A) \[{{C}_{2}}{{H}_{5}}I\]
B) \[{{C}_{2}}{{H}_{5}}OH\]
C) \[CH{{I}_{3}}\]
D) \[C{{H}_{3}}CHO\]
Correct Answer: C
Solution :
\[C{{H}_{2}}=C{{H}_{2}}\xrightarrow{HBr}{{H}_{3}}C-\underset{\underset{(X)}{\mathop{Br}}\,}{\mathop{\underset{|}{\mathop{C}}\,}}\,{{H}_{2}}\xrightarrow{aq.\,KOH}C{{H}_{3}}-\underset{\underset{(Y)}{\mathop{OH}}\,}{\mathop{\underset{|}{\mathop{C}}\,}}\,{{H}_{2}}\]\[C{{H}_{3}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,{{H}_{2}}\xrightarrow[{{I}_{2}}\,excess]{N{{a}_{2}}C{{O}_{3}}}\underset{(Z)}{\mathop{\underset{Iodoform}{\mathop{CH{{I}_{3}}}}\,}}\,\]
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