MCQ
$HO - (CH_2)_6 - OH,$ this conversion can be achieved by


- A$O_3 , Zn,$ then $LiAlH_4$
- B$O_3 /H_2O_2,$ then $LiAlH_4$
- Ccold dil. $KMnO_4 , HIO_4$ then $LiAlH_4$
- ✓all of these


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$(i)\,\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus - C{H_3}} \\
| \\
{\,\,\,\,\,\,OC{H_3}}
\end{array}$
$(ii)\,\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus - C{H_3}} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iii)\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus {H_2} - NH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iv)\,C{H_3} - \mathop C\limits^ \oplus {H_2}$