Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Textbook Question
Chapter 19.SE, Problem 80AP
Compound B is isomeric with A (Problem 19-79) and shows an IR peak at 1715 cm-1. The 1H NMR spectrum of B has peaks at 2.4 δ (1 H, septet, 1=7 Hz), 2.1 δ (3 H, singlet), and 1.2 δ (6 H, doublet, 1=7 Hz). What is the structure of B?
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Propose a structure for each of the following two isomers with formula C6H14 given their 1H-NMR spectra.
Isomer A: δ = 0.84 (d, 12 H), 1.39 (septet, 2H) ppm
Isomer B: δ = 0.84 (t, 3 H), 0.86 (t, 9H), 1.22 (q, 2H) ppm
Reaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed bytreatment 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 thestructure of R?
Compound B of molecular formula C9H19N shows a noteworthy infrared absorption at 3300 cm-1. Its 1H-NMR spectrum shows three singlets – δ 1.0 (6H), 1.1 (12H), 1.4 (1H) ppm. Its 13C-NMR spectrum has four signals – δ 25, 28, 41, 64 ppm. Suggest a structure for this compound.
Chapter 19 Solutions
Organic Chemistry
Ch. 19.1 - Prob. 1PCh. 19.1 - Draw structures corresponding to the following...Ch. 19.2 - Prob. 3PCh. 19.2 - How would you carry out the following reactions?...Ch. 19.4 - Treatment of an aldehyde or ketone with cyanide...Ch. 19.4 - p-Nitrobenzaldehyde is more reactive toward...Ch. 19.5 - Prob. 7PCh. 19.5 - The oxygen in water is primarily (99.8) 16O, but...Ch. 19.6 - Prob. 9PCh. 19.8 - Show the products you would obtain by...
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