I have tried to rationalize the questions with resonance in their intermediate forms, but have not been able to see a difference. Does it have to do with steric interactions based on ortho, meta, and para positions? How do I approach this question?enter image description here

  • $\begingroup$ Do you know about Hammett plots and linear free energy relationships (LFER)s? $\endgroup$ – Zhe Mar 13 '17 at 2:34

RBr + MeOH → ROMe + H2O ?? I. don't. think. so. There is one reaction which the proximity of the ortho MeO on the ring might cause steric hindrance. The two other effects you should be considering are: i) the electron withdrawing/donating nature of the substituents on the ring and ii) their location (ortho, meta, or para) because their location has an electronic (not steric) effect.

Since these considerations are (obviously) the reason you are being asked this question, I leave it to you to digest what those effects will be. You should understand that their effects are not only on the reactant but on the transition states as well.

  • $\begingroup$ I totally did not think about activating groups, thank you! $\endgroup$ – Biomed Boy Mar 13 '17 at 0:41

Think about which direction the electrons will flow based on each group.

In A, the ester puts a positive charge on the carbonyl carbon, thereby drawing electrons away from the benzene ring, and increasing its electron withdrawing effect on the reaction site. With electrons pulled away from the reaction site, bromine will have a stronger bond to the reactive-carbon.

In B, the ethers are donating electrons. Thereby decreasing the strength of the bromine bond.

In C, the benzene ring has electron withdrawing effects, similar to in A.

In D, the ether is donating electrons, like in B (but to a lesser extent).

I trust you can figure it out from here!


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