Question

At what position, and on what ring, is bromination of phenyl benzoate expected to occur? Show your product and explain your answer.

Answer 1

Interesting question. First of all, you have to decide which ring is more strongly activated (or less strongly deactivated, as the case may be). Bromination will take place in that ring. Then, you have to decide whether the substituent is ortho-para director meta-directing to determine which position(s) in that ring will be brominated.

The ring on the left has a -COOPh substituent, and the ring on the right has a -OCOPh substituent (even though they are each other substituents, you can regard them individually.) Refer to your textbook for these two substituents' polar characters—find the ring that is more activated to electrophilic aromatic substitution.

Hint: In general, Atoms with lone pairs tend to be electron-donating and activating (e.g., OH group); an atom without lone pair, but bonded strongly electronegative atom (e.g.-NO2 group), tends to be electron-attracting and deactivating. An atom without a lone pair but bonded to a less electronegative atom (e.g., CH3 group) tends to be weakly electron-donating and activating. Halogens and the nitroso group are exceptions - they are electron-attracting but activating.

The left ring has the substituent -COOR attached; the carbon has no lone pairs and is attached to more electronegative atoms; therefore, we find that the left ring is deactivated. The right ring has the substituent -OCOR attached; the oxygen has two lone pairs. As a result, it is electron-donating and activating; bromination is therefore expected to occur in the right ring. Finally, the -OCR is ortho-para directing, so we expect a mixture of the ortho- and para-substituted products.