A) \[AlC{{l}_{3}}\]
B) \[\text{anhyd}\text{.}\,\,\text{AlC}{{\text{l}}_{\text{3}}}\]
C) \[{{N}_{2}}\]
D) \[He\]
Correct Answer: B
Solution :
Friedel-Crafts reaction: In this reaction alkyl group is introduced in the benzene ring in presence of anhy. \[\text{AlC}{{\text{l}}_{\text{3}}}\]
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