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