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
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Chapter 19, Problem 19.41P
Interpretation Introduction

(a)

Interpretation:

The organometallic compound that would be produced by the given reaction is to be drawn.

Concept introduction:

To form a bond between two carbons of like charge, a separate reaction that first reverses the charge (or polarity) at one of the carbons is carried out. Thus, one carbon atom would become electron rich while the other would remain electron poor. This general idea of reversing a charge at a particular atom is called umpolung. In the alkyl halide reactant, the C atom bonded to the halogen atom bears a partial positive charge and is relatively electron poor. By contrast, that C atom has to become electron rich in the organometallic compound produced. An alkyllithium reagent (RLi) can be synthesized from an alkyl bromide by treating it with solid lithium in ether.

Expert Solution
Check Mark

Answer to Problem 19.41P

The organometallic compound that would be produced by the given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  1

Explanation of Solution

The given reaction is,

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  2

In the above reaction, the C atom of the C-Br bond is electron-poor. When the C-Br bond is bonded to the metal, the carbon will become electron-rich. An alkyllithium reagent (RLi) can be synthesized from an alkyl bromide by treating it with solid lithium in ether. Therefore, the organometallic compound that would be produced by the given reaction is as shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  3

Conclusion

The organometallic compound that would be produced by the given reaction is drawn by converting the electron-poor carbon of C-Br bond to an electron-rich carbon.

Interpretation Introduction

(b)

Interpretation:

The organometallic compound that would be produced by the given reaction is to be drawn.

Concept introduction:

To form a bond between two carbons of like charge, a separate reaction that first reverses the charge (or polarity) at one of the carbons is carried out. Thus, one carbon atom would become electron rich while the other would remain electron poor. This general idea of reversing a charge at a particular atom is called umpolung. In the alkyl halide reactant, the C atom bonded to the halogen atom bears a partial positive charge and is relatively electron poor. By contrast, that C atom has to become electron rich in the organometallic compound produced. An alkyl bromide can be converted to a Grignard reagent (RMgBr) simply by treating it with solid magnesium in an ether solvent such as tetrahydrofuran (THF).

Expert Solution
Check Mark

Answer to Problem 19.41P

The organometallic compound that would be produced by the given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  4

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  5

In the above reaction, the C atom of the C-Br bond is electron-poor. When the C-Br bond is bonded to the metal, the carbon will become electron-rich. An alkyl bromide can be converted to a Grignard reagent (RMgBr) simply by treating it with solid magnesium in an ether solvent such as tetrahydrofuran (THF). So the C-Br bond will become C-Mg bond. Therefore, the organometallic compound that would be produced by the given reaction is as shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  6

Conclusion

The organometallic compound that would be produced by the given reaction is drawn by converting the electron-poor carbon of C-Br bond to an electron-rich carbon.

Interpretation Introduction

(c)

Interpretation:

The organometallic compound that would be produced by the given reaction is to be drawn.

Concept introduction:

To form a bond between two carbons of like charge, a separate reaction that first reverses the charge (or polarity) at one of the carbons is carried out. Thus, one carbon atom would become electron rich while the other would remain electron poor. This general idea of reversing a charge at a particular atom is called umpolung. In the alkyl halide reactant, the C atom bonded to the halogen atom bears a partial positive charge and is relatively electron poor. By contrast, that C atom has to become electron rich in the organometallic compound produced. An alkyl bromide can be converted to a Grignard reagent (RMgBr) simply by treating it with solid magnesium in an ether solvent such as tetrahydrofuran (THF).

Expert Solution
Check Mark

Answer to Problem 19.41P

The organometallic compound that would be produced by the given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  7

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  8

In the above reaction, the C atom of the C-Br bond is electron-poor. When the C-Br bond is bonded to the metal, the carbon will become electron-rich. An alkyl bromide can be converted to a Grignard reagent (RMgBr) simply by treating it with solid magnesium in an ether solvent such as tetrahydrofuran (THF). So the C-Br bond will become C-Mg bond. Therefore, the organometallic compound that would be produced by the given reaction is as shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  9

Conclusion

The organometallic compound that would be produced by the given reaction is drawn by converting the electron-poor carbon of C-Br bond to an electron-rich carbon.

Interpretation Introduction

(d)

Interpretation:

The organometallic compound that would be produced by the given reaction is to be drawn.

Concept introduction:

To form a bond between two carbons of like charge, a separate reaction that first reverses the charge (or polarity) at one of the carbons is carried out. Thus, one carbon atom would become electron rich while the other would remain electron poor. This general idea of reversing a charge at a particular atom is called umpolung. In the alkyl halide reactant, the C atom bonded to the halogen atom bears a partial positive charge and is relatively electron poor. By contrast, that C atom has to become electron rich in the organometallic compound produced. An alkyl bromide can be converted to a Grignard reagent (RMgX) simply by treating it with solid magnesium in an ether solvent.

Expert Solution
Check Mark

Answer to Problem 19.41P

The organometallic compound that would be produced by the given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  10

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  11

In the above reaction, the C atom of the C-Cl bond is electron-poor. When the C-Br bond is bonded to the metal, the carbon will become electron-rich. An alkyl chloride can be converted to a Grignard reagent (RMgBr) simply by treating it with solid magnesium in an ether. So the C-Cl bond will become C-Mg bond. Therefore, the organometallic compound that would be produced by the given reaction is as shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  12

Conclusion

The organometallic compound that would be produced by the given reaction is drawn by converting the electron-poor carbon of C-Br bond to an electron-rich carbon.

