Question

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

• A. 1-butene \(+\mathrm{HF}\)
• B. 2-butanol \(+\mathrm{H_2SO_4}\)
• C. Butanoyl chloride \(+\mathrm{AlCl_3}\) then \(\mathrm{Zn, HCl}\)
• D. Butyl choloride \(+\mathrm{AlCl_3}\)

Hint:

2-Phenylbutane requires the sec-butyl carbocation (or its equivalent) to attack benzene. Check which route gives a straight-chain butyl group instead.

Answer 1

The Correct Option is C

2-Phenylbutane is \(\mathrm{C_6H_5{-}CH(CH_3)CH_2CH_3}\); it forms when the sec-butyl cation attacks benzene.
(1) 1-Butene + HF follows Markovnikov addition to give 2-fluorobutane, which with a Lewis acid gives the sec-butyl cation, so 2-phenylbutane forms.
(2) 2-Butanol + \(\mathrm{H_2SO_4}\) is protonated and loses water to give the sec-butyl cation, so 2-phenylbutane forms.
(4) n-Butyl chloride + \(\mathrm{AlCl_3}\) gives the primary n-butyl cation, which undergoes a 1,2-hydride shift to the more stable sec-butyl cation, so 2-phenylbutane is the major product.
(3) Butanoyl chloride + \(\mathrm{AlCl_3}\) is Friedel-Crafts acylation giving \(\mathrm{C_6H_5COCH_2CH_2CH_3}\); Clemmensen reduction (Zn, HCl) converts C=O to \(\mathrm{CH_2}\), giving n-butylbenzene (no rearrangement), not 2-phenylbutane.
Answer: (3) Butanoyl chloride + AlCl₃ then Zn, HCl