A) \[PC{{l}_{5}}\]
B) Reduction
C) Oxidation with potassium dichromate
D) Ozonolysis
Correct Answer: C
Solution :
n-propyl alcohol and isopropyl alcohol give different product on oxidation with \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}.\] \[\underset{\begin{smallmatrix} n-propyl \\ alcohol \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}}}\,-C{{H}_{2}}OH\xrightarrow[{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}]{[O]}\] \[\underset{\text{propionaldehyde}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CHO}}\,\] \[\underset{\text{isopropyl}\,\text{alcohol}}{\mathop{C{{H}_{3}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}}}\,\xrightarrow[{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}]{[O]}\] \[C{{H}_{3}}-\underset{\text{acetone}}{\mathop{\underset{O}{\mathop{\underset{|\,|}{\mathop{C}}\,}}\,}}\,-C{{H}_{3}}\]
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