A) butyric acid
B) isobutyric acid
C) propanoic acid
D) none of these
Correct Answer: B
Solution :
\[\underset{\underset{(A)}{\mathop{C{{H}_{3}}}}\,}{\mathop{\underset{|}{\mathop{{{H}_{3}}C-CH-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-OAg}}\,}}\,\xrightarrow[-C{{O}_{2}},AgBr]{CC{{l}_{4}}}\] \[\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{{{H}_{3}}C-CH-BR}}\,}}\,\xrightarrow[EtOH]{KOH}\underset{\Pr opene}{\mathop{{{H}_{3}}CCH=C{{H}_{2}}}}\,\]
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