Carboxylic Acids and their Derivatives - Result Question 40
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.
$(2006,3 \times 4 M=12 M)$
26. Which is the rate determining step in Hofmann’s bromamide degradation?
(a) Formation of (i)
(b) Formation of (ii)
(c) Formation of (iii)
(d) Formation of (iv)
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Answer:
Correct Answer: 26. (d)
Solution:
Rearrangement of (iii) to (iv) is the rate determining step :
BETA
