- A$1-$ butane
- B$1-$ butene
- ✓$1-$ butyne
- D$2-$ butene
$\underset{\text{Ethyl bromide}}{\mathop{{{C}_{2}}{{H}_{5}}Br}}\,+\underset{\text{ sodium}}{\mathop{NaC}}\,\equiv \underset{\text{acetylide}}{\mathop{CH}}\,\to $ $\underset{\text{ }1\text{ -butyne}}{\mathop{{{C}_{2}}{{H}_{5}}C\equiv CH}}\,+\underset{\begin{smallmatrix}
\text{sodium } \\
\text{bromide}
\end{smallmatrix}}{\mathop{NaBr\text{ }}}\,$
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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 $3.$
