Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Aromatic substitution and it's effect on proton NMR?

  1. Nov 25, 2011 #1
    I can't figure out how aromatic substituents effect aromatic rings. I can think of it in a couple ways:

    Is the group ortho or meta directing? When we learned about aromatic electrophilic substitution, we obviously learned about these groups. When an aromatic proton is lost, it will be preferentially lost in a certain position due to the additional resonance of whatever group is attached to the ring. So this tells you where you can find extra positive charge on the ring (and thus where it would be more deshielded). For example, an ortho directing group would have more positive charge on the meta positions. This reasoning seems to work when I am doing problems.

    However you aren't actually pulling off protons, you are just looking at their chemical environments. So take an ortho directing group like OH. The O is electronegative, so it should have a partial negative charge. If it has a partial negative charge then that means it drew electrons out of the aromatic system. So the whole thing would be deshielded, but maybe slightly less deshielded the further away you get from oxygen. This reasoning doesn't seem to work when doing problems.

    So basically, if you take a neutral aromatic ring with a substituent on it such as an ortho/para director like -OH, is there actually deshielding occuring at the meta positions? We only learned about it in terms of resonance that could be drawn after you pulled a proton off and the ring had an overall +1 charge.

    I guess this boils down to me not understanding when the inductive effect applies and when resonance stabilization applies.
    Last edited: Nov 25, 2011
  2. jcsd
  3. Nov 25, 2011 #2


    User Avatar
    Science Advisor
    2017 Award

    Consider the case of methoxybenzene shown below:

    You can see that resonance will delocalize the lone pair on the oxygen into the aromatic ring. In this case, the oxygen is actually donating electron density to the aromatic ring. Hence, this and other ortho-/para-directing groups are also consider activating groups because they increase the nucleophilicity of the aromatic ring. While oxygen is electronegative and should remove electron density from the aromatic ring through the inductive effect, the resonance effect is stronger than the inductive effect, so the methoxy group overall donates electron density to the aromatic ring.

    The increased electron density goes primarily to the ortho and para positions on the ring giving these positions a partial negative charge and increasing the shielding of the protons at those locations.
  4. Nov 26, 2011 #3
    Thank you that was very helpful.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook