A) ethanoic acid
B) methanoic acid
C) propanoic acid
D) Either (a) or (c)
Correct Answer: C
Solution :
\[C{{H}_{3}}COOH+2NaOH\xrightarrow[630\,\,K]{CaO}\underset{Methane}{\mathop{C{{H}_{4}}}}\,\] \[+N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O\] \[C{{H}_{3}}C{{H}_{2}}COOH+2NaOH\xrightarrow[630\,\,K]{CaO}\] \[C{{H}_{3}}-\underset{ethan}{\mathop{C{{H}_{3}}}}\,+N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O\] Note: Methanoic acid does not undergo such kind of reaction.
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