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