Problem 125 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following reaction, \(\mathrm{X}\) is and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}+4 \mathrm{H} \stackrel{\text { Red } \mathrm{P}+\mathrm{HI}}{\longrightarrow} 2 \mathrm{X}+\mathrm{H}_{2} \mathrm{O}\) (a) ethylene (b) propane (c) ethane (d) butane

Step-by-Step Solution

Answer

The reaction's product \(\mathrm{X}\) is ethane (option (c)).

Step 1: Understand the Reaction Components

Identify that the given reaction involves the compound \(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5\) being reacted with hydrogen in the presence of red phosphorus and hydroiodic acid (HI). The product of interest labeled as \(\mathrm{X}\) is involved in the creation of \(2\,\mathrm{X}\) molecules.

Step 2: Determine the Type of Chemical Reaction

Realize that the reaction is a reduction process where a mixed ether (diethyl ether, \(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5\)) is reduced, typically resulting in the cleavage of the ether linkage to form two separate hydrocarbons and water as byproducts.

Step 3: Evaluate Possible Reaction Pathways

In the presence of HI and red phosphorus, diethyl ether is likely to be cleaved to produce ethyl iodide \(\text{C}_2\text{H}_5\text{I}\). Subsequent reactions with excess HI could further reduce these iodine-bound hydrocarbons to their respective alkanes.

Step 4: Hypothesize Possible Products

Considering the stoichiometry and typical outcomes from similar chemical reactions, deduce the possible formation of ethane \(\text{C}_2\text{H}_6\) when ethyl iodide undergoes further reduction and dehalogenation.

Step 5: Match the Hypothesized Product with Given Options

Identify the transformation of each ethyl fragment into ethane due to the nature of hydrogenation and the presence of red phosphorus and HI. Thus, the molecule \(\mathrm{X}\) corresponds to ethane.

Key Concepts

Diethyl Ether Reduction

Diethyl ether is a common ether with the chemical formula \(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5\). In reduction reactions, it often undergoes cleavage into smaller hydrocarbons. Reduction, in chemistry, generally means the addition of hydrogen or the removal of oxygen from a molecule.
When diethyl ether is subjected to reduction conditions, such as the presence of red phosphorus and hydroiodic acid, the ether linkage (C-O-C bond) is broken. This process is known as ether cleavage.
The cleavage of diethyl ether specifically produces smaller alkanes and sometimes halide intermediates, depending on the reagents used. In the specific reaction described, diethyl ether is reduced to form ethyl iodide as an intermediate, which is later converted to an alkane by the action of further reducing agents.

Hydroiodic Acid Reactions

Hydroiodic acid (HI) is a powerful reducing agent widely used in organic synthesis to cleave various chemical bonds. In the context of ether cleavage reactions, HI plays a crucial role in breaking down ethers into simpler compounds.
When HI interacts with ethers like diethyl ether, it facilitates the cleavage of the ether's C-O-C bond, forming alkyl iodides. The reaction mechanism typically involves the protonation of the ether oxygen by HI, leading to the formation of a good leaving group, which then undergoes nucleophilic substitution.
In context, HI not only cleaves the ether but also reduces the resulting alkyl halides to corresponding alkanes by further hydrogenation, leveraging its strong reducing capacity.

Chemical Reaction Mechanisms

Understanding chemical reaction mechanisms is key to predicting the products of complex reactions, like the cleavage of ethers. A reaction mechanism details the step-by-step process through which reactants are transformed into products, including all intermediary species formed along the way.
In the case of ether cleavage using HI, the mechanism involves several steps starting with protonation. This step is followed by nucleophilic attack, where iodide ions replace the ether linkage, forming an alkyl iodide.
Subsequent steps involve reduction of the alkyl iodide back to an alkane. Electron-rich iodide acts as a nucleophile throughout the reaction, with hydrogen ions facilitating reductions in the presence of HI. Understanding these detailed pathways helps chemists predict the types and proportions of products formed in such reactions.