A) 1-butene + HF
B) 2-butanol+ \[{{\operatorname{H}}_{2}}{{\operatorname{SO}}_{4}}\]
C) Butanoykhloride \[+AlC{{l}_{3}}\] then \[Zn,\,HCl\]
D) Butyl chloride \[+AlC{{l}_{3}}\]
Correct Answer: C
Solution :
The Friedal-crafts alkylation reaction will give propyi phenyl ketone which further on Clemmenson 's reduction will give butyl benzene \[{{C}_{6}}{{H}_{6}}+C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COCl\xrightarrow{AlC{{l}_{3}}}\] \[{{C}_{6}}{{H}_{5}}COC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\xrightarrow{Zn-Hg/HCl}\] \[\underset{Bultyl\,bezene}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\]
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