Question

In Friedel-Craft's synthesis of toluene, the reactants in addition to anhydrous $AlCl_3$ are

• A. $ C_6 H_5 Cl + CH_4 $
• B. $ C_6 H_5 CI + CH_3 Cl $
• C. $ C_6 H_6 + CH_4 $
• D. $ C_6 H_6 + CH_3 Cl $

Answer 1

The Correct Option is D

The Friedel-Crafts alkylation is an organic reaction catalyzed by a Lewis acid such as anhydrous aluminum chloride (AlCl_3). This reaction involves the alkylation of an aromatic ring (such as benzene) with an alkyl halide. For the synthesis of toluene, the reactants need to be benzene (C_6H_6) and an alkyl halide containing a methyl group, such as methyl chloride (CH_3Cl).

Let's discuss why the combination C_6H_6 + CH_3Cl is the correct choice for the Friedel-Crafts synthesis of toluene:

1. Mechanism: In the presence of AlCl_3, the methyl chloride (CH_3Cl) forms a stable carbocation, which then reacts with the benzene ring (C_6H_6) to form toluene (C_6H_5CH_3).
2. Reagents: The reaction specifically requires an aromatic compound like benzene for the alkylation process. The methyl group from CH_3Cl is necessary to form the toluene product.
3. Reaction Condition: Anhydrous AlCl_3 is used as the catalyst, which helps in the formation of the carbocation from the alkyl halide.

Other options do not meet these criteria:

• C_6H_5Cl + CH_4: Chlorobenzene and methane will not react under Friedel-Crafts conditions to form toluene.
• C_6H_5Cl + CH_3Cl: Although CH_3Cl is present, C_6H_5Cl (Chlorobenzene) is less reactive toward electrophilic substitution than benzene.
• C_6H_6 + CH_4: Methane is not an alkylating agent in Friedel-Crafts alkylation.

Thus, the correct answer for the synthesis of toluene via Friedel-Crafts alkylation is using benzene (C_6H_6) and methyl chloride (CH_3Cl).