Which compound is more acidic?

[FROGGBUSTER]

Which compound is more acidic? m-chlorophenol or m-nitrophenol?

We would only look at inductive withdrawing effects, so which is more electronegative? Chlorine or a Nitro group?

Please don't say Nitro "because it can accept electrons into itself thru resonance", people kept on insisting that was correct before when I asked something similar but I know it's wrong; the negative charges can't get onto that meta carbon if you draw the resonance structures.

Thanks.

---

Members don't see this ad.

 

[FROGGBUSTER]

This one is probably too tough for the DAT huh?

 

[bud1010]

the nitro group is very withdrawing, more withdrawing than chlorine/halogen group. i think when people were saying "it can accept electrons into itself thru resonance" they meant that the NO2 can actually become part of the resonance structure becuase of the two oxygen atoms (one gets double bond one gets single bond) idk if that makes sense but i tried ha.

 

[Toothfairy525]

i had the same question, go to chad's O3 outline, and he has them ordered very nicely.

 

[LetsGo2DSchool]

The nitro is a much much much stronger withdrawing group than a halogen. This is due to the oxygens attached to nitrogen allowing the electron density from the ring to be distributed over these atoms. This extra conjugated resonance will stabilize the ring and thus making the compound more acidic.

As a rule of thumb, whenever you have an O attached to an aromatic ring providing an extra conjugated system outside the ring (i.e. COOH), it will provide resonance and stability and thus more acidity.

 

[woox]

Cl is a O/P director btw

 

[FROGGBUSTER]

The nitro is a much much much stronger withdrawing group than a halogen. This is due to the oxygens attached to nitrogen allowing the electron density from the ring to be distributed over these atoms. This extra conjugated resonance will stabilize the ring and thus making the compound more acidic.

As a rule of thumb, whenever you have an O attached to an aromatic ring providing an extra conjugated system outside the ring (i.e. COOH), it will provide resonance and stability and thus more acidity.

But this distribution of electron density is thru the sigma network correct?

 

[LetsGo2DSchool]

But this distribution of electron density is thru the sigma network correct?

Not exactly sure what you're asking. Maybe this can help clarify:

All resonance and delocalization of electrons involve the conjugated pi bond network. Electrons in sigma bonds are permanent and cannot freely move around the molecule unlike with pi bonds. Electrons in pi bonds are in p-orbitals which are perpendicular to sigma bonds and thus they can participate in the conjugated system with other pi electrons.

 

[FROGGBUSTER]

I drew something out.

If the nitro and -OH are meta, you cannot donate into the nitro group thru the pi network thru resonance. If the nitro and -OH are ortho or para, then yes you can.

That was what I thought but people continue to insist to me nitro is more electron-withdrawing because of its ability to allow electrons to be donated into itself. I would agree in an ortho or para case, but not meta.

This is what I'm asking:

Because you cannot donate electrons to NO2 thru resonance/pi network, the main criteria for evaluating acidity/basicity in a meta-situated case is the inductive withdrawing power of the substituents.

So which would be more electron-withdrawing INDUCTIVELY?? Is it nitro or chloro?

 

[LetsGo2DSchool]

Ok, I see what you mean. I checked my textbook too and that seems to be the case that the nitro must be in the para position to participate in resonance.

As for inductive effects, the nitro should be stronger than the chloro because of two combined reasons (which I think is correct but not 100% sure):

1) the two O have a stronger pull on the pi electrons from the ring than would the chloro, when combined with the reason below.....

2) because the N has a (+) charge, it is electro-positive meaning that it would attract the pi electrons from the ring even though it's not true resonance

 

[FROGGBUSTER]

Ok, I see what you mean. I checked my textbook too and that seems to be the case that the nitro must be in the para position to participate in resonance.

As for inductive effects, the nitro should be stronger than the chloro because of two combined reasons (which I think is correct but not 100% sure):

1) the two O have a stronger pull on the pi electrons from the ring than would the chloro, when combined with the reason below.....

2) because the N has a (+) charge, it is electro-positive meaning that it would attract the pi electrons from the ring even though it's not true resonance

Yeah, that's why I'm unsure. 1N + 2 O's would seemingly outweigh 1 Chloro, but the 2 O's are 2 bonds away from the ring while the Chloro is directly connected.

But you're right, the + charge does seem to indicate that the answer would be NO2.

Thanks. Hopefully, this sort of stuff won't come up on the DAT.