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
Question
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Chapter 26, Problem 26.37P
Interpretation Introduction

(a)

Interpretation:

Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. The mechanism and products for these two competing steps are to be drawn.

Concept introduction:

In any free radical polymerization reaction generally, three steps are involved viz. initiation, propagation, and termination. In initiation step, free radicals are generated from either monomer or from another compound such as hydrogen peroxide or benzoyl peroxide called a radical initiator. In propagation, step radicals react with the monomer to grow the chain to build up polymer and this process is called a chain-growth polymerization or addition polymerization. Finally, in terminating step, the growth of polymer is stopped by two ways: combination or disproportionation. In combination step, two growing polymer radicals combine to form the uncharged product, polymer. The termination occurs when the one radical abstracts a hydrogen atom from another radical is called disproportionation. The breaking of a covalent bond, whereby the electrons making up that bond are distributed equally to the atoms which are disconnected is known as the homolytic bond dissociation or homolysis. So in homolysis generally radicals are formed. In homolysis, a covalent bond is a breakdown equally and each atom acquires a single electron which is called a radical and a single barbed arrow (Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  1) is used to represents the movement of a single electron in a homolysis process.

Expert Solution
Check Mark

Answer to Problem 26.37P

Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. The mechanism and products for these two competing steps are given below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  2

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  3

Explanation of Solution

The mechanism for the propagation and its corresponding product for a given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  4

So in the propagation step, secondary radical is formed as a product.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  5

The mechanism for the hydrogen atom abstraction and its corresponding product in a given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  6

So in the hydrogen atom abstraction step, allylic radical is formed as a product.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  7

Conclusion

Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. The mechanism and products for these two competing steps are drawn.

Interpretation Introduction

(b)

Interpretation:

Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is to be explained which step produces a more stable product.

Concept introduction:

In any free radical polymerization reaction generally, three steps are involved viz. initiation, propagation, and termination. In initiation step, free radicals are generated from either monomer or from another compound such as hydrogen peroxide or benzoyl peroxide called a radical initiator. In propagation, step radicals react with the monomer to grow the chain to build up polymer and this process is called a chain-growth polymerization or addition polymerization. Finally, in terminating step, the growth of polymer is stopped by two ways: combination or disproportionation. In combination step, two growing polymer radicals combine to form the uncharged product, polymer. The termination occurs when the one radical abstracts a hydrogen atom from another radical is called disproportionation. The breaking of a covalent bond, whereby the electrons making up that bond are distributed equally to the atoms which are disconnected is known as the homolytic bond dissociation or homolysis. So in homolysis generally radicals are formed. In homolysis, a covalent bond is a breakdown equally and each atom acquires a single electron which is called a radical and a single barbed arrow (Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  8) is used to represents the movement of a single electron in a homolysis process. The weakest bond possesses the smallest bond dissociation energy. The stability order for radicals is methyl radical < 1oradical < 2oradical < 3oradical. Since radicals are electron-poor like carbocation, the electron-donating groups increases the stability of radical.

Expert Solution
Check Mark

Answer to Problem 26.37P

Due to resonance stabilization, an allylic radical is more stable than the secondary radical.

Explanation of Solution

The products of both these steps are:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  9 Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  10

The stability order of the radical is methyl radical < 1oradical < 2oradical < Allylic, benzylic radical < 3oradical. Therefore, an allylic radical is more stable than secondary radical due to the resonance stabilization as shown below.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  11

Conclusion

Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is explained which step produce a more stable product on the basis of the stability of radicals.

Interpretation Introduction

(c)

Interpretation:

Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is to be explained why propylene is poor reactive in free radical polymerization.

Concept introduction:

In any free radical polymerization reaction generally, three steps are involved viz. initiation, propagation, and termination. In initiation step, free radicals are generated from either monomer or from another compound such as hydrogen peroxide or benzoyl peroxide called a radical initiator. In propagation, step radicals react with the monomer to grow the chain to build up polymer and this process is called a chain-growth polymerization or addition polymerization. In a such case for propagation step less stable radical must be formed. Finally, in terminating step, the growth of polymer is stopped by two ways: combination or disproportionation. In combination step, two growing polymer radicals combine to form the uncharged product, polymer. The termination occurs when the one radical abstracts a hydrogen atom from another radical is called disproportionation. The breaking of a covalent bond, whereby the electrons making up that bond are distributed equally to the atoms which are disconnected is known as the homolytic bond dissociation or homolysis. So in homolysis generally radicals are formed. In homolysis, a covalent bond is a breakdown equally and each atom acquires a single electron which is called a radical and a single barbed arrow (Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  12) is used to represents the movement of a single electron in a homolysis process. The weakest bond possesses the smallest bond dissociation energy. The stability order for radicals is methyl radical < 1oradical < 2oradical < 3oradical. Since radicals are electron-poor like carbocation, the electron-donating groups increase the stability of radical.

