
- A$A = HCl ;$ Anhydrous $AlCl _{3}$
- B$A = HCl , ZnCl _{2}$
- ✓$A = Cl _{2} ; UV$ light
- D$A = Cl _{2}$; dark, Anhydrous $AlCl _{3}$

The reaction is free radical halogenation.
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$ {I_2} + 2{e^ - } \to \,2{I^ - }\,;\,\,{E^o}\, = \,\,0.54\,\,V $
$ MnO_4^ - \, + \,8{H^ + }\, + \,5{e^ - } \to \,M{n^{2 + }}\, + \,4{H_2}O\,;\,{E^{o\,}} = 1.52\,\,V $
$ F{e^{3 + }} + {e^ - } \to \,\,F{e^{2 + }}\,;\,\,{E^o}\, = \,\,0.77\,\,V $
$ S{n^{4 + }} + 2{e^ - } \to \,\,S{n^{2 + }}\,;\,\,{E^o}\, = \,\,0.1\,\,V $
The strongest reducant and oxidant respectively are
$Xe{F_6}\,\xrightarrow{{ + {H_2}O}}A\,\xrightarrow{{ + {H_2}O}}B\,\xrightarrow{{ + {H_2}O}}C$

$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\
{C{H_3} - CH = CH - C{H_2} - CH - C{H_3}}
\end{array}$ $\longrightarrow $ $C{H_3} - CH = CH - C{H_2}C{O_2}H$

Considering the above chemical reaction, identify the product $"X"$