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

Question

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

Step-by-Step Solution

Answer

Product C is ethyl methyl ether.

Step 1: Identify Reaction Type for A

The first step involves the reaction of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\) with \(\mathrm{aq} \cdot \mathrm{NaOH}\). This is a nucleophilic substitution reaction where \(\mathrm{Br}^-\) is replaced by \(\mathrm{OH}^-\). Thus, product \(A\) is ethanol, \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\).

Step 2: Reaction of A to Form B

In the second step, ethanol \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH})\) is treated with sodium \((\mathrm{Na})\). This leads to the formation of sodium ethoxide \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{ONa})\) (Product B) and hydrogen gas is released.

Step 3: Reaction of B to Form C

In the final step, sodium ethoxide \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{ONa})\) reacts with methyl iodide \((\mathrm{CH}_{3}\mathrm{I})\). This is a Williamson ether synthesis that produces ethyl methyl ether \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OCH}_{3})\) as product C.

Key Concepts

Nucleophilic Substitution Reaction

A nucleophilic substitution reaction is a fundamental type of organic reaction where one nucleophile replaces another attached to a carbon atom. This type of reaction plays a key role in organic chemistry transformations.
In the given exercise, the reaction of ethyl bromide (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\)) with aqueous sodium hydroxide (\(\mathrm{aq} \cdot \mathrm{NaOH}\)) is an example of a nucleophilic substitution reaction. Here, the bromide ion (\(\mathrm{Br}^-\)) is substituted by a hydroxide ion (\(\mathrm{OH}^-\)).
This reaction is facilitated by the polar nature of the carbon-bromine bond in ethyl bromide, which makes the carbon atom susceptible to nucleophilic attack.

• The hydroxide ion, being a strong nucleophile, attacks the electrophilic carbon atom.
• This attack leads to the release of the bromide ion and the formation of ethanol (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\)), which is product A.

Understanding this substitution process is essential for grasping how nucleophiles can replace leaving groups in organic molecules.

Williamson Ether Synthesis

Williamson ether synthesis is an important method for forming ethers from an alkoxide ion and a primary alkyl halide. This reaction utilizes two key components: the nucleophilic alkoxide and the electrophilic alkyl halide.
In the exercise, sodium ethoxide (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{ONa}\)) acts as the nucleophile. It is formed from ethanol and sodium in the previous reaction.
Nucleophilic sodium ethoxide then reacts with methyl iodide (\(\mathrm{CH}_{3}\mathrm{I}\)) to form ethyl methyl ether (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OCH}_{3}\)), product C.

• This reaction involves an SN2 mechanism, where the nucleophile directly attacks the electrophile while the leaving group (iodide) departs.
• The transition state is reached quickly, and the formation of the ether is complete in one step.

This synthesis route showcases how the choice of nucleophile and leaving group determines the success and efficiency of ether formation.

Ethanol Formation

Ethanol formation in organic reactions often involves the conversion of haloalkanes through nucleophilic substitution. In our example, ethanol is product A which results from reacting ethyl bromide with aqueous sodium hydroxide.
The formation of ethanol involves substituting the bromide ion with a hydroxide ion.

• This nucleophilic attack breaks the carbon-bromine bond and forms a new carbon-oxygen bond.
• Ethanol thus formed is a simple alcohol with industrial and lab importance.

Ethanol can serve as a precursor for further reactions, as seen with the subsequent formation of sodium ethoxide in the exercise. Awareness of ethanol's formation pathways helps in understanding its role as both a product and a reagent in synthesis processes.