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Which of the following can be used as the halide component for Friedel-Crafts reaction?
A
Chloroethene
B
Isopropyl chloride
C
Chlorobenzene
D
Bromobenzene
Explanation
Aryl and vinyl halides cannot form carbocations easily.
Detailed Solution
Benzene $+ CH_3-CH(Cl)-CH_3 \xrightarrow{Anhy.\ AlCl_3}$ cumene
In chlorobenzene, bromobenzene and chloroethene, the lone pair of the halogen is delocalised with $\pi$ bonds, so the C–X bond attains double bond character and they do not form carbocations.
In chlorobenzene, bromobenzene and chloroethene, the lone pair of the halogen is delocalised with $\pi$ bonds, so the C–X bond attains double bond character and they do not form carbocations.