A) \[C{{H}_{3}}C{{H}_{3}}\]
B)
C) \[C{{H}_{3}}CH=CHC{{H}_{2}}C{{H}_{3}}\]
D) \[\underset{\,\,\,\,\,C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{\,\,\,\,\,\,C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C{{H}_{3}}-C-C{{H}_{3}}}}\,}}\,}}\,}}\,\]
Correct Answer: C
Solution :
\[C{{H}_{3}}-CH=CH-C{{H}_{2}}-C{{H}_{3}}\] it decolourizes \[KMn{{O}_{4}}\] solution because double bond is present.
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