A) \[C{{H}_{3}}C{{H}_{2}}CHO\]
B) \[~C{{H}_{2}}=CHC{{H}_{2}}OH\]
C) \[C{{H}_{3}}COC{{H}_{3}}\]
D) \[C{{H}_{2}}=CH-O-C{{H}_{3}}\]
Correct Answer: D
Solution :
Since, the compound does not give a precipitate with 2, 4-dinitrophenylhydrazine, it cannot be an aldehyde or a ketone. Further since it does not react with metallic sodium, therefore, it cannot be an alcohol. Thus, the compound \[{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}\text{O}\]must be an ether \[i.e.,\]option (d) is correct.
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