(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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EP ORGANIC CHEMISTRY-OWL V2 ACCESS
- 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_forwardDraw a structural formula for the alkene with the molecular formula C5H10 that reacts with Br2 to give each product.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_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 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_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 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_forwardSelect the reagent that you would use to convert an alkene to a Markovnikov (more substituted) alcohol without possibility of carbocation rearrangement. o H2SO4, H2O OBH3-THF, then H2O2, H2O, NaOH Hg(OAc)2, H2O, then NaBH4 o OsO4, TBHP, H2Oarrow_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 structure of the four allylic halides formed when 3-methylcyclohexene undergoes allylic halogenation with NBS + hv.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
- Ignoring stereochemistry, draw the alkylborane formed from the addition of one equivalent of BH3 to the alkene. The alkylborane formed in Part 1 is further treated with H2O2 and HO−. Draw the two stereoisomers of the final product of the reaction. Include stereochemistry where relevant. How many stereocenters are formed from the reaction? What is the relationship between the stereoisomers?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_forwardAnswer the following questions related to organic reactions a) Identify the nucleophile, electrophile and leaving group when bromocyclohexane reacts with sodium cyanide. b) Draw the expected product(s) resulting from the following SN2 reaction: 3-iodoheptane and KOH c) Draw the expected product(s) resulting from the following SN1 reaction: 1-chloro-1-phenylethane and (CH3)2NHarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning