Concept explainers
(a)
Interpretation:
The
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
Answer to Problem 19.41P
The organometallic compound that would be produced by the given reaction is
Explanation of Solution
The given reaction is,
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:
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.
(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).
Answer to Problem 19.41P
The organometallic compound that would be produced by the given reaction is shown below:
Explanation of Solution
The given reaction is
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:
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.
(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).
Answer to Problem 19.41P
The organometallic compound that would be produced by the given reaction is shown below:
Explanation of Solution
The given reaction is
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:
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.
(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.
Answer to Problem 19.41P
The organometallic compound that would be produced by the given reaction is shown below:
Explanation of Solution
The given reaction is
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:
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.
(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.
Answer to Problem 19.41P
The organometallic compound that would be produced by the given reaction is shown below:
Explanation of Solution
The given reaction is
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:
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.
(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.
Answer to Problem 19.41P
The organometallic compound that would be produced by the given reaction is shown below:
Explanation of Solution
The given reaction is
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:
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.
(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.
Answer to Problem 19.41P
The organometallic compound that would be produced by the given reaction is shown below:
Explanation of Solution
The given reaction is
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:
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)
- Three of these reactions proceed using the same intermediate. Draw the structure of that intermediate.arrow_forwardDraw the organic product(s) for each of the following reactions. If more than one product is expected indicate the major product.arrow_forwardExplain why this reaction is not possible that it cannot yield the provided productarrow_forward
- Predict the major product of the reaction.arrow_forwardGiven the general reaction coordinate diagram for an El reaction, what does the 1st transition state refer to? TS E TS, SM Reaction Coordinate An activated complex between the leaving group and carbocationic species. An activated complex between the nucleophile and carbocationic species. None of these O An activated complex between the base and carbocationic species.arrow_forwardWhich reaction intermediate (A or B) is more likely to form in the epoxide ring opening reaction? Reactions prefer to react through lower energy intermediates. How might you justify one structure being lower in energy than the other? Draw structures as part of your explanation why.arrow_forward
- First, draw the structure of the alkyl bromide that will ONLY give the alkene shown in an elimination reaction; then, draw the structure of the alkyl bromide that will give a mixture of alkene products in an elimination reaction.arrow_forwardWhen this compound is treated with a base, elimination occurs. What is the product of this elimination? Explain your reasoning.arrow_forwardDraw a reaction coordinate diagram in which the structure of the transition site is more similar to the product of the reaction than to the reactant.arrow_forward
- Based on Markovnikov’s rule, draw only the Major product for the following asymmetrical addition (hydration) reactionarrow_forward. One of these reactions occurs rapidly while the other is so slow that substitution products are not observed. Determine which reaction is which and explain the difference in rate using structural drawing and a few words. Br + 'Br +arrow_forwardNo reaction occurs when benzaldehyde and propenenitrile (acrylonitrile) are combined. In the presence of a catalytic amount of NaCN, however, the reaction shown here takes place. Draw a complete, detailed mechanism to account for these results. Hint: See Problem 18.70. NaCN TH. + CN `CN HOH,0, ELOHarrow_forward
- EBK A SMALL SCALE APPROACH TO ORGANIC LChemistryISBN:9781305446021Author:LampmanPublisher:CENGAGE LEARNING - CONSIGNMENTOrganic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning