Concept explainers
Interpretation:
For the given reaction that produces both substitution and elimination products at room temperature, it is be determined which product will be the major product if the temperature at which the reaction is run is raised and which product will be the major product if the temperature of the reaction is lowered.
Concept introduction:
Most of the reaction often produces both substitution as well as elimination products. When substitution and elimination reactions are both favored under a specific temperature, it is often possible to change the outcome by changing the temperature under which the reactions are carried out. Substitution and elimination products are formed in roughly equal amounts at the lower temperatures. Increasing the temperature of the reaction tends to promote elimination more than substitution because there is significantly more disorder in the products of an elimination reaction than in the products of a competing substitution reaction.
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Chapter 9 Solutions
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- The overall dehydration reaction we are performing (the dehydration of cyclohexanol) is reversible and actually slightly endothermic. How can we shift the equilibrium to the product side, assuming we can not limitlessly increase the temperature? Increased magnetic stirring le Chatelier's principle Increased amount of catalyst Extended reaction timearrow_forwardBy taking into account electronegativity differences, draw the products formed by heterolysis of the carbon–heteroatom bond in each molecule. Classify the organic reactive intermediate as a carbocation or a carbanion.arrow_forwardWhat about the second step in the electrophilic addition of HCl to an alkene-the reaction of chloride ion with the carbocation intermediate? Is this step exergonic or endergonic? Does the transition state for this second step resemble the reactant (carbocation) or the product (alkyl chloride)? Make a rough diagram of what the transition state may look like?arrow_forward
- 2. The following carbocation is generated as an intermediate in the addition of H-Br to an alkene. Draw the structure of all possible alkenes that could have formed this intermediate.arrow_forwardFor the given SN2 reaction, draw the organic and inorganic products of the reaction, and identify the nucleophile, substrate, and leaving group. Organic product Inorganic product Brarrow_forwardFor a reversible reaction at an equilibrium: The rate of the reaction forward is the same as that of the reaction in reverse. The rate constant for the reaction forward is the same as that for the reverse. The free energy of substrate is the same as that of product. No conversion of substrate to product, and product to substrate occurs. Position of an equilibrium does not depend on substrate and product concentration , only on the magnitude of equilibrium constant.arrow_forward
- Write the reagents and reaction conditions that are necessary for each of the following transformations.arrow_forwardThe equilibrium constant for reversible reactions is equal to a) [Products]/[Reagents] [Reagents] [Reagents] [Productsb) [Reagents]/[Products] [Products]/[Reagents].c) [Products]/[Products].arrow_forwardH3C CH3 H3C NA C→XT Br Br₂ CH₂Cl₂ H3C Electrophilic addition of bromine, Br₂, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermediate known as a bromonium ion. The reaction occurs in an anhydrous solvent such as CH₂Cl₂. CH3 Br In the second step of the reaction, bromide is the nucleophile and attacks at one of the carbons of the bromonium ion to yield the product. Due to steric clashes, the bromide ion always attacks the carbon from the opposite face of the bromonium ion so that a product with anti stereochemistry is formed. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions Br CH3 H3C CH3arrow_forward
- CH3 CH3 Br- Br2 .CH3 CH2Cl2 CH3 H3C H3C Br Electrophilic addition of bromine, Brɔ, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermediate known as a bromonium ion. The reaction occurs in an anhydrous solvent such as CH,Cl,. In the second step of the reaction, bromide is the nucleophile and attacks at one of the carbons of the bromonium ion to yield the product. Due to steric clashes, the bromide ion always attacks the carbon from the opposite face of the bromonium ion so that a product with anti stereochemistry is formed. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions CH3 CH3 CH3 CH3 H3C H3C :Br: :Br:arrow_forwardwrite the main product that will be formed as a result of each reactionarrow_forwardSynthesize the products by drawing out reagents and intermediates along the way.arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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