Among the following reaction(s) Which gives (give) tert-butylbenzene as the major product
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → B, C and D
tert-Butylbenzene is formed by Friedel–Crafts alkylation of benzene when the electrophile is the tert-butyl carbocation, \( \ce{(CH3)3C+} \). Any reagent that produces this cation in the presence of benzene gives the product; a primary cation formed on the way rearranges (1,2-hydride shift) to the more stable tertiary cation.
\( \ce{C6H6 + (CH3)3C+ -> C6H5C(CH3)3 + H+} \)
(A) tert-Butyl bromide with \( \ce{C2H5ONa} \) — Does not give tert-butylbenzene. Ethoxide is a strong base; with a tertiary halide it causes elimination to 2-methylpropene, and no carbocation is formed for the ring to attack.
\( \ce{(CH3)3CBr + C2H5ONa -> (CH3)2C=CH2 + C2H5OH + NaBr} \)
(B) Isobutyl chloride with \( \ce{AlCl3} \) — Gives tert-butylbenzene. \( \ce{AlCl3} \) removes the chloride; the primary isobutyl cation rearranges by a 1,2-hydride shift to the tertiary cation.
\( \ce{(CH3)2CHCH2Cl + AlCl3 -> (CH3)2CHCH2+ + AlCl4-} \)
\( \ce{(CH3)2CHCH2+} \xrightarrow{\text{1,2-H shift}} \ce{(CH3)3C+} \)
(C) 2-Methylpropene with \( \ce{H2SO4} \) — Gives tert-butylbenzene. The acid protonates the alkene at the \( \ce{CH2} \) end (Markovnikov), giving the tertiary cation directly.
\( \ce{(CH3)2C=CH2 + H+ -> (CH3)3C+} \)
(D) 2-Methylpropan-1-ol with \( \ce{BF3.OEt2} \) — Gives tert-butylbenzene. The Lewis acid removes the –OH group; the primary cation rearranges to the tertiary cation.
\( \ce{(CH3)2CHCH2OH + BF3 -> (CH3)2CHCH2+ + [HOBF3]-} \)
\( \ce{(CH3)2CHCH2+} \xrightarrow{\text{1,2-H shift}} \ce{(CH3)3C+} \)
(E) 1-Chloropropane (n-propyl chloride) with \( \ce{H2SO4} \) — Does not give tert-butylbenzene. A three-carbon unbranched chain cannot give a tert-butyl cation; at most it could give an isopropyl group.