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
✓
Answer
Correct option: D.
all of these
d
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$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{C{H_2} - OH\,\,} \\
{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
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{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
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