A) \[{{C}_{6}}{{H}_{5}}Br+C{{H}_{2}}=CH-C{{H}_{2}}-ONa\]
B) \[C{{H}_{2}}=CH-C{{H}_{2}}-Br-{{C}_{6}}{{H}_{5}}ONa\]
C) \[{{C}_{6}}{{H}_{5}}-CH=CH-Br-C{{H}_{3}}-ONa\]
D) \[C{{H}_{2}}=CH-Br-{{C}_{6}}{{H}_{5}}-C{{H}_{2}}-ONa\]
Correct Answer: A
Solution :
\[{{C}_{6}}{{H}_{5}}Br+NaO-C{{H}_{2}}-CH=C{{H}_{2}}\xrightarrow[{}]{{}}\] \[\underset{Allyl\,phenyl\,ether}{\mathop{{{C}_{6}}{{H}_{5}}-O-C{{H}_{2}}-CH=C{{H}_{2}}+NaBr}}\,\]
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