A) \[{{C}_{2}}{{H}_{6}}\]
B) \[{{C}_{2}}{{H}_{5}}-{{C}_{2}}{{H}_{5}}\]
C) \[{{C}_{2}}{{H}_{5}}-O-{{C}_{2}}{{H}_{5}}\]
D) \[{{C}_{2}}{{H}_{5}}-C{{H}_{3}}\]
Correct Answer: C
Solution :
\[{{C}_{2}}{{H}_{5}}I\]and\[A{{g}_{2}}O\]reacts as below \[2{{C}_{2}}{{H}_{5}}I+A{{g}_{2}}O\xrightarrow[{}]{{}}\underset{Ether}{\mathop{{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}}}\,+{{I}_{2}}\] \[+2Ag\] Thus,\[{{C}_{2}}{{H}_{5}}-O-{{C}_{2}}{{H}_{5}}\]is produced.
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