Question

Starting with the compounds chloromethane, chloroethane, sodium azide, sodium cyanide, and a reducing agent, suggest how the following compounds could be synthesized. (a) \(N\) -methylpropanamide (b) ethylethanoate (c) methylethylamine (d) tetramethylammonium chloride

Short answer

(a) N-methylpropanamide can be synthesized from chloroethane using sodium cyanide and a reducing agent, followed by reaction with methyl chloroformate. \n(b) Ethylethanoate can be synthesized from chloroethane by performing an esterification reaction with acetic acid. \n(c) Methylethylamine can be synthesized from chloromethane and chloroethane, followed by formation of ethyl azide, its reduction to ethylamine, and reaction with chloromethane. \n(d) Tetramethylammonium chloride can be synthesized through a successive methylation of ammonia using an excess of chloromethane.

Step-by-step solution

Synthesis of N-methylpropanamide

N-methylpropanamide can be synthesized from chloroethane via a two-step process, involving the transformation of the halide into an amine, followed by acid-catalysed amide formation: \n\n1.1 - Reaction of chloroethane with sodium cyanide to produce ethanenitrile: \n\n\(CH3CH2Cl + NaCN \rightarrow CH3CH2CN + NaCl\) \n\n1.2 - Reduction of the nitrile group into an amine using a reducing agent: \n\n\(CH3CH2CN + H2 \rightarrow CH3CH2NH2\) \n\n1.3 - Acylation of the amine with methyl chloroformate to yield N-methylpropanamide: \n\n\(CH3CH2NH2 + CH3OCOCl \rightarrow CH3CH2NHCOOCH3 + HCl\)

Synthesis of Ethylethanoate

Ethylethanoate can be synthesized from chloroethane by performing an esterification reaction with acetic acid:\n\n \(CH3CH2OH + CH3COOH \rightarrow CH3COOCH2CH3 + H2O\)

Synthesis of Methylethylamine

Methylethylamine can be synthesized from chloromethane and chloroethane. By treating chloroethane with sodium azide, we get ethyl azide, which upon reduction gives ethylamine. The ethylamine can then be reacted with chloromethane to give methylethylamine:\n\n3.1 - Formation of Ethyl Azide: \n\n \(CH3CH2Cl + NaN3 \rightarrow CH3CH2N3 + NaCl\)\n\n3.2 - Reduction of Ethyl Azide to Ethylamine: \n\n \(CH3CH2N3 + 3H2 \rightarrow CH3CH2NH2 + 3H2O\)\n\n3.3 - Reaction of Ethylamine with Chloromethane: \n\n \(CH3CH2NH2 + CH3Cl \rightarrow CH3CH2NHCH3 + HCl\)

Synthesis of Tetramethylammonium Chloride

Tetramethylammonium chloride can be synthesized through a reaction of chloromethane with ammonia, followed by successive methylations using excessive chloromethane. The use of excessive chloromethane ensures that all the hydrogen atoms on the nitrogen get replaced by methyl groups:\n\n \(NH3 + 4CH3Cl \rightarrow [CH3]4N+Cl- + 3HCl \)

Key concepts

Nitrile Reduction

The process of nitrile reduction plays a crucial role in organic synthesis, as it allows the transformation of a nitrile group into an amine. Amines are fundamental building blocks in the synthesis of various organic compounds, including pharmaceuticals and polymers.

In the provided exercise, the nitrile reduction is part of synthesizing N-methylpropanamide. The process begins by reacting chloroethane with sodium cyanide to form ethanenitrile ( \(CH3CH2Cl + NaCN \rightarrow CH3CH2CN + NaCl\) ). The nitrile group is then reduced using a reducing agent, commonly hydrogen in the presence of a metal catalyst, to form an amine ( \(CH3CH2CN + H2 \rightarrow CH3CH2NH2\) ). This process involves the addition of hydrogen across the carbon-nitrogen triple bond of the nitrile, leading to the formation of a primary amine.

To aid student understanding, it is important to stress that the selection of the appropriate reducing agent and reaction conditions are crucial aspects of successfully conducting a nitrile reduction. Common reducing agents include hydrogen gas with a transition metal catalyst or chemical hydrides like LiAlH4 or NaBH4.

Esterification Reaction

Esterification is a fundamental organic chemistry reaction where a carboxylic acid reacts with an alcohol to form an ester and water. It's a key reaction in synthesizing perfumes, flavors, and plastics.

The exercise leads students through the creation of ethylethanoate ( \(CH3CH2OH + CH3COOH \rightarrow CH3COOCH2CH3 + H2O\) ). To demystify this process for learners, it's helpful to highlight that esterification is an equilibrium reaction usually catalyzed by an acid. This means that to drive the reaction forward, one may need to remove the produced water or use an excess of one of the reactants.

When educating students on this topic, discuss why the reactants (an alcohol and a carboxylic acid) need to be carefully chosen based on the desired ester. Also, explore the role of catalysts, such as sulfuric acid, to speed up the reaction without being consumed in the process.

Amine Acylation

Amine acylation is an invaluable tool in organic synthesis for creating amides from amines. This reaction involves the introduction of an acyl group onto the nitrogen atom of an amine compound. In pharmaceuticals and agrochemicals production, this reaction is pivotal.

In our textbook problem, amine acylation is used to convert ethylamine into N-methylpropanamide through the reaction with methyl chloroformate ( \(CH3CH2NH2 + CH3OCOCl \rightarrow CH3CH2NHCOOCH3 + HCl\) ). It's essential to inform students that acylation typically requires activation of the carboxylic acid derivative (like chloroformates or anhydrides) to make it more reactive.

Clear explanations on the choice of acylating agents and the stoichiometry of the reaction can enhance understanding. Additionally, covering the mechanism of amine acylation, including the formation of a tetrahedral intermediate and subsequent rearrangement, will provide students with a more comprehensive grasp of the concept.

Azide to Amine Conversion

Converting azides to amines is a significant transformation in organic chemistry, especially useful in synthesizing drugs and agrochemicals. This transformation is often achieved through reduction reactions.

The exercise solution demonstrates the synthesis of methylethylamine from ethyl azide, an intermediate produced by reacting chloroethane with sodium azide ( \(CH3CH2Cl + NaN3 \rightarrow CH3CH2N3 + NaCl\) ). The ethyl azide is then reduced, typically with hydrogen and a catalyst, to yield ethylamine ( \(CH3CH2N3 + 3H2 \rightarrow CH3CH2NH2 + 3H2O\) ).

For students to fully comprehend this process, it’s crucial to discuss the nature of azides and their high reactivity. The choice of reducing agent, for example, Staudinger reduction or catalytic hydrogenation, can vary based on the desired reaction conditions and scale of the reaction. Clarifying these points would significantly assist students in understanding the route from azides to amines.