A) \[\alpha \]-chlorodiethyl ether
B) \[\alpha \], \[\alpha \]'-dichlorodiethyl ether
C) perchlorodiethyl ether
D) none of these.
Correct Answer: C
Solution :
In the presence of light and excess of chlorine, all the hydrogen atoms of diethyl ether are substituted to give perchlorodiethyl ether. \[C{{H}_{3}}C{{H}_{2}}-O-C{{H}_{2}}C{{H}_{3}}+\underset{(excess)}{\mathop{10C{{l}_{2}}}}\,\xrightarrow{hv}\] \[CC{{l}_{3}}CC{{l}_{2}}-O-CC{{l}_{2}}-CC{{l}_{3}}3+10HCl\] Perchlorodiethyl ether
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