A) Methanol
B) Methanal
C) Propanol-1
D) Propanol-2
Correct Answer: D
Solution :
\[CaOC{{l}_{2}}+{{H}_{2}}O\to Ca{{(OH)}_{2}}+C{{l}_{2}}\] \[\underset{\text{2-propanol}}{\mathop{C{{H}_{3}}-\underset{\underset{OH}{\mathop{|\,\,\,\,}}\,}{\mathop{CH}}\,-C{{H}_{3}}+C{{l}_{2}}}}\,\to \]\[C{{H}_{3}}-\underset{\underset{O}{\mathop{||}}\,}{\mathop{C}}\,-C{{H}_{3}}+2HCl\] \[C{{H}_{3}}-\underset{\underset{O}{\mathop{||}}\,}{\mathop{C\,}}\,-C{{H}_{3}}+3C{{l}_{2}}\to CC{{l}_{3}}-CO-C{{H}_{3}}+3HCl\] \[\begin{matrix} CC{{l}_{3}}COC{{H}_{3}} \\ CC{{l}_{3}}COC{{H}_{3}} \\ \end{matrix}+Ca<\begin{matrix} OH \\ OH \\ \end{matrix}\to 2CHC{{l}_{3}}+\begin{matrix} C{{H}_{3}}COO \\ C{{H}_{3}}COO \\ \end{matrix}>Ca\]
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