A) \[C{{H}_{3}}I\]
B) \[C{{H}_{3}}OH\]
C) \[C{{H}_{3}}C{{H}_{2}}I\]
D) \[{{C}_{2}}{{H}_{5}}OH\]
Correct Answer: A
Solution :
\[C{{H}_{3}}I+2[H]\xrightarrow[or\,Zn-Cu\,or\,{{C}_{2}}{{H}_{5}}OH]{Zn/HCl}\underset{Methane}{\mathop{C{{H}_{4}}+HI}}\,\] \[2C{{H}_{3}}I+2Na\xrightarrow[{}]{dry\,ether}\underset{Ethane}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,+2NaI\] Therefore, from CH3l both methane and ethane can be prepared.You need to login to perform this action.
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