Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 23.2, Problem 14LC

a.

Interpretation Introduction

Interpretation: To predict the product that is obtained by the following addition reactions.

Concept Introduction: The conversion of double and triple bonds to different functional groups is generally done through addition reactions. The reaction converts the unsaturated compounds to saturated compounds.

a.

Expert Solution
Check Mark

Answer to Problem 14LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 23.2, Problem 14LC , additional homework tip  1

Explanation of Solution

In this reaction, the nucleophile which is OH will gets attached by breaking the double bond which creates an electrophilic center that helps the nucleophile to attack at that center resulting in the formation of alcohol in this case.

b.

Interpretation Introduction

Interpretation: To predict the product that is obtained by the following addition reactions.

Concept Introduction: The conversion of double and triple bonds to different functional groups is generally done through addition reactions. The reaction converts the unsaturated compounds to saturated compounds.

b.

Expert Solution
Check Mark

Answer to Problem 14LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 23.2, Problem 14LC , additional homework tip  2

Explanation of Solution

In this reaction, the chlorine will act as a nucleophile. The nucleophile gets attached by breaking the double bond which creates an electrophilic center that helps the nucleophile to attack at that center resulting in the formation of a haloalkane in this case.

c.

Interpretation Introduction

Interpretation: To predict the product that is obtained by the following addition reactions.

Concept Introduction: The conversion of double and triple bonds to different functional groups is generally done through addition reactions. The reaction converts the unsaturated compounds to saturated compounds.

c.

Expert Solution
Check Mark

Answer to Problem 14LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 23.2, Problem 14LC , additional homework tip  3

Explanation of Solution

In this reaction, the HBr will add following anti-Markovnikov’s rule. This reaction will take place in the presence of peroxide which results in the free radical mechanism and initiate the reaction and add bromine radical at the terminal carbon. This mechanism results in the product which is 2-bromobutane in this case.

d.

Interpretation Introduction

Interpretation: To predict the product that is obtained by the following addition reactions.

Concept Introduction: The conversion of double and triple bonds to different functional groups is generally done through addition reactions. The reaction converts the unsaturated compounds to saturated compounds.

d.

Expert Solution
Check Mark

Answer to Problem 14LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 23.2, Problem 14LC , additional homework tip  4

Explanation of Solution

The hydrogen will add to an alkene by a process called hydrogenation. This process will take place in the presence of a catalyst. The catalyst used in the process of hydrogenation is generally metals. In this process, two hydrogens are added across the double bond resulting in the saturated compound. The double bond in butene is replaced with the two hydrogens and this results in the saturated compound called butane.

Chapter 23 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 23.2 - Prob. 11LCCh. 23.2 - Prob. 12LCCh. 23.2 - Prob. 13LCCh. 23.2 - Prob. 14LCCh. 23.2 - Prob. 15LCCh. 23.3 - Prob. 16LCCh. 23.3 - Prob. 17LCCh. 23.3 - Prob. 18LCCh. 23.3 - Prob. 19LCCh. 23.3 - Prob. 20LCCh. 23.3 - Prob. 21LCCh. 23.3 - Prob. 22LCCh. 23.3 - Prob. 23LCCh. 23.4 - Prob. 24LCCh. 23.4 - Prob. 25LCCh. 23.4 - Prob. 26LCCh. 23.4 - Prob. 27LCCh. 23.4 - Prob. 28LCCh. 23.4 - Prob. 29LCCh. 23.4 - Prob. 30LCCh. 23.4 - Prob. 31LCCh. 23 - Prob. 32ACh. 23 - Prob. 33ACh. 23 - Prob. 34ACh. 23 - Prob. 35ACh. 23 - Prob. 36ACh. 23 - Prob. 37ACh. 23 - Prob. 38ACh. 23 - Prob. 39ACh. 23 - Prob. 40ACh. 23 - Prob. 41ACh. 23 - Prob. 42ACh. 23 - Prob. 43ACh. 23 - Prob. 44ACh. 23 - Prob. 45ACh. 23 - Prob. 46ACh. 23 - Prob. 47ACh. 23 - Prob. 48ACh. 23 - Prob. 49ACh. 23 - Prob. 50ACh. 23 - Prob. 51ACh. 23 - Prob. 52ACh. 23 - Prob. 53ACh. 23 - Prob. 54ACh. 23 - Prob. 55ACh. 23 - Prob. 56ACh. 23 - Prob. 57ACh. 23 - Prob. 58ACh. 23 - Prob. 59ACh. 23 - Prob. 60ACh. 23 - Prob. 61ACh. 23 - Prob. 62ACh. 23 - Prob. 63ACh. 23 - Prob. 64ACh. 23 - Prob. 65ACh. 23 - Prob. 66ACh. 23 - Prob. 67ACh. 23 - Prob. 68ACh. 23 - Prob. 69ACh. 23 - Prob. 70ACh. 23 - Prob. 71ACh. 23 - Prob. 72ACh. 23 - Prob. 73ACh. 23 - Prob. 74ACh. 23 - Prob. 75ACh. 23 - Prob. 76ACh. 23 - Prob. 77ACh. 23 - Prob. 78ACh. 23 - Prob. 79ACh. 23 - Prob. 80ACh. 23 - Prob. 81ACh. 23 - Prob. 82ACh. 23 - Prob. 83ACh. 23 - Prob. 84ACh. 23 - Prob. 85ACh. 23 - Prob. 86ACh. 23 - Prob. 87ACh. 23 - Prob. 88ACh. 23 - Prob. 89ACh. 23 - Prob. 90ACh. 23 - Prob. 91ACh. 23 - Prob. 92ACh. 23 - Prob. 93ACh. 23 - Prob. 94ACh. 23 - Prob. 95ACh. 23 - Prob. 1STPCh. 23 - Prob. 2STPCh. 23 - Prob. 3STPCh. 23 - Prob. 4STPCh. 23 - Prob. 5STPCh. 23 - Prob. 6STPCh. 23 - Prob. 7STPCh. 23 - Prob. 8STPCh. 23 - Prob. 9STPCh. 23 - Prob. 10STPCh. 23 - Prob. 11STPCh. 23 - Prob. 12STPCh. 23 - Prob. 13STPCh. 23 - Prob. 14STPCh. 23 - Prob. 15STPCh. 23 - Prob. 16STPCh. 23 - Prob. 17STP
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