Friedel craft's reaction is :

C_{6} H_{5} B r + 2 N a + {B r C H}_{3} ⟶ C_{6} H_{5} {C H}_{3} + 2 N a B r

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C_{6} H_{6} + {C H}_{3} C O C l {A l C l}_{3} - --- \to C_{6} H_{5} {C O C H}_{3} + H C l

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C_{6} H_{5} O H + Z n a n h y d r o u s - ------- \to C_{6} H_{6} + Z n O

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None of above

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Solution

The correct option is B $C_{6} H_{6} + {C H}_{3} C O C l {A l C l}_{3} - --- \to C_{6} H_{5} {C O C H}_{3} + H C l$

Friedel–Crafts alkylation involves the alkylation of an aromatic ring with an alkyl halide using a strong Lewis acid $A l {C l}_{3}$ as the catalyst.With anhydrous aluminium chloride as a catalyst, the alkyl group attaches at the former site of the chloride ion.

Only the second option is undergoing aromatic alkylation with $A l {C l}_{3}$ as a catalyst.

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