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 25, Problem 25.40P
Interpretation Introduction

(a)

Interpretation:

The major product of the given reaction is to be predicted. A complete, detailed mechanism for the reaction is to be provided.

Concept introduction:

Halogen molecules undergo homolysis when irradiated with UV light to produce two radical atoms. Homolysis is the type of bond dissociation where each atom gets one of the bond electrons. An unpaired electron present in the radical of atom or molecule. It is a highly reactive and unstable species. This homolytic dissociation is called initiation.

Being reactive, a radical will react with other species present. This produces another radical, which can continue the reaction is a step called propagation. The reaction is generally branched as each initiation step produces two radicals. Propagation steps continue as long as radicals are present and may be terminated by a reaction between two radicals.

The reaction generally results in substitution, often of a hydrogen atom from the hydrocarbon skeleton. When there are distinct types of hydrogen atoms in the reactant, the hydrogen removed is one that will lead to the formation of the most stable radical. The relative stability of radicals changes in the same way as the stability of the carbocation. This decides the major product of the reaction.

Expert Solution
Check Mark

Answer to Problem 25.40P

The major product of the reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  1

The complete mechanism of the reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  2

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  3

Irradiation of the reaction mixture with UV light leads to the formation of two bromine radicals through homolysis in the initiation step.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  4

One of the bromine radicals reacts with the hydrocarbon and extracts a hydrogen atom (radical). There are three distinct types of – hydrogen atoms in the substrate; therefore three types of radicals can be produced in this step – 1o, 2o or 3o. The most stable radical of these is the 3o radical; therefore, it is produced preferentially in the first of the propagation steps.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  5

The 3o alkyl radical produced in this step reacts with a molecule of bromine to produce the brominated product and another bromine radical through homolysis of the bromine molecule.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  6

The bromine radical can continue the radical reaction chain or combine with another alkyl or bromine radical, in which case, the particular chain is terminated.

Thus, the complete reaction mechanism for the reaction can be drawn as

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  7

The major product of reaction will be

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  8

Conclusion

The major product of the reaction is determined on the basis of the stability of the alkyl radical produced after the initiation step.

Interpretation Introduction

(b)

Interpretation:

The major product of the given reaction is to be predicted. A complete, detailed mechanism for the given reaction is to be provided.

Concept introduction:

Halogen molecules undergo homolysis when irradiated with UV light to produce two radical atoms. Homolysis is the type of bond dissociation where each atom gets one of the bond electrons. An unpaired electron present in the radical of atom or molecule. It is a highly reactive and unstable species. This homolytic dissociation is called initiation.

Being reactive, a radical will react with other species present. This produces another radical, which can continue the reaction is a step called propagation. The reaction is generally branched as each initiation step produces two radicals. Propagation steps continue as long as radicals are present and may be terminated by a reaction between two radicals.

The reaction generally results in substitution, often of a hydrogen atom from the hydrocarbon skeleton. When there are distinct types of hydrogen atoms in the reactant, the hydrogen removed is one that will lead to the formation of the most stable radical. The relative stability of radicals changes in the same way as the stability of the carbocation. This decides the major product of the reaction.

Expert Solution
Check Mark

Answer to Problem 25.40P

The major product of the reaction is a mixture of four isomers:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  9

The complete mechanism for the reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  10

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  11

Irradiation of NBS (N-bromosuccinimide) produces bromine radicals in the initiation step.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  12

The bromine radical reacts with the substrate in the first of the propagation steps to produce the most stable radical. The most stable alkyl radical that can be produced on the extraction of a hydrogen atom is the allylic radical. Therefore, the hydrogen in the allylic position is extracted preferentially. The allylic radical is resonance stabilized.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  13

In the next propagation step, each resonance form of the allylic radical reacts with another bromine radical to produce the product. Since the cyclic radical is unsymmetric, each one produces a mixture of two isomers, giving a total of four isomeric products.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  14

The complete mechanism for the reaction can be drawn as

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  15

Thus the major product of the reaction is a mixture of four isomers.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  16

Conclusion

The major product of the reaction is determined on the basis of the stability of the alkyl radical produced after the initiation step.

Interpretation Introduction

(c)

Interpretation:

The major product of the given reaction is to be predicted. A complete, detailed mechanism for the given reaction is to be provided.

Concept introduction:

Under conditions that do not result in homolysis, halogen molecules, particularly chlorine and bromine, behave as electrophiles if electron-rich species such as unsaturated hydrocarbons are present. Unsaturated hydrocarbons react with the halogen molecule in an electrophilic addition reaction. The halogen molecule is added across the double bond, replacing the CC π bond with two CBr σ bonds.

The addition is a two-step reaction. In the first step, a cyclic halonium intermediate is produced. In the second step, the halide ion formed in the first step acts as a nucleophile and adds to one of the carbons of the halonium intermediate to produce the vicinal dihalide product. Since the cyclic halonium intermediate is planar, the halide addition in the second step can occur from above or below the plane of the ring, giving a mixture of two isomers if the alkene is unsymmetric.

