A) \[{{C}_{2}}{{H}_{5}}OH\]
B) \[{{C}_{2}}{{H}_{5}}I\]
C) \[C{{H}_{3}}CHO\]
D) \[CH{{I}_{3}}\]
Correct Answer: D
Solution :
\[C{{H}_{2}}=C{{H}_{2}}+HBr\xrightarrow{{}}\underset{X}{\mathop{C{{H}_{3}}-C{{H}_{2}}Br}}\,\] \[\underset{X}{\mathop{C{{H}_{3}}C{{H}_{2}}Br}}\,\xrightarrow[-KBr]{Aq.KOH}\underset{Y}{\mathop{\underset{ethyl\,alcohol}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,}}\,\] \[C{{H}_{3}}C{{H}_{2}}OH+N{{a}_{2}}C{{O}_{3}}+{{I}_{2}}\xrightarrow[(idoform\,test)]{}\underset{Z}{\mathop{\underset{iodoform}{\mathop{CH{{I}_{3}}}}\,}}\,\] \[+HCOONa+NaI+C{{O}_{2}}+{{H}_{2}}O\]
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