EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 16, Problem 16.40AP
Interpretation Introduction
Interpretation:
The differences between ethylbenzene,
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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From the following 1H-NMR and 13C-NMR spectra, assign a structure for a compound with a formula of C4H9Br. Give reasons (explanations) for the choosen structure.
There are several isomeric alkanes of molecular formula C6H14.Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers.
Isomer A: δ = 0.84 (d, 12 H), 1.39 (septet, 2H) ppm
Isomer B: δ = 0.84 (t, 3 H), 0.86 (s, 9H), 1.22 (q, 2H) ppm
Following are 1H-NMR spectra for compounds B (C6H12O2) and C (C6H10O). Upon warming in dilute acid, compound B is converted to compound C. Deduce the structural formulas for compounds B and C.
Chapter 16 Solutions
EBK ORGANIC CHEMISTRY
Ch. 16 - Prob. 16.1PCh. 16 - Prob. 16.2PCh. 16 - Prob. 16.3PCh. 16 - Prob. 16.4PCh. 16 - Prob. 16.5PCh. 16 - Prob. 16.6PCh. 16 - Prob. 16.7PCh. 16 - Prob. 16.8PCh. 16 - Prob. 16.9PCh. 16 - Prob. 16.10P
Ch. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - Prob. 16.29PCh. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35APCh. 16 - Prob. 16.36APCh. 16 - Prob. 16.37APCh. 16 - Prob. 16.38APCh. 16 - Prob. 16.39APCh. 16 - Prob. 16.40APCh. 16 - Prob. 16.41APCh. 16 - Prob. 16.42APCh. 16 - Prob. 16.43APCh. 16 - Prob. 16.44APCh. 16 - Prob. 16.45APCh. 16 - Prob. 16.46APCh. 16 - Prob. 16.47APCh. 16 - Prob. 16.48APCh. 16 - Prob. 16.49APCh. 16 - Prob. 16.50APCh. 16 - Prob. 16.51APCh. 16 - Prob. 16.52APCh. 16 - Prob. 16.53APCh. 16 - Prob. 16.54APCh. 16 - Prob. 16.55APCh. 16 - Prob. 16.56APCh. 16 - Prob. 16.57APCh. 16 - Prob. 16.58APCh. 16 - Prob. 16.59APCh. 16 - Prob. 16.60APCh. 16 - Prob. 16.61APCh. 16 - Prob. 16.62APCh. 16 - Prob. 16.63APCh. 16 - Prob. 16.64APCh. 16 - Prob. 16.65APCh. 16 - Prob. 16.66APCh. 16 - Prob. 16.67APCh. 16 - Prob. 16.68AP
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- There are several isomeric alkanes of molecular formula C6H14. Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers. 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) ppmarrow_forwardIn Section 15.5c, we learned that the frequency of the C=0 stretch for an amide is lower than it is for a ketone, suggesting that the C=0 bond is weaker in an amide. This weakening of the C=0 bond can be explained by the significant contribution from one of its resonance structures toward the resonance hybrid. Draw that resonance structure for N,N-dimethylacetamide [CH3CON(CH3)2] and explain how it accounts for the weakening of the C=0 bond.arrow_forwardCompound 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.arrow_forward
- 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. Please show work.arrow_forwardCompound P has molecular formula C5H9ClO2. Deduce the structure of P from its 1H and 13C NMR spectra.arrow_forward08) The NMR spectra of the two isomeric compounds with formula C3H5ClO2 are shown in letters a and b. Low-field protons appearing in the NMR spectrum around 12.1 and 11.5 ppm, respectively, are shown highlighted. Draw the structures of the isomers.arrow_forward
- There 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_forwardGive reasons for the following: (i) p-nitrophenol is more acidic than p-methylphenol. (ii) Bond length of C—O bond in phenol is shorter than that in methanol. (iii) (CH3)3C—Br on reaction with sodium methoxide (Na+ _OCH3) gives alkene as the main product and not an ether.arrow_forwardCompounds 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)arrow_forward
- Use the 1H NMR and IR spectra given below to identify the structures of two isomers (A and B) having molecular formula C4H8O2.arrow_forwardDescribe how utlraviolet (UV) spectroscopy could distinguish propiophenone from 1-phenyl-2-propanone and p-ethylbenzaldehyde. Consider their respective wavelengths at their maximum UV absorbances, (λ max).arrow_forwardSolve the structure of this compound by looking at the 1H, 13C NMR data provided. You must fill all of the parts to this question in the given spaces. Molecular formula: C6H14N2O2arrow_forward
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