The Ritter reaction is especially valuable for the synthesis of 3° alkanamines. In fact, there are few alternative routes to them. This reaction is illustrated by the first step in the following sequence. In the second step, the Ritter product is hydrolyzed to the amine. HCN H,O HO- N-C-H H NH, H,SO4 КОН Ritter product

The Ritter reaction is especially valuable for the synthesis of 3° alkanamines. In fact, there are few alternative routes to them. This reaction is illustrated by the first step in the following sequence. In the second step, the Ritter product is hydrolyzed to the amine. HCN H,O HO- N-C-H H NH, H,SO4 КОН Ritter product

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

Chapter23: Amines

Section: Chapter Questions

Problem 23.54P

See similar textbooks

Similar questions

Treatment of 1-aminoadamantane, C10H17N, with methyl 2,4-dibromobutanoate in the presence of a nonnucleophilic base, R3N, involves two successive SN2 reactions and gives compound A. Propose a structural formula for compound A.
Amines are converted into alkenes by a two-step process called Hofmann elimination. SN2 reaction of the amine with an excess of CH3I in the first step yields an intermediate that undergoes E2 reaction when treated with silver oxide as base. Pentylamine, for example, yields 1-pentene. Propose a structure for the intermediate, and explain why it readily undergoes elimination.
Undecanal is commercially available, but 6-methyl-1-bromoheptane is not. Propose two methods for the synthesis of 6-methyl-1-bromoheptane using precursors containing fewer than five carbon atoms (Cn n < 5). A protecting group will probably be required.
The formula for the danger sign pheromone for a species of ant is C7H7O. When treated with I2 NaOH, this pheromone produces iodoform and n-hexanoic acid. How is this pheromone structure?
What is the major product of the reaction of 1 mol of propyne with each of the following reagents? a. HBr (1 mol) e. aqueous H2SO4, HgSO4 h. H2/Lindlar catalyst b. HBr (2 mol) f. R2BH in THF followed by i. sodium amide c. Br2 (1 mol)/CH2Cl2 H2O2/HO-/H2O j. the product of part i followed by d. Br2 (2 mol)/CH2Cl2 g. excess H2, Pd/C 1-chloropropane
The reaction of a nitrile with an alcohol in the presence of a strong acid forms an N-substituted amide. This reaction, known as the Ritter reaction, doesnot work with primary alcohols. a. Propose a mechanism for the Ritter reaction.b. Why does the Ritter reaction not work with primary alcohols?c. How does the Ritter reaction differ from the acid-catalyzed hydrolysis of a nitrile to form an amide?
Predict the product of the following reactions. Part A Part B NaOEt,EtOH, NaOMe,MeOH, heat H30+quench
(2R, 3S) -3-phenylbutan-2-ol tosylate undergoes E2 removal when treated with sodium ethoxide.Draw the structures of the products obtained (one majority and one minority). Use Newman's projections for this reactive in this deduction.
Predict the products of reaction of pent-1-yne with the following reagents.(a) 1 equivalent of HCl (b) 2 equivalents of HCl (c) excess H2, Ni(d) H2, Pd>BaSO4, quinoline
Several sulfonylureas, a class of compounds containing RSO2NHCONHR, are useful drugs as orally active replacements for injected insulin in patients with adult-onset diabetes. These drugs decrease blood glucose concentrations by stimulating b cells of the pancreas to release insulin and by increasing the sensitivity of insulin receptors in peripheral tissues to insulin stimulation. Tolbutamide is synthesized by the reaction of the sodium salt of p-toluenesulfonamide and ethyl N-butylcarbamate . Propose a mechanism for this step.
Boron tribromide (BBr3) cleaves ethers to give alkyl halides and alcohols.R¬O¬R′ + BBr3 ¡ R¬O¬BBr2 + R′BrR¬O¬BBr2 + 3 H2O ¡ ROH + B(OH)3 + 2 HBrThe reaction is thought to involve attack by a bromide ion on the Lewis acid–base adductof the ether with BBr3 (a strong Lewis acid). Propose a mechanism for the reaction ofbutyl methyl ether with BBr3 to give (after hydrolysis) butan-1-ol and bromomethane.
The base-promoted rearrangement of an a-haloketone to a carboxylic acid, known as the Favorskii rearrangement, is illustrated by the conversion of 2-chlorocyclohexanone to cyclopentanecarboxylic acid. Jo Han Sow La NaOH, THF O Nat H3O+ OH NaOH, THF The mechanism involves the following 5 steps: -ō cyclopropanone intermediate 1. Abstraction of a proton to form enolate anion 1; 2. Formation of a cyclopropanone intermediate 2 with expulsion of chloride ion; 3. Addition of hydroxide ion to form tetrahedral intermediate 3; 4. Collapse of the tetrahedral intermediate and breakage of the three-membered ring to form carbanion intermediate 4; 5. Proton transfer to form the rearranged carboxylic acid. For the following reaction, draw the reaction out on paper, and then draw the structure of cyclopropanone intermediate 2 in the window. 1. NaOH, THF 2. H3O* OH