Question

Action of $AlC{l_3}$ in Friedel Crafts reaction is:
A) To absorb ${\text{HCl}}$
B) To release ${\text{HCl}}$
C) To generate electrophile
D) To produce nucleophile

Answer 1

Hint: The reagent aluminium chloride i.e. $AlC{l_3}$ acts as a lewis acid which means the reagent accepts an electron or accepts negatively charged species. One can think about what possibilities can arise by using this Lewis acid and choose the correct choice.

Complete step by step answer:

1) First of all let's learn about the reagent $AlC{l_3}$ which is called aluminium chloride. This reagent is electrons deficient as there are vacant d-orbitals present in the central aluminium atom.
2) Hence, the reagent aluminium chloride will try to accept the electron pair from other atoms or reagents to make itself stabilized.
3) As the aluminium chloride is always used as a reagent with the presence of chlorine then the aluminium chloride will accept the one chlorine atom from the chlorine molecule and the reaction will go as follows,
$AlC{l_3} + {\text{ Cl - Cl }} \to {\text{ }}{\left[ {{\text{AlC}}{{\text{l}}_4}} \right]^ - } + C{l^ + }$
4) The chlorine atom produced is a positively charged species which is an electrophile. This electrophile is further used in Friedel Crafts reaction.
5) Therefore, the action of $AlC{l_3}$ in Friedel Crafts reaction is to generate electrophile

which shows option C as a correct choice.

Additional information: In the Friedel-Craft reaction the formed electrophile later gets added to the benzene ring. This reaction is a type of electrophilic aromatic substitution reaction which is the synthesis of monoacetylated products by the reaction between the components arenes and acyl anhydrides or chlorides. In this reaction, a stoichiometric amount of the Lewis acid catalyst is required for both the substrate and the product form complexes.

Note:
The reagent $AlC{l_3}$ gets coordinated with halogens and favours the breaking of the bond present between them. This will result in the increase in the electrophilicity of its binding partner i.e. positive halogen making it much more reactive electrophile.