Question

Wurtz reaction of methyl iodide yields an organic compound \(\mathrm{X}\). Which one of he following reactions also yields \(\mathrm{X}\) ? (a) \(\mathrm{CHCl}_{3} \stackrel{\text { Ag powder, } \triangle}{\longrightarrow}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa} \longrightarrow\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{LiAlH}_{4}\) (d) \(\mathrm{C}_{2}^{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{Mg}\) Dry ether

Short answer

Option (d) yields ethane, the same compound as the Wurtz reaction of methyl iodide.

Step-by-step solution

Determine Wurtz Reaction Product

The Wurtz reaction involves halogenated compounds, usually alkyl halides, and sodium metal, producing an alkane. For methyl iodide, the general equation is: \[ 2 ext{CH}_3I + 2 ext{Na} ightarrow ext{CH}_3- ext{CH}_3 + 2 ext{NaI} \]This indicates that ethane (\( \text{CH}_3\text{CH}_3 \)) is the organic compound \( X \).

Analyze Option (a)

In reaction (a), heating chloroform (\(\text{CHCl}_3 \)) with silver powder typically does not yield ethane. This reaction can lead to other types of products, not related to simple hydrocarbons like ethane.

Analyze Option (b)

This reaction between ethyl chloride \((\text{C}_2\text{H}_5\text{Cl})\) and sodium ethoxide \((\text{C}_2\text{H}_5\text{ONa})\) generally leads to ether formation, not ethane, via a Williamson synthesis, which can produce diethyl ether.

Analyze Option (c)

Lithium aluminium hydride \((\text{LiAlH}_4)\) is a reducing agent used in hydride transfer reactions that typically reduce aldehydes, ketones, or carboxylic acid derivatives. It's unlikely to simply yield ethane from ethyl chloride.

Analyze Option (d)

In this Grignard reaction, \(\text{C}_2\text{H}_5\text{Cl}\) with magnesium in dry ether forms an ethyl magnesium chloride \((\text{C}_2\text{H}_5\text{MgCl})\). Reacting with water, Grignard reagents can yield the corresponding alkane: \[ \text{C}_2\text{H}_5\text{MgCl} + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_6 + \text{MgCl(OH)} \]This matches the product \( X \), ethane.

Key concepts

Organic Compound Formation

Organic compounds are built on the backbones of carbon and hydrogen atoms. Converting simpler molecules into complex organic compounds is the essence of organic chemistry. Various reactions synthesize these compounds by forming specific bonds between atoms.
Take the Wurtz reaction for instance. It's an important method in the formation of alkanes. By reacting alkyl halides with sodium metal, we unite two alkyl groups into a single alkane, releasing a byproduct of sodium halide.
In the case of methyl iodide, Wurtz reaction leads to the formation of ethane, \( \text{CH}_3\text{CH}_3 \). This showcases a conversion from a simple haloalkane to a basic hydrocarbon, indispensable in numerous synthesis sequences.

Grignard Reaction

A standout in organic synthesis, the Grignard reaction uses Grignard reagents to form carbon-carbon bonds. It's a flexible method contributing to diverse chemical synthesis, notably for forming alcohols but also simple hydrocarbons like in our examined problem.
The formation of Grignard reagents involves the reaction of an alkyl halide with magnesium, creating a compound like \( \text{C}_2\text{H}_5\text{MgCl} \). Despite their high reactivity, careful handling spearheads successful synthetic schemes.
When a Grignard reagent reacts with water, it results in an alkane. In our example, reacting \( \text{C}_2\text{H}_5\text{MgCl} \) with water produces the sought-after ethane (\( \text{C}_2\text{H}_6 \)). This capability to introduce larger groups makes Grignard reactions a go-to in constructing complex molecules.

Williamson Synthesis

Williamson Ether Synthesis is a classic reaction for creating ethers. It involves reacting an alkyl halide with a metal alkoxide.
This versatile reaction allows the linkage of alkyl groups, forming an ether bond critical in molecules for fuel, solvents, and other applications.
In the original problem, the reaction between ethyl chloride and sodium ethoxide is a classic Williamson synthesis example. Instead of forming a simple alkane like ethane, it creates diethyl ether. This illustrates how selectivity in reaction conditions alters product outcomes.

Lithium Aluminium Hydride Reduction

Lithium aluminium hydride (LiAlH₄) is a robust reducing agent pivotal in organic transformations. Unlike mild reagents, it's capable of reducing a wide array of carbonyl compounds, ranging from aldehydes to acids. However, it doesn’t typically yield simple alkanes.
In our context, using LiAlH₄ with ethyl chloride would not just strip away the chlorine but also potentially transform any carbonyl derivatives. This powerful reduction may not create ethane but instead illustrates the breadth of transformation possible when reducing agents are applied.
The specific reactivity of LiAlH₄ showcases its usage beyond merely transforming single bonds, but in altering the oxidation states of entire functional groups in complex molecules.