A) \[C{{H}_{3}}C{{H}_{2}}OH+{{C}_{6}}{{H}_{6}}\]
B) \[C{{H}_{3}}CH=C{{H}_{2}}+{{C}_{6}}{{H}_{6}}\]
C) \[C{{H}_{2}}=C{{H}_{2}}+{{C}_{6}}{{H}_{6}}\]
D) \[C{{H}_{3}}-C{{H}_{3}}+{{C}_{6}}{{H}_{6}}\]
Correct Answer: A
Solution :
(a, c) \[{{C}_{6}}{{H}_{5}}+C{{H}_{2}}=C{{H}_{2}}\xrightarrow[\Delta ,P]{AlC{{l}_{2}}}\] \[\underset{(See\,Friedel-Crafts\,reaction)}{\mathop{{{C}_{6}}{{H}_{5}}{{C}_{2}}{{H}_{5}}}}\,\]
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