MCQ
Which product is obtained on heating sodium acetate with sodalime?
- AButane
- BPropane
- CEthane
- ✓Methane
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$\xrightarrow{{Mg/THF}}A\mathop {\xrightarrow{{(i)\,C{H_3}CHO}}}\limits_{(ii)\,aq.N{H_4}Cl} B$
$ClF _{3}, IF _{7}, BrF _{5}, BrF _{3}, I _{2} Cl _{6}, IF _{5}, ClF , ClF _{5}$
$(I)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow[{5\% \,\,water}]{{95\% \,\,\,acetone}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(II)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow[{10\% \,\,water}]{{90\% \,\,\,acetone}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(III)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow[{20\% \,\,water}]{{80\% \,\,\,acetone}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(IV)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow{{100\% \,water}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
Arrange these reactions in decreasing order of greater proportion of inverted product and select correct answer from the codes given below :