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The reaction of toluene with $Cl_2$ in presence of $FeCl_3$ gives X and reaction in presence of light gives Y. Thus, X and Y are
A
X = Benzyl chloride, Y = m-chlorotoluene
B
X = Benzal chloride, Y = o-chlorotoluene
C
X = m-chlorotoluene, Y = p-chlorotoluene
D
X = o- and p-chlorotoluene, Y = Trichloromethyl benzene
Detailed Solution
With $Cl_2$ in the presence of the Lewis acid $FeCl_3$, the electrophile $Cl^+$ is generated and electrophilic substitution occurs on the ring.
The $-CH_3$ group is ortho/para directing, so X is a mixture of o-chlorotoluene and p-chlorotoluene.
With $Cl_2$ in the presence of light, chlorine free radicals are formed and substitution occurs on the side chain (benzylic hydrogens) by a free radical mechanism.
Successive substitution gives benzyl chloride, benzal chloride and finally benzotrichloride: $C_6H_5CH_3 \xrightarrow{Cl_2/h\nu} C_6H_5CH_2Cl \rightarrow C_6H_5CHCl_2 \rightarrow C_6H_5CCl_3$. So Y is trichloromethyl benzene.
Hence X = o- and p-chlorotoluene and Y = trichloromethyl benzene.
Note: the source prints the last option as 'Y = o- and p-chlorotoluene, Y = ...'; the first 'Y' is a misprint for X.
The $-CH_3$ group is ortho/para directing, so X is a mixture of o-chlorotoluene and p-chlorotoluene.
With $Cl_2$ in the presence of light, chlorine free radicals are formed and substitution occurs on the side chain (benzylic hydrogens) by a free radical mechanism.
Successive substitution gives benzyl chloride, benzal chloride and finally benzotrichloride: $C_6H_5CH_3 \xrightarrow{Cl_2/h\nu} C_6H_5CH_2Cl \rightarrow C_6H_5CHCl_2 \rightarrow C_6H_5CCl_3$. So Y is trichloromethyl benzene.
Hence X = o- and p-chlorotoluene and Y = trichloromethyl benzene.
Note: the source prints the last option as 'Y = o- and p-chlorotoluene, Y = ...'; the first 'Y' is a misprint for X.