[a confused chiral girl]

Hi!

I want to ask a question about which of the following is the most basic. the four choices are aniline, 4-nitroaniline, 4-methylaniline, 4-methoxyaniline.

I thought it would be 4-methylaniline, because the anime (NH2) would pull the electron density away from CH3 since the amine is more eletronegative. However, I'm also contemplating just aniline is most basic, because it wouldn't have any resonance to stabilize it.

thanks!

[refid]

basicity is determined by availability of lone pairs on the nitrogen..my guess would be 4-methoxyaniline would be most basic, since the methoxy group has a +R effect donating electrons into the benzene, making the lone pairs on the Amino group available..

[a confused chiral girl]

thanks for your reply! 

I feel a bit hesistant about this, because 4 methoxyaniline has a CH3O group attached to the ring. Since the oxygen is quite electronegative, it will pull electrons toward itself as well, thus fighiting with the N on the other side on the ring. so wouldn't that make the N less basic instead?

[english]

For this question you will need to look at the opposite of what you would look for in finding the most acidic aniline.  A more detailed explanation will follow anyway. 

The most basic aniline is the one in which electron density is supplied to it from one of the three substituents (methyl, nitro, methoxy).


Since an amino group's basic character is determined largely by its lone pairs, and these lone pairs can be delocalized into the ring, you will need a substituent that will "re-supply" this lone pair through resonance.

Alkyl groups cannot donate electrons through resonance (review the conditions required for electron delocalization).  

You've got the reverse thinking.  You're not trying to find which substituent "feels" withdrawal more by the amino group, but how the amino group "feels" electron density by each substituent.

Inductive effects are too weak to really matter compared to resonance effects.  Since alkyl groups cannot participate through electron resonance donation, we can rule out methyl.  

Nitro groups, as you know, deactivate benzene rings.  If a nitro group is placed para to NH₂, the lone pairs of NH₂ can be delocalized onto it.  So you know that a nitro group would increase acidity, not basicity.  

Basicity will follow the opposite trend to acidity.  So find the most activating group for the most basic aniline.  


Think of it this way.  You know that in order to have a strong acid, its conjugate base must be stable (much weaker), and vice versa (think of water auto-ionizing; 2H₂O --> HO^- + H₃O⁺.  HO is a stonger base because water is a weaker acid).  

So first, protonate the NH₂, giving NH₃⁺, it's conjugate acid.  You know that any time you protonate something, you increase its acidity (protonating water gives H₃O⁺, a much stronger acid than water).  

So we've now set down the following rule:  the stronger the base, the more stable its conjugate acidic form.  We've just increased the Ka of NH₂ by protonating it, so now we must find which substituent (methyl, methoxy, or nitro) will stabilize the acid the best, giving the stonger conjugate base (NH₂).  

Placing MeO para to NH₃⁺, see what happens when you move a lone pair of the O of MeO into the ring.  The C atom connecting the protonated amino group has a negative charge (an extra lone pair), and it is this extra lone pair that can conjugate with the positively charged N to stabilize its charge.

More stable acidic form, the stronger is its basic form.

Now the question you must ask yourself is, can a nitro group do this?

[Ψ×Ψ]

Oxygen is pretty electronegative, but alkoxy groups are electron donating substituents.

[a confused chiral girl]

For this question you will need to look at the opposite of what you would look for in finding the most acidic aniline.  A more detailed explanation will follow anyway. 

The most basic aniline is the one in which electron density is supplied to it from one of the three substituents (methyl, nitro, methoxy).


Since an amino group's basic character is determined largely by its lone pairs, and these lone pairs can be delocalized into the ring, you will need a substituent that will "re-supply" this lone pair through resonance.

Alkyl groups cannot donate electrons through resonance (review the conditions required for electron delocalization).  

You've got the reverse thinking.  You're not trying to find which substituent "feels" withdrawal more by the amino group, but how the amino group "feels" electron density by each substituent.

Inductive effects are too weak to really matter compared to resonance effects.  Since alkyl groups cannot participate through electron resonance donation, we can rule out methyl.  

Nitro groups, as you know, deactivate benzene rings.  If a nitro group is placed para to NH₂, the lone pairs of NH₂ can be delocalized onto it.  So you know that a nitro group would increase acidity, not basicity.  

Basicity will follow the opposite trend to acidity.  So find the most activating group for the most basic aniline.  


Think of it this way.  You know that in order to have a strong acid, its conjugate base must be stable (much weaker), and vice versa (think of water auto-ionizing; 2H₂O --> HO^- + H₃O⁺.  HO is a stonger base because water is a weaker acid).  

So first, protonate the NH₂, giving NH₃⁺, it's conjugate acid.  You know that any time you protonate something, you increase its acidity (protonating water gives H₃O⁺, a much stronger acid than water).  

So we've now set down the following rule:  the stronger the base, the more stable its conjugate acidic form.  We've just increased the Ka of NH₂ by protonating it, so now we must find which substituent (methyl, methoxy, or nitro) will stabilize the acid the best, giving the stonger conjugate base (NH₂).  

Placing MeO para to NH₃⁺, see what happens when you move a lone pair of the O of MeO into the ring.  The C atom connecting the protonated amino group has a negative charge (an extra lone pair), and it is this extra lone pair that can conjugate with the positively charged N to stabilize its charge.

More stable acidic form, the stronger is its basic form.

Now the question you must ask yourself is, can a nitro group do this?

G_ENglish,
thank you so much for your detailed explanation. I have been trying to go along with this reasoning and absorbing it for the past couple of days. You said that we can rule out CH3 right away, because alkyl groups cannot donate electrons through resonance. I didn't think of that right away, because I didn't think of resonance.

I was contemplating between methyl, and methoxy because they are both electron donating groups. However, before reading your explanation and thinking of resonance, I was going to choose methyl, because I thought the Oxygen on the methoxy is going to withdraw a bit of the electrons because of it's electronegativity. but now after your explanation, instead of thinking that O withdraws a bit of the electron, I should actually think of it as donating electrons through resonance. I understand this now, just a bit weird to accept because O is electronegative, whereas in methyl, nothing is electronegative and can donate through induction.

also, to answer your question, nitro group would be unable to do that because nitro would withdraw electrons from the amine, so it would'nt contribute to resonance and donate electrons.

thank you once again!! 

[english]

also, to answer your question, nitro group would be unable to do that because nitro would withdraw electrons from the amine, so it would'nt contribute to resonance and donate electrons.

thank you once again!! 

Yes, but because nitro will withdraw the ring electrons through resonance.

Your welcome.