Question

In Williamson synthesis, ethoxyethane is prepared by (a) heating sodium ethoxide with ethyl bromide (b) passing ethanol over heated alumina (c) heating ethanol with dry \(\mathrm{Ag}_{2} \mathrm{O}\) (d) treating ethyl alcohol with excess of at \(430-440 \mathrm{~K}\)

Short answer

The answer is (a) heating sodium ethoxide with ethyl bromide.

Step-by-step solution

Understanding Williamson Synthesis

Williamson synthesis is a chemical reaction that is used to prepare ethers, such as ethoxyethane, through the reaction of an alkoxide ion with a primary alkyl halide.

Identifying the Reaction Components

Ethoxyethane is obtained by combining the alkoxide ion (like sodium ethoxide, \( \text{C}_2\text{H}_5\text{ONa} \)) with an alkyl halide such as ethyl bromide (\( \text{C}_2\text{H}_5\text{Br} \)).

Analyzing Each Option

Evaluate each given option to see which fits the Williamson synthesis criterion: - (a) Involves sodium ethoxide and ethyl bromide, matching Williamson synthesis requirements. - (b), (c), and (d) involve different reactions or conditions that do not align with Williamson synthesis.

Selecting the Correct Option

The correct method for preparing ethoxyethane through Williamson synthesis is (a), where sodium ethoxide reacts with ethyl bromide.

Key concepts

Ether preparation

Ether preparation is one of the fascinating transformations in organic chemistry that involves creating a molecule from two building blocks. Ethers, such as ethoxyethane, are formed through a reaction called Williamson synthesis. This method is renowned for its effectiveness in synthesizing complex molecules.

In the process, a nucleophile called an alkoxide ion reacts with an electrophile known as a primary alkyl halide. The reaction takes place under mild conditions, making it attractive for laboratory and industrial settings. What makes this method so special? It allows for the creation of asymmetric ethers, which are crucial in various applications, including pharmaceuticals and perfumes.

To prepare an ether like ethoxyethane, the reaction involves heating reactants such as sodium ethoxide and ethyl bromide. This typically takes place in an organic solvent under controlled temperatures, ensuring a high yield of the desired ether. Understanding this preparation process is essential for anyone delving deeper into organic synthesis.

Alkoxide ion

An alkoxide ion serves as the reactive partner in the formation of ethers like ethoxyethane. This ion forms when an alcohol loses its hydrogen, gaining a negative charge. For instance, sodium ethoxide is an example of an alkoxide ion derived from ethanol.

Key characteristics of alkoxide ions include their strong basicity and nucleophilicity. This means they are keen to donate electrons and react with positively charged species such as primary alkyl halides. This reactivity makes them perfect for Williamson synthesis, where they attack the carbon atom in the alkyl halide, leading to the formation of an ether.

During the synthesis, maintaining the strong basic nature of the alkoxide ion is crucial. It’s typically prepared in an inert atmosphere to prevent any unwanted reactions with moisture or air. This step ensures that when the alkoxide ion meets the alkyl halide, the reaction proceeds smoothly, forming the desired ether efficiently.

Primary alkyl halide

Primary alkyl halides are essential components in the Williamson synthesis and play a significant role in ether preparation. These are molecules where a halogen atom is attached to a carbon that is only connected to one other carbon. In the case of ethoxyethane, ethyl bromide acts as the primary alkyl halide.

Unique characteristics of primary alkyl halides in this reaction include their role as excellent electrophiles. They lure nucleophiles like alkoxide ions for the synthesis to be successful. The halogen atom in these molecules is replaced by the alkoxide, forming the ether linkage.

It’s essential to note the type of halide selected. Primary alkyl halides ensure that the Williamson synthesis proceeds efficiently, as they are less hindered by steric factors, unlike secondary or tertiary halides. Therefore, choosing the right type of alkyl halide is crucial for the success and yield of the ether being synthesized. This concept is a cornerstone in understanding the intricacies of organic reactions.