A) sodium ethanoate
B) sodium propanoate
C) sodium methanoate
D) sodium ethoxide
Correct Answer: B
Solution :
Hypohalite\[(X{{O}^{-}},\text{ }X=Cl,Br,I)\]group oxidises ketones with terminal \[(C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,)\]group into acid salt. Hence, ethyl methyl ketone on treatment with a solution of sodium hypochlorite gives chloroform and sodium propanoate. \[\underset{ethyl\,methyl\,ketone}{\mathop{{{C}_{2}}{{H}_{5}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{3}}}}\,\xrightarrow{NaOCl}\underset{\begin{smallmatrix} \,\,\,\,\,\,\,sodium \\ propanoate \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{5}}COONa}}\,+\underset{chloroform}{\mathop{CHC{{l}_{3}}}}\,\]
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