Toluene when treated with $B r_{2} / F e$ , given p-bromotoluene as the major product because the $- C H_{3}$ group

is p-directing

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deactives the ring

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is m - directing

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activates the ring by hyperconjugation

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Solution

The correct options are
A activates the ring by hyperconjugation
B is p-directing
Due to hyperconjugation, the methyl group of toluene donates electron density to the benzene ring which causes extra electron density on ortho and para positions. In these positions, the attack of electrophile is more susceptible. Hence, $C H_{3}$ is an ortho-para directing group.

Hence, Options "A" and "D" are correct answers.

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C H_{3} - C H_{3} | C H - C H_{2} - C H_{3} + B r_{2} h v - \to

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