Question

The reaction of toluene with $Cl_2$ in the presence of $FeCl_3$ gives 'X' and reaction in presence of light gives 'Y' Thus, 'X' and 'Y' are

• A. X = benzal chloride, Y = o-chlorotoluene
• B. X = m-chlorotoluene, Y = p-chlorotoluene
• C. X = o and p-chlorotoluene, Y = trichloromethyl benzene
• D. X = benzyl chloride, Y = m-chlorotoluene

Answer 1

The Correct Option is C

To solve this question, we need to understand the reactions of toluene (C_6H_5CH_3) with chlorine (Cl_2) under different conditions:

1. Reaction with Cl_2 in the presence of FeCl_3:

   This reaction is an electrophilic aromatic substitution. The Lewis acid FeCl_3 helps in the generation of a strong electrophile, Cl^+, which attacks the aromatic ring. Toluene being an activated benzene ring due to the electron-donating methyl group, undergoes chlorination preferentially at the ortho and para positions. Therefore, the products formed are ortho-chlorotoluene and para-chlorotoluene.

2. Reaction with Cl_2 in the presence of light:

   This condition favors free radical substitution reactions, mainly at the benzylic position (the carbon adjacent to the aromatic ring). In toluene, this results in the substitution of hydrogen atoms of the methyl group with chlorine atoms. Finally, this leads to the formation of benzyl chloride, dichloromethyl benzene, and trichloromethyl benzene (commonly known as benzotrichloride) as products. The major product when the substitution goes to completion is trichloromethyl benzene.

Based on these interpretations, the correct answer is:

X = o and p-chlorotoluene, Y = trichloromethyl benzene

This option correctly reflects the major products for each reaction condition: electrophilic substitution (o and p-chlorotoluene) and free radical substitution (trichloromethyl benzene).