Organic Chemistry: Principles And Mechanisms: Study Guide/solutions Manual (second)
Organic Chemistry: Principles And Mechanisms: Study Guide/solutions Manual (second)
2nd Edition
ISBN: 9780393655551
Author: KARTY, Joel
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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How might you prepare each of the following using a nucleophilic substitution reaction at some step? (a) (b) (c) (d) CH3 CH3C CCHCH3 CH3 CH3 0 CCH3 ☐ CH3 CH3CH2CH2CH2CN CH3CH2CH2NH2
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Manganese(II) Arsenate is insoluble in water at room temperature. (Note: Arsenate = AsO4³-) In the presence of aqueous ammonia, solid Manganese(II) Arsenate becomes more soluble and aqueous tetraamminemanganese (II) ion forms. When solid Manganese(II) arsenate was placed in a 2.00 M solution of ammonia, at equilibrium, 0.308 M of ammonia remains. If the Kf of tetraamminemanganese (II) ion is 250.0, Determine the Ksp of Manganese(II) arsenate Hint: You will have to figure out the Kspf of the overall chemical equation first, then solve for Ksp by using Kf and Kspf

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Organic Chemistry: Principles And Mechanisms: Study Guide/solutions Manual (second)

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