Problem 11
Question
When acetamide is treated with \(\mathrm{NaOBr}\), the product formed is (a) \(\mathrm{CH}_{3} \mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CN}\) (d) \(\mathrm{CH}_{3} \mathrm{COBr}\)
Step-by-Step Solution
Verified Answer
(b) \(\mathrm{CH}_3 \mathrm{NH}_2\) is formed.
1Step 1: Identify the Reaction
The reaction in question involves the treatment of acetamide with sodium hypobromite (NaOBr). This reaction is known as the Hofmann Bromamide Degradation Reaction.
2Step 2: Understand the Hofmann Bromamide Degradation
In the Hofmann Bromamide Degradation, an unsubstituted amide reacts with \(\mathrm{NaOBr}\)\ (derived from bromine and base) to produce a primary amine. During this reaction, the carbonyl group (-C=O) in the amide is lost, leading to a one-carbon shorter chain in the amine product.
3Step 3: Determine the Product
Acetamide, \(\mathrm{CH}_3\mathrm{CONH}_2\), under Hofmann degradation, will lose the carbonyl group and form methylamine, \(\mathrm{CH}_3\mathrm{NH}_2\). This occurs as the reaction proceeds with the elimination of \(\mathrm{CO}_2\).
4Step 4: Match the Product to the Options
Comparing the product \(\mathrm{CH}_3\mathrm{NH}_2\) to the given choices, option (b) corresponds to methylamine, which is the product formed in this reaction.
Key Concepts
Acetamide ReactionSodium HypobromiteAmide to Amine Conversion
Acetamide Reaction
The acetamide reaction is a key process in organic chemistry known for transforming an acetamide molecule, such as \( \mathrm{CH}_3\mathrm{CONH}_2 \), into other products through chemical reactions. This particular reaction is interesting because it is part of a special type of chemical transformation called the Hofmann Bromamide Degradation. In this process, the acetamide molecule undergoes a sequence of changes when treated with sodium hypobromite (\( \mathrm{NaOBr} \)).
- This reaction is important for reducing the length of the carbon chain in organic compounds.
- During the transformation, the carbonyl group \( (-C=O) \) in acetamide is effectively removed, converting amide into a primary amine.
- It also produces valuable by-products such as carbon dioxide \( (\mathrm{CO}_2) \), which is released during the reaction.
Sodium Hypobromite
Sodium hypobromite \( (\mathrm{NaOBr}) \) plays a vital role in the Hofmann Bromamide Degradation reaction. It acts as a brominating agent, usually formed by the reaction of bromine with a base such as sodium hydroxide (\( \mathrm{NaOH} \)). In the context of the acetamide reaction, sodium hypobromite is essential as it facilitates the transformation of amides into amines.
- \( \mathrm{NaOBr} \) is often used because it can efficiently remove the carbonyl group from amides.
- This chemical behaves as an oxidizing agent, which helps in cleaving the bond between the carbon and the rest of the amide structure.
- The use of sodium hypobromite simplifies the process, making it an effective choice for amide to amine conversion.
Amide to Amine Conversion
The conversion of an amide to an amine is a significant chemical transformation. During this process, an amide compound, such as acetamide \((\mathrm{CH}_3\mathrm{CONH}_2)\), is transformed into an amine, specifically methylamine \((\mathrm{CH}_3\mathrm{NH}_2)\). This conversion is achieved through the Hofmann Bromamide Degradation, leveraging the reactivity of sodium hypobromite.
- The reaction breaks the carbon-to-nitrogen double bond, effectively removing the carbonyl group.
- As a result, the chain length of the molecule reduces by one carbon atom, since \((\mathrm{CO}_2)\) is expelled from the structure.
- This conversion is not only efficient but also quite fascinating as it allows for the synthesis of simpler and often more reactive organic compounds.
Other exercises in this chapter
Problem 9
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Which of the following compounds cannot by identified by carbylamine test? (a) \(\mathrm{CHCl}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{NH}-\mathrm{C}_
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Which is formed when \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{NOH}\) is heated? (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) (b) \(\left(\mathrm{CH}_
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