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Write a stepwise mechanism for each of the following reactions. Explain why a more stable
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- If phenoxide ion is allowed to react with 1-bromopentane, pentyl phenyl ether is obtained. However, if cyclohexane is used as the alkyl halide, the major products are phenol and cyclohexene. Explain how these products were formed.arrow_forwardNAME and DRAW the STRUCTURE of ALL the alkenes which could undergo catalytic hydrogenation at 900°C to form methylcyclopentane. Circle the alkene with the HIGHEST stability and X the alkene with the HIGHEST heat of hydrogenation. Give reasons for your choice.arrow_forwardDraw the structure for an alkene that gives the following reaction product. CH3 -CHCH₂CH3 ? CH₂l2, Zn/Cuarrow_forward
- CCH H20, H2SO4 H9SO4 CH3 Alkynes do not react directly with aqueous acid as do alkenes, but will do so in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the OH group adds to the more highly substituted carbon and the H adds to the less highly substituted carbon. The initial product of the reaction is a vinyl alcohol, also called an enol. The enol immediately rearranges to a more stable ketone via tautomerization. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions Hjö: -CH3 -CH3 H3O*arrow_forwardC=CH H20, H2SO4 H9SO4 CH3 Alkynes do not react directly with aqueous acid as do alkenes, but will do so in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the OH group adds to the more highly substituted carbon and the H adds to the less highly substituted carbon. The initial product of the reaction is a vinyl alcohol, also called an enol. The enol immediately rearranges to a more stable ketone via tautomerization. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions H-OH HO: Hjö: C=CH c=CH Hö Hg Hgarrow_forwardIdentify two alkenes that react with HBr to form 1-bromo-1-methylcyclohexane without undergoing a carbocation rearrangement.arrow_forward
- Which represents an efficient synthetic route to go from an alkane to an alkene? O elimination with NaNH2, followed by a water workup O anti-Markovnikov hydrohalogenation, followed by elimination O radical bromination, followed by elimination O hydration, followed by elimination O hydration, followed by ozonolysis of the double bondarrow_forwardIntermidiates and reagents of the following two final products from a Acetylene Note: the final products are different reactions but use the same starting material.arrow_forwardDraw the alkene that would react with the reagent given to account for the product formed. CH3 HCI CH3 CHCCH, ? + Či CH3 You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one.arrow_forward
- Nucleophilic substitution happens on compounds having nucleophilic groups as leaving groups. The rule is, the weaker the basicity of a group of the substrate, the better is its leaving ability. In these substitution reactions, the basicity of leaving group must be less than the incoming nucleophilic group. Nucleophilic substitution reaction at sp3-hybridized carbon is either bimolecular (SN2) or unimolecular. Bimolecular reaction takes place in single step, involving transition state intermediate. In SN2 reaction, inversion in configuration occurs. In case of optically active alkyl halides, the inversion in configuration is called Walden inversion. SN2 reaction is preferred if the compound has less steric hindrance. On the other hand, unimolecular (SN1) reaction involves two steps and a carbonium ion intermediate. Optically active substrates give racemic mixture in these type of reactions. Which of the following will produce enantiomeric pair on treatment with HOH? " I ÇH, C,Hs-C-Br…arrow_forwardChoose the correct answer on the following questions: Zaitsev’s rule states that: the alkene with less alkyl substituents is the major product in base-induced elimination reactions the more highly substituted carbocation is formed as the intermediate rather than the less highly substituted one the alkene with more alkyl substituents is the major product in base-induced elimination reactions the less highly substituted carbocation is formed as the intermediate rather than the more highly substituted one Markovnikov rule states that: the alkene with less alkyl substituents is the major product in base-induced elimination reactions the more highly substituted carbocation is formed as the intermediate rather than the less highly substituted one the alkene with more alkyl substituents is the major product in base-induced elimination reactions the less highly substituted carbocation is formed as the intermediate rather than the more highly substituted one The Hoffman rule states that:…arrow_forwardWrite a stepwise mechanism for each of the following reactions. Explain why a more stable alkyne (but-2-yne) is isomerized to a less stable alkyne (but-1-yne), but under similar conditions. 2.5-dimethylhex-3-yne forms 2,5-dimethylhexa-2.3-diene.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