Which will give Fe^ 3+ ions in solution ?

Benzene $(C_6H_6)$ $+\,C{{H}_{3}}-CH=C{{H}_{2}}\,\xrightarrow{{{H}^{\oplus }}}A\xrightarrow{{{O}_{2}}}B$ $\xrightarrow{{{H}^{\oplus }}\,/\,\Delta }C+$$\begin{matrix}
O \\
|| \\
C{{H}_{3}}-C-C{{H}_{3}} \\
\end{matrix}$

The structure of intermediate compound $'B'$ will be

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The major product in the given reaction is :

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The atomic weight of an element is double its atomic number. If there are four electrons in $2p$ orbital, the element is

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Out of all the known elements the number of transitional elements is

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Lanthanum is grouped with $f-$ block elements because

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$\mathrm{RCONH}_2$ is converted into $\mathrm{RNH}_2$ by means of Hofmann bromamide degradation.

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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 $3.$