Question

Using resonance structures as part of your answer, explain why Br in bromobenzene is: a. A mild ring deactivator b. An o-, p-director. [Note: It is not enough to use resonance structures from part "a" to "prove" part "b". One must show that o-, p-attack is faster, i.e. has lower Ea (AGa) by actually showing an electrophile, e.g. the generalized E+, attacking at the o-, m-& p-positions and, using Hammond's Principle, explain why the o-, p-attacks are favored.] Explain what polar (inductive, ?-system) effect you would expect Br to have on the reactivity of the benzene ring. From a Lewis structure of bromobenzene would you expect Br to donate or remove electron density [ep] from benzene's ?-system? That is, would you expect Br to activate or deactivate benzene? C. d. e. Based on your answer to parts "c" and "d" and the reactivity of bromobenzene, f. What must be the hybrid state of Br in bromobenzene? Why? How do your g. If we had looked at fluorobenzene instead why would F also be a ring deactivator. how strong is Br's resonance (T-system) effect versus its polar effect? Explain resonance structure from part "a" of this question help support this conclusion? It's p-orbitals are about the same size as the ring C's p-orbitals.

Answer 1

A)Bromine belongs to halogen family in periodic table.Though bromine have 3 loan pair of electron,it is strong electronegative atom.Because of strong electronegativity bromine pulls the electron away from the ring so it is mild ring deactivator. B)because of loan pair of electron and delocalisation of pi electron it forms resonance structure which has more electron density on ortho and para position so bromobenzene gives electrophilic substitution at ortho and para position.