Problem 146

Question

In the following sequence of reactions: $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{KMnO}_{4}}{\longrightarrow}(\mathrm{a}) \stackrel{\mathrm{SOCl}_{2}, \mathrm{NH}_{3}}{\longrightarrow} $$. (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

Verified

Answer

The end product (c) is methylamine.

1Step 1: Identify the First Reaction

The substance starts as ethanol, which reacts with potassium permanganate ($ \mathrm{KMnO}_4 $). This is an oxidation reaction that converts ethanol ($ \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} $) to acetic acid ($ \mathrm{CH}_3\mathrm{COOH} $).

2Step 2: Second Reaction

The acetic acid ($ \mathrm{CH}_3\mathrm{COOH} $) from the first reaction is then treated with $ \mathrm{SOCl}_2 $, which converts carboxylic acids to acyl chlorides, resulting in acetyl chloride ($ \mathrm{CH}_3\mathrm{COCl} $). Next, it reacts with $ \mathrm{NH}_3 $ (ammonia) to produce an amide, specifically acetamide ($ \mathrm{CH}_3\mathrm{CONH}_2 $).

3Step 3: Final Reaction

In the final step, acetamide ($ \mathrm{CH}_3\mathrm{CONH}_2 $) is reacted with $ \mathrm{Br}_2 $ and $ \mathrm{NaOH} $ through the Hofmann rearrangement. This process removes a carbonyl group from an amide, resulting in a primary amine. Therefore, acetamide ($ \mathrm{CH}_3\mathrm{CONH}_2 $) is converted to methylamine ($ \mathrm{CH}_3\mathrm{NH}_2 $).

Key Concepts

Ethanol to Acetic AcidAcyl Chloride FormationHofmann Rearrangement

Ethanol to Acetic Acid

Ethanol ($ \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} $) is commonly converted to acetic acid ($ \mathrm{CH}_3\mathrm{COOH} $) through an oxidation process. This transformation is a fundamental reaction in organic chemistry, often carried out using oxidizing agents like potassium permanganate ($ \mathrm{KMnO}_4 $).

### The Chemistry Behind the ReactionIn this reaction, $ \mathrm{KMnO}_4 $ acts as a strong oxidizing agent. It oxidizes ethanol, which is a primary alcohol, to acetic acid, a carboxylic acid. During the reaction, the hydroxyl group (OH) in ethanol is transformed into a carbonyl group (C=O). This is accompanied by the loss of hydrogen, facilitating the formation of acetic acid.

Understanding the oxidation of ethanol to acetic acid is crucial for recognizing broader concepts such as redox reactions and their significance in transforming functional groups.

This process is an essential step in producing vinegar, where fermentation further helps in acetic acid production.
It illustrates the basic principles of alcohol oxidation and carboxyl group chemistry.

Acyl Chloride Formation

The conversion of acetic acid ($ \mathrm{CH}_3\mathrm{COOH} $) to acyl chloride is achieved using thionyl chloride ($ \mathrm{SOCl}_2 $). This reaction is vital because acyl chlorides are highly reactive and serve as important intermediates in organic synthesis.

### How Acyl Chlorides are FormedWhen acetic acid is treated with thionyl chloride, it undergoes a substitution reaction. The hydroxyl group (OH) is replaced by a chlorine atom ($ \mathrm{Cl} $) to form acetyl chloride ($ \mathrm{CH}_3\mathrm{COCl} $).
The reaction proceeds as follows: \[ \mathrm{CH}_3\mathrm{COOH} + \mathrm{SOCl}_2 \rightarrow \mathrm{CH}_3\mathrm{COCl} + \mathrm{SO}_2 + \mathrm{HCl} \]

The products include acetyl chloride, sulfur dioxide ($ \mathrm{SO}_2 $), and hydrogen chloride ($ \mathrm{HCl} $), with the latter two being gases that easily leave the reaction mixture. This makes the reaction particularly efficient, driving it to completion.

Acyl chlorides, like acetyl chloride, are pivotal in forming other organic compounds such as amides, esters, and anhydrides.
Recognizing this reaction helps in understanding broader synthetic pathways and manipulation of functional groups in organic chemistry.

Hofmann Rearrangement

The Hofmann rearrangement is an important reaction that involves the conversion of an amide to an amine. In this context, acetamide ($ \mathrm{CH}_3\mathrm{CONH}_2 $) is converted into methylamine ($ \mathrm{CH}_3\mathrm{NH}_2 $) using bromine ($ \mathrm{Br}_2 $) and sodium hydroxide ($ \mathrm{NaOH} $).

### Understanding the MechanismIn the Hofmann rearrangement, $ \mathrm{Br}_2 $ and $ \mathrm{NaOH} $ together create a powerful basic and oxidative environment. The mechanism, although complex, essentially leads to the loss of a carbon atom within the carbonyl group and results in the formation of a primary amine.

The reaction steps can be simplified as follows:

Bromination: The amide is first brominated to form an N-bromoamide intermediate.
Rearrangement: The N-bromoamide undergoes rearrangement, losing the carbon in the form of carbon dioxide, and forming an isocyanate intermediate.
Hydrolysis: The isocyanate is then hydrolyzed by the basic solution, yielding the primary amine.

This process is notable for its utility in providing a straightforward method to produce primary amines from amides. It highlights not only the reaction conditions but also the mechanism by which the carbon skeleton of molecules can be altered.

Problem 145

Problem 147

Other exercises in this chapter

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Problem 145

The end product (c) in this reaction, $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{COOH} \stackrel{\mathrm{CaCO}_{3}}{\longrightarrow \mathrm{A}} \stackrel{\text

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Problem 147

Which of the following will give yellow precipitate with $\mathrm{I}_{2} / \mathrm{NaOH} ?$ 1\. $\mathrm{ICH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}$ 2. \(\m

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Problem 148

Consider the following substances: 1\. HCHO 2\. $\mathrm{CH}_{3} \mathrm{CHO}$ 3\. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}$ 4\. \(\mathrm{CH}_{3}

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