Substituents on phenol

[inaccensa]

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Imagine a nitro group in ortho position in phenol a and meta postion in phenol b. Which is more acidic? If i drawn the resonance structures, I don't see much of a difference. Can somebody please explain this

 

[97spooncivic]

The nitro group is electron withdrawing. When the hydroxyl group is deprotonated, it will bare a negative charge. The ortho position is closer the hydroxyl and therefore will better stabilize the negative charge making it more acidic.

 

[inaccensa]

Thanks for replying. I actually drew the resonance structures again and saw the obvious error I had made. When the NO2 is in meta, there aren't as many resonance structures as when its in the o/p position.

 

[combatwombat]

The ortho position is closer the hydroxyl and therefore will better stabilize the negative charge making it more acidic.

I think this is basically right, except that i'd point out that "closeness" of the NO2 group to the OH group is not what makes it acidity per se. It's the NO2 group being in the ortho or para positions that allow for extra resonance structures. The meta position, while "closer" than the para position, doesn't allow for as many resonance structures and therefore makes the OH group less acidic. Best to think of it in terms of resonance structures.

 

[eric51]

I think this is basically right, except that i'd point out that "closeness" of the NO2 group to the OH group is not what makes it acidity per se. It's the NO2 group being in the ortho or para positions that allow for extra resonance structures. The meta position, while "closer" than the para position, doesn't allow for as many resonance structures and therefore makes the OH group less acidic. Best to think of it in terms of resonance structures.

Correct. It has absolutely nothing to do with how "close" the group is. Ortho/para EWGs will increase the acidity because of resonance. A meta EWG will still increase the acidity a little bit because of it's sigma withdrawing nature, but not nearly as much as the ortho/para EWG can because of the pi withdrawing.

 

[loveoforganic]

You contradicted yourself eric 😛 It has a little bit to do with closeness, just nothing compared to the resonance effect. In this case, if I had to guess, acidity ranking would prob be para > ortho > meta. I'd put ortho behind para due to increasing electrostatic repulsion that would happen following deprotonation.

 

[eric51]

You contradicted yourself eric 😛 It has a little bit to do with closeness, just nothing compared to the resonance effect. In this case, if I had to guess, acidity ranking would prob be para > ortho > meta. I'd put ortho behind para due to increasing electrostatic repulsion that would happen following deprotonation.

No, I did not contradict myself. It really does have nothing to do with closeness. Even an EWG that has no EWG capability because of resonance will still be way more effective in the ortho/para position because it makes the ortho/para carbons more electron deficient, and that is where the partial negative charge will lie. An EWG in the meta position will only make a slight difference in the electron density of the overall sigma framework, and make a slight difference in acidity in this case.

 

[loveoforganic]

So, if there's some inductive EWG effect, o- or p- position would be equivalent?

 

[eric51]

So, if there's some inductive EWG effect, o- or p- position would be equivalent?

That's a lot harder to predict, as having a EWG is the ortho vs. para position can really change the MOs. You'd probably have to do some HF or DFT calculations on your system to know for sure the differences, or do some actual laboratory experiments. That is definetely out of the scope of the MCAT though.