When excess C_6H_6 is reacted with CH_2Cl_2 i | vedclass

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(D) The reaction of benzene $(C_6H_6)$ with dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction

Vedclass · Accepted Answer

Diphenylmethane

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(D) The reaction of benzene $(C_6H_6)$ with dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.Since benzene is in excess,the reaction proceeds in two steps:$1$. $C_6H_6 + CH_2Cl_2 \xrightarrow{AlCl_3} C_6H_5-CH_2Cl + HCl$$2$. $C_6H_5-CH_2Cl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5-CH_2-C_6H_5 + HCl$The final product obtained is diphenylmethane $(C_6H_5-CH_2-C_6H_5)$.

When excess $C_6H_6$ is reacted with $CH_2Cl_2$ in the presence of anhydrous $AlCl_3$,which of the following products is obtained?

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