Solved: CHEMISTRY Question for NEET | Sushrut

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NEET CHEMISTRYEasy

Benzene reacts with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form:

A

Toluene

B

Chlorobenzene

C

Benzylchloride

D

Xylene

Step-by-Step Solution

Identify the Reaction: The reaction of benzene with an alkyl halide (like methyl chloride, $CH_3Cl$ ) in the presence of a Lewis acid catalyst (like anhydrous aluminium chloride, $AlCl_3$ ) is known as Friedel-Crafts Alkylation .
Mechanism: It involves Electrophilic Aromatic Substitution.

Step 1 (Generation of Electrophile): The Lewis acid $AlCl_3$ reacts with $CH_3Cl$ to generate the electrophile, the methyl carbocation ( $CH_3^+$ ) . $CH_3Cl + AlCl_3 \rightarrow CH_3^+ + [AlCl_4]^-$
Step 2 (Attack): The benzene ring attacks the electrophile ( $CH_3^+$ ) to form a sigma complex (arenium ion).
Step 3 (Deprotonation): Loss of a proton restores aromaticity, yielding methylbenzene, commonly known as Toluene.

Reaction Equation: $C_6H_6 + CH_3Cl \xrightarrow{\text{anhyd. } AlCl_3} C_6H_5CH_3 \text{ (Toluene)} + HCl$
Other Options: Chlorobenzene is formed by electrophilic substitution with $Cl_2/FeCl_3$ . Benzyl chloride is formed by free-radical substitution of toluene. Xylene involves di-alkylation.

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Solved: CHEMISTRY Question for NEET | Sushrut