Expert Solution
Check Mark

Answer to Problem 26.37P

In free radical polymerization reaction less stable radical must be formed to propagate the chain. But in given case more stable, allylic carbocation is formed which affects the propagation step results in rate of polymerisation is very low.

Explanation of Solution

The products of the both steps in free radical polymerization of propylene are given below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  13 Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  14

Due to the formation of more stable, allylic radical in hydrogen atom abstraction step it will interfere with the formation of polymer.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 26, Problem 26.37P , additional homework tip  15

In free radical polymerization reaction less stable radical must be formed to propagate the chain. But in given case more stable, allylic carbocation is formed which affects the propagation step results in rate of polymerisation is very low.

Conclusion

Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is explained why propylene is poor reactive in free radical polymerization.

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Chapter 26 Solutions

Organic Chemistry: Principles and Mechanisms (Second Edition)

Ch. 26 - Prob. 26.11PCh. 26 - Prob. 26.12PCh. 26 - Prob. 26.13PCh. 26 - Prob. 26.14PCh. 26 - Prob. 26.15PCh. 26 - Prob. 26.16PCh. 26 - Prob. 26.17PCh. 26 - Prob. 26.18PCh. 26 - Prob. 26.19PCh. 26 - Prob. 26.20PCh. 26 - Prob. 26.21PCh. 26 - Prob. 26.22PCh. 26 - Prob. 26.23PCh. 26 - Prob. 26.24PCh. 26 - Prob. 26.25PCh. 26 - Prob. 26.26PCh. 26 - Prob. 26.27PCh. 26 - Prob. 26.28PCh. 26 - Prob. 26.29PCh. 26 - Prob. 26.30PCh. 26 - Prob. 26.31PCh. 26 - Prob. 26.32PCh. 26 - Prob. 26.33PCh. 26 - Prob. 26.34PCh. 26 - Prob. 26.35PCh. 26 - Prob. 26.36PCh. 26 - Prob. 26.37PCh. 26 - Prob. 26.38PCh. 26 - Prob. 26.39PCh. 26 - Prob. 26.40PCh. 26 - Prob. 26.41PCh. 26 - Prob. 26.42PCh. 26 - Prob. 26.43PCh. 26 - Prob. 26.44PCh. 26 - Prob. 26.45PCh. 26 - Prob. 26.46PCh. 26 - Prob. 26.47PCh. 26 - Prob. 26.48PCh. 26 - Prob. 26.49PCh. 26 - Prob. 26.50PCh. 26 - Prob. 26.51PCh. 26 - Prob. 26.52PCh. 26 - Prob. 26.53PCh. 26 - Prob. 26.54PCh. 26 - Prob. 26.55PCh. 26 - Prob. 26.56PCh. 26 - Prob. 26.57PCh. 26 - Prob. 26.58PCh. 26 - Prob. 26.59PCh. 26 - Prob. 26.60PCh. 26 - Prob. 26.61PCh. 26 - Prob. 26.62PCh. 26 - Prob. 26.63PCh. 26 - Prob. 26.64PCh. 26 - Prob. 26.65PCh. 26 - Prob. 26.66PCh. 26 - Prob. 26.67PCh. 26 - Prob. 26.68PCh. 26 - Prob. 26.69PCh. 26 - Prob. 26.70PCh. 26 - Prob. 26.71PCh. 26 - Prob. 26.72PCh. 26 - Prob. 26.73PCh. 26 - Prob. 26.74PCh. 26 - Prob. 26.75PCh. 26 - Prob. 26.76PCh. 26 - Prob. 26.77PCh. 26 - Prob. 26.78PCh. 26 - Prob. 26.1YTCh. 26 - Prob. 26.2YTCh. 26 - Prob. 26.3YTCh. 26 - Prob. 26.4YTCh. 26 - Prob. 26.5YTCh. 26 - Prob. 26.6YTCh. 26 - Prob. 26.7YTCh. 26 - Prob. 26.8YTCh. 26 - Prob. 26.9YTCh. 26 - Prob. 26.10YTCh. 26 - Prob. 26.11YTCh. 26 - Prob. 26.12YTCh. 26 - Prob. 26.13YTCh. 26 - Prob. 26.14YTCh. 26 - Prob. 26.15YTCh. 26 - Prob. 26.16YTCh. 26 - Prob. 26.17YTCh. 26 - Prob. 26.18YTCh. 26 - Prob. 26.19YTCh. 26 - Prob. 26.20YTCh. 26 - Prob. 26.21YTCh. 26 - Prob. 26.22YTCh. 26 - Prob. 26.23YT
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