Answer
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Hint: We know that Hoffmann bromamide reaction is the reaction which produces alkyl amine on reaction of amide with bromine in presence of ethanolic solution of sodium or potassium hydroxide.
Complete step by step answer:
In the question, it is given that reaction ‘C’ with bromine in presence of potassium hydroxide solution gives ethaylamine that means, the reaction is Hoffmann Bromoamide reaction. Then, C must be an amide. ‘C’ forms by heating of B. Then, ‘B’ must be a carboxylate ion. Then ‘A’ must be a carboxylic acid. Now, we have to determine, which carboxylic acid among the given options is correct. We know that, in Hoffmann Bromamide degradation reaction, the loss of one carbon atom takes place. In the question product formed is ethylamine, that means, the amide composed of three carbon atoms, which indicates that the carboxylic acid ‘A’ also contains three carbon atoms, that is propanoic acid.
Now, we write the whole reaction.
Thus, the correct answer is option D, i.e., ${\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{COOH}}$..
Additional Information:
Hoffmann bromoamide reaction is a degradation reaction, where migration of an aryl or alkyl group from carbonyl carbon of amide to nitrogen atom takes place. So, the amine forms in the reaction contain one less carbon atom than that of the amide.
Note:
Students might get confused in choosing the correct carboxylic acid from the given options. They might think that, as the product of the reaction (ethylamine) consists of two carbon atoms, so the carboxylic acid also has two carbon atoms. But they must not forget that Hoffmann bromoamide reaction is a degradation reaction.
Complete step by step answer:
In the question, it is given that reaction ‘C’ with bromine in presence of potassium hydroxide solution gives ethaylamine that means, the reaction is Hoffmann Bromoamide reaction. Then, C must be an amide. ‘C’ forms by heating of B. Then, ‘B’ must be a carboxylate ion. Then ‘A’ must be a carboxylic acid. Now, we have to determine, which carboxylic acid among the given options is correct. We know that, in Hoffmann Bromamide degradation reaction, the loss of one carbon atom takes place. In the question product formed is ethylamine, that means, the amide composed of three carbon atoms, which indicates that the carboxylic acid ‘A’ also contains three carbon atoms, that is propanoic acid.
Now, we write the whole reaction.
Thus, the correct answer is option D, i.e., ${\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{COOH}}$..
Additional Information:
Hoffmann bromoamide reaction is a degradation reaction, where migration of an aryl or alkyl group from carbonyl carbon of amide to nitrogen atom takes place. So, the amine forms in the reaction contain one less carbon atom than that of the amide.
Note:
Students might get confused in choosing the correct carboxylic acid from the given options. They might think that, as the product of the reaction (ethylamine) consists of two carbon atoms, so the carboxylic acid also has two carbon atoms. But they must not forget that Hoffmann bromoamide reaction is a degradation reaction.
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