Question

What is the product of the reaction between Benzene and \(CH_3Cl\) in the presence of anhydrous \(AlCl_3\)?

• A. Toluene
• B. Phenol
• C. Chlorobenzene
• D. Benzaldehyde

Hint:

Benzene reacting with an alkyl halide in the presence of anhydrous AlCl3 is a standard signal for Friedel-Crafts alkylation. Recall that this reaction attaches an alkyl group to the ring by replacing one hydrogen atom.

Answer 1

The Correct Option is A

Concept: This reaction is known as Friedel–Crafts Alkylation, where an alkyl group substitutes a hydrogen atom of the benzene ring in the presence of a Lewis acid catalyst such as \(AlCl_3\).

Step 1: Formation of electrophile. In the presence of \(AlCl_3\), methyl chloride forms a methyl carbocation. \[ CH_3Cl + AlCl_3 \rightarrow CH_3^+ + AlCl_4^- \]

Step 2: Electrophilic substitution on benzene. The methyl carbocation attacks the benzene ring, replacing one hydrogen atom. \[ C_6H_6 + CH_3^+ \rightarrow C_6H_5CH_3 \]

Step 3: Product formation. The product formed is Toluene.

Answer 2

Concept:

• An alkyl halide reacting with benzene in the presence of anhydrous $AlCl_3$ is the standard reagent combination for Friedel-Crafts alkylation, a well known named reaction in aromatic chemistry.
• Each of the other three options is the expected product of a different named reaction, so checking the reagents needed for each option rules them out directly.

Step 1: Recognise the reagent pattern.
Benzene plus an alkyl halide plus anhydrous $AlCl_3$ (a Lewis acid catalyst) is the defining condition for Friedel-Crafts alkylation, which substitutes an alkyl group onto the ring.

Step 2: Rule out the other options by their actual reagents.
Phenol needs the cumene process or hydrolysis of chlorobenzene, not an alkyl halide.
Chlorobenzene needs $Cl_2$ with $FeCl_3$ (halogenation), not $CH_3Cl$ used as an alkylating agent.
Benzaldehyde needs $CO$ and $HCl$ with $AlCl_3$, known as the Gattermann-Koch reaction, not simple alkylation.

Step 3: Confirm the product.
Since the given reagents match Friedel-Crafts alkylation, the methyl group from $CH_3Cl$ replaces a hydrogen atom on the benzene ring, giving methylbenzene.

Final Answer: The product is Toluene.