Benzene + CH_3-CH_2-CH_2-Cl xrightarrow{AlCl_ | vedclass

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Question

(C) The reaction of benzene with $n$-propyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.In this reaction,th

Vedclass · Accepted Answer

Isopropylbenzene

Vedclass · Answer

(C) The reaction of benzene with $n$-propyl chloride in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.In this reaction,the primary carbocation $(CH_3-CH_2-CH_2^+)$ formed initially is unstable and undergoes a $1,2-$hydride shift to form a more stable secondary carbocation $(CH_3-CH^+-CH_3)$.This secondary carbocation then attacks the benzene ring to form isopropylbenzene (cumene) as the major product.Therefore,the correct option is $C$.

Benzene + $CH_3-CH_2-CH_2-Cl$ $\xrightarrow{AlCl_3}$ the product is:

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