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EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 13, Problem 13.59AP
Interpretation Introduction
Interpretation:
The structure of compound X with molecular formula
Concept introduction:
Many nuclei and electrons have spin. Due to this spin magnetic moment arises. The energy of this magnetic moment depends on the orientation of the applied magnetic field. In NMR spectroscopy, every nucleus has a spin. There is an
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Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.
Chapter 13 Solutions
EBK ORGANIC CHEMISTRY
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Prob. 13.4PCh. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10P
Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - Prob. 13.24PCh. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27PCh. 13 - Prob. 13.28PCh. 13 - Prob. 13.29PCh. 13 - Prob. 13.30PCh. 13 - Prob. 13.31PCh. 13 - Prob. 13.32PCh. 13 - Prob. 13.33PCh. 13 - Prob. 13.34PCh. 13 - Prob. 13.35PCh. 13 - Prob. 13.36APCh. 13 - Prob. 13.37APCh. 13 - Prob. 13.38APCh. 13 - Prob. 13.39APCh. 13 - Prob. 13.40APCh. 13 - Prob. 13.41APCh. 13 - Prob. 13.42APCh. 13 - Prob. 13.43APCh. 13 - Prob. 13.44APCh. 13 - Prob. 13.45APCh. 13 - Prob. 13.46APCh. 13 - Prob. 13.47APCh. 13 - Prob. 13.48APCh. 13 - Prob. 13.49APCh. 13 - Prob. 13.50APCh. 13 - Prob. 13.51APCh. 13 - Prob. 13.52APCh. 13 - Prob. 13.53APCh. 13 - Prob. 13.54APCh. 13 - Prob. 13.55APCh. 13 - Prob. 13.56APCh. 13 - Prob. 13.57APCh. 13 - Prob. 13.58APCh. 13 - Prob. 13.59APCh. 13 - Prob. 13.60APCh. 13 - Prob. 13.61APCh. 13 - Prob. 13.62APCh. 13 - Prob. 13.63APCh. 13 - Prob. 13.64APCh. 13 - Prob. 13.65APCh. 13 - Prob. 13.67AP
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- The 'H NMR spectrum of compound A (C3H100) has four signals: a multiplet at 8 = 7.25-7.32 ppm (5 H), a singlet at d = 5.17 ppm (1 H), a quartet at d = 4.98 ppm (1 H), and a doublet at ô = 1.49 ppm (3 H). There are 6 signals in its 13C NMR spectrum. The IR spectrum has a broad absorption in the -3200 cm-1 region. Compound A reacts with KMNO4 in a basic solution followed by acidification to give compound B with the molecular formula C7H6O2. Draw structures for compounds A and B.arrow_forward5. A Compound (C7H140) has a strong infrared absorption at 1715 cm³¹. The ¹HNMR spectrum has two signals at 8 1.10 (doublet) and 8 2.77 (septet), ratio 6:1. The 13CNMR spectrum shows three lines at 8 218, 39, and 18. Which of the following structure best fits with this spectroscopic data? X H ?arrow_forwardGive the structure that corresponds to the following molecular formula and H1 NMR spectrum:: C7H16O4: δ 1.93 (triplet); δ 3.35 (s); δ 4.49 (triplet); relative integral 1:6:1.arrow_forward
- An unknown compound has the formula C5H10O2. Elucidate its structure by scrutinizing its 1H NMR spectra, shown. Specifically, label each different type of H atom in the final structure with its NMR chemical shift in ppm.arrow_forwardIdentify the structures of isomers H and I (molecular formula C8H11N).a.Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm−1b.Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm−1arrow_forwardDraw the structure of molecular formula C8H10O that produced the 1H NMR spectra shown below. The IR spectrum does not show a broad absorbance at 3300 cm–1 or a strong absorbance at 1710 cm–1.arrow_forward
- The 1H NMR spectra of two carboxylic acids with molecular formula C3H5O2Cl are shown below. Identify the carboxylic acids. (The “offset” notation means that the farthest-left signal has been moved to the right by the indicated amount to fit on the spectrum; thus, the signal at 9.8 ppm offset by 2.4 ppm has an actual chemical shift of 9.8 + 2.4 = 12.2 ppm.)arrow_forwardCompound X of the molecular formula C7H10 has the 13C NMR spectrum (5 signals) shown below. On treatment with excess H2/Pt (catalytic hydrogenation), X is converted to methylcyclohexane. Propose a structure for X and justify your reasoning by clearly labeling each carbon signal and write out the reaction. 200 180 160 140 120 100 80 60 40 20arrow_forwardWhat IR and H-NMR peaks would you expect to see in methyl 3-oxohexanoate (see p559, attached image)? Include band position, chemical shift, splitting in your response.arrow_forward
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