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(D) The reaction of anisole $(C_6H_5OCH_3)$ with isobutyl bromide $((CH_3)_2CHCH_2Br)$ in the presence of $AlCl_3$ is a Friedel-Crafts alkylation reac

Vedclass · Accepted Answer

$1-$methoxy$-4-$tert-butylbenzene

Vedclass · Answer

(D) The reaction of anisole $(C_6H_5OCH_3)$ with isobutyl bromide $((CH_3)_2CHCH_2Br)$ in the presence of $AlCl_3$ is a Friedel-Crafts alkylation reaction.$AlCl_3$ is a Lewis acid that coordinates with the bromine atom of the alkyl halide to form a carbocation intermediate.Initially,a primary carbocation $(CH_3)_2CHCH_2^+$ is formed,which undergoes a $1,2-$hydride shift to form a more stable tertiary carbocation $(CH_3)_3C^+$.This tertiary carbocation then acts as an electrophile and attacks the benzene ring of anisole.The $-OCH_3$ group is an ortho/para directing group.Due to steric hindrance at the ortho position,the para-substituted product is the major product.Therefore,the major product is $1-$methoxy$-4-$tert-butylbenzene.

The major product formed in the reaction is:
$C_6H_5OCH_3 + (CH_3)_2CHCH_2Br \xrightarrow{AlCl_3}$

Similar Questions

Which of the following undergoes $S_{N^{1}}$ reaction?

Reaction $(A)$:
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Correct statement is:

Which of the following chloro-compounds shows complete inversion of configuration during a reaction?

Identify the reagent $A$ in the following conversion.
Alkyl halide $\xrightarrow{A}$ Alkyl nitrite

The major product $(A)$ formed in the reaction given below is:

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The major product formed in the reaction is:$C_6H_5OCH_3 + (CH_3)_2CHCH_2Br \xrightarrow{AlCl_3}$