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 12, Problem 47P
Anthracene undergoes a Diels–Alder reaction with maleic anhydride to give a cycloadduct
with the formula
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Chapter 12 Solutions
Organic Chemistry - Standalone book
Ch. 12.2 - Write structural formulas for toluene (C6H5CH3)...Ch. 12.3 - Prob. 2PCh. 12.5 - Prob. 3PCh. 12.5 - Prob. 4PCh. 12.6 - Prob. 5PCh. 12.6 - Chrysene is an aromatic hydrocarbon found in coal...Ch. 12.8 - Prob. 7PCh. 12.9 - As measured by their first-order rate constants,...Ch. 12.9 - Give the structure of the principal organic...Ch. 12.9 - Prob. 10P
Ch. 12.10 - Prob. 11PCh. 12.11 - Prob. 12PCh. 12.12 - Prob. 13PCh. 12.13 - Prob. 14PCh. 12.13 - Prob. 15PCh. 12.15 - The regioselectivity of Birch reduction of...Ch. 12.16 - Prob. 17PCh. 12.17 - Both cyclooctatetraene and styrene have the...Ch. 12.17 - Prob. 19PCh. 12.18 - Give an explanation for each of the following...Ch. 12.19 - Prob. 21PCh. 12.19 - What does a comparison of the heats of combustion...Ch. 12.20 - Prob. 23PCh. 12.20 - Prob. 24PCh. 12.20 - Prob. 25PCh. 12.20 - Prob. 26PCh. 12.20 - Prob. 27PCh. 12.20 - Prob. 28PCh. 12.21 - Prob. 29PCh. 12.21 - Prob. 30PCh. 12.22 - Prob. 31PCh. 12.22 - Prob. 32PCh. 12 - Write structural formulas and give the IUPAC names...Ch. 12 - Prob. 34PCh. 12 - Prob. 35PCh. 12 - Prob. 36PCh. 12 - Prob. 37PCh. 12 - Acridine is a heterocyclic aromatic compound...Ch. 12 - Prob. 39PCh. 12 - Prob. 40PCh. 12 - Prob. 41PCh. 12 - Evaluate each of the following processes applied...Ch. 12 - Prob. 43PCh. 12 - Prob. 44PCh. 12 - Prob. 45PCh. 12 - Prob. 46PCh. 12 - Anthracene undergoes a DielsAlder reaction with...Ch. 12 - Prob. 48PCh. 12 - Prob. 49PCh. 12 - The relative rates of reaction of ethane, toluene,...Ch. 12 - Both 1,2-dihydronaphthalene and...Ch. 12 - Prob. 52PCh. 12 - Prob. 53PCh. 12 - Prob. 54PCh. 12 - Prob. 55PCh. 12 - Prob. 56PCh. 12 - Each of the following reactions has been described...Ch. 12 - Prob. 58PCh. 12 - A compound was obtained from a natural product and...Ch. 12 - Prob. 60PCh. 12 - Suggest reagents suitable for carrying out each of...Ch. 12 - Prob. 62PCh. 12 - Prob. 63DSPCh. 12 - Prob. 64DSPCh. 12 - Prob. 65DSPCh. 12 - Prob. 66DSP
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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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