Organic Chemistry, 12e Study Guide/Student Solutions Manual
12th Edition
ISBN: 9781119077329
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 14, Problem 11PP
Practice Problem 14.11
In addition to a signal downfield, the
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(b) One isomer of dimethoxybenzoic acid has 1H NMR spectrum dµ (ppm) 3.85 (6H, s), 6.63 (1H, t, J 2 Hz),
and 7.17 (2H, d, J 2Hz), One isomer of coumalic acid has 'H NMR spectrum dн (ppm) 6.41 (1H, d, J 10
Hz), 7.82 (1H, dd, J 2 Hz, 10 Hz) and 8.51 (1H, d, J 2Hz). In each case, which isomer is represented
here? The bonds sticking into the centre of the ring can be to any carbon atom.
Note: COOH proton is not indicated in the spectra
MeO
HO₂C
CO2H
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dimethoxybenzoic acid
coumalic acid
Draw the structure of a compound, CaH8O3 that exhibits IR absorptions at 1710 and 2500-3000 cm³¹ and the
following ¹H NMR signals: 11.1 (1H, singlet), 4.14 (2H, singlet), 3.63 (2H, quartet), 1.26 (3H, triplet) ppm.
For each compound shown below,(1) sketch the 13C NMR spectrum (totally decoupled, with a singlet for each type of carbon), showing approximatechemical shifts.(2) show the multiplicity expected for each signal in the off-resonance-decoupled spectrum.(3) sketch the spectra expected using the DEPT-90 and DEPT-135 techniques.(a)OCH3 C O CH2 CH3ethyl acetate
Chapter 14 Solutions
Organic Chemistry, 12e Study Guide/Student Solutions Manual
Ch. 14 - PRACTICE PROBLEM 14.1 Provide a name for each of...Ch. 14 - Prob. 2PPCh. 14 - Prob. 3PPCh. 14 - Practice Problem 14.4 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.5 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.6 1,3,5-Cycloheptatriene is...Ch. 14 - Prob. 7PPCh. 14 - Prob. 8PPCh. 14 - Practice Problem 14.9 In 1967 R. Breslow (of...Ch. 14 - Prob. 10PP
Ch. 14 - Practice Problem 14.11 In addition to a signal...Ch. 14 - PRACTICE PROBLEM 14.12
Azulene has an appreciable...Ch. 14 - Practice Problem 14.13 (a) The -Sh group is...Ch. 14 - Practice Problem 14.14
Explain how NMR...Ch. 14 - PRACTICE PROBLEM 14.15 Four benzenoid compounds,...Ch. 14 - Prob. 16PCh. 14 - Write structural formulas and give acceptable...Ch. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Which of the hydrogen atoms shown below is more...Ch. 14 - 14.22 The rings below are joined by a double bond...Ch. 14 - Prob. 23PCh. 14 - 14.24 (a) In 1960 T. Katz (Columbia University)...Ch. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - 14.27 5-Chloro-1,3-cyclopentadiene (below)...Ch. 14 - Prob. 28PCh. 14 - Furan possesses less aromatic character than...Ch. 14 - 14.30 For each of the pairs below, predict...Ch. 14 - Assign structures to each of the compounds A, B,...Ch. 14 - Prob. 32PCh. 14 - Give a structure for compound F that is consistent...Ch. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - The IR and 1H NMR spectra for compound X(C8H10)...Ch. 14 - Prob. 37PCh. 14 - Prob. 38PCh. 14 - 14.39 Given the following information, predict the...Ch. 14 - Consider these reactions: The intermediate A is a...Ch. 14 - Prob. 41PCh. 14 - Compound E has the spectral features given below....Ch. 14 - Draw all of the molecular orbitals for...Ch. 14 - Prob. 1LGPCh. 14 - Prob. 2LGPCh. 14 - 3. The NMR signals for the aromatic hydrogens of...Ch. 14 - Prob. 4LGPCh. 14 - Prob. 5LGP
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- Draw the structure of the compound identified by the simulated 'H NMR and C NMR spectra. The molecular formula of the compound is C,0H,O. 12 (Blue numbers next to the lines in the 'H NMR spectra indicate the integration values.) H NMR 1H 2H 2H 2H 2H|| 3H 10 8. 4 (dd) g 13C NMR 220 200 180 160 140 120 100 80 60 40 20 8 (ppm) Deduce the structure from the spectra. Select Draw Rings More Erasearrow_forwardcould you help me out with these question and explain to me step by step thanksarrow_forwardThe data for four 13C NMR spectra are given below. Match each spectrum with one of the following structures. Briefly justify your reasoning by labeling each struture. ÓH -NH2 А C Spectrum 1: 26, 29, 32, 39, 112, 127, 128, 144 ppm • Spectrum 2: 28, 68 ppm • Spectrum 3: 12, 23, 33, 70 ppm • Spectrum 4: 16, 17, 38, 52 ppmarrow_forward
- In the presence of a small amount of acid, a solution of acetaldehyde (CH3CHO) in methanol (CH3OH) was allowed to stand and a new compound L was formed. L has a molecular ion in its mass spectrum at 90 and IR absorptions at 2992 and 2941 cm−1. L shows three signals in its 13C NMR at 19, 52, and 101 ppm. The 1H NMR spectrum of L is given below. What is the structure of L?arrow_forwardPredict the H-NMR spectra of Diphenylacetylene including number of signals, chemical shift of each signal, multiplicity (splitting) of each signal and integration of each signal.arrow_forwardFollowing is the 1H-NMR spectrum of compound O, molecular formula C7H12. Compound O reacts with bromine in carbon tetrachloride to give a compound with the molecular formula C7H12Br2. The 13C-NMR spectrum of compound O shows signals at d 150.12, 106.43, 35.44, 28.36, and 26.36. Deduce the structural formula of compound O.arrow_forward
- Compound K, molecular formula C6H14O, readily undergoes acid-catalyzed dehydration when warmed with phosphoric acid to give compound L, molecular formula C6H12, as the major organic product. The 1H-NMR spectrum of compound K shows signals at d 0.90 (t, 6H), 1.12 (s, 3H), 1.38 (s, 1H), and 1.48 (q, 4H). The 13C-NMR spectrum of compound K shows signals at d 72.98, 33.72, 25.85, and 8.16. Deduce the structural formulas of compounds K and L.arrow_forwardWhy does the 13C-NMR spectrum of the product only show six signals for the 1,4-di-tert.-butyl-2,5-dimethoxybenzene? Structure and 13C are attachedarrow_forwardThe 1H- and 13C-NMR data of an ester of molecular formula C6H10O2 are given below. Also shown are the COSY and HETCOR NMR spectra of the ester. Draw the structure of the ester, explaining how you reach your conclusion. 1H-NMR: 7.20-6.90 (1H), 5.85 (1H), 4.16 (2H), 1.88 (3H), 1.31 (3H) ppm 13C-NMR: 166.7, 144.5, 123.0 , 60.2, 18.0, 14.3 ppmarrow_forward
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