A) \[{{C}_{2}}{{H}_{6}}\]\[{{C}_{2}}{{H}_{5}}OH\]\[C{{H}_{4}}\]
B) \[C{{H}_{4}}\]\[{{C}_{2}}{{H}_{5}}OH\]\[{{C}_{2}}{{H}_{6}}\]
C) \[{{C}_{2}}{{H}_{6}}\]\[C{{H}_{3}}COC{{H}_{3}}\]\[{{C}_{3}}{{H}_{8}}\]
D) \[{{(C{{H}_{3}}CO)}_{2}}O\]\[{{C}_{2}}{{H}_{6}}\]\[{{C}_{2}}{{H}_{6}}\]
Correct Answer: B
Solution :
\[\underset{\text{sodium}\,\text{acetate}}{\mathop{C{{H}_{3}}COONA}}\,\xrightarrow[-N{{a}_{2}}C{{O}_{3}}]{\text{Sodalime}/\Delta }\underset{\begin{smallmatrix} methane \\ A \end{smallmatrix}}{\mathop{C{{H}_{4}}}}\,\] \[\underset{\text{sodium}\,\text{acetate}}{\mathop{C{{H}_{3}}COONA}}\,\xrightarrow[-N{{a}_{2}}C{{O}_{3}}]{\text{Sodalime}/\Delta }\underset{\begin{smallmatrix} methane \\ A \end{smallmatrix}}{\mathop{C{{H}_{4}}}}\,\] \[\underset{\text{acetic}\,\text{acide}}{\mathop{C{{H}_{3}}COOH}}\,\xrightarrow{LiAl{{H}_{4}}}\underset{\begin{smallmatrix} \text{ethanol} \\ B \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,\] \[C{{H}_{3}}COONa\xrightarrow[\begin{smallmatrix} -C{{O}_{2}} \\ -NaOH \end{smallmatrix}]{\text{Kolbe }\!\!\!\!\text{ s}\,\text{electrolysis}}\underset{\begin{smallmatrix} \text{ethane} \\ C \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{6}}}}\,\] Thus A, B, and C are respectively \[C{{H}_{4}},{{C}_{2}}{{H}_{5}}OH\]and \[{{C}_{2}}{{H}_{6}}\]
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