Benzene xrightarrow[{Anhy. AlCl_3}]{{CH_3Cl}} | vedclass

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Question

(D) $1$. Benzene reacts with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) to form Toluene $(A)$: $C_6H_6 + CH_3Cl \xrigh

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$C = 	ext{Benzyl isocyanide}$

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(D) $1$. Benzene reacts with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) to form Toluene $(A)$: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 (A) + HCl$$2$. Toluene reacts with $Cl_2$ in the presence of light $(hv)$ via free radical substitution at the side chain to form Benzyl chloride $(B)$: $C_6H_5CH_3 + Cl_2 \xrightarrow{hv} C_6H_5CH_2Cl (B) + HCl$$3$. Benzyl chloride reacts with $AgCN$ (an ionic cyanide source) to form Benzyl isocyanide $(C)$ as the major product,because $AgCN$ is covalent and the nitrogen atom is more nucleophilic: $C_6H_5CH_2Cl + AgCN ightarrow C_6H_5CH_2NC (C) + AgCl$Thus,the correct statement is $C = 	ext{Benzyl isocyanide}$.

Benzene $\xrightarrow[{Anhy. AlCl_3}]{{CH_3Cl}} A \,$ $\xrightarrow[{hv}]{{Cl_2 (1 \, mole)}} B$ $\xrightarrow{{AgCN}} C$
Which is correct for the above reaction?

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Benzene $\xrightarrow[{Anhy. AlCl_3}]{{CH_3Cl}} A \,$ $\xrightarrow[{hv}]{{Cl_2 (1 \, mole)}} B$ $\xrightarrow{{AgCN}} C$Which is correct for the above reaction?