$(i)\,CH_3CH_2OH$
$(ii)\ CH_3COCH_3$
$(iii)\ \begin{array}{*{20}{c}}
{C{H_3} - CHOH} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iv) \,CH_3OH$
Which of the above compound(s), on being warmed with iodine solution and $NaOH,$ will give iodoform ?
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$(I)\,\,Zn\,+\,Conc. HNO_3\,(hot) \to Zn(NO_3)_2\,+\,X\,+\,H_2O$
$(II)\,\,Zn\,+\,dil. HNO_3\,(Cold) \to Zn(NO_3)_2\,+\,Y\,+\,H_2O$
Compound $X$ and $Y$ are respectively
$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} O{\mkern 1mu} } \\
{\,\,\,|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} ||} \\
{C{H_3} - CH - C - OH}
\end{array}$ $+ C{H_3} - N{H_2} \to 'A'\xrightarrow[\Delta ]{}'B'\xrightarrow{\begin{subarray}{l}
{\text{LiAl}}{{\text{H}}_4} \\
{\text{(excess)}}
\end{subarray} }'C'$
The final product $‘C’$ will be