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