• # question_answer Methyl benzene can be prepared by reacting  benzene with bromomethane in the presence of A) $B{{r}_{2}}/CC{{l}_{4}}$                        B) Dil.${{H}_{2}}S{{O}_{4}}$                     C)  Anhydrous$AlC{{l}_{3}}$D) $Ni/{{H}_{2}}.70{{\,}^{o}}C$

${{C}_{6}}{{H}_{6}}+C{{H}_{3}}Br\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{5}}C{{H}_{3}}+HBr$ This reaction is known as Friedal-Craft's alkylatiol Hence, the correct option is (c).