Two linkage isomers possible $NO _2 \rightarrow$ Ambidentate ligand
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$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Z}^{-}$$\xrightarrow[{{\text{Sublimation}}}]{{{k_s}}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Z}+\mathrm{Br}^{-}$
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Z}^{-}$$\xrightarrow[{{\text{elimination}}}]{{{k_e}}}\mathrm{CH}_{3} \mathrm{CH}= \mathrm{CH}_{2} +\mathrm{HZ}+\mathrm{Br}^{-}$
where
$\mathrm{Z}^{-}=\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(\mathrm{A})$ or $\begin{array}{*{20}{c}}
{\,C{H_3}} \\
{|\,\,\,\,\,} \\
{C{H_3} - C - {O^ - }(B)} \\
{|\,\,\,\,} \\
{\,\,C{H_3}}
\end{array}$
$\mathrm{k}_{\mathrm{s}}$ and $\mathrm{k}_{\mathrm{e}},$ are $,$ respectively, the rate constants for the substitution and elimination, and $\mu=\frac{\mathrm{k}_{\mathrm{s}}}{\mathrm{k}_{\mathrm{e}}},$ the correct options is
$\underset{{{C}_{3}}{{H}_{6}}C{{l}_{2}}}{\mathop{A}}\,\,\xrightarrow[2.\,\,{{H}^{+}}\,\,(\min d)]{\begin{smallmatrix}
1\,.\,\,alcoholic\,\,KOH\,/\,\Delta \\
followed\,\,by\,\,NaN{{H}_{2}}\,/\,\Delta
\end{smallmatrix}}$ $B\,\xrightarrow[{{H}_{2}}{{O}_{2}}/\overset{\Theta }{\mathop{O}}\,H]{{{B}_{2}}{{H}_{6}}}C$
$A\,\xrightarrow{aq.\,\,KOH\,(excess)}C$
