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