Organic Chemistry Study Guide and Solutions
6th Edition
ISBN: 9781936221868
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 17, Problem 17.53AP
Interpretation Introduction
Interpretation:
The compounds,
Concept introduction:
Diels Alder reaction is the
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A 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?
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)
Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The 1H-NMR spectrum of B shows two singlets, one at 2.52 pm and one at 8.13 ppm. The 13C-NMR spectrum of B shows five signals. The mass spectrum of B shows a peak at m/z = 260 and another peak at m/z = 262; the relative height of the two peaks is 1:1 respectively.
- Identify compound B, explaining your reasoning
Chapter 17 Solutions
Organic Chemistry Study Guide and Solutions
Ch. 17 - Prob. 17.1PCh. 17 - Prob. 17.2PCh. 17 - Prob. 17.3PCh. 17 - Prob. 17.4PCh. 17 - Prob. 17.5PCh. 17 - Prob. 17.6PCh. 17 - Prob. 17.7PCh. 17 - Prob. 17.8PCh. 17 - Prob. 17.9PCh. 17 - Prob. 17.10P
Ch. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22APCh. 17 - Prob. 17.23APCh. 17 - Prob. 17.24APCh. 17 - Prob. 17.25APCh. 17 - Prob. 17.26APCh. 17 - Prob. 17.27APCh. 17 - Prob. 17.28APCh. 17 - Prob. 17.29APCh. 17 - Prob. 17.30APCh. 17 - Prob. 17.31APCh. 17 - Prob. 17.32APCh. 17 - Prob. 17.33APCh. 17 - Prob. 17.35APCh. 17 - Prob. 17.36APCh. 17 - Prob. 17.37APCh. 17 - Prob. 17.38APCh. 17 - Prob. 17.39APCh. 17 - Prob. 17.40APCh. 17 - Prob. 17.41APCh. 17 - Prob. 17.42APCh. 17 - Prob. 17.43APCh. 17 - Prob. 17.44APCh. 17 - Prob. 17.45APCh. 17 - Prob. 17.46APCh. 17 - Prob. 17.47APCh. 17 - Prob. 17.48APCh. 17 - Prob. 17.49APCh. 17 - Prob. 17.50APCh. 17 - Prob. 17.51APCh. 17 - Prob. 17.52APCh. 17 - Prob. 17.53APCh. 17 - Prob. 17.54APCh. 17 - Prob. 17.55APCh. 17 - Prob. 17.56APCh. 17 - Prob. 17.57APCh. 17 - Prob. 17.58APCh. 17 - Prob. 17.59AP
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- Propose a structure for a compound of molecular formula C3H8O with an IR absorption at 3600–3200 cm−1 and the following NMR spectrum:arrow_forwardCompound A has molecular formula C7H7X. Its 1H-NMR spectrum shows a singlet at 2.26 ppm and two doublets, one at 6.95 ppm and one at 7.28 ppm. The singlet has an integral of three and the doublets each have an integral of two. Its 13C-NMR shows five signals. The mass spectrum of A shows a peak at m/z = 170 and another peak at m/z = 172; the relative height of the two peaks is 1:1 respectively. - Identify what atom X is, explaining your reasoning - Identify Compound A, explaining your reasoning Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The 1H-NMR spectrum of B shows two singlets, one at 2.52 pm and one at 8.13 ppm. The 13C-NMR spectrum of B shows five signals. The mass spectrum of B shows a peak at m/z = 260 and another peak at m/z = 262; the relative height of the two peaks is 1:1 respectively. - Identify compound B, explaining your reasoning Compound B is treated with sodium ethoxide to generate compound C. The 1H-NMR spectrum of C shows…arrow_forwardThree compounds, A, B, and C, have the same molecular formula, C5H6. In the presence of a platinum catalyst, all three compounds absorb 3 molar equivalents of hydrogen and yield pentane. Compounds B and C give a precipitate when treated with ammoniacal silver nitrate; compound A give no reaction. Compounds A and B show an absorption maximum near 250 nm. Compound C shows no absorption maximum beyond 200 nm. Propose a structure for A, B, and C.arrow_forward
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