Question
Which of the following would not give 2-phenylbutane as the major product in a Friedel-
Crafts alkylation reaction on benzene ring ?
A.
$${\text{1 - butene}} + HF$$
B.
$${\text{2 - butanol}} + {H_2}S{O_4}$$
C.
$${\text{Butanoyl chloride}} + AlC{l_3}\,{\text{then}}\,Zn,HCl$$
D.
$${\text{Butyl chloride}} + AlC{l_3}$$
Answer :
$${\text{Butanoyl chloride}} + AlC{l_3}\,{\text{then}}\,Zn,HCl$$
Solution :
The Friedal-crafts acylation reaction will give propyl phenyl ketone which further on Clemmenson’s reduction will give butyl benzene
$${C_6}{H_6} + C{H_3}C{H_2}C{H_2}COCl$$ \[\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{5}}COC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\] \[\xrightarrow{Zn-Hg/HCl}\underset{\text{Butyl benzene}}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\]