A) Ethane
B) Acetylene
C) Ethyl alcohol
D) None of these
Correct Answer: A
Solution :
\[{{C}_{2}}{{H}_{5}}-COONa\underset{\text{Propionate ion}}{\mathop{\underset{\downarrow }{\mathop{{{C}_{2}}{{H}_{5}}CO{{O}^{-}}}}\,}}\,+N{{a}^{+}}\] At anode: \[{{C}_{2}}{{H}_{5}}CO{{O}^{-}}\xrightarrow{{}}\underset{\text{Propionate}\,free\,radical}{\mathop{{{C}_{2}}{{H}_{5}}-COO+{{e}^{-}}}}\,\] \[{{\overset{\centerdot }{\mathop{C}}\,}_{2}}{{H}_{5}}-{{\overset{\centerdot }{\mathop{C}}\,}_{2}}{{H}_{5}}\xrightarrow{{}}{{C}_{4}}{{H}_{10}}\text{butane}\] \[{{\overset{\centerdot }{\mathop{C}}\,}_{2}}{{H}_{5}}-{{\overset{\centerdot }{\mathop{C}}\,}_{2}}{{H}_{5}}\xrightarrow{{}}\underset{Ethylene}{\mathop{{{C}_{4}}{{H}_{4}}}}\,+\underset{Ethane}{\mathop{{{C}_{2}}{{H}_{6}}}}\,\] \[{{C}_{2}}{{H}_{5}}\overset{\centerdot }{\mathop{C}}\,OO+{{\overset{\centerdot }{\mathop{C}}\,}_{2}}{{H}_{5}}\xrightarrow{{}}\underset{(ester)}{\mathop{\underset{Ethyl\,propionate}{\mathop{{{C}_{2}}{{H}_{5}}COO{{C}_{2}}{{H}_{5}}}}\,}}\,\]
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