Benzene reacts with n-propyl chloride in the | vedclass

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(B) This reaction is known as Friedel-Crafts alkylation. In the presence of anhydrous $AlCl_3$,$n-$propyl chloride acts as an alkylating agent. It is

Vedclass · Accepted Answer

Isopropyl benzene

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(B) This reaction is known as Friedel-Crafts alkylation. In the presence of anhydrous $AlCl_3$,$n-$propyl chloride acts as an alkylating agent. It is an electrophilic aromatic substitution reaction.Step $1$: Generation of electrophile:$CH_3-CH_2-CH_2-Cl + AlCl_3 \rightarrow CH_3-CH_2-CH_2^+ + AlCl_4^-$Step $2$: Carbocation rearrangement:The primary carbocation $(CH_3-CH_2-CH_2^+)$ formed in step $1$ rearranges to form a more stable secondary carbocation $(CH_3-CH^+-CH_3)$ via a $1,2-$hydride shift.Step $3$: Electrophilic attack:This secondary carbocation attacks the benzene ring to form the intermediate.Step $4$: Deprotonation:The final step is the abstraction of a proton from the intermediate to restore aromaticity,yielding isopropyl benzene (cumene) as the major product.

Benzene reacts with $n-$propyl chloride in the presence of anhydrous $AlCl_3$ to give predominantly

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