Question

Ethoxy ethane is obtained as a major product in the reaction: (A) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{443 \mathrm{~K}}\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}-\frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{413 \mathrm{~K}}\) (C) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{(\mathrm{i}) \mathrm{Na}}{(\mathrm{i}) \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Cl}}\) (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OCH}_{2}-\mathrm{CH}_{3} \underset{\mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow}\)

Short answer

The correct reaction that produces ethoxy ethane as a major product is \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{\mathrm{H}_{2}\mathrm{SO}_{4}}{413 \mathrm{~K}}\) (option B). The lower temperature of 413 K promotes the formation of ethoxy ethane through intermolecular dehydration.

Step-by-step solution

(Step 1: Identify Ethoxy Ethane structure)

Ethoxy ethane, also known as diethyl ether, is an ether with the molecular formula \(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}\). Its structure is shown like this: \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\). Now let's examine each reaction option given in the exercise:

(Step 2: Analyze Reaction A)

In Reaction (A) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{\mathrm{H}_{2}\mathrm{SO}_{4}}{443 \mathrm{~K}}\), ethanol is treated with sulfuric acid, H2SO4, at 443 K. Under these conditions, dehydration of ethanol occurs and results in the formation of ethylene, not ethoxy ethane. So, option (A) is incorrect.

(Step 3: Analyze Reaction B)

In Reaction (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{\mathrm{H}_{2}\mathrm{SO}_{4}}{413 \mathrm{~K}}\), ethanol is treated with sulfuric acid, H2SO4, at 413 K. This set of reaction conditions promotes the formation of ethoxy ethane through intermolecular dehydration. So, option (B) is correct.

(Step 4: Analyze Reaction C)

In Reaction (C) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{(\mathrm{i})\mathrm{Na}}{(\mathrm{i}) \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Cl}}\), ethanol is treated with sodium metal and an alkyl halide, ethyl chloride. This reaction is known as the Williamson ether synthesis, which can produce ethoxy ethane, but it's not the major product because of the formation of a competing byproduct, ethane. So, option (C) is incorrect.

(Step 5: Analyze Reaction D)

In Reaction (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OCH}_{2}-\mathrm{CH}_{3}\underset{\mathrm{H}_{3}\mathrm{O}^{+}}{\longrightarrow}\), ethoxy ethane is treated with a hydronium ion (H3O+). This reaction will lead to the cleavage of the ether and not the formation of ethoxy ethane. So, option (D) is incorrect. Based on the above analysis, we can conclude that Reaction (B) is the correct answer, as it produces ethoxy ethane as the major product.

Key concepts

Dehydration of Alcohols

Dehydration of alcohols is a fundamental chemical reaction where an alcohol loses a molecule of water to form a new product. This process requires the presence of a strong acid, such as sulfuric acid, to act as a catalyst, and heat to drive the reaction forward. There are two main types of dehydration reactions that can occur with alcohols:

• Intermolecular Dehydration: This involves two alcohol molecules combining with the removal of water to form an ether. A typical condition for this type of reaction is heating ethanol with sulfuric acid at a lower temperature, such as 413 K, leading to the formation of ethoxy ethane.
• Intramolecular Dehydration: Here, a single alcohol molecule loses a water molecule, usually at higher temperatures (around 443 K), resulting in the formation of an alkene. This is what happens when ethanol undergoes dehydration at 443 K, yielding ethylene instead of an ether.

In the context of the given exercise, reaction B involves intermolecular dehydration, correctly resulting in the formation of ethoxy ethane, as the major product. It demonstrates the importance of controlling reaction conditions to obtain the desired product.

Williamson Ether Synthesis

The Williamson ether synthesis is a highly versatile method to produce ethers from alcohols. This reaction involves the conversion of an alcohol to an alkoxide ion by using a strong base, such as sodium or potassium. The alkoxide ion then acts as a nucleophile and attacks an alkyl halide to form the ether. In the exercise, reaction C is an example of this synthesis:

• First, ethanol reacts with sodium metal to form sodium ethoxide ( ext{C}_2 ext{H}_5 ext{ONa}).
• The sodium ethoxide then attacks ethyl chloride ( ext{C}_2 ext{H}_5 ext{Cl}), leading to the formation of ethoxy ethane ( ext{C}_4 ext{H}_{10} ext{O}).

However, this reaction can also result in the formation of byproducts. In this case, the presence of sodium ethoxide also leads to the formation of ethane, which makes ethoxy ethane not the major product in this reaction. This highlights the power of this method to create ethers, but also the necessity to consider competing side reactions.

Reaction Analysis

Analyzing a chemical reaction involves identifying all possible products and determining the main outcome based on the reaction conditions. For Ethoxy Ethane Formation analysis in the exercise, a systematic approach was needed:

• Option A: Ethanol treated with sulfuric acid at high temperatures (443 K) favors the formation of ethylene, an alkene, through intramolecular dehydration. Thus, not producing ethoxy ethane.
• Option B: Ethanol at a slightly lower temperature (413 K) undergoes intermolecular dehydration. This successfully produces ethoxy ethane as a major product by combining two ethanol molecules.
• Option C: Involves a Williamson Ether Synthesis method where sodium and ethyl chloride form ethoxy ethane, but with possible byproduct creation, compromising the yield.
• Option D: Considers the hydrolysis of an ether, breaking down ethoxy ethane rather than forming it.

Understanding these reactions and their conditions is crucial for correctly predicting reaction outcomes, particularly in determining which method most efficiently produces a desired chemical like ethoxy ethane.