Under which of the following conditions would toluene \(C_6H_5–CH_3\), be converted into bromomethyl benzene, \(C_6H_5–CH_2Br\)?

Reaction with \(Br_2\) in sunlight

The correct answer is option 3. Reaction with \(Br_2\) in sunlight.

Toluene can be converted into bromomethyl benzene, \(C_6H_5–CH_2Br\), by reaction with \(Br_2\) in sunlight. This reaction is a free radical substitution reaction, and the presence of sunlight provides the necessary energy to initiate the reaction. The reaction with \(Br_2\) in dark or with \(HBr\) will not produce bromomethyl benzene.

The reaction with \(Br_2/FeBr_3\) will produce a mixture of ortho- and para-bromotoluene. This is because \(FeBr_3\) is a catalyst that activates the benzene ring, making it more susceptible to attack by \(Br_2\). The reaction with HBr will produce benzyl bromide, \(C_6H_5CH_2Br\). This is because HBr is a strong nucleophile that attacks the benzene ring at the ortho and para positions.