Connect Online Access 1-Semester for Organic Chemistry
Connect Online Access 1-Semester for Organic Chemistry
6th Edition
ISBN: 9781260475609
Author: SMITH, Janice
Publisher: Mcgraw-hill Higher Education (us)
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Chapter 6, Problem 26P

Draw the products of homolysis or heterolysis of each indicated bond. Use electronegativity

differences to decide on the location of charges in the heterolysis reaction. Classify each

carbon reactive intermediate as a radical, carbocation, or carbanion.

a.Chapter 6, Problem 26P, Draw the products of homolysis or heterolysis of each indicated bond. Use electronegativity , example  1b. Chapter 6, Problem 26P, Draw the products of homolysis or heterolysis of each indicated bond. Use electronegativity , example  2

Expert Solution
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Interpretation Introduction

(a)

Interpretation: The products of homolysis or heterolysis of the indicated bond is to be drawn by using the electronegativity differences. Each carbon reactive intermediate is to be classified as a radical, carbocation, or carbanion.

Concept introduction: In organic chemistry, the formation of carbocation or carbanion occurs due to the heterolysis or homolysis process. Carbocations possess six electrons around them, whereas carbanions possess the lone pair of electrons. Carbocation behaves as electrophile due to lack of electrons and incomplete octet.

Answer to Problem 26P

In the given indicated bond, homolysis takes place that results in the formation of the radicals.

The first product is,

Connect Online Access 1-Semester for Organic Chemistry, Chapter 6, Problem 26P , additional homework tip  1

The second product is,

Connect Online Access 1-Semester for Organic Chemistry, Chapter 6, Problem 26P , additional homework tip  2

Explanation of Solution

The heterolysis does not take place in the given compound due to the less electronegativity difference between atoms.

The product of homolysis is shown below.

Connect Online Access 1-Semester for Organic Chemistry, Chapter 6, Problem 26P , additional homework tip  3

Figure 1

In the above reaction, cyclohexane forms cyclohexyl radical and hydrogen radical by homolysis. Homolysis is opposite to the heterolysis. It forms radical with unpaired electron because the electrons are not attracted toward one element in the homolysis.

Conclusion

In the given indicated bond, homolysis takes place that results in the formation of the radical. The product of homolysis is shown in Figure 1.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation: The products of homolysis or heterolysis of the indicated bond is to be drawn by using the electronegativity differences. Each carbon reactive intermediate is to be classified as a radical, carbocation, or carbanion.

Concept introduction: In organic chemistry, the formation of carbocation or carbanion occurs due to the heterolysis or homolysis process. Carbocations possess six electrons around them, whereas carbanions possess the lone pair of electrons. Carbocation behaves as electrophile due to lack of electrons and incomplete octet.

Carbanion behaves as a nucleophile in the chemical reaction due to the presence of excess electrons. Heterolysis is the process in which unequal sharing of electrons results in the breaking of the bond.

Answer to Problem 26P

In the given indicated bond, heterolysis takes place that results in the formation of the carbocation.

The first product is,

Connect Online Access 1-Semester for Organic Chemistry, Chapter 6, Problem 26P , additional homework tip  4

The second product is,

Connect Online Access 1-Semester for Organic Chemistry, Chapter 6, Problem 26P , additional homework tip  5

Explanation of Solution

Heterolysis in the compound takes place due to the more electronegativity difference. The product of heterolysis is shown below.

Connect Online Access 1-Semester for Organic Chemistry, Chapter 6, Problem 26P , additional homework tip  6

Figure 2

In the above reaction, ethanol forms ethyl carbocation and hydroxide ion by heterolysis. The heterolysis in the chemical reaction leads to the formation of ionic species because electrons are attracted toward more electronegative atom. Oxygen is more electronegative than carbon. Therefore, heterolysis and the formation of carbocation take place in the reaction.

Conclusion

In the given indicated bond, heterolysis takes place that results in the formation of the carbocation. The product of heterolysis is shown in Figure 2.

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