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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 $3.$

$Toluene\,\, \xrightarrow[{2.\,{H^ + }}]{{1.\,KMn{O_4}/O{H^ - }}}I\xrightarrow{{SOC{l_2}}}II\xrightarrow{{N{H_3}}}III\xrightarrow{{OB{r^ - }}}IV$ is