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
The product is (b) 9{CH}_3 9{NH}_2.
1Step 1: Identify the Reaction
The reaction of acetamide with sodium hypobromite (NaOBr) is an example of the Hofmann bromamide reaction (also known as Hofmann rearrangement). This reaction involves the conversion of an amide to a primary amine with one less carbon atom.
2Step 2: Recognize the Reactants and Products
Acetamide ({CH}__{3} 9{CONH}_2) is the starting compound. When it undergoes the Hofmann rearrangement with 9{NaOBr}9, the carbonyl (9{C=O}9) is removed, resulting in the primary amine, methylamine (9{CH}_3 9{NH}_2).
3Step 3: Choose the Correct Product
From the reaction, we know the product is a primary amine. The options given include
(a) 9{CH}_3 9{OH}
(b) 9{CH}_3 9{NH}_2
(c) 9{CH}_3 9{CN}
(d) 9{CH}_3 9{COBr}.
The correct product of the Hofmann rearrangement is methylamine, which corresponds to option (b).
Key Concepts
Amide to Amine ConversionPrimary AmineMethylamineChemical Reaction Mechanism
Amide to Amine Conversion
The Hofmann Bromamide Reaction exemplifies the conversion of an amide to an amine. This reaction is particularly intriguing because it involves a rearrangement. During this process, a carbon atom is removed from the structure, leading to a simpler amine. In the case of acetamide treated with sodium hypobromite (NaOBr), the reaction results in the loss of the carbonyl group (C=O) present in acetamide.
This transforms the molecule into a primary amine with one fewer carbon atom. It's a valuable reaction in organic chemistry due to its ability to simplify complex molecules. Keeping the reagents, mechanism, and structural changes in mind can greatly assist in understanding this conversion.
This transforms the molecule into a primary amine with one fewer carbon atom. It's a valuable reaction in organic chemistry due to its ability to simplify complex molecules. Keeping the reagents, mechanism, and structural changes in mind can greatly assist in understanding this conversion.
Primary Amine
Understanding primary amines is essential when studying the Hofmann Bromamide Reaction. A primary amine is an organic compound with one amino group \(\text{(-NH}_2)\) attached to a carbon atom, and two hydrogen atoms coming off the nitrogen.
The simplicity of a primary amine's structure makes it a foundational unit in biochemical processes and pharmaceuticals. In the context of our original exercise, the reaction of acetamide with sodium hypobromite (OBr) yields methylamine, which is a primary amine. Recognizing the formation of primary amines in organic reactions helps chemists predict product outcomes and design synthesis pathways.
The simplicity of a primary amine's structure makes it a foundational unit in biochemical processes and pharmaceuticals. In the context of our original exercise, the reaction of acetamide with sodium hypobromite (OBr) yields methylamine, which is a primary amine. Recognizing the formation of primary amines in organic reactions helps chemists predict product outcomes and design synthesis pathways.
Methylamine
Methylamine (CH extsubscript{3}NH extsubscript{2}) becomes the product when acetamide reacts through the Hofmann Bromamide Reaction. This small and simple molecule plays significant roles in many fields.
Methylamine serves as a building block in organic synthesis and is especially noted for its use in pharmaceuticals and agricultural chemicals. Furthermore, its presence as an output of the reaction indicates a successful conversion and signifies the breakdown of more complex amides to simpler structures. Recognizing methylamine’s importance can anchor one's understanding of the broader applications of this reaction.
Methylamine serves as a building block in organic synthesis and is especially noted for its use in pharmaceuticals and agricultural chemicals. Furthermore, its presence as an output of the reaction indicates a successful conversion and signifies the breakdown of more complex amides to simpler structures. Recognizing methylamine’s importance can anchor one's understanding of the broader applications of this reaction.
Chemical Reaction Mechanism
The chemical reaction mechanism in the Hofmann Bromamide Reaction involves several intricate steps. Initially, the reaction begins with the formation of NaOBr. Then, this reagent interacts with the amide, resulting in the formation of an N-bromoamide.
Subsequently, this bromoamide undergoes rearrangement, a crucial step where the nitrogen migrates to replace the carbonyl group, allowing the elimination of a carbon atom. This sequence efficiently converts the substance into a primary amine.
Subsequently, this bromoamide undergoes rearrangement, a crucial step where the nitrogen migrates to replace the carbonyl group, allowing the elimination of a carbon atom. This sequence efficiently converts the substance into a primary amine.
- The rearrangement is unique because it involves migration and simultaneous changes in the molecular framework.
- It's essential to note the intermediate products, which form and react swiftly, making the understanding of each step fundamental to mastering organic chemistry.
Other exercises in this chapter
Problem 9
Which of the following reacts with \(\mathrm{NaNO}_{2}+\mathrm{HCl}\) at \(273-278 \mathrm{~K}\) to give alcohol/phenol? (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)
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Hinsberg reagent is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{Cl}\) (c) \(\mathrm{C}_{6} \ma
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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}_
View solution Problem 13
The compound which one reaction with aqueous nitrous acid at low temperature produces an oily nitrosamine is (a) methylamine (b) ethylamine (c) triethylamine (d
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