Problem 78
Question
In the mechanism of Hoffimann reaction, which intermediate rearranges to alkyl isocyanate? (a) nitrene (b) amide (c) nitroso (d) bromamide
Step-by-Step Solution
Verified Answer
The intermediate that rearranges to alkyl isocyanate is bromamide (d).
1Step 1: Understanding the Hoffmann Reaction
The Hoffmann rearrangement is a chemical reaction in which a primary amide is converted to a primary amine with the loss of one carbon atom. This reaction typically involves stages including the formation of a bromamide intermediate.
2Step 2: Identifying the Key Intermediate
In the Hoffmann reaction, a primary amide reacts with bromine in an alkaline solution to initially form a bromamide. This bromamide then rearranges.
3Step 3: Recognizing the Rearrangement Process
The bromamide intermediate undergoes a rearrangement where the carbonyl group excides from the nitrogen, leading to the formation of an alkyl isocyanate intermediate before hydrolysis to a primary amine.
4Step 4: Conclusion on Rearranging Intermediate
Based on the mechanism of the Hoffmann rearrangement, the bromamide is the key intermediate that rearranges to form the alkyl isocyanate.
Key Concepts
Understanding Primary AminesFormation and Role of Bromamide IntermediateAlkyl Isocyanate: A Transitional Stage
Understanding Primary Amines
Primary amines are organic compounds that contain a nitrogen atom bonded to one hydrocarbon group and two hydrogen atoms. They are essential in various biological processes and synthetic chemistry.
Some basic characteristics of primary amines include:
Some basic characteristics of primary amines include:
- The nitrogen atom in primary amines has a lone pair of electrons, making them nucleophilic and reactive.
- They are generally basic, meaning they can accept a proton to form an ammonium ion.
- Primary amines have a distinctive fishy odor and are typically polar due to the N-H bond.
Formation and Role of Bromamide Intermediate
The bromamide intermediate plays a critical role in the Hoffmann rearrangement mechanism. It is a pivotal step that involves significant transformation in the conversion process from a primary amide to primary amine.
In the initial step of the Hoffmann reaction, a primary amide reacts with bromine in an alkaline solution, typically sodium or potassium hydroxide, to form a bromamide. This reaction introduces the bromine atom to the amide nitrogen, forming a haloamide structure.
In the initial step of the Hoffmann reaction, a primary amide reacts with bromine in an alkaline solution, typically sodium or potassium hydroxide, to form a bromamide. This reaction introduces the bromine atom to the amide nitrogen, forming a haloamide structure.
- Bromamide intermediate is prone to rearrangement due to its reactive structure.
- The carbonyl group excides from the nitrogen, paving the way for subsequent steps.
- This transformation is crucial as it sets the stage for the formation of an alkyl isocyanate.
Alkyl Isocyanate: A Transitional Stage
During the Hoffmann rearrangement, one intermediary stage involves forming an alkyl isocyanate from the bromamide intermediate. Alkyl isocyanates are highly reactive compounds characterized by an -N=C=O group.
This formation is a crucial rearrangement process where the carbonyl group is eliminated, allowing the remaining structure to reorganize into an isocyanate.
This formation is a crucial rearrangement process where the carbonyl group is eliminated, allowing the remaining structure to reorganize into an isocyanate.
- Alkyl isocyanates are reactive and generally short-lived, swiftly proceeding to further reactions.
- In the Hoffmann sequence, isocyanates hydrolyze to yield a primary amine and carbon dioxide as by-products.
- The transition through an alkyl isocyanate is pivotal as it symbolizes the conversion stage that enables carbon elimination.
Other exercises in this chapter
Problem 76
A fruity smell is produced by the reaction of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) with (a) \(\mathrm{PCl}_{5}\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}\)
View solution Problem 77
Which of the following formula is more reactive for hydrolysis? (a) \(-\mathrm{COOR}\) (b) \(-\mathrm{CONH}_{2}\) (c) \(-\mathrm{COOH}\) (d) \(-\mathrm{COCl}\)
View solution Problem 80
Here the strongest acid is (a) \(\mathrm{CH}_{3}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{COOH}\) (c) \(\mathrm{Cl}-\mathrm{CH}_{2}-\mathrm
View solution Problem 81
Identify the wrong statement from the following: 8 (a) methyl salicylate is an ester (b) salicylic acid is a monobasic acid (c) methyl salicylate does not occur
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