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Toluene when treated with $Br_2/Fe$ gives p-bromotoluene as the major product because $CH_3$ group

$\begin{array}{1 1}(a)\;\text{deactivates the ring}\\(b)\;\text{is meta directing}\\(c)\;\text{activates the ring by hyperconjugation}\\(d)\;\text{is para directing and activates the ring by hyperconjugation}\end{array}$

Can you answer this question?

Methyl group activates the benzene nucleus due to its electron releasing nature and also shows hyperconjugation.It is an ortho and para-directing group(i.e) on substitution it forms always a mixture of ortho and para derivatives.
Hence (d) is the correct option.
answered Mar 20, 2014