
Organic Chemistry - Standalone book
10th Edition
ISBN: 9780073511214
Author: Francis A Carey Dr., Robert M. Giuliano
Publisher: McGraw-Hill Education
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Chapter 14, Problem 52P
NMR and IR spectra for a compound with the formula
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Chapter 14 Solutions
Organic Chemistry - Standalone book
Ch. 14.3 - Prob. 1PCh. 14.3 - Prob. 2PCh. 14.4 - The 1H NMR signal for bromoform (CHBr3) appears at...Ch. 14.5 - identify the most shielded and least shielded...Ch. 14.5 - (a) Assign the chemical shifts 1.6, 2.2, and 4.8...Ch. 14.5 - Assign the chemical shifts 1.1, 1.7, 2.0, and 2.3...Ch. 14.5 - Assign the chemical shifts 1.6, 4.0, 7.5, 8.2, and...Ch. 14.6 - The 300-MHz 1H NMR spectrum of 1,4-dimethylbenzene...Ch. 14.6 - Prob. 9PCh. 14.6 - How many signals would you expect to find in the...
Ch. 14.7 - Describe the appearance of the 1H NMR spectrum of...Ch. 14.8 - Describe the appearance of the 1H NMR spectrum of...Ch. 14.11 - Prob. 13PCh. 14.11 - Prob. 14PCh. 14.12 - Hydrogen bonding between the oxygen of dimethyl...Ch. 14.14 - Prob. 16PCh. 14.15 - The 13C NMR spectrum of 1-bromo-3-chloropropane...Ch. 14.15 - Consider carbons x, y, and z in p-methylanisole....Ch. 14.15 - Prob. 19PCh. 14.16 - To which of the compounds of Problem 14.16 does...Ch. 14.18 - DEPT spectra for a compound with the formula...Ch. 14.20 - Vibrational frequencies are sensitive to isotopic...Ch. 14.21 - Prob. 23PCh. 14.22 - Prob. 24PCh. 14.23 - Prob. 25PCh. 14.23 - Which one of the C5H8 isomers shown has its max at...Ch. 14.24 - Knowing what to look for with respect to isotopic...Ch. 14.24 - The base peak appears at m/z105 for one of the...Ch. 14.24 - Mass spectra of 1-bromo-4-propylbenzene and...Ch. 14.25 - Prob. 30PCh. 14 - Each of the following compounds is characterized...Ch. 14 - Deduce the structure of each of the following...Ch. 14 - From among the isomeric compounds of molecular...Ch. 14 - The H1NMR spectrum of fluorene has signals at 3.8...Ch. 14 - Prob. 35PCh. 14 - H1NMR spectra of four isomeric alcohols with...Ch. 14 - Prob. 37PCh. 14 - We noted in Section 14.13 that an NMR spectrum is...Ch. 14 - Identify each of the C4H10O isomers on the basis...Ch. 14 - A compound (C3H7ClO2) exhibited three peaks in its...Ch. 14 - Label nonequivalent carbons in the following...Ch. 14 - Compounds A and B are isomers of molecular formula...Ch. 14 - C13 NMR spectra for four isomeric alkyl bromides...Ch. 14 - Prob. 44PCh. 14 - Prob. 45PCh. 14 - Identify the C3H5Br isomers on the basis of the...Ch. 14 - Prob. 47PCh. 14 - A compound (C8H10O) has the IR and H1NMR spectra...Ch. 14 - Deduce the structure of a compound having the...Ch. 14 - Figure 14.53 presents IR, H1NMR, C13NMR and mass...Ch. 14 - H1NMR, C13NMR, IR, and mass spectra are shown for...Ch. 14 - 1H NMR and IR spectra for a compound with the...Ch. 14 - FriedelCraftsalkylation of benzene with...Ch. 14 - Prob. 54DSPCh. 14 - Prob. 55DSPCh. 14 - Prob. 56DSPCh. 14 - Prob. 57DSPCh. 14 - Prob. 58DSP
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- am I right? Does this molecule not undergo beta-oxidation?arrow_forwardPredict the reagents (nucleophile and solvent) for each of the followingarrow_forward1. Answer the following questions about molecule 1. a) circle the most basic atom in molecule b) Provide the structure of the conjugate acid and a resonance structure of the conjugate acid that explain its bascity. HO H+ molecule 1 c) The pKa of the conjugate acid of the most basic atom in molecule is approximately -1. Which of the following acids can completely protonate the most basic atom? (Keq>10³). HI H₂O IZ Cl3C OH (+) NH3arrow_forward
- Question 3 Some photophysical parameters (lifetime, t, and quantum yield, $) for fluorescence (n = 370 nm) and phosphorence (p = 580 nm) of pyrene, 1-chloropyrene and 1-bromopyrene are given in the table below, as measured at room temperature (RT) and at 77 K in a frozen ethanol glass. фе τη Фл Фр Tp (RT) (RT) (77 K) (77 K) (77 K) (ns) (s) X = H 0.72 530 0.9 < 0.001 0.39 X X = CI 0.22 75 0.59 0.058 0.10 X = Br 0.032 2 0.17 0.085 0.004 fl = fluorescence; p = phosphorescence (a) Construct a Jablonski diagram for pyrene (X = H) at 77K. (b) Pyrene (X = H) has an absorbance maximum, Amax, at 330 nm and a fluorescence maximum, 11, at 370 nm. Why does this difference in wavelengths occur? (c) Explain why the lifetime for phosphorescence is longer than that for fluorescence. (d) Why does the fluorescence quantum yield increase with decreasing temperature? (e) Explain the trend in phosphorescence quantum yield as X is varied.arrow_forwardQuestion 4 The photoisomerization of alkenes is a photochemical transformation between the E- and Z-stereoisomers. The irradiation of the E-isomer (shown below) with radiation at 340 nm gives an E:Z ratio of 5:95. Some relevant information for each compound is shown in the table below. Amax 340nm E290 E340 (L mol¹ cm¹) (L mol¹ cm-1) E 340 nm 8000 20000 PE-Z = 0.60 Z 290 nm 16000 2000 Oz-E = 0.30 (a) The reaction proceeds through an excited state. Explain the nature of this excited state, and explain how it allows formation of the E- and Z-isomers. Explain why this isomerisation is unlikely to occur thermally. (b) The product of the equilibrium shown above gives a final ratio with substantially more of one isomer. Explain why this occurs. (c) Explain why the starting concentration of isomers does not affect the final ratio after irradiation. (d) If the irradiating wavelength used was changed to 290 nm and you started with Z- rather than E-isomer, use the data in the table above to…arrow_forwardQuestion 5 The photoisomerisation of cinnamonitriles (shown below) has been used as a model for the molecular transformations that lead to vision in mammals. The outcome of the reaction under various conditions is shown in the table below. CN sensitizer visible light 忌 CN СОН HO.. OH OH N NH 'N' Riboflavin Starting isomer E Sensitizer Riboflavin Radiation No light Final Z:E ratio 0:100 E No sensitizer 402 nm 4:96 E Riboflavin 402 nm 99:1 Z Riboflavin 402 nm 99:1 (a) Explain what is happening in the photoisomerisation reaction above. (b) Give the structure of the intermediate that allows the reaction to occur, and explain how it forms. (c) Explain why each reaction gives the particular E:Z ratio that it does, and why it is considered photostationary. (d) Explain the role of the sensitizer, riboflavin, and why it is a requirement for this reaction. (e) Explain why the irradiation of both the E- and Z-isomers in the presence of riboflavin gives the same result. What would be the final…arrow_forward
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