Starting with benzene and methane, and suitable inorganic reagents, suggest how the following compound might be synthesized. See Exercise 37 for a description of how an alkyl group can be added to the benzene ring.

Exercise 37

Alkylation of benzene can be accomplished by treating benzene with haloalkane (RX) in the presence of AlCl₃. The reaction is known as a Friedel–Crafts alkylation reaction. (The reaction is named after Charles Friedel, a French chemist, and James M. Crafts, an American chemist, who discovered this method of making alkylbenzenes in 1877.) An example of a Friedel–Crafts alkylation reaction is shown below:

The mechanism for this reaction involves the following steps. First, (CH₃)₂CHCl and AlCl₃ react in a Lewis acid–base reaction to form an adduct, (CH₃)₂CH—Cl—AlCl₃, in which a chlorine atom is bonded to both carbon and aluminum. The adduct then dissociates to (CH₃)₂CH⁺ a carbocation, and AlCl₄^–.The carbocation acts as an electrophile in a reaction with benzene, forming an arenium ion. Finally, a proton is removed from the arenium ion by AlCl₄^–, yielding an alkylbenzene, HCl, and AlCl₃. Write chemical equations for the elementary processes involved in forming cumene and HCl from (CH₃)₂CHCl and benzene. Use curved arrows to show the movement of electrons.

Transcribed Image Text:

+ (CH3)2CHCI AICI 3 CH(CH3)2 Cumene + HCI