Question

What is the major product of the reaction between benzene and iso-butyl bromide in the presence of AlCl3?

• A. Cumene
• B. Toluene
• C. n-butyl benzene
• D. Tert-butyl benzene

Hint:

Check whether the carbocation formed from iso-butyl bromide is stable, or if it can rearrange.

Answer 1

The Correct Option is D

Step 1: This is a Friedel-Crafts alkylation. Iso-butyl bromide, $(CH_3)_2CHCH_2Br$, reacts with the Lewis acid AlCl3, which abstracts the bromide ion to generate a carbocation. \[ (CH_3)_2CHCH_2Br + AlCl_3 \rightarrow (CH_3)_2CHCH_2^+ + AlCl_3Br^- \] Step 2: The carbocation formed first is a primary carbocation, which is highly unstable. Step 3: A hydride shift occurs from the adjacent carbon to the positively charged carbon, converting the unstable primary carbocation into a much more stable tertiary carbocation. \[ (CH_3)_2CHCH_2^+ \rightarrow (CH_3)_3C^+ \] Step 4: This rearranged tert-butyl cation is the electrophile that actually attacks the benzene ring, since Friedel-Crafts alkylations always proceed through the most stable available carbocation. Step 5: Benzene attacks this tertiary carbocation, and loss of a proton from the arenium ion intermediate restores aromaticity, giving tert-butylbenzene. \[ C_6H_6 + (CH_3)_3C^+ \rightarrow C_6H_5C(CH_3)_3 + H^+ \] \[\boxed{\text{Tert-butyl benzene}}\]