A) \[{{C}_{2}}{{H}_{5}}Cl+Mg\xrightarrow{\text{dry}\,\text{ether}}\]
B) \[{{C}_{2}}{{H}_{5}}Cl+LiAl{{H}_{4}}\xrightarrow{{}}\]
C) \[{{C}_{2}}{{H}_{5}}Cl+{{C}_{2}}{{H}_{5}}ONa\xrightarrow{{}}\]
D) \[CHC{{l}_{3}}\xrightarrow{\text{Ag }\,\text{powder}}\]
Correct Answer: B
Solution :
Wurtz reaction: In this reaction alkyl halide reacts with sodium in presence of ether forms alkane. \[C{{H}_{3}}\underset{(Ether)}{\mathop{}}\,\,C{{H}_{3}}\xrightarrow{{}}\underset{\begin{smallmatrix} \text{Ethane} \\ \,\,\,\,\text{(X)} \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{6}}}}\,+2NaI\] So, the compound X will be ethane.
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