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 Friedal - crafts alkylation reaction will give propyl phenyl ketone which on further Clemmenson’s reduction will give butyl benzene.

$\text{C}_{\text{6}} \text{H}_{\text{6}} \, \text{+} \, \text{CH}_{\text{3}} \text{CH}_{\text{2}} \text{CH}_{\text{2}} \text{COCl} \overset{\text{AlCl}_{\text{3}}}{ \rightarrow } \text{C}_{\text{6}} \text{H}_{\text{5}} \text{COCH}_{\text{2}} \text{CH}_{\text{2}} \text{CH}_{\text{3}} \overset{\text{Zn-Hg/HCl}}{ \rightarrow } \underset{\text{Butylbenzene}}{\text{C}_{\text{6}} \text{H}_{\text{5}} \text{CH}_{\text{2}} \text{CH}_{\text{2}} \text{CH}_{\text{2}} \text{CH}_{\text{3}}}$

Rest other will give 2-phenylbutane as in them rearrangement of carbocation is possible.