Organic Chemistry: Principles and Mechanisms (Second Edition)
Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
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Chapter 7, Problem 7.41P
Interpretation Introduction

Interpretation:

It is to be explained why the equilibrium percentage of the first given molecule in its keto form is lower than that of the second given molecule, referring to Table 7-1.

Concept introduction:

In aqueous basic or acidic conditions, ketones and aldehydes exist in rapid equilibrium with a rearranged form called enol form. As a ketone or aldehyde, the species is called the keto form. In the enol form, the species has a carbon atom that is simultaneously a part of a C = C bond, which is a characteristic of alkenes, and is bonded to OH, which is a characteristic of alcohol. These two forms are always in equilibrium with each other, and this is called keto-enol tautomerization. In enol form, the alkyl substitution of the alkene group provides more stability to the molecule.

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Which statement is false regarding the following system? + HX Addition Elimination H -X O A positive AG means that the equilibrium will favor the reactants because the entropy term is larger than the enthalpy term representing the gain of entropy when one molecule becomes two. O A negative AG means that the equilibrium will favor the products because the enthalpy term is larger than the entropy term representing the loss of entropy when two molecules become one. Elimination reactions are thermodynamically favored at high temperatures, where the entropy of the reaction will be larger than the enthalpy, causing a positive AG and the reactants to be favored over the products. Addition reactions are thermodynamically favored at high temperatures, where the entropy of the reaction will be larger than the enthalpy, causing a positive AG and the reactants to be favored over the products.
Use the SN1 reaction shown below to answer questions 4-7. H2S -S-CH3 acetone Which option gives the correct order for bonds broken and bonds formed during the arrow-pushing mechanism for this reaction? Step 1: C-O bond breaks Step 2: S-C bond forms O A. Step 3: S-H bond breaks at the same time as an O-H bond forms Step 1: S-H bond breaks at the same time as an O-H bond forms O B. Step 2: C-O bond breaks at the same time as an S-C bond forms Step 1: S-C bond breaks OC. Step 2: C-O bond forms Step 1: S-H bond breaks at the same time as a C-O bond breaks O D. Step 2: S-C bond forms Step 3: O-H bond forms
6-53 List the following carbocations in decreasing order of their stability. CH₂ *CH₂ 520 +CH₂ ADDED CH, CH, TEA

Chapter 7 Solutions

Organic Chemistry: Principles and Mechanisms (Second Edition)

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