Problem 104 - A Complete Resource Book in Chemistry for JEE Main

Question

Product \(\mathrm{C}\) in the following reaction, will be \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\text { aq. } \mathrm{NaOH}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{Na}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{CH}_{3} \mathrm{I}}{\longrightarrow} \mathrm{C}\) (a) ethane (b) ethyl methyl ether (c) ethyl iodide (d) propane

Step-by-Step Solution

Answer

The product \( \mathrm{C} \) is ethyl methyl ether (b).

Step 1: Understanding the Reaction

The given reaction involves multiple steps where each step involves a chemical transformation. It starts with \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \). This is ethyl bromide, which is going to undergo reactions in sequence.

Step 2: First Reaction - Aqueous NaOH

The first reaction is \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \) reacting with aqueous \( \mathrm{NaOH} \). This is a nucleophilic substitution reaction where \( \mathrm{Br} \) is replaced by \( \mathrm{OH} \), forming \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \), which is ethanol (\( \mathrm{A} \)).

Step 3: Second Reaction - Sodium Metal

The product \( \mathrm{A} \) (ethanol) is then reacted with sodium metal (\( \mathrm{Na} \)). This reaction forms sodium ethoxide (\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa} \)) which is compound \( \mathrm{B} \).

Step 4: Third Reaction - Methyl Iodide

In the final step, compound \( \mathrm{B} \) (sodium ethoxide) reacts with methyl iodide \( \mathrm{CH}_{3} \mathrm{I} \). The ethoxide ion (\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} \)) acts as a nucleophile and attacks the methyl group, displacing the iodide ion, forming ethyl methyl ether (\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OCH}_{3} \)), which is compound \( \mathrm{C} \).

Step 5: Conclusion: Identify Product C

From the steps above, the final product of the reactions, compound \( \mathrm{C} \), is ethyl methyl ether.

Key Concepts

Nucleophilic Substitution

Nucleophilic substitution is a fascinating and fundamental reaction in organic chemistry. It involves the replacement of an atom or group of atoms, known as the leaving group, by a nucleophile. Let's break it down a bit further.
This type of reaction is characterized by the nucleophile, which is a chemical species that donates an electron pair to an electrophile. In simpler terms, a nucleophile is like a chemical that "loves" positively charged particles.

• The reaction begins with a substrate, which contains a good leaving group. Common examples of leaving groups include halogens like bromine, chlorine, and iodine.
• During the reaction, the nucleophile attacks the substrate, forming a new bond while the leaving group departs.

The first step in our exercise is a nucleophilic substitution where ethyl bromide ( C_2H_5Br ) reacts with aqueous NaOH. Here, the hydroxide ion ( OH^- ) is the nucleophile that replaces the bromine atom. As a result, ethanol ( C_2H_5OH ) is formed. This process is crucial in transforming the reactants into the intermediate compounds needed for further reactions.

Sodium Ethoxide

Sodium ethoxide (C_2H_5ONa) is a strong base commonly used in organic synthesis, especially in ether formation and other nucleophilic substitution reactions. Understanding its formation and role is essential for grasping many chemical transformations.
Sodium ethoxide is formed when ethanol (C_2H_5OH), a product from our previous reaction, reacts with metallic sodium (Na). This process is a simple displacement reaction whereby sodium displaces the hydrogen in the hydroxyl group of ethanol. This reaction can be summarized as:
\[ \text{C}_2\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_2\text{H}_5\text{ONa} + \frac{1}{2}\text{H}_2 \]

• Sodium ethoxide is robust and acts as a good nucleophile because of the largely electronegative oxygen atom.
• In the following step, it plays a critical role in the formation of ethers, as it can readily react with alkyl halides.

This makes sodium ethoxide a versatile reagent in organic chemistry, particularly due to its dual function as a strong base and potent nucleophile, facilitating the formation of new bonds leading to product C.

Ethyl Methyl Ether

Ethyl methyl ether ( C_2H_5OCH_3 ) serves as the target product in our discussed reaction sequence. Ethers like this one are compounds characterized by an oxygen atom between two alkyl groups, in this case, ethyl and methyl.
In the final step, sodium ethoxide comes into play again by reacting with methyl iodide ( CH_3I ). This step further exemplifies nucleophilic substitution, where the ethoxide ion ( C_2H_5O^{-} ) from sodium ethoxide acts as a nucleophile. It attacks the methyl group of methyl iodide, displacing the iodide ion and forming the new ether linkage.
Understanding the formation of ethers is essential in organic chemistry as they are not only prevalent in natural products but also serve as solvents and intermediates in various chemical syntheses.

• Ethyl methyl ether is relatively simple and versatile.
• Knowledge of its synthesis is pivotal given its applicability in creating more complex molecules.

Thus, the comprehension of these reactions is not just academic but also practical in real-world organic synthesis, emphasizing the significance of learning the mechanisms and conditions that facilitate these transformations.