A) Ether
B) Alcohol
C) Alkylhalide
D) Ketone
Correct Answer: A
Solution :
\[{{C}_{2}}{{H}_{5}}ONa+I{{C}_{2}}{{H}_{5}}\xrightarrow{{}}\] \[\underset{\text{Di ethyl ether}}{\mathop{{{C}_{2}}{{H}_{5}}-O-{{C}_{2}}{{H}_{5}}+NaI}}\,\] Ether form by Williamson's synthesis only.
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