Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 13, Problem 13.53AP
Interpretation Introduction
Interpretation:
The structure of compound A 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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Give 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.
A compound, C7H12O2, exhibits the following 13C NMR shifts and substitution, determined by DEPT.
13C NMR: δ25.5 (2o), 25.9 (2o), 29.0 (2o), 43.1 (3o), 183.0 (4o)
Draw the structure
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
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Chapter 13 Solutions
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 1H-NMR spectrum of Compound D of molecular formula C10H12O shows three singlets – δ 2.20 (6H, s), 4.86 (4H), 7.10 (2H) ppm. Its 13C-NMR spectrum has five signals – 20, 74, 127, 135, 146 ppm. Suggest a structure for this compound.arrow_forwardAn unknown compound C3H2NCl shows moderately strong IR absorptions around 1650 cm-1 and 2200 cm-1. Its NMR spectrum consists of two doublets (J = 14 Hz) at δ 5.9 and δ 7.1. Propose a structure consistent with this data?arrow_forwardCompound K, molecular formula C6H14O, readily undergoes acid-catalyzed dehydration when warmed with phosphoric acid to give compound L, molecular formula C6H12, as the major organic product. The 1H-NMR spectrum of compound K shows signals at d 0.90 (t, 6H), 1.12 (s, 3H), 1.38 (s, 1H), and 1.48 (q, 4H). The 13C-NMR spectrum of compound K shows signals at d 72.98, 33.72, 25.85, and 8.16. Deduce the structural formulas of compounds K and L.arrow_forward
- 2) A compound having molecular formula C3H60 gave the following spectral data: ( 1 ) υν : λax 292 nm, εman 21 (ii) IR: 2720 cm1 (w) and 1738 cm(s) . Deduce the structure of the compound.arrow_forwardThere are several isomeric alcohols and ethers of molecular formula C5H12O. Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers. Isomer A: δ = 0.92 (t, 7.8 Hz, 3 H), 1.20 (s, 6H), 1.49 (q, 7.8 Hz, 2H), 1.85 (s, 1H) ppm Isomer B: δ = 1.19 (s, 9 H), 3.21 (s, 3H) ppmarrow_forwardThe 1H-NMR spectrum of Compound C shows five signals – δ 2.38 (1H, dt), 2.72 (1H, dt), 5.34 (1H, t), 5.49 (2H, ddd), 6.27 (2H, dd) ppm. Its 13C-NMR spectrum has four signals – δ 26, 58, 127, 129 ppm. In the compound’s mass spectrum, the M+1 peak appears at m/z = 115. An M+2 peak, whose intensity is roughly one-third that of the M+1 peak, also appears. Suggest a structure for this compound.arrow_forward
- The two isomeric compounds with the formula C3H5ClO2 have NMR spectra shown in 4a and 4b below. The downfield protons appearing in the NMR spectra at about 12.1 and 11.5 ppm, respectively, are shown as insets. Determine the structure of the two isomersarrow_forwardThe only organic compound obtained when compound Z undergoes the following sequence of reactions gives the 1H NMR spectrum shown. Identify compound Z.arrow_forwardThe only organic compound obtained when compound Z undergoes the following sequence of reactions gives the 1H NMR spectrum shown.Identify compound Z.arrow_forward
- A and B are isomeric dicarbonyl compounds of the molecular formula C5H&O2. The 'H NMR spectrum of A contains a singlet at 2.05 ppm and another singlet at 5.40 ppm. The 'H NMR spectrum of B contains three signals: a singlet at 2.3 ppm, a triplet at 1.10 ppm and a quartet at 2.70 ppm. Suggest structures for A and B and draw them in their respective boxes below. 1st attemptarrow_forwardIndicate two basic differences that exist between the spectra of 1H y 13C in NMR.arrow_forwardHow could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?arrow_forward
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