Which Is More Acidic?
The Answer Is Orth/para.
Anybody Explain?
The Answer Is Orth/para.
Anybody Explain?
[orgo]acidity
- Thread starter ippie
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I believe that the ortho is more acidic because firstly NO is an electron withdrawing group and will increase acidity do to the inductive effect. The closer the electro withdrawing group the more acidic is the molecule. Secondly, after the phenol loses its hydrogen the negative charge will be able to rest on the ortho position do to resonance. This also increases the acicdty having the electron withrdawing group directly on top of the negative charge.
The NO2 is more acidic but I think NO withdraws electron density via inductive effects as stated but also donates electron density to the ring via resonance. The main example of an o/p director that deactivates the ring is the halogens but additionally NO fits that bill too (less known) so on that basis you could probably answer the question alone but its even easier to see that the NO2 can stabalize the negative charge via resonance and would be responsible for the one with the NO2 being more acidic.
I sorta meant that, but i dont think it depends on the strength of the withdrawing group as much as it depends on the location. I think the location is most important and The ortho would be stronger which like you said is the NO2. I didnt realize that one was NO2 and the other NO i thought the were both NO. My mistake. Either way what i said before is true just that is isnt an NO rather an NO2. I didnt look at the attachment carefully enough.
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The answer is ortho para because NO is like the halogens, it directs ortho/para. It is however electron withdrawing, therefore increasing the acidity. I hope that helps-
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The Ortho/Para form is always more acidic ecause when you pull an H off of the O, you generate more resonance on the anion. Since you have a lot of resonance on the anion, the anion is considered more stable. The more stable your anion the more acidic is your compound because stability of the anion will allow this compound to dissociate in H2O.
Hope this helps.
Hope this helps.
It has a lone pair on the N and it donates via resonance just like any other atom that has a lone pait on the benzylic position.
There is a lone pair on the Nitrogen. Lets look at nitrosobenzene. The lone pair on the nitrogen donates electrons into the ring and then you get a negative charge on a carbon that can move around via resonance. Sorry the lone pair is not shown in the picture.
http://en.wikipedia.org/wiki/Image:Nitrosobenzene.png
The aromatic ring is deactivated toward electrophillic aromatic substitution by the combined electron-withdrawing inductive effect of the electronegative nitrogen and the oxygen. The lone pair on the nitrogen can however stabalize by resonance the ortho and para substituted intermediates. The stabilization is more effective for intermediates of the ortho and para than for the meta attack.
so we get a rare nonhalogen ortho-para directing deactivating group!!
Either way the prtho would be more acididc regardless of whether that was a NO2 or NO
