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 35P
Interpretation Introduction
Interpretation:
The structure for compound
Concept Introduction:
▸ In
▸ The splitting of the molecules is determined by the
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Compound A has molecular formula C5H8Br4 but shows only one singlet in the 1H-NMR spectrum. Suggest a structure for A and explain your reasoning.
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Compound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning.
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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- Compounds A and B are isomers having the molecular formula C4H8O3. Identify A and B on the basis of their 1H NMR spectra.Compound A: δ 1.3 (3H, triplet); 3.6 (2H, quartet); 4.1 (2H, singlet); 11.1 (1H, broad singlet)Compound B: δ 2.6 (2H, triplet); 3.4 (3H, singlet); 3.7 (2H triplet); 11.3 (1H, broad singlet)arrow_forwardDraw the structure of a compound with the formula C5H10O2 (along with the reasons of choosing it) which, upon analysis, gave key peaks in an infrared spectrum at 3450 cm-1 and 1713 cm-1, as well as the following 1H-NMR spectrum.arrow_forwardPropose a structure for a compound of molecular formula C3H8O with an IR absorption at 3600–3200 cm−1 and the following NMR spectrum:arrow_forward
- Compound X (molecular formula C10H120) was treated with NH2NH2, ¯OH to yield compound Y (molecular formula C10H14). Match the 1H NMR spectra of X and Y to the corresponding structures of X and Y. Compound NH2NH2 Compound 'H NMR of X 6 H OH Y 1 H 5H 8. 6. 4 ppm or H NMR of Y 6 H 2H 5H 1 H multiplet multiplet 8. 6. 4. 3. 1 nnm 2. 2. 3, O:arrow_forwardTreatment of compound C (molecular formula C4H8O) with C6H5MgBr, followed by H2O, affords compound D (molecular formula C10H14O). Compound D has a strong peak in its IR spectrum at 3600–3200 cm−1. The 1H NMR spectral data of C and D are given. What are the structures of C and D? Compound C signals at 1.3 (singlet, 6 H) and 2.4 (singlet, 2 H) ppm Compound D signals at 1.2 (singlet, 6 H), 1.6 (singlet, 1 H), 2.7 (singlet, 2 H), and 7.2 (multiplet, 5 H) ppmarrow_forwardA hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?arrow_forward
- Compound I (C11H14O2) is insoluble in water, aqueous acid, and aqueous NaHCO3, but dissolves readily in 10% Na2CO3 and 10% NaOH. When these alkaline solutions are acidified with 10% HCl, compound I is recovered unchanged. Given this information and its 1H-NMR spectrum, deduce the structure of compound I.arrow_forwardwhat is is the structure of a compound of molecular formula c 10 h 14 o 2 that shows a strong ir absorption at 3150-2850 cm − 1 and give the following 1 h nmr absorptions: 1.4 (triplet, 6 h), 4.0 (quarter, 4h), and 6.8 (singlet, 4h) ppm?arrow_forwardName the following compounds A and B. How could you distinguish these two molecules by using 1H NMR and IR techniques? Propose an analytical technique to determine the iron content of these compounds. Calculate the mass percentages of C and H of compound B (C: 12.01 g/mol; H: 1.008 g/mol; Fe: 55.845 g/mol).arrow_forward
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