
What is the role of $AlC{{l}_{3}}$ in Friedel crafts acylation reaction?
(A) Primary attack on the reactant to form intermediate.
(B) Removal of water
(C) Stabilization of the intermediate.
(D) None of these.
Answer
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Hint: $AlC{{l}_{3}}$ acts as a catalyst in this reaction. Catalyst is the substance which increases the rate of the reaction by altering the reaction mechanism. The catalyst will get regenerated at the end of the reaction.
Complete step by step solution:
First, what is Friedel crafts acylation reaction? Friedel crafts acylation reaction is an electrophilic aromatic substitution reaction. It is the reaction in which the acyl group is attached to the benzene ring in presence of the anhydrous $AlC{{l}_{3}}$ catalyst.
$C{{H}_{3}}COCl+{{C}_{6}}{{H}_{6}}\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{{}}}COC{{H}_{3}}+HCl$
- Friedel crafts reaction uses acetyl chloride as the acylating agent. Acid anhydride can also be used as an acylating agent. The catalyst used here is$AlC{{l}_{3}}$, in order to allow the attachment of the acetyl group to the aromatic ring. In Friedel crafts reaction the aromatic ring gets converted into a ketone, after the attachment of the acyl group to the ring.
- Acyl halide forms a highly electrophilic acylium ion with the strong Lewis acid$AlC{{l}_{3}}$ and later it is stabilized by resonance.
- Therefore, the role of $AlC{{l}_{3}}$ in Friedel crafts reaction is to attack the reactant to form an intermediate and later, stabilization of that intermediate.
So, the correct answer is “Option A and C”.
Additional Information:
-$AlC{{l}_{3}}$ is a strong Lewis acid.
- It can be used as a catalyst for various reactions such as halogenation, Friedel craft alkylation and Merwin-Pondorf-Verley reduction.
- $AlC{{l}_{3}}$ is anhydrous, non-explosive, non-inflammable but corrosive solid.
So, the correct answer is “Option A and C”.
Note: -In acylation reaction, the product formed is only ketones. Because when we use formyl chloride as the acylating agent, it decomposes into $CO$ and $HCl$.
- Aryl amines cannot be used in the Friedel crafts reaction because they produce highly unreactive complexes with $AlC{{l}_{3}}$.
-$FeC{{l}_{3}}$ can also be used as a catalyst for Friedel crafts reactions.
Complete step by step solution:
First, what is Friedel crafts acylation reaction? Friedel crafts acylation reaction is an electrophilic aromatic substitution reaction. It is the reaction in which the acyl group is attached to the benzene ring in presence of the anhydrous $AlC{{l}_{3}}$ catalyst.
$C{{H}_{3}}COCl+{{C}_{6}}{{H}_{6}}\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{{}}}COC{{H}_{3}}+HCl$
- Friedel crafts reaction uses acetyl chloride as the acylating agent. Acid anhydride can also be used as an acylating agent. The catalyst used here is$AlC{{l}_{3}}$, in order to allow the attachment of the acetyl group to the aromatic ring. In Friedel crafts reaction the aromatic ring gets converted into a ketone, after the attachment of the acyl group to the ring.
- Acyl halide forms a highly electrophilic acylium ion with the strong Lewis acid$AlC{{l}_{3}}$ and later it is stabilized by resonance.
- Therefore, the role of $AlC{{l}_{3}}$ in Friedel crafts reaction is to attack the reactant to form an intermediate and later, stabilization of that intermediate.
So, the correct answer is “Option A and C”.
Additional Information:
-$AlC{{l}_{3}}$ is a strong Lewis acid.
- It can be used as a catalyst for various reactions such as halogenation, Friedel craft alkylation and Merwin-Pondorf-Verley reduction.
- $AlC{{l}_{3}}$ is anhydrous, non-explosive, non-inflammable but corrosive solid.
So, the correct answer is “Option A and C”.
Note: -In acylation reaction, the product formed is only ketones. Because when we use formyl chloride as the acylating agent, it decomposes into $CO$ and $HCl$.
- Aryl amines cannot be used in the Friedel crafts reaction because they produce highly unreactive complexes with $AlC{{l}_{3}}$.
-$FeC{{l}_{3}}$ can also be used as a catalyst for Friedel crafts reactions.
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