- A$4$
- B$2$
- C$5$
- ✓$6$
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Given $\mathrm{R}=8.31\, \mathrm{~J} \,\mathrm{~K}^{-1} \,\mathrm{~mol}^{-1} ; \log 6.36 \times 10^{-3}=-2.19$ $\left.10^{-4.79}=1.62 \times 10^{-5}\right]$
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In this reaction, $\mathrm{RCONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.
$1.$ How can the conversion of $(i)$ to $(ii)$ be brought about?
$(A)$ $\mathrm{KBr}$ $(B)$ $\mathrm{KBr}+\mathrm{CH}_3 \mathrm{ONa}$ $(C)$ $\mathrm{KBr}+\mathrm{KOH}$ $(D)$ $\mathrm{Br}_2+\mathrm{KOH}$
$2.$ Which is the rate determining step in Hofmann bromamide degradation?
$(A)$ Formation of $(i)$ $(B)$ Formation of $(ii)$ $(C)$ Formation of $(iii)$ $(D)$ Formation of $(iv)$
$3.$ What are the constituent amines formed when the mixture of $(i)$ and $(ii)$ undergoes Hofmann bromamide degradation?
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give the answer question $1$, $2$, and $3.$