Anhydrous $ AlCl_3 $ is used in Friedel-Crafts reaction because it is (A) Soluble in ether (B) Ionisable to Al and Cl ions (C) Electron rich (D) Electron deficient molecule
Hint : $ AlCl_3 $ behaves as a Lewis acid in the Friedel-Crafts reaction, which enables the reaction to go forward. It is a catalyst in this reaction. The Friedel-Crafts reaction is a set of reactions where an alkyl or acyl group is added to a benzene molecule by an electrophilic aromatic substitution.
Complete Step By Step Answer: The electrophile attacks the pi electrons to form a nonaromatic carbocation. For this reaction to occur, an electron acceptor is required. Anhydrous $ AlCl_3 $ ionises to form $ Al^(3+) $ and $ 3Cl^- $ ions. Therefore, it is capable of accepting electrons and acts as a Lewis acid. A Lewis acid is any substance that can accept a pair of nonbonding electrons. It is an electron-pair acceptor. Such as $ AlCl_3 $ in this question. Being an electron-acceptor implies that there are vacant spots available in the shell of $ AlCl_3 $ for it to accept electrons. Thus, it is electron deficient. $ AlCl_3 $ accepts Cl and converts to $ AlCl_4^- $ therefore the organic compound from which Cl had been removed becomes an electrophile. An electrophile is a molecule that forms a bond to its nucleophile by accepting both bonding electrons from that reaction partner (nucleophile). Therefore, for this question, the correct answer is (D). $ AlCl_3 $ is used in Friedel-Crafts reaction because it is an electron deficient molecule.
Note : In contrast, Lewis bases are substances that donate a pair of nonbonding electrons. Such questions can be tricky to answer as they are very specific. But, by knowing the organic compound reactions – such as Friedel-Crafts reaction – such questions can be answered. Also, thoroughly learn about the various reagents used/recommended for such specific organic compound reactions mentioned in the text.