What is the role of $A l C l_{3}$ in Friedel crafts acylation reaction?

Primary attack on reactant to form intermediate

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Removal of water

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Stabilization of intermediate

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None of these

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Solution

The correct option is C Primary attack on reactant to form intermediate
In Friedel crafts acylation reaction, $A l C l_{3}$ attacks reactant to form intermediate. $A l C l_{3}$ reacts with acetyl chloride to form donor-acceptor complex. This complex then produces acyl cation (acylium ion). Then benzene ring attacks this acyl cation (acylium ion).

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Similar questions

In Friedel crafts alkylation and acylation the attacking reagent is an:

C H_{3} M g B r

(C_{2} H_{5})_{2} O

H_{2} O

H_{2} S O_{4}

0^{\circ} C

C H_{3} C O C l

A l C l_{3}

C H_{2} C l_{2}

H_{2} O

A l C l_{3}

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