A) Butane
B) Ethyl ethanoate
C) Ethyl propanoate
D) Ethene
Correct Answer: B
Solution :
\[C{{H}_{3}}C{{H}_{2}}COOK\xrightarrow[{}]{Electrolysis}\underset{I}{\mathop{C{{H}_{3}}C{{H}_{2}}CO{{O}^{\bullet }}}}\,\] \[\xrightarrow[{}]{{}}\underset{II}{\mathop{CH_{3}^{\bullet }C{{H}_{2}}}}\,\] \[2II\xrightarrow[{}]{{}}\]Butane;\[I+II\xrightarrow[{}]{{}}\]ethyl propanoate \[II\xrightarrow[{}]{disproportionation}\]ethene + ethane.
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