[JoshP]

Hi guys,

I'd really appreciate any help with this question! I've tried to make a stab at it but I'm not really sure I'm getting anywhere. This is what I thought, in decreasing order of pKaH:

meta-NO2, para-chlorine, unsubstituted, para-OMe, para-NO2.

I figured that the electron-withdrawing NO2 in the meta position is going to destabilise the +ve charge. Then and similar argument for the chlorine. O-Me is electron-donating, so is likely to stabilise said +ve charge, and the para NO2 can delocalise the +ve charge over itself, lending stability. That leaves the unsubstituted analine in the middle.

However, from having a look at some pKaH values, there seems to be something a bit wrong with my answer! So yeah, I'd really appreciate any help.

Cheers, Josh

[Dan]

The equilibrium you need to consider is:

AH⁺ <----> A + H⁺

So you can approach this problem from two angles, either by explaining the relative stabilities of AH⁺ or by considering the relative basicities of A - I think the latter is far easier.

So the more basic A is, the less acidic AH⁺ is.

Try and rank p-nitro-, m-nitro, p-chloro, p-methoxy and unsubstituted aniline in order of basicity, and use that to work out the order of acidity of their conjugate acids. Think about the availability of the N lone pair.

[JoshP]

The equilibrium you need to consider is:

AH⁺ <----> A + H⁺

So you can approach this problem from two angles, either by explaining the relative stabilities of AH⁺ or by considering the relative basicities of A - I think the latter is far easier.

So the more basic A is, the less acidic AH⁺ is.

Try and rank p-nitro-, m-nitro, p-chloro, p-methoxy and unsubstituted aniline in order of basicity, and use that to work out the order of acidity of their conjugate acids. Think about the availability of the N lone pair.

Ok, well in terms of basicity, this is what I came up with:

I thought methyoxy is going to be the most basic due to it's electron-donating effect, followed by unsubstituted because it has no electron-withdrawing substituents. NO2 in the meta position won't be as withdrawing as a chlorine in para, which is followed by NO2 in para because is going to withdraw by inductive + conjugation effects.

So: p-NO2, p-Cl, m-NO2, unsub., O-Me, in order of decreasing acidity.

How does that look now? Cheers, J

[Dan]

Looking better, but I think you're off on the p-chloro vs m-nitro. Remember that while chloro is inductively withdrawing it is mesomerically donating. Also, inductive effects fall off with distance, it is mesomeric effects that manifest strongly in the ortho/para positions - don't confuse the two. Inductive effects are stronger in the ortho position, then meta, then para.

Case in point: Note that pK_aH for o-, m-, and p-chloroaniline are 2.7, 3.5 & 4.2 respectively. Inductive effects dominate, and fall off with distance.

pK_aH for o-, m-, and p-nitroaniline are -0.3, 2.5 & 1.0 respectively. Mesomeric (resonance) effects dominate, mesomerically withdrawing in the ortho/para positions.