Carboxylic Acids and their Derivatives - Result Question 41
Passage
$R \mathrm{CONH}_2$ is converted into $R \mathrm{NH}_2$ by means of Hofmann’s bromamide degradation.
In this reaction, $R \mathrm{CONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann’s degradation reaction is an intramolecular reaction.
$(2006,3 \times 4 M=12 M)$
27. What are the constituent amines formed when the mixture of (1) and (2) undergoes Hofmann’s bromamide degradation?
Show Answer
Answer:
Correct Answer: 27. (b)
Solution:
The rate determining step of Hofmann’s bromamide reaction is unimolecular rearrangement of bromamide anion (iii) and no cross-products are formed when mixture of amides are taken.
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