(a)
Interpretation:
The mechanism of given reaction has to be described.
Concept Introduction:
The
The first step of
Frist step is the slow step also rate determining step so the rate of the reaction is depends on the concentration of substrate only.
Nucleophile attacks the both front and back side of carbocation in
Order of the substrate that favored in
Poor nucleophiles are favored the
Polar aprotic or good ionizing solvent are favored the
Good leaving group is favored the
(b)
Interpretation:
The mechanism of given reaction has to be described.
Concept Introduction:
The rate of the reaction is depends on a single reactant in reaction is known as
The first step of
Frist step is the slow step also rate determining step so the rate of the reaction is depends on the concentration of substrate only.
Nucleophile attacks the both front and back side of carbocation in
Order of the substrate that favored in
Poor nucleophiles are favored the
Polar aprotic or good ionizing solvent are favored the
Good leaving group is favored the
(c)
Interpretation:
The mechanism of given reaction has to be described.
Concept Introduction:
The rate of the reaction is depends on a single reactant in reaction is known as
The first step of
Frist step is the slow step also rate determining step so the rate of the reaction is depends on the concentration of substrate only.
Nucleophile attacks the both front and back side of carbocation in
Order of the substrate that favored in
Poor nucleophiles are favored the
Polar aprotic or good ionizing solvent are favored the
Good leaving group is favored the
(d)
Interpretation:
The mechanism of given reaction has to be described.
Concept Introduction:
The rate of the reaction is depends on a single reactant in reaction is known as
The first step of
Frist step is the slow step also rate determining step so the rate of the reaction is depends on the concentration of substrate only.
Nucleophile attacks the both front and back side of carbocation in
Order of the substrate that favored in
Poor nucleophiles are favored the
Polar aprotic or good ionizing solvent are favored the
Good leaving group is favored the
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OWLv2 with MindTap Reader, 1 term (6 months) Printed Access Card for Brown/Iverson/Anslyn/Foote's Organic Chemistry, 8th Edition
- Alcohols are important for organic synthesis, especially in situations involving alkenes. The alcohol might be the desired product, or the OH group might be transformed into another functional group via halogenation, oxidation, or perhaps conversion to a sulfonic ester derivative. Formation of an alcohol from an alkene is particularly powerful because conditions can be chosen to produce either the Markovnikov or non-Markovnikov product from an unsymmetrical alkene. Using your reaction roadmap as a guide, show how to convert 4-methyl-1-pentene into 5-methylhexanenitrile. You must use 4-methyl-1-pentene and sodium cyanide as the source of all carbon atoms in the target molecule. Show all reagents needed and all molecules synthesized along the way.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_forwardDraw a structural formula for the alkene with the molecular formula C5H10 that reacts with Br2 to give each product.arrow_forward
- Ethers can be prepared by reaction of an alkoxide or phenoxide ion with a primary alkyl halide. Draw the structure of the expected organic product of the reaction of iodomethane with the following alkoxide ion: CH3 H3C O Na You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. • Do not include lone pairs in your answer. They will not be considered in the grading. • Do not include counter-ions, e.g., Na", I, in your answer. орy вste ChemDoodlearrow_forwardFour alkenes are formed from the E1 reaction of 3-bromo-2,3-dimethylpentane and methanol. Draw the structures of the alkenes and rank them according to the amount that would be formed.arrow_forwardEthers can be prepared by reaction of an alkoxide or phenoxide ion with a primary alkyl halide. Draw the structure of the expected organic product of the reaction of iodoethane with the following alkoxide ion: H3C CH3 + Na You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading. • Do not include counter-ions, e.g., Na", I, in your answer. P opy aste [*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_forwardDraw the line-angle formula of the enol formed in the following alkyne hydration reaction and then draw the structural formula of the carbonyl compound with which this enol is in equilibrium. -C=CH 1. (sia)2BH 2. NaOH/H₂O₂ an enol carbonyl compound • You do not have to consider stereochemistry. • Draw both the enol and the carbonyl forms. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. • Separate structures using the sign from the drop-down menu. ChemDoodleⓇ <arrow_forwardThe compound below is treated with chlorine in the presence of light. H3C CH3 H3C° `CH2CH3 Draw the structure for the organic radical species produced by reaction of the compound with a chlorine atom. Assume reaction occurs at the weakest C-H bond. • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms.arrow_forward
- 5A In the following reactions, mixtures of alkenes and ethyl ethers are formed. Draw their structures. Explain which is or are likely to be the main product(s) in each reaction. In case of formation of two isomers of alkenes, explain which is formed in greater proportion CH3 ofi H3C- -Br CH3 EtOHarrow_forwardAlkynes 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 -X티 Hö: H-O -CH3 -CH3 H30*arrow_forwardWhen 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1- butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base? base Br A. Less than 45% B. C. 45% Between 45% and 70% D. More than 70%arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning