Problem 235 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction, \(\mathrm{CH}_{3} \mathrm{COOH} \stackrel{\mathrm{LiAH}_{4}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{PCl}_{5}}{\longrightarrow} \mathrm{B} \stackrel{\text { AIC.KOH }}{\longrightarrow} \mathrm{C}\), the product \(\mathrm{C}\) is: (a) Ethylene (b) Acetyl chloride (c) Acetaldehyde (d) Acetylene

Step-by-Step Solution

Answer

The product \(\mathrm{C}\) is ethylene, corresponding to option (a).

Step 1: Identify the first reaction

The first reaction involves the reduction of acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) with lithium aluminium hydride (\(\mathrm{LiAlH}_{4}\)), a strong reducing agent. This will reduce the carboxylic acid to the corresponding primary alcohol. Thus the product \(\mathrm{A}\) is ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)).

Step 2: Understand the second reaction

In the second step, ethanol is treated with phosphorus pentachloride (\(\mathrm{PCl}_{5}\)), resulting in the substitution of the hydroxyl group (\(-OH\)) with chlorine (\(\mathrm{-Cl}\)). Therefore, \(\mathrm{B}\) is chloroethane (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Cl}\)).

Step 3: Analyze the final reaction

Finally, chloroethane is treated with alcoholic potassium hydroxide (KOH). This is an elimination reaction where the chlorine atom is eliminated as KCl and the adjacent hydrogen as water, resulting in the formation of an alkene. The product \(\mathrm{C}\) is ethylene (\(\mathrm{CH}_{2} = \mathrm{CH}_{2}\)).

Key Concepts

Reduction with LiAlH4

In organic chemistry, reduction reactions are essential for transforming carboxylic acids into alcohols. Lithium aluminium hydride (\(\mathrm{LiAlH}_4\)) is a powerful reducing agent often used for this purpose. When \(\mathrm{LiAlH}_4\) is added to acetic acid (\(\mathrm{CH}_3\mathrm{COOH}\)), it donates electrons to reduce the carbonyl group (C=O) of the carboxylic acid to a primary alcohol.

As a result, the carboxylic acid (\(\mathrm{CH}_3\mathrm{COOH}\)) is transformed into ethanol (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\)). Key points to remember:

• The carboxylic acid group (-COOH) is reduced to a hydroxyl group (-OH).
• The oxygen from the acid's carbonyl group becomes a part of the hydroxyl group of the alcohol.
• This reaction is specific because \(\mathrm{LiAlH}_4\) cannot be used with water or alcohols directly, as it reacts violently.

Understanding this reaction is crucial as it lays the groundwork for converting more complex functional groups and synthesizing various organic compounds.

Substitution with PCl5

Substitution reactions involve the replacement of one atom or group with another. Phosphorus pentachloride (\(\mathrm{PCl}_5\)) is widely used for substituting hydroxyl groups (-OH) with chlorine atoms (-Cl) to form alkyl chlorides.

In the task, ethanol (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\)) reacts with \(\mathrm{PCl}_5\). During this substitution, the hydroxyl group in ethanol is replaced by chlorine, resulting in chloroethane (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl}\)). Important aspects of this reaction:

• \(\mathrm{PCl}_5\) provides the chlorine atom that replaces the hydroxyl group.
• This reaction produces additional by-products like POCl3 and HCl.
• Substitution reactions with \(\mathrm{PCl}_5\) are often useful in preparing alkyl chlorides from alcohols, which are important intermediates in many chemical syntheses.

This step highlights how substitution reactions can modify molecules for further reactions or applications in chemical processes.

Elimination reactions

Elimination reactions play a critical role in forming alkenes by removing elements from a molecule. In these reactions, two atoms or groups are removed from a substrate, often resulting in the formation of a double bond. With chloroethane (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl}\)) as the starting material, an elimination reaction is achieved using alcoholic potassium hydroxide (KOH).

In this context, the chlorine atom and an adjacent hydrogen atom are removed:

• The chlorine atom is eliminated as part of potassium chloride (KCl).
• The hydrogen atom combines with a hydroxide ion from KOH to form water.
• The removal of these atoms results in the formation of a carbon-carbon double bond, creating ethylene (\(\mathrm{CH}_2=\mathrm{CH}_2\)).

Elimination reactions are significant for synthesizing alkenes, which serve as basic building blocks in organic synthesis and industrial applications. Recognizing and applying these reactions can open pathways for creating various complex organic structures.