Concept explainers
Videos
(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.
Want to see more full solutions like this?
Chapter 19 Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
- О δα HO- H -Br δα HO-- + + -Br [B] 8+ HO- -Br δα नarrow_forward1/2 - 51% + » GAY Organic Reactions Assignment /26 Write the type of reaction that is occurring on the line provided then complete the reaction. Only include the major products and any byproducts (e.g. H₂O) but no minor products. Please use either full structural diagrams or the combination method shown in the lesson. Skeletal/line diagrams will not be accepted. H3C 1. 2. CH3 A Acid OH Type of Reaction: NH Type of Reaction: + H₂O Catalyst + HBr 3. Type of Reaction: H3C 4. Type Reaction: 5. H3C CH2 + H2O OH + [0] CH3 Type of Reaction: 6. OH CH3 HO CH3 + Type of Reaction: 7. Type of Reaction: + [H]arrow_forwardhumbnai Concentration Terms[1].pdf ox + New Home Edit Sign in Comment Convert Page Fill & Sign Protect Tools Batch +WPS A Free Trial Share Inter Concreting Concentration forms. Hydrogen peroxide is a powerful oxidizing agent wed in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous sulation of H2O2 is 30% by mass and has density of #liligime calculat the Ⓒmolality ⑥mole fraction of molarity. 20 9. B. A sample of Commercial Concentrated hydrochloric ETarrow_forward
- If a reaction occurs, what would be the major products? Please include a detailed explanation as well as a drawing showing how the reaction occurs and what the final product is.arrow_forwardWould the following organic synthesis occur in one step? Add any missing products, required catalysts, inorganic reagents, and other important conditions. Please include a detailed explanation and drawings showing how the reaction may occur in one step.arrow_forward(a) Sketch the 'H NMR of the following chemical including the approximate chemical shifts, the multiplicity (splitting) of all signals and the integration (b) How many signals would you expect in the 13C NMR? CH3arrow_forward
- Draw the Show the major and minor product(s) for the following reaction mechanisms for both reactions and show all resonance structures for any Explain why the major product is favoured? intermediates H-Brarrow_forwardChoose the right answerarrow_forward8. What is the major product of the following reaction? KMnO4 b a TOH OH OH C d OH "OH HO OH OHarrow_forward
- Choose the right answerarrow_forward3. Draw ALL THE POSSBILE PRODUCTS AND THE MECHANISMS WITH ALL RESONANCE STRUCTURES. Explain using the resonance structures why the major product(s) are formed over the minor product(s). H₂SO4, HONO CHarrow_forward7. Provide the product(s), starting material(s) and/or condition(s) required for the No mechanisms required. below reaction HO + H-I CI FO Br2, FeBr3 O I-Oarrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
EBK A SMALL SCALE APPROACH TO ORGANIC LChemistryISBN:9781305446021Author:LampmanPublisher:CENGAGE LEARNING - CONSIGNMENT