$C{H_3} - \mathop {\mathop {\mathop {\mathop C\limits^| }\limits^{Cl} - }\limits_{|\,\,\,\,} }\limits_{OH} C{H_2}CH = CHC{H_3}$
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${I^ - }(aq.) + MnO_4^ - (aq.)\xrightarrow[{weakly\,\,O{H^ - }}]{{Neutral\,\,or}}Y + Mn{O_2}$
$MnO_4^ - (aq.) + M{n^{2 + }}(aq.)\xrightarrow{{ZnS{O_4}}}Z + 4{H^ + }$
Products $X,\,Y$ and $Z$ are respectively :
Consider the above reaction and identify the intermediate ' $X$ '
$\begin{array}{*{20}{c}}
{HO - C{H_2} - CH - CH = C - C{H_2} - C - C - OH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,||} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,O\,}
\end{array}$