Question

Which of the following would not give $2$ -phenylbutane as the major product in a Friedel-Crafts alkylation reaction with benzene?

• A. $1$ -butene + $HF$
• B. $2$ -butanol + $H_{2}SO_{4}$
• C. Butanoylchloride + $AlCl_{3}$ then $\text{Zn-Hg/HCl}$
• D. Butyl chloride + $AlCl_{3}$

Answer 1

Answer: (C)

The Friedel - crafts alkylation reaction will give propyl phenyl ketone which on further
Clemmensen's reduction will give butyl benzene.
$C _6 H _6+ CH _3 CH _2 CH _2 COCl \xrightarrow{ AlCl _3} C _6 H _5 COCH _2 CH _2 CH _3 \xrightarrow{ Zn - Hg / HCl } \underset{\text{Butylbenzene}}{C _6 H _5 CH _2 CH _2 CH _2 CH _3}$
Rest other will give 2-phenylbutane as in them rearrangement of carbocation is possible.