Interpretation Introduction

(e)

Interpretation:

The organometallic compound that would be produced by the given reaction is to be drawn.

Concept introduction:

To form a bond between two carbons of like charge, a separate reaction that first reverses the charge (or polarity) at one of the carbons is carried out. Thus, one carbon atom would become electron rich while the other would remain electron poor. This general idea of reversing a charge at a particular atom is called umpolung. In the alkyl halide reactant, the C atom bonded to the halogen atom bears a partial positive charge and is relatively electron poor. By contrast, that C atom has to become electron rich in the organometallic compound produced. A lithium dialkylcuprate is synthesized from the corresponding alkyllithium reagent by treating it with copper(I) iodide, CuI.

Expert Solution
Check Mark

Answer to Problem 19.41P

The organometallic compound that would be produced by the given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  13

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  14

In the above reaction, the C atom of the C-Br bond is electron-poor. When the C-Br bond is bonded to the metal, the carbon will become electron-rich. Lithium dialkylcuprate is synthesized from the corresponding alkyllithium reagent by treating it with copper(I) iodide, CuI. So the C-Br bond will become C-CuLi bond. Therefore, the organometallic compound that would be produced by the given reaction is as shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  15

Conclusion

The organometallic compound that would be produced by the given reaction is drawn by converting the electron-poor carbon of C-Br bond to an electron-rich carbon.

Interpretation Introduction

(f)

Interpretation:

The organometallic compound that would be produced by the given reaction is to be drawn.

Concept introduction:

To form a bond between two carbons of like charge, a separate reaction that first reverses the charge (or polarity) at one of the carbons is carried out. Thus, one carbon atom would become electron rich while the other would remain electron poor. This general idea of reversing a charge at a particular atom is called umpolung. In the alkyl halide reactant, the C atom bonded to the halogen atom bears a partial positive charge and is relatively electron poor. By contrast, that C atom has to become electron rich in the organometallic compound produced. An alkyllithium reagent (RLi) can be synthesized from an alkyl bromide by treating it with solid lithium in ether solvent such as THF.

Expert Solution
Check Mark

Answer to Problem 19.41P

The organometallic compound that would be produced by the given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  16

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  17

In the above reaction, the C atom of the C-Br bond is electron-poor. When the C-Br bond is bonded to the metal, the carbon will become electron-rich. An alkyllithium reagent (RLi) can be synthesized from an alkyl bromide by treating it with solid lithium in ether solvent such as THF. So the C-Br bond will become C-Li bond. Therefore the organometallic compound that would be produced by the given reaction is as shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  18

Conclusion

The organometallic compound that would be produced by the given reaction is drawn by converting the electron-poor carbon of C-Br bond to an electron-rich carbon.

Interpretation Introduction

(g)

Interpretation:

The organometallic compound that would be produced by the given reaction is to be drawn.

Concept introduction:

To form a bond between two carbons of like charge, a separate reaction that first reverses the charge (or polarity) at one of the carbons is carried out. Thus, one carbon atom would become electron rich while the other would remain electron poor. This general idea of reversing a charge at a particular atom is called umpolung. In the alkyl halide reactant, the C atom bonded to the halogen atom bears a partial positive charge and is relatively electron poor. By contrast, that C atom has to become electron rich in the organometallic compound produced. A lithium dialkylcuprate is synthesized from the corresponding alkyllithium reagent by treating it with copper(I) iodide, CuI.

Expert Solution
Check Mark

Answer to Problem 19.41P

The organometallic compound that would be produced by the given reaction is shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  19

Explanation of Solution

The given reaction is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  20

In the above reaction, the C atom of the C-Cl bond is electron-poor. When the C-Br bond is bonded to the metal, the carbon will become electron-rich. A lithium dialkylcuprate is synthesized from the corresponding alkyllithium reagent by treating it with copper(I) iodide, CuI. Therefore, the organometallic compound that would be produced by the given reaction is as shown below:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 19, Problem 19.41P , additional homework tip  21

Conclusion

The organometallic compound that would be produced by the given reaction is drawn by converting the electron-poor carbon of C-Br bond to an electron-rich carbon.

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Chapter 19 Solutions

Organic Chemistry: Principles and Mechanisms (Second Edition)

Ch. 19 - Prob. 19.11PCh. 19 - Prob. 19.12PCh. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Prob. 19.21PCh. 19 - Prob. 19.22PCh. 19 - Prob. 19.23PCh. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - Prob. 19.29PCh. 19 - Prob. 19.30PCh. 19 - Prob. 19.31PCh. 19 - Prob. 19.32PCh. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Prob. 19.37PCh. 19 - Prob. 19.38PCh. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.41PCh. 19 - Prob. 19.42PCh. 19 - Prob. 19.43PCh. 19 - Prob. 19.44PCh. 19 - Prob. 19.45PCh. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Prob. 19.53PCh. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Prob. 19.66PCh. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - Prob. 19.70PCh. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - Prob. 19.73PCh. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Prob. 19.77PCh. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.1YTCh. 19 - Prob. 19.2YTCh. 19 - Prob. 19.3YTCh. 19 - Prob. 19.4YTCh. 19 - Prob. 19.5YTCh. 19 - Prob. 19.6YTCh. 19 - Prob. 19.7YTCh. 19 - Prob. 19.8YTCh. 19 - Prob. 19.9YTCh. 19 - Prob. 19.10YT
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