Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following sequence of reactions: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{KMnO}_{4}}{\longrightarrow}\) (a) \(\stackrel{\mathrm{SOCl}_{2}, \mathrm{NH}_{3}}{\longrightarrow}\) (b) \(\mathrm{Br}_{2}+\mathrm{NaOH}\) (c) the end product (c) is (a) Acetone (b) Ethylamine (c) Acetic acid (d) Methyl amine

Step-by-Step Solution

Answer

The end product (c) is methyl amine (d).

Step 1: Reaction with KMnO4

The first step involves the oxidation of ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)) using potassium permanganate (\(\mathrm{KMnO}_{4}\)). This reaction oxidizes the primary alcohol to acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)), forming compound (a).

Step 2: Reaction with SOCl2 and NH3

Acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) is then treated with thionyl chloride (\(\mathrm{SOCl}_{2}\)) to form acetyl chloride (\(\mathrm{CH}_{3}\mathrm{COCl}\)). Upon further reaction with ammonia (\(\mathrm{NH}_{3}\)), acetamide (\(\mathrm{CH}_{3}\mathrm{CONH}_{2}\)) is obtained, which is compound (b) in the sequence.

Step 3: Hofmann Bromamide Reaction

In the final step, acetamide (\(\mathrm{CH}_{3}\mathrm{CONH}_{2}\)) undergoes the Hofmann bromamide reaction, where it is treated with bromine (\(\mathrm{Br}_{2}\)) and a base like sodium hydroxide (\(\mathrm{NaOH}\)). This reaction results in the formation of methyl amine (\(\mathrm{CH}_{3}\mathrm{NH}_{2}\)), which is compound (c), the end product.

Key Concepts

Oxidation of Alcohols

Oxidation of alcohols is a fundamental chemical process. It involves the transformation of alcohols into aldehydes, ketones, or carboxylic acids, depending on the type of alcohol and the conditions.

• Primary alcohols are typically oxidized first to aldehydes and then to carboxylic acids.
• Secondary alcohols are often oxidized to ketones.
• Tertiary alcohols generally do not undergo oxidation without the breakdown of the carbon skeleton.

In the described reaction, ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)) is a primary alcohol. Potassium permanganate (\(\mathrm{KMnO}_{4}\)) is a potent oxidizing agent, which means it can effectively promote the oxidation process. Upon oxidizing ethanol, we get acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)). This transformation is key, as the acetic acid is a stepping-stone to the next reaction stage. Remember, the power of an oxidizing agent is critical as it determines how far the oxidation proceeds.

Thionyl Chloride Reaction

The thionyl chloride reaction is a process where thionyl chloride (\(\mathrm{SOCl}_{2}\)) converts carboxylic acids to acyl chlorides. This reaction is particularly useful in organic synthesis as acyl chlorides are versatile intermediates.

• Thionyl chloride (\(\mathrm{SOCl}_{2}\)) is a chlorinating agent that replaces the hydroxyl group (-OH) of carboxylic acids with a chlorine atom, forming an acyl chloride.
• This reaction is beneficial because it can be done at mild conditions, and it often desolves releasing gases like \(\mathrm{SO}_{2}\) and HCl, aiding in the purification process.

In the provided sequence, acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) reacts with thionyl chloride to form acetyl chloride (\(\mathrm{CH}_{3}\mathrm{COCl}\)). Acetyl chloride is reactive and further reacts with ammonia (\(\mathrm{NH}_{3}\)) to yield acetamide (\(\mathrm{CH}_{3}\mathrm{CONH}_{2}\)). Acetamide plays a crucial role as it transitions into the final step of the sequence, showcasing the interconnectedness of chemical reactions.

Amide Formation

Amide formation is a reaction where an acyl chloride or carboxylic acid reacts with an amine. The convergence of these two entities results in an amide and, often, a side product such as \(\mathrm{HCl}\) if acyl chloride is the starting material.

• Acyl chlorides are highly reactive and excellent candidates for forming amides when treated with ammonia or amines.
• This reaction is central in making peptides, as proteins are merely long chains of amides.
• In this chemical path, the reactivity of acyl chloride towards ammonia is the driving factor for efficient amide formation.

For our study, acetyl chloride reacts with ammonia to form acetamide (\(\mathrm{CH}_{3}\mathrm{CONH}_{2}\)). This compound is then subject to the Hofmann bromamide reaction. This reaction involves \(\mathrm{Br}_{2}\) and NaOH to produce methyl amine (\(\mathrm{CH}_{3}\mathrm{NH}_{2}\)), a smaller amine. Through these series of transformations, ammonia's reactivity is pivotal, operating efficiently in the transition from an acid chloride to an amide and finally a primary amine.