Show how one could synthesize-and resolve methylethylallylamine oxide from allylamine, with the knowledge that amine oxides are somewhat basic substances having \(\mathrm{K}_{\mathrm{B}}\) values at about \(10^{-11}\).

Short Answer

Expert verified
To synthesize methylethylallylamine oxide from allylamine, perform two reactions: 1. Alkylation: React allylamine with methyl iodide (CH3I) and sodium ethoxide (NaOEt) under anhydrous conditions and an inert solvent, such as THF or DCM, at low temperature. The reaction is: \( H_2C=CH-CH_2NH_2 + CH_3I + NaOEt \rightarrow CH_3CH_2NHCH_2CH=CH_2 + NaI + H_2O \). 2. Oxidation: React methylethylallylamine with hydrogen peroxide (H2O2) and a weak base like sodium bicarbonate (NaHCO3) to maintain the pH. The reaction is: \( CH_3CH_2NHCH_2CH=CH_2 + H_2O_2 \rightarrow CH_3CH_2N(CH_2CH=CH_2)O + H_2O \). Amine oxides are somewhat basic (\(\mathrm{K}_{\mathrm{B}}\) values around \(10^{-11}\)), but their basicity is weaker than that of the corresponding amines due to the partial positive charge on the nitrogen atom.

Step by step solution

01

Identify the chemical structures of allylamine and methylethylallylamine oxide

First, we need to determine the chemical structures of both compounds: - Allylamine (\( H_2C=CH-CH_2NH_2 \)) is an organic compound with a double bond between the first and second carbons and an amine group on the third carbon. - Methylethylallylamine oxide (\( CH_3CH_2NHCH_2CH=CH_2 \)) has a methyl group (CH3) and ethyl group (CH2CH3) attached to the nitrogen atom and is an oxide.
02

Determine the difference between the two structures

The main difference between the two structures is that allylamine (\(H_2C=CH-CH_2NH_2\)) contains only an amine group, while methylethylallylamine oxide (\( CH_3CH_2NHCH_2CH=CH_2\)) has an amine group with a methyl group and an ethyl group attached to the nitrogen atom.
03

Find a suitable reaction to convert allylamine into methylethylallylamine oxide

To convert allylamine into methylethylallylamine oxide, we need to perform two reactions: 1. Alkylation: The first reaction is the alkylation of allylamine, in which we add both a methyl group and an ethyl group to the nitrogen atom. Following are the reagents and conditions required for this reaction: - Allylamine - Methyl iodide (CH3I) and sodium ethoxide (NaOEt) as alkylating agents - Anhydrous conditions and an inert solvent, such as THF or DCM - Low temperature to control the extent of alkylation The reaction will be as follows: \( H_2C=CH-CH_2NH_2 + CH_3I + NaOEt \rightarrow CH_3CH_2NHCH_2CH=CH_2 + NaI + H_2O \) 2. Oxidation: The second reaction is the oxidation of the amine to form the amine oxide: - Methylethylallylamine as the substrate - Hydrogen peroxide (H2O2) as an oxidizing agent - A weak base such as sodium bicarbonate (NaHCO3) to maintain the pH of the reaction mixture The reaction will be as follows: \( CH_3CH_2NHCH_2CH=CH_2 + H_2O_2 \rightarrow CH_3CH_2N(CH_2CH=CH_2)O + H_2O \)
04

Discuss the basicity of amine oxides and their \(\mathrm{K}_{\mathrm{B}}\) values

Amine oxides are somewhat basic substances, with \(\mathrm{K}_{\mathrm{B}}\) values around \(10^{-11}\). The basicity of amine oxides arises from the lone pair of electrons on the nitrogen atom, which can act as a Lewis base and accept a proton (H+) from an acid, forming a conjugate acid. However, since amine oxides have a partial positive charge on the nitrogen atom due to the oxygen atom's higher electronegativity, their basicity is weaker than that of the corresponding amines.

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