Question

What are A, B and C in the following reactions? (1) $ C{{H}_{3}}C{{O}_{2}}Na\xrightarrow{\text{Sodalime}/\Delta }A $ (2) $ C{{H}_{3}}C{{O}_{2}}H\xrightarrow{LiAI{{H}_{4}}}B $ (3) $ C{{H}_{3}}C{{O}_{2}}Na\xrightarrow{\text{Kolbe }\!\!\!\!\text{ s}\,\text{electrolysis}}C $

• 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}} $

Answer 1

Answer: (B)

$ \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}} $