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 Baeyer's reagent. It decolourises by alkene and alkyne both compound which decolourises Baeyer's reagent shows the presence of multiple bonds. \[\underset{C{{H}_{2}}}{\overset{C{{H}_{2}}}{\mathop{||}}}\,+{{H}_{2}}O+[O]\xrightarrow[KMn{{O}_{4}}]{1%\,\,Alkaline}\underset{ethylene\,glycol}{\mathop{\underset{C{{H}_{2}}-OH}{\overset{C{{H}_{2}}-OH}{\mathop{|}}}\,}}\,\] \[\underset{CH}{\overset{CH}{\mathop{|||}}}\,+2[O]\xrightarrow[KMn{{O}_{4}},\,\,{{H}_{2}}O]{1%\,\,Alkaline}\underset{glyoxal}{\mathop{\underset{H}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{\underset{C=O}{\overset{C=O}{\mathop{|}}}\,}}}\,}}}\,}}\,\]
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