A) Alkane
B) Alkene
C) Alkene and alkyne both
D) None of the above
Correct Answer: C
Solution :
\[1%\] alkaline \[KMn{{O}_{4}}\] solution is known as Baeyers reagent. It decolourises by nikene and alkyne both compound which decolourises Baeyers reagent shows the presence of multiple bonds. \[\begin{matrix} \begin{align} & C{{H}_{2}} \\ & || \\ & C{{H}_{2}} \\ \end{align} & +{{H}_{2}}O+[O]\xrightarrow[KMn{{O}_{4}}]{1%Alkaline} \\ \end{matrix}\underset{ethylene\,\,glycol}{\mathop{\begin{matrix} C{{H}_{2}}-OH \\ \begin{align} & | \\ & C{{H}_{2}}-OH \\ \end{align} \\ \end{matrix}}}\,\] \[\begin{matrix} \begin{align} & CH \\ & ||| \\ & CH \\ \end{align} & +2[O] \\ \end{matrix}\xrightarrow[KMn{{O}_{4}}\cdot {{H}_{2}}O]{1%\,\,Alkaline}\underset{\begin{align} & | \\ & C=O \\ & | \\ & C=O \\ & | \\ & H \\ & glyoxal \\ \end{align}}{\mathop{H}}\,\]
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