Organic Chemistry
Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 20.SE, Problem 38AP
Interpretation Introduction

a) lactic acid, Ka = 8.4 x 10-4

Interpretation:

The pKa of lactic acid with Ka = 8.4 x 10-4 is to be calculated.

Concept introduction:

pKa = - log Ka. By taking the inverse of log of Ka the pKa value can be obtained.

To calculate:

The pKa of lactic acid with Ka = 8.4 x 10-4.

Interpretation Introduction

b) Acrylic acid, Ka = 5.6 x 10-6

Interpretation:

The pKa of acrylic acid with Ka = 5.6 x 10-6 is to be calculated.

Concept introduction:

pKa = - log Ka. By taking the inverse of log of Ka the pKa value can be obtained.

To calculate:

The pKa of acrylic acid with Ka = 5.6 x 10-6.

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Students have asked these similar questions
(a) Given that Ka for acetic acid is 1.8 x 10-5 and that forhypochlorous acid is 3.0 x 10-8, which is the stronger acid?(b) Which is the stronger base, the acetate ion or the hypochloriteion? (c) Calculate Kb values for CH3COO- and ClO-.
(a) Predict the products of the following acid-base reactions using curved-arrow mechanisms to indicate electron flow. (b) Indicate the acid, base, conjugate acid, and conjugate base of each reaction. (c) Indicate whether the reactants or products are favored at equilibrium a) CH,COOH CH3O b) CH,CH,OH H2N
(a) Given that Kb for ammonia is 1.8 x10-5 and that forhydroxylamine is 1.1 x 10-8, which is the stronger base?(b) Which is the stronger acid, the ammonium ion or thehydroxylammonium ion? (c) Calculate Ka values for NH4+and H3NOH+.

Chapter 20 Solutions

Organic Chemistry

Ch. 20.6 - Prob. 11PCh. 20.6 - How might you carry out the following...Ch. 20.7 - Prob. 13PCh. 20.7 - Prob. 14PCh. 20.8 - Cyclopentanecarboxylic acid and...Ch. 20.8 - Prob. 16PCh. 20.SE - Prob. 17VCCh. 20.SE - Prob. 18VCCh. 20.SE - The following carboxylic acid can’t be prepared...Ch. 20.SE - Electrostatic potential maps of anisole and...Ch. 20.SE - Predict the product(s) and provide the mechanism...Ch. 20.SE - Predict the product(s) and provide the mechanism...Ch. 20.SE - Prob. 23MPCh. 20.SE - Predict the product(s) and provide the complete...Ch. 20.SE - Acid-catalyzed hydrolysis of a nitrile to give a...Ch. 20.SE - Prob. 26MPCh. 20.SE - Naturally occurring compounds called cyanogenic...Ch. 20.SE - 2-Bromo-6, 6-dimethylcyclohexanone gives 2,...Ch. 20.SE - Naturally occurring compounds called terpenoids,...Ch. 20.SE - In the Ritter reaction, an alkene reacts with a...Ch. 20.SE - Give IUPAC names for the following compounds:Ch. 20.SE - Prob. 32APCh. 20.SE - Prob. 33APCh. 20.SE - Prob. 34APCh. 20.SE - Prob. 35APCh. 20.SE - Prob. 36APCh. 20.SE - Prob. 37APCh. 20.SE - Prob. 38APCh. 20.SE - Calculate the Ka's for the following acids: (a)...Ch. 20.SE - Thioglycolic acid, HSCH2CO2H, a substance used in...Ch. 20.SE - Prob. 41APCh. 20.SE - Prob. 42APCh. 20.SE - How could you convert butanoic acid into the...Ch. 20.SE - How could you convert each of the following...Ch. 20.SE - How could you convert butanenitrile into the...Ch. 20.SE - How would you prepare the following compounds from...Ch. 20.SE - Prob. 47APCh. 20.SE - Using 13CO2 as your only source of labeled carbon,...Ch. 20.SE - Prob. 49APCh. 20.SE - Which method-Grignard carboxylation or nitrile...Ch. 20.SE - Prob. 51APCh. 20.SE - Prob. 52APCh. 20.SE - Propose a structure for a compound C6H12O2 that...Ch. 20.SE - Prob. 54APCh. 20.SE - How would you use NMR (either 13C or 1H) to...Ch. 20.SE - Prob. 56APCh. 20.SE - A chemist in need of 2,2-dimethylpentanoic acid...Ch. 20.SE - Prob. 58APCh. 20.SE - Prob. 59APCh. 20.SE - Prob. 60APCh. 20.SE - Prob. 61APCh. 20.SE - Prob. 62APCh. 20.SE - Prob. 63APCh. 20.SE - The following pKa values have been measured....Ch. 20.SE - Identify the missing reagents a-f in the following...Ch. 20.SE - Propose a structure for a compound, C4H7N, that...Ch. 20.SE - Prob. 67APCh. 20.SE - The 1H and 13C NMR spectra below belong to a...Ch. 20.SE - Propose structures for carboxylic acids that show...Ch. 20.SE - Carboxylic acids having a second carbonyl group...
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