Organic Chemistry, 12e Study Guide/Student Solutions Manual
12th Edition
ISBN: 9781119077329
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 14, Problem 4LGP
Interpretation Introduction
Interpretation:
The structure of adenine is to be drawn, and the nonbonding electron pairs that are not part of the
Concept introduction:
The electrons on nitrogen atom that take part in the aromatic system are not available for protonation, thereby making the nitrogen less basic.
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Chapter 14 Solutions
Organic Chemistry, 12e Study Guide/Student Solutions Manual
Ch. 14 - PRACTICE PROBLEM 14.1 Provide a name for each of...Ch. 14 - Prob. 2PPCh. 14 - Prob. 3PPCh. 14 - Practice Problem 14.4 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.5 Apply the polygon-and-circle...Ch. 14 - Practice Problem 14.6 1,3,5-Cycloheptatriene is...Ch. 14 - Prob. 7PPCh. 14 - Prob. 8PPCh. 14 - Practice Problem 14.9 In 1967 R. Breslow (of...Ch. 14 - Prob. 10PP
Ch. 14 - Practice Problem 14.11 In addition to a signal...Ch. 14 - PRACTICE PROBLEM 14.12
Azulene has an appreciable...Ch. 14 - Practice Problem 14.13 (a) The -Sh group is...Ch. 14 - Practice Problem 14.14
Explain how NMR...Ch. 14 - PRACTICE PROBLEM 14.15 Four benzenoid compounds,...Ch. 14 - Prob. 16PCh. 14 - Write structural formulas and give acceptable...Ch. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Which of the hydrogen atoms shown below is more...Ch. 14 - 14.22 The rings below are joined by a double bond...Ch. 14 - Prob. 23PCh. 14 - 14.24 (a) In 1960 T. Katz (Columbia University)...Ch. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - 14.27 5-Chloro-1,3-cyclopentadiene (below)...Ch. 14 - Prob. 28PCh. 14 - Furan possesses less aromatic character than...Ch. 14 - 14.30 For each of the pairs below, predict...Ch. 14 - Assign structures to each of the compounds A, B,...Ch. 14 - Prob. 32PCh. 14 - Give a structure for compound F that is consistent...Ch. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - The IR and 1H NMR spectra for compound X(C8H10)...Ch. 14 - Prob. 37PCh. 14 - Prob. 38PCh. 14 - 14.39 Given the following information, predict the...Ch. 14 - Consider these reactions: The intermediate A is a...Ch. 14 - Prob. 41PCh. 14 - Compound E has the spectral features given below....Ch. 14 - Draw all of the molecular orbitals for...Ch. 14 - Prob. 1LGPCh. 14 - Prob. 2LGPCh. 14 - 3. The NMR signals for the aromatic hydrogens of...Ch. 14 - Prob. 4LGPCh. 14 - Prob. 5LGP
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- 4-Methylphenol is more acidic than ethanol (pKa 10.36 vs 16.0) , even though both contain an OH group and a methyl group. Draw the structures of the anions formed from loss of the alcoholic protons from both compounds. Use resonance to explain the difference in their respective acidities.arrow_forwardExplain why phenol is more acidic than alcohols by considering the resonance effect anddrawing the resonance structures of phenoxide ions.arrow_forward2. Will a 0.10 molar solution of CH3NH3OCN (methylamine cyanate) produce a basic or acidic solution. Explain quantitatively.arrow_forward
- In the following acid - base reactions, a) Draw Lewis structures of the reactants and the products. b) Determine which species are acting as electrophiles (acids) and which are acting as nucleophiles (bases). c) Use the curved - arrow formalism to show the movement of electron pairs in these reactions and the imaginary movement in the resonance hybrids of the products. d) Indicate which reactions are best termed Brønsted-Lowry acid - base reactions i. CH3CHO + HCI-- > CH3CH2O + + Cl- ii. CH3CHO + OH- - - > CH3CO-(OH) Harrow_forward3) In the mid-1930s, the German theoretical chemist Erich Hückel developed a rule that dealt with the aromaticity of various compounds, which became known as the Hückel Rule. Which (parts) of the compounds listed below are aromatic? Justify your answer based on Hückel's rule. You can treat the rings separately or together as you wish.image iv: colchicine: a highly poisonous alkaloid,obtained from autumn turmeric and used to treat gout.arrow_forwardConsider cyclohexane and benzene (draw the structures below). Draw the conjugate base of each molecule and then explain why cyclohexane is the weaker acid.arrow_forward
- The presence of a pi bond also makes a compound a base. With this in mind, draw the conjugate acid of ethylene, CH2=CH2.arrow_forwardThe pKa of ascorbic acid (vitamin C, page 55) is 4.17, showing that it is slightly more acidic than acetic acid (CH3COOH, pKa 4.74). Compare the most stable conjugate base of ascorbic acid with the conjugate base of acetic acid, and suggest why these two compounds have similar acidities, even though ascorbic acid lacks the carboxylic acid (COOH) group.arrow_forwardThe C-H bond in acetone, (CH,),C=0, has a pk, of 19.2. Draw two resonance structures for its conjugate base. Then, explain why acetone is much more acidic than propane, CH,CH,CH, (pK,= 50). Use skeletal structures for the resonance structures. Resonance structures: draw structure draw structure (negative charge on C) (negative charge on O) Acetone is much more acidic than propane because: Acetone's conjugate base has fewer resonance structures than propane's conjugate base. The negative charge on acetone's conjugate base is stabilized by resonance. The negative charge on acetone's conjugate base is stabilized by inductance, Acetone's higher molecular mass makes it more acidic than propane.arrow_forward
- The C-H bond in acetone, (CH3)C=O, has a pką of 19.2. Draw two resonance structures for its conjugate base. Then, explain why acetone is much more acidic than propane, CH;CH,CH3 (pKa = 50). %3Darrow_forwardDoarrow_forwardThe pKa of ascorbic acid (vitamin C, page 55) is 4.17, showing that it is slightly more acidic than acetic acid (CH3COOH, pKa 4.74).Compare the stabilities of these four conjugate bases, and predict which OH group of ascorbic acid is the most acidicarrow_forward
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