A) \[B{{r}_{2}}\]
B) \[{{O}_{3}}\]
C) conc.\[{{H}_{2}}S{{O}_{4}}\]
D) \[KMn{{O}_{4}}\]
Correct Answer: D
Solution :
Key Idea: On oxidation with \[KMn{{O}_{4}}\] n-pentane and iso-pentane give different oxidation products. So, by identifying these products we can distinguish n-pentane and iso-pentane. \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\xrightarrow[[O]]{KMn{{O}_{4}}}\]\[\underset{{{\text{1}}^{o}}\text{ }alcohol}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH}}\,\] \[C{{H}_{3}}C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ \underset{iso-pentane}{\mathop{C{{H}_{3}}}}\, \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}\xrightarrow[[O]]{KMn{{O}_{4}}}\]\[C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \\ {{3}^{o}}\text{ }alcohol \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{3}}\] n-pentane gives \[{{1}^{o}}\] alcohol and iso-pentane gives \[{{3}^{o}}\] alcohol and the alcohols can be distinguished by Lucas test \[\therefore \] n-pentane and iso-pentane can be distinguished by \[KMn{{O}_{4}}\].
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