Anisole undergoes bromination with bromine in ethanoic acid even in the absence of iron (III)bromide catalyst(a) Due to the activation of benzene ring by the methoxy group.(b) Due to the de-activation of benzene ring by the methoxy group.(c) Due to the increase in electron density at ortho and para positions(d) Due to the formation of stable carbocation.

The correct answer is (a) and (c).

(a) The methoxy group (-OCH3) is an activating group, meaning it increases the reactivity of the benzene ring towards electrophilic aromatic substitution. This is because the oxygen atom in the methoxy group is electron-rich and can donate electron density to the benzene ring, making it more nucleophilic and thus more reactive towards electrophiles like bromine.

(c) The methoxy group is also an ortho/para-directing group. This means that it directs incoming electrophiles to the ortho and para positions relative to itself on the benzene ring. This is again due to the electron-donating nature of the methoxy group. The extra electron density it provides to the benzene ring is most concentrated at the ortho and para positions, making these positions more nucleophilic and thus more likely to react with electrophiles.

(b) is incorrect because the methoxy group is an activating, not deactivating group.

(d) is incorrect because the reaction does not involve the formation of a carbocation. The reaction mechanism for electrophilic aromatic substitution involves the formation of a sigma complex (or arenium ion), which is a resonance-stabilized intermediate, not a carbocation.