A) \[B{{r}_{2}}\]
B) \[{{O}_{3}}\]
C) cone, \[{{H}_{2}}S{{O}_{4}}\]
D) \[KMn{{O}_{4}}\]
Correct Answer: D
Solution :
On oxidation, with \[KMn{{O}_{4}}\], they give different alcohols. \[\underset{\text{n-pentane}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,-\xrightarrow[[O]]{KMn{{O}_{4}}}\] \[\underset{{{1}^{o}}alcohol}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH}}\,\] \[\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C{{H}_{3}}C{{H}_{2}}-CH-C{{H}_{3}}\xrightarrow[[O]]{KMn{{O}_{4}}}}}\,}}\,\] \[\underset{{{3}^{\mathbf{o}}}-alcohol}{\mathop{\overset{OH}{\mathop{\overset{|}{\mathop{\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C{{H}_{3}}C{{H}_{2}}-C-C{{H}_{3}}}}\,}}\,}}\,}}\,}}\,\]
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