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 abillity. 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,Hg-C-Br C,Hs-C-CI CH3 IA H CH3 C,Hs-C-I CH3-C-Br CH C,H,

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 abillity. 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,Hg-C-Br C,Hs-C-CI CH3 IA H CH3 C,Hs-C-I CH3-C-Br CH C,H,

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter10: Alcohols

Section: Chapter Questions

Problem 10.52P: Alcohols are important for organic synthesis, especially in situations involving alkenes. The...

See similar textbooks

Similar questions

If anhydrides react like acid chlorides with the nucleophiles, draw the products formed when each of the following nucleophiles reacts with benzoic anhydride [(C6H5CO)2O]: (a) CH3MgBr (2 equiv), then H2O; (b) LiAlH4, then H2O; (c) LiAlH[OC(CH3)3]3, then H2O.
H3C-C O: H ethanal :CN: and the electrophile is cyanide ion In the step you have just written, the nucleophile is slow rate-determining Ö—0—1 H3C-C-C=N; adduct
What are the relative rates of the following nucleophilic substitution reaction below.
An enolate is a very strong nucleophile. Bromine is a strong electrophile, so it can react withmuch weaker nucleophiles. Give mechanisms for the reactions of bromine with cyclopenteneand with phenol, which are both much weaker nucleophiles than an enolate
Which of the following would be a better starting material for a nucleophilic aromatic substitution reaction? Explain your choice using structures. Then provide the complete arrow pushing mechanism for the reaction of the better starting material and excess ethanol. CI CI O₂N NO₂ NO₂ NO₂
Fill in the missing reagents a-e in the following scheme:
Propose a pericyclic mechanism to account for the following transformation:
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.
For the given molecule, highlight any nucleophilic sites in red and any electrophilic sites in blue. O: 439 0-H
.Which one of the following molecules can react as electrophile in reactions with alkenes? CH3OH NaCI HCI KCN NaCI O HCI O KCN O CH3OH
H₂C=CHCCH3 then H3O+ The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an a,ß-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the ẞ-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions N 9a
Rank the set of compounds in order of reactivity (fastest to slowest) in a nucleophilic addition-elimination reaction with a nucleophile such as CH3NH2.