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EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 20, Problem 20.60AP
Interpretation Introduction
Interpretation:
The structure of compound A is to be identified and two peaks in the mass spectrum are to be rationalized.
Concept Introduction:
In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the
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Reaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 18.
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.
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Chapter 20 Solutions
EBK ORGANIC CHEMISTRY
Ch. 20 - Prob. 20.1PCh. 20 - Prob. 20.2PCh. 20 - Prob. 20.3PCh. 20 - Prob. 20.4PCh. 20 - Prob. 20.5PCh. 20 - Prob. 20.6PCh. 20 - Prob. 20.7PCh. 20 - Prob. 20.8PCh. 20 - Prob. 20.9PCh. 20 - Prob. 20.10P
Ch. 20 - Prob. 20.11PCh. 20 - Prob. 20.12PCh. 20 - Prob. 20.13PCh. 20 - Prob. 20.14PCh. 20 - Prob. 20.15PCh. 20 - Prob. 20.16PCh. 20 - Prob. 20.17PCh. 20 - Prob. 20.18PCh. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Prob. 20.21PCh. 20 - Prob. 20.22PCh. 20 - Prob. 20.23PCh. 20 - Prob. 20.24PCh. 20 - Prob. 20.25PCh. 20 - Prob. 20.26PCh. 20 - Prob. 20.27APCh. 20 - Prob. 20.28APCh. 20 - Prob. 20.29APCh. 20 - Prob. 20.30APCh. 20 - Prob. 20.31APCh. 20 - Prob. 20.32APCh. 20 - Prob. 20.33APCh. 20 - Prob. 20.34APCh. 20 - Prob. 20.35APCh. 20 - Prob. 20.36APCh. 20 - Prob. 20.37APCh. 20 - Prob. 20.38APCh. 20 - Prob. 20.39APCh. 20 - Prob. 20.40APCh. 20 - Prob. 20.41APCh. 20 - Prob. 20.42APCh. 20 - Prob. 20.43APCh. 20 - Prob. 20.44APCh. 20 - Prob. 20.45APCh. 20 - Prob. 20.46APCh. 20 - Prob. 20.47APCh. 20 - Prob. 20.48APCh. 20 - Prob. 20.49APCh. 20 - Prob. 20.50APCh. 20 - Prob. 20.51APCh. 20 - Prob. 20.52APCh. 20 - Prob. 20.53APCh. 20 - Prob. 20.54APCh. 20 - Prob. 20.55APCh. 20 - Prob. 20.56APCh. 20 - Prob. 20.57APCh. 20 - Prob. 20.58APCh. 20 - Prob. 20.59APCh. 20 - Prob. 20.60AP
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- Treatment of ketone A with ethynyllithium (HC≡CLi) followed by D3O+ afforded a compound B of molecular formula C12H13DO3, which gave an IR absorption at approximately 1715 cm−1. What is the structure of B and how is it formed?arrow_forwardCompound 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 reasoningarrow_forward. Compound A, MW = 86, shows an IR absorption at 1730 cm' and a very simple 'H NMR spectrum with peaks at 9.7 8 (1 H, singlet) and 1.2 8 (9 H, singlet). Propose a structure for A.arrow_forward
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