Benzene + CH_3CH_2CH_2CH_2Cl xrightarrow{AlCl | vedclass

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Question

(A) The reaction is a Friedel-Crafts alkylation of benzene with $n$-butyl chloride in the presence of a Lewis acid catalyst,$AlCl_3$.$1$. The reaction

Vedclass · Accepted Answer

sec-Butylbenzene

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(A) The reaction is a Friedel-Crafts alkylation of benzene with $n$-butyl chloride in the presence of a Lewis acid catalyst,$AlCl_3$.$1$. The reaction proceeds through the formation of a carbocation intermediate.$2$. Initially,$n$-butyl chloride reacts with $AlCl_3$ to form a primary carbocation,$CH_3CH_2CH_2CH_2^+$.$3$. Primary carbocations are unstable and undergo rearrangement to form more stable carbocations. The primary carbocation undergoes a $1,2$-hydride shift to form a secondary carbocation,$CH_3CH_2CH^+CH_3$.$4$. This secondary carbocation then undergoes electrophilic aromatic substitution with benzene to form $sec$-butylbenzene as the major product.

Benzene + $CH_3CH_2CH_2CH_2Cl \xrightarrow{AlCl_3}$ hydrocarbon $(X)$. The major product $X$ is:

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