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