Can NH2 with no formal charge on it be protonated?

[ngu9997]

So I have this question that asks which is the strongest base. I have one structure with a benzene ring with an NH2 group on one side and a methoxy group on another side. Automatically I knew that the NH2 group would be what is protonated in the situation that this structure acts as a base and becomes protonated. But the more I think about it, the less it makes sense. N already has 2 H bonds on it and is attached to the benzene ring and has a lone pair giving it a formal charge of 0. How could it form another H bond in the case it becomes protonated (I assume there has to be some logic as to why the N is what I'm looking at to become protonated rather than the oxygen on the methoxy group).

 

[mjc123]

How could it form another H bond in the case it becomes protonated

How does any amine (or ammonia) get protonated? Just the same.

 

[ngu9997]

Oh wow , I forgot that it was H+ so it wouldn't violate any rules of having more than 8 electrons in N's octet. That makes sense, but how would we determine that the NH2 group gains the hydrogen in protonation rather than the oxygen on the methoxy group?

 

[snorkack]

Oh wow , I forgot that it was H+ so it wouldn't violate any rules of having more than 8 electrons in N's octet. That makes sense, but how would we determine that the NH2 group gains the hydrogen in protonation rather than the oxygen on the methoxy group?

Because N is usually a stronger base than O.

If you have NH₃ and H₂O in ammonia water solution and add acid, then you can protonate either water - forming H₃O⁺ - or ammonia - forming NH₄⁺.
But ammonia is much stronger base. H₃O⁺ has pKa of -1,5, while NH₄⁺ has pKa of +9,25.
Now, in aromatic amines, the aromatic ring actually withdraws some of the free electron pair and makes the amine a weaker base. Aniline, C₆H₅NH