A) \[C{{H}_{3}}C{{H}_{2}}Cl\]
B) \[C{{H}_{3}}ClC{{H}_{2}}OH\]
C) \[CHC{{l}_{2}}C{{H}_{2}}OH\]
D) \[CC{{l}_{3}}CHO\]
Correct Answer: D
Solution :
Primary alcohols are oxidized by halogens to form aldehydes. When \[{{C}_{2}}{{H}_{5}}OH\] reacts with \[C{{l}_{2}}\], it gets oxidized and to form chloral. \[\underset{Ethanol}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,\xrightarrow[\begin{smallmatrix} oxidation \\ (-2HCl) \end{smallmatrix}]{C{{l}_{2}}}\underset{Aldehyde}{\mathop{C{{H}_{3}}CHO}}\,\] \[\xrightarrow[chorination]{C{{l}_{2}}(excess)}\underset{Chloral}{\mathop{CC{{l}_{3}}CHO}}\,\]
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