A) \[PC{{l}_{5}}\]
B) reduction
C) oxidation with potassium dichromate
D) ozonolysis
Correct Answer: C
Solution :
| [c] n-propyl alcohol and isopropyl alcohol gives different product on oxidation with \[{{K}_{2}}C{{r}_{7}}\]. |
| \[\underset{\begin{smallmatrix} \text{n-Propyl} \\ \text{ alcohol} \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}}}\,-C{{H}_{2}}OH\xrightarrow[{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}]{[O]}\]\[\underset{\text{Prophionaldehyde}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CHO}}\,\] |
| \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \\ \text{Isopropyl alcohol} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\xrightarrow[{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}]{[O]}\]\[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ O \\ \text{Acetone} \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}\] |
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