1. CNBr2. LiAlH43. H₂ONH22aPrimary amines can be prepared from nitriles by reduction with LiAlH4. The two-step sequence involves SN2 displacement of halide ion by cyanide ion followed by reduction, and yields a primary aminewith one more carbon than was present in the alkyl halide.Draw curved arrows to show the movement of electrons in this step of the mechanism.Arrow-pushing InstructionsBrCEN:-C=N:Br

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Answered: 1. CN Br 2. LiAlH4 3. H₂O NH2 2a…

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

Chapter20: Dienes, Conjugated Systems, And Pericyclic Reactions

Section: Chapter Questions

Problem 20.47P

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1. CN Br 2. LIAIH4 3. H₂O σ NH2 Primary amines can be prepared from nitriles by reduction with LiAlH4. The two-step sequence involves SN2 displacement of halide ion by cyanide ion followed by reduction, and yields a primary amine with one more carbon than was present in the alkyl halide. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions X 2- H-OH NH + OH 2 C
5. The acyl group is a protecting group for amines which can be deprotected by treatment with sodium hydroxide. Please write mechanisms for both the protection and deprotection steps. NaOH CH30-H NH2 protect deprotect R NH2 amine acetic anhydride N-acyl amine
Amoxycillin is often adminisstered as the sodium salt. It is more soluble form that results from the reaction: Amoxycillin + NaOH --> Sodium-amoxycillin + H2O What functional grouo omn amoxycillin allows it to act as an acid in the reaction?
Br 1. N3 2. LiAlH4 3. H₂O NH₂ 26 Preparation of primary amines using azide ion avoids the problem of over-alkylation since the alkyl azide is not nucleophilic. The alkyl azide is prepared from an alkyl halide via an SN2 reaction with N3 as the nucleophile. Subsequent reduction of the azide with LiAlH4 leads to the desired amine. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions H |- H-AI-H H H 1 Al. H
1. Show the products of the reductive amination reactions below; a) Acetone + NH3, followed by H2/Ni2 b) Acetone + CH3NH2, followed by LIBH3CN c) Acetone + (CH3)2NH, followed by LIBH3CN 2. Write the reaction and the names of the products formed in the reaction of aniline and N-methylaniline with: a) NaNO3+HCI followed by CuClI b) NaNO3+HCI followed by Cu20, Cu2*, H20
14. What products result from the reaction between water and the following amine: a. butyl ethyl methyl ammonium ion + OH+ b. butyl ethyl methyl ammonium ion + H3O+ c. butyl ethyl methyl ammonium ion + H3O d. butyl ethyl methyl ammonium ion + OH- e. butyl ethyl methyl ammonium ion + H2O
Give the major product for the reaction. Pyrrolidine is a secondary amine. CH CH O CH CH! O OH CH2 O CH. I CH.CH Ї TCH CHE OH CH CH₂CH₂ CH.CH CH.OH2 0= RO pyrrolidine. 2 GH CH2 3 HCI H,0 TOH OH: CH₂CH₂ TCH CH: CH₂CH. CH₂CH₂
13
Several additional amine syntheses are effectively limited to making primary amines. The reduction of azides and nitrocompounds and the Gabriel synthesis leave the carbon chain unchanged. Formation and reduction of a nitrile adds onecarbon atom. Show how these amine syntheses can be used for the following conversions. (c) 1@bromo@3@phenylheptane S 3@phenylheptan@1@amine (d) 1@bromo@3@phenylheptane S 4@phenyloctan@1@amine
H3C OH 1. Br2, PBг3 2. H₂O H3C OH Br The a-bromination of carbonyl compounds by Br₂ in acetic acid is limited to aldehydes and ketones because acids, esters, and amides don't enolize to a sufficient extent. Carboxylic acids, however, can be a-brominated by first converting the carboxylic acid to an acid bromide by treatment with PBr3. Following enolization of the acid bromide, Br2 reacts in an a-substitution reaction. Hydrolysis of the acid bromide completes the reaction. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions Br Br зь P-Br Br H₂C H₂C SH Br Br
H₂C 0 NH₂ B₂, NaOH 98 H₂C OH When an a-hydroxy amide is treated with try in aqueous NaOH under Hofmann rearrangement conditions, loss of CO, occurs and a chain-shortened aldehyde is formed. The mechanhum involves the following steps: Base abstracts an acidic amide proton, yielding amide anion The amide anion reacts with bromine in an e-substitution reaction to give N-bromoamide 2. Abstraction of the remaining amide proton by base gives a resonance-stabilized bromoamide anion Rearangement occurs to yield isocyanate Water adds to the isocyanate to yield carbamic acid Elimination of CD, yields carbinolamine Following proton transfer, expubion of ammonia yields the final product aldehyde Write out the mechanism on a separate sheet of paper, and then draw the structures of carbamic acid 5 and isocyanate 4. H 400, ANH, You do not have to consider stereochemistry. You do not have to explicitly draw Hatoms De not include lone pairs in your answer. They will not be considered in the grading…
State all the products formed from Hofmann elimination of the following amine. Choose the major product. Show the mechanism to obtain the major product. а. b. с. NH2 CH,CHCH3 NH2 NH2

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