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 :
\[C{{H}_{2}}=C{{H}_{2}}\xrightarrow{HBr}C{{H}_{3}}-\underset{\underset{Br\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}}}\,\xrightarrow{\text{Hydrolysis}}\]\[C{{H}_{3}}-\underset{\underset{OH\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}}}\,\underset{{{I}_{2}}\,\text{excess}}{\mathop{\xrightarrow{N{{a}_{2}}C{{O}_{3}}}}}\,\underset{\begin{smallmatrix} \text{Yellow}\,\text{ppt} \\ \left( \text{Iodoform} \right) \end{smallmatrix}}{\mathop{\,CH{{I}_{3}}}}\,\]
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