The usual acid-base

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In a question, it has been asked whether o-nitrophenol or m-nitrophenol is more acidic. It has something to do with the electron denstiy being at the meta position because of the nitro group, therefore o- is more acidic.

Why don't we consider what the phenol group does? Also, can somebody explain it in better terms (I just don't understand the language Kaplan put it in)? Will is be the same if there were any other meta directing group and how would it change if there were 2 electron withdrawing groups? i know that would be more acidic, but which one (o, m or P) would be more acidic?

Thanks a ton to everybody who answers my question.
 
In a question, it has been asked whether o-nitrophenol or m-nitrophenol is more acidic. It has something to do with the electron denstiy being at the meta position because of the nitro group, therefore o- is more acidic.

Why don't we consider what the phenol group does? Also, can somebody explain it in better terms (I just don't understand the language Kaplan put it in)? Will is be the same if there were any other meta directing group and how would it change if there were 2 electron withdrawing groups? i know that would be more acidic, but which one (o, m or P) would be more acidic?

Thanks a ton to everybody who answers my question.

The order of acidity would be p>o>m:
both the o- and p- will have anions formed after deprotonation that are directly resonance stabilized by the strong W -NO2 group
So, both would be expected to be much stronger acids than the m-.
Also the question of H-bonding in the acid is interesting, the usual argument is that H-bonding will weaken the acidity of the O-H bond since it is "held" by the -NO2, and the o- can have such H-bonding, but neither the m- or p- can.
So para- is the strongest since the anion is stabilized by resonance, followed by ortho since it stabilized by resonance but the -OH is H-bonded followed by meta- for which the anion is not stabilized.

Hope it helps!
 
The order of acidity would be p>o>m:
both the o- and p- will have anions formed after deprotonation that are directly resonance stabilized by the strong W -NO2 group
So, both would be expected to be much stronger acids than the m-.
Also the question of H-bonding in the acid is interesting, the usual argument is that H-bonding will weaken the acidity of the O-H bond since it is "held" by the -NO2, and the o- can have such H-bonding, but neither the m- or p- can.
So para- is the strongest since the anion is stabilized by resonance, followed by ortho since it stabilized by resonance but the -OH is H-bonded followed by meta- for which the anion is not stabilized.

Hope it helps!

Hmm..pooyah I don't really get what you're saying. Are we suppose to deprotonate it??? That sound's familiar but I never went to my orgo class to know shortcuts, but heres my way, can you check it.

goes like this:
1. Lets start with phenol
2. Test add NO2 in o, m, and p -----> they will all produce a carbocation
3. Only o and p can resonance stabilize with the OH
4. meta position can NOT stabilize thru resonance..(draw it and try it)
5. So now you have a fight of O vs P, who will win?
6. Para can H-bond intermolecularly with other stuff while Ortho gets distracted by the NO2 and H-bonds with that..intramolecular therefore it decreases acidity, MP, and BP.
7. Para wins.
 
Hmm..pooyah I don't really get what you're saying. Are we suppose to deprotonate it??? That sound's familiar but I never went to my orgo class to know shortcuts, but heres my way, can you check it.

goes like this:
1. Lets start with phenol
2. Test add NO2 in o, m, and p -----> they will all produce a carbocation
3. Only o and m can resonance stabilize with the OH
4. Para position can NOT stabilize thru resonance..(draw it and try it)
5. So now you have a fight of O vs P, who will win?
6. Para can H-bond intermolecularly with other stuff while Ortho gets distracted by the NO2 and H-bonds with that..intramolecular therefore it decreases acidity, MP, and BP.
7. Para wins.

I think you're missing a point that para can definitely resonance stabilize and meta absolutely can not lol I wish I could draw it, but para would make resonance structures and so does ortho!
Simply put, look at the resonance structures, para is more resonance stabilized than ortho, thus it makes a stronger acid!
 
Also, do we always deprotonate something connected to the ring when we look for trends?
 
lol yeh. I reread it and was like wtf???? I got every o m p letter mixed up. Arite its ready now, can you read it again.

Yes you're on the right track! As I said in my first post intermolecular hydrogen bonding actually weakens acidity, and ortho would be involved in intermolecular H-bonding since it's closer to -OH, but I would say para wins mainly because it is more resonance stabilized than ortho!
 
Also, do we always deprotonate something connected to the ring when we look for trends?

Normally yes, once it gives up the hydrogen you need to determine whether it is resonance stabilized or not! But that's only if you want to determine whether -p, -m, or -o positions have higher acidity strength!
 
Normally yes, once it gives up the hydrogen you need to determine whether it is resonance stabilized or not! But that's only if you want to determine whether -p, -m, or -o positions have higher acidity strength!


K got it. By the way, when are you taking your test?
 
Thanks guys!! Hey, I take mine on the 9th!!!
Keep us updated about your scores and the breakdown
Goodluck... Crack it!!!👍
 
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