Expert Solution
Check Mark

Answer to Problem 25.40P

The major product of the reaction is a mixture of vicinal dibromo isomers.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  17

The whole mechanism of the reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  18

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  19

In the absence of UV irradiation, bromine will not undergo homolysis to produce radicals. Therefore, the reaction will be a simple electrophilic addition.

The π pair of the double bond will act as a nucleophile and attack one of the bromine atoms from the molecule, leading to heterolysis of the BrBr bond and formation of a cyclic bromonium cation intermediate.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  20

In the second step, the bromide ion produced in the first step will add to one of the carbons of the bromonium ring. Since the bromonium ring is planar, this addition can be from above or below the plane of the ring. This will lead to the formation of a mixture of isomers.

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  21

Thus, the whole mechanism for the reaction can be drawn as

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  22

The major product of the reaction will be

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 25, Problem 25.40P , additional homework tip  23

Conclusion

The major product of the reaction is determined on the basis of an electrophilic addition of the bromine molecule to the double bond.

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Chapter 25 Solutions

Organic Chemistry: Principles and Mechanisms (Second Edition)

Ch. 25 - Prob. 25.11PCh. 25 - Prob. 25.12PCh. 25 - Prob. 25.13PCh. 25 - Prob. 25.14PCh. 25 - Prob. 25.15PCh. 25 - Prob. 25.16PCh. 25 - Prob. 25.17PCh. 25 - Prob. 25.18PCh. 25 - Prob. 25.19PCh. 25 - Prob. 25.20PCh. 25 - Prob. 25.21PCh. 25 - Prob. 25.22PCh. 25 - Prob. 25.23PCh. 25 - Prob. 25.24PCh. 25 - Prob. 25.25PCh. 25 - Prob. 25.26PCh. 25 - Prob. 25.27PCh. 25 - Prob. 25.28PCh. 25 - Prob. 25.29PCh. 25 - Prob. 25.30PCh. 25 - Prob. 25.31PCh. 25 - Prob. 25.32PCh. 25 - Prob. 25.33PCh. 25 - Prob. 25.34PCh. 25 - Prob. 25.35PCh. 25 - Prob. 25.36PCh. 25 - Prob. 25.37PCh. 25 - Prob. 25.38PCh. 25 - Prob. 25.39PCh. 25 - Prob. 25.40PCh. 25 - Prob. 25.41PCh. 25 - Prob. 25.42PCh. 25 - Prob. 25.43PCh. 25 - Prob. 25.44PCh. 25 - Prob. 25.45PCh. 25 - Prob. 25.46PCh. 25 - Prob. 25.47PCh. 25 - Prob. 25.48PCh. 25 - Prob. 25.49PCh. 25 - Prob. 25.50PCh. 25 - Prob. 25.51PCh. 25 - Prob. 25.52PCh. 25 - Prob. 25.53PCh. 25 - Prob. 25.54PCh. 25 - Prob. 25.55PCh. 25 - Prob. 25.56PCh. 25 - Prob. 25.57PCh. 25 - Prob. 25.58PCh. 25 - Prob. 25.59PCh. 25 - Prob. 25.60PCh. 25 - Prob. 25.61PCh. 25 - Prob. 25.62PCh. 25 - Prob. 25.63PCh. 25 - Prob. 25.64PCh. 25 - Prob. 25.65PCh. 25 - Prob. 25.66PCh. 25 - Prob. 25.67PCh. 25 - Prob. 25.68PCh. 25 - Prob. 25.69PCh. 25 - Prob. 25.70PCh. 25 - Prob. 25.71PCh. 25 - Prob. 25.72PCh. 25 - Prob. 25.73PCh. 25 - Prob. 25.74PCh. 25 - Prob. 25.75PCh. 25 - Prob. 25.76PCh. 25 - Prob. 25.77PCh. 25 - Prob. 25.1YTCh. 25 - Prob. 25.2YTCh. 25 - Prob. 25.3YTCh. 25 - Prob. 25.4YTCh. 25 - Prob. 25.5YTCh. 25 - Prob. 25.6YTCh. 25 - Prob. 25.7YTCh. 25 - Prob. 25.8YTCh. 25 - Prob. 25.9YTCh. 25 - Prob. 25.10YTCh. 25 - Prob. 25.11YTCh. 25 - Prob. 25.12YTCh. 25 - Prob. 25.13YTCh. 25 - Prob. 25.14YTCh. 25 - Prob. 25.15YTCh. 25 - Prob. 25.16YTCh. 25 - Prob. 25.17YTCh. 25 - Prob. 25.18YTCh. 25 - Prob. 25.19YTCh. 25 - Prob. 25.20YTCh. 25 - Prob. 25.21YTCh. 25 - Prob. 25.22YTCh. 25 - Prob. 25.23YTCh. 25 - Prob. 25.24YT
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