and IR spectra recorded below. Provide a structural formula for this compound. 1 NMR 1 3 3 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 13 C NMR 2 225 200 175 1 125 113 75 25 0
IR 90 80 70 %Transmittance 60 40 30 20 10 0 4000 30 3000 20 2000 Wavenumbers (cm-1) 10 0 0 114
13. A chemically pure compound with molecular formula C 9 10 O 2 has the following 1 NMR and IR spectra. Provide its structural formula. 1 NMR 1 2 2 2 3 10 9 8 7 6 5 4 3 2 1 IR 110 90 80 70 %Transmittance 60 40 30 20 10 0 4000 30 3000 20 2000 Wavenumbers (cm-1) 10 0 115
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