Benzene + (CH_3)_3C-CH_2Cl xrightarrow{AlCl_ | vedclass

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Question

(D) The reaction is a Friedel-Crafts alkylation. The alkyl halide is $(CH_3)_3C-CH_2Cl$ (neopentyl chloride). When $AlCl_3$ acts on this primary alkyl

Vedclass · Accepted Answer

tert-Pentylbenzene

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(D) The reaction is a Friedel-Crafts alkylation. The alkyl halide is $(CH_3)_3C-CH_2Cl$ (neopentyl chloride). When $AlCl_3$ acts on this primary alkyl halide,it forms a carbocation intermediate. Initially,a primary carbocation $(CH_3)_3C-CH_2^+$ is formed,which is highly unstable. It immediately undergoes a $1,2-$methyl shift to form a more stable tertiary carbocation $(CH_3)_2C^+-CH_2-CH_3$. This tertiary carbocation then attacks the benzene ring to form the major product,which is tert-pentylbenzene ($2$-methyl$-2-$phenylbutane).

Benzene $ + (CH_3)_3C-CH_2Cl \xrightarrow{AlCl_3} \text{Major product}$

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