Organic Chemistry Study Guide and Solutions
6th Edition
ISBN: 9781936221868
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 19, Problem 19.67AP
Interpretation Introduction
Interpretation:
The compound that is in accordance with the given spectroscopic data is to be stated.
Concept introduction:
Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and
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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)
As reaction of (CH3)2CO with LIC≡CH followed by H2O affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 3600−3200, 3303, 2938, and 2120 cm−1. D shows the following 1H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?
How could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?
Chapter 19 Solutions
Organic Chemistry Study Guide and Solutions
Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - Prob. 19.4PCh. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Prob. 19.9PCh. 19 - Prob. 19.10P
Ch. 19 - Prob. 19.11PCh. 19 - Prob. 19.12PCh. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Prob. 19.21PCh. 19 - Prob. 19.22PCh. 19 - Prob. 19.23PCh. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - Prob. 19.29PCh. 19 - Prob. 19.30PCh. 19 - Prob. 19.31PCh. 19 - Prob. 19.32PCh. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Prob. 19.37PCh. 19 - Prob. 19.38PCh. 19 - Prob. 19.39PCh. 19 - Prob. 19.40APCh. 19 - Prob. 19.41APCh. 19 - Prob. 19.42APCh. 19 - Prob. 19.43APCh. 19 - Prob. 19.44APCh. 19 - Prob. 19.45APCh. 19 - Prob. 19.46APCh. 19 - Prob. 19.47APCh. 19 - Prob. 19.48APCh. 19 - Prob. 19.49APCh. 19 - Prob. 19.50APCh. 19 - Prob. 19.51APCh. 19 - Prob. 19.53APCh. 19 - Prob. 19.54APCh. 19 - Prob. 19.55APCh. 19 - Prob. 19.56APCh. 19 - Prob. 19.57APCh. 19 - Prob. 19.58APCh. 19 - Prob. 19.59APCh. 19 - Prob. 19.60APCh. 19 - Prob. 19.61APCh. 19 - Prob. 19.62APCh. 19 - Prob. 19.63APCh. 19 - Prob. 19.64APCh. 19 - Prob. 19.65APCh. 19 - Prob. 19.66APCh. 19 - Prob. 19.67APCh. 19 - Prob. 19.68APCh. 19 - Prob. 19.69APCh. 19 - Prob. 19.70APCh. 19 - Prob. 19.71APCh. 19 - Prob. 19.72AP
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- Reaction of butanenitrile (CH3CH2CH2CN) with methylmagnesium bromide (CH3MgBr), followed by treatment with aqueous acid, forms compound G. G has a molecular ion in its mass spectrum at m/z = 86 and a base peak at m/z = 43. G exhibits a strong absorption in its IR spectrum at 1721 cm−1 and has the 1H NMR spectrum given below. What is the structure of G?arrow_forwardA compound (C7H14O) has a strong peak in its IR spectrum at 1710 cm–1. Its 1H NMR spectrum consists of three singlets in the ratio 9:3:2 at δ 1.0, 2.1, and 2.3, respectively. Identify the compound.arrow_forwardTreatment of compound D with LiAlH4 followed by H2O forms compound E. D shows a molecular ion in its mass spectrum at m/z = 71 and IR absorptions at 3600–3200 and 2263 cm−1. E shows a molecular ion in its mass spectrum at m/z = 75 and IR absorptions at 3636 and 3600–3200 cm−1. Propose structures for D and E from these data and the given 1H NMR spectra.arrow_forward
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- Please give the correct answer to the following NMR spectroscopy question and provide detailed solutions of the answer.arrow_forwardCompound 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.arrow_forward(a) The 'H-NMR spectrum of cyclobutanone shows two signals - signal A at 3.00 ppm and signal B at 1.95 ppm. Give the multiplicity of each signal. cyclobutanone (b) When cyclobutanone is treated with D20 and NaOD, the only signal observable in the 1H-NMR is a singlet at 2.00 ppm. Explain why this is the case. [Note: Deuterium atoms do not display signals in the TH-NMR spectrum]arrow_forward
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