Pushing Electrons
4th Edition
ISBN: 9781133951889
Author: Weeks, Daniel P.
Publisher: Cengage Learning
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Question
Chapter 3, Problem 5EQ
Interpretation Introduction
Interpretation:
The leaving group in the given molecule is to be determined in order identify the positively charged and neutral fragments produced during mechanism.
Concept Introduction:
Heterolytic cleavage of sigma bond occurs under a variety of conditions. The mechanism of bond cleavage is as shown below,
The arrow in the above mechanism indicates that the sigma electrons form A-B bond are leaving A and becoming the exclusive property of B. Since the fragment A is formally losing one electron, it must become positively charged and B must become negatively charged since it gains an electron.
Often, heterolytic cleavage occurs from a charged intermediate that was formed in a previous step. The process is the same, but the charge on the products is different.
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Students have asked these similar questions
Consider this reaction:
Br
CH;OH
Br-Br
H3CO
The mechanism proceeds through a first cationic intermediate, intermediate 1. Nucleophilic attack leads to intermediate 2, which goes on to form the final product. In cases that involve a
negatively charged nucleophile, the attack of the nucleophile leads directly to the product.
Br
+ CH3OH
Br
Intermediate 1
Intermediate 2
(product)
In a similar fashion, draw intermediate 1 and intermediate 2 (final product) for the following reaction.
OH
+ Br2
+ HBr
Br
racemic mixture
What is the slow, rate-determining step, in the acid-catalyzed dehydration of 2-
butanol?
Loss of a b-hydrogen from the carbocation to form an alkene.
Protonation of the alcohol to form an oxonium ion.
Loss of water from the oxonium ion to form a carbocation.
The simultaneous loss of a B-hydrogen and water from the oxonium ion.
Consider the given reaction in which NC−NC− is the nucleophile and CH3CNCH3CN is the solvent. The reactant molecule has a structure with solid and dashed wedge bonds. A solid wedge () is used to show the bond that is above the plane of the paper, and a dashed wedge () is used to show the bond that is behind the plane of the paper.
Draw the product of the following reaction:
Chapter 3 Solutions
Pushing Electrons
Ch. 3 - Prob. 1EQCh. 3 - Prob. 2EQCh. 3 - Prob. 3EQCh. 3 - Prob. 4EQCh. 3 - Prob. 5EQCh. 3 - Prob. 6EQCh. 3 - Here are some exercises in sigma bond breaking....Ch. 3 - Prob. 8EQCh. 3 - Prob. 9EQCh. 3 - Prob. 10EQ
Ch. 3 - Prob. 11EQCh. 3 - Prob. 12EQCh. 3 - Prob. 13EQCh. 3 - Prob. 14EQCh. 3 - Prob. 15EQCh. 3 - Prob. 16EQCh. 3 - Prob. 17EQCh. 3 - Prob. 18EQCh. 3 - Prob. 19EQCh. 3 - Prob. 20EQCh. 3 - Prob. 21EQCh. 3 - Prob. 22EQCh. 3 - Prob. 23EQCh. 3 - Prob. 24EQCh. 3 - Prob. 25EQCh. 3 - Prob. 26EQCh. 3 - Prob. 27EQCh. 3 - Prob. 28EQCh. 3 - Prob. 29EQCh. 3 - Prob. 30EQCh. 3 - The reaction just described is reversible....Ch. 3 - Prob. 32EQCh. 3 - Prob. 59EQ
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