2 text{ Benzene} xrightarrow[{AlCl_3}]{{CH_2C | vedclass

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Question

(B) The reaction of $2$ moles of benzene with dichloromethane $(CH_2Cl_2)$ in the presence of a Lewis acid catalyst like $AlCl_3$ is a Friedel-Crafts

Vedclass · Accepted Answer

Diphenylmethane

Vedclass · Answer

(B) The reaction of $2$ moles of benzene with dichloromethane $(CH_2Cl_2)$ in the presence of a Lewis acid catalyst like $AlCl_3$ is a Friedel-Crafts alkylation reaction. In this reaction,the $CH_2Cl_2$ acts as an alkylating agent. First,one chlorine atom is replaced by a phenyl group to form benzyl chloride $(C_6H_5CH_2Cl)$. Then,the second chlorine atom is replaced by another phenyl group to form diphenylmethane $(C_6H_5CH_2C_6H_5)$. The overall reaction is: $2 C_6H_6 + CH_2Cl_2 \xrightarrow{AlCl_3} C_6H_5CH_2C_6H_5 + 2 HCl$ Thus,the product is diphenylmethane.

$2 \text{ Benzene} \xrightarrow[{AlCl_3}]{{CH_2Cl_2}} \text{ Product}$

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