A) \[{{C}_{2}}{{H}_{5}}Cl+LiAl{{H}_{4}}\xrightarrow{{}}\]
B) \[{{C}_{2}}{{H}_{5}}Cl+Mg\xrightarrow{Dry\text{ }ether}\]
C) \[{{C}_{2}}{{H}_{5}}Cl+{{C}_{2}}{{H}_{5}}ONa\xrightarrow{{}}\]
D) \[CHC{{l}_{5}}Cl+Ag(powder)\xrightarrow{\Delta }\]
Correct Answer: A
Solution :
\[2C{{H}_{3}}+2Na\xrightarrow{Dry\text{ }ether}\underset{ethane}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,+2Nal\] \[4{{C}_{2}}{{H}_{5}}Cl+LiAl{{H}_{4}}\xrightarrow{{}}\underset{ethane}{\mathop{4{{C}_{2}}{{H}_{6}}}}\,+AlC{{l}_{3}}+LiCl\]
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