Why is para nitrophenol more acidic than meta nitrophenol? I worked out the resonance and even in meta, the electrons do move around? Is it a matter of the electrons being able to move all around the ring in a para setting vs. meta?

I'm a bit confused. Thanks!

does it have anything to do with the -OH being ortho/para directing & activating and thus more stable?

and the nitro is meta-deactivating, isn't it?

are ortho, meta, para ring systems on the exam?

are ortho, meta, para ring systems on the exam?
you might be asked which product is predicted to be most abundant

I can not remember exactly but it has to do with an electron withdrawing (NO2) group at the ortho/para position and resonance abilities.

Try drawing out the resonance forms for both the meta and para. With the para notice that you can take the a lone pair from the oxygen that has lost a hydrogen, drop them into the carbon they are attached to, move the double bond between carbons 1 and 2 to between carbons 2 and 3, and then take the double bond from between carbons 3 and 4 and move these electrons onto the nitrogen then moving one of the electron pairs between oxygen and nitrogen and move it onto the oxygen. You cannot do this with the meta postion. I hope this helps...

So, in the para nitrophenol, the electron is basically able to move around from the O that has lost a proton all the way to the nitrogen, which is not the case in meta nitrophenol?

Also, here is a general question about resonance:

If there is a pi bond between carbons 1 and 2, that pi bond can move to carbons 2 and 3. What about lone pairs? if there is a lone pair on carbon 1, it can move to carbon 2 and then 3 and so forth? so basically, lone pairs can travel one atom at a time?

Ummm....lemme think about it. Here's a reference to similar question
http://forums.studentdoctor.net/showpost.php?p=5423585&postcount=623


Edit:
Why is para nitrophenol more acidic than meta nitrophenol? I worked out the resonance and even in meta, the electrons do move around? Is it a matter of the electrons being able to move all around the ring in a para setting vs. meta?

I'm a bit confused. Thanks!Remember the relative acidity of a molecule is related to the stability of the conjugate base formed when deprotonated. In this case, the para-structure has more resonance contributors relative to the meta structure indicating the the para-structure conjugate base is more stable than meta conjugate base. As you've indicated it's a matter of moving electrons around after deprotonation.

Oops...double post.

you might be asked which product is predicted to be most abundant

Maybe choose b/w ortho/para and meta. If you can sort out ortho/para ratios, I have a prize for you to pick up in Stockholm

I didn't read other ppls response, so I will just assume that the answer was given but I wanted to just ask another note. I remember an experiment that I did a long time ago with dyes and we had to convert it to some sort of quinone and I remember that the para position for a benzoquinone was unusually stabilized due to some reason that pertained to the delocalized electrons moving straight through the beneze (or something like that). I don't know if this is relevant but if you get the double bound on the nitrogen and one on the oxygen for a para-nitrophenol, it could be a similar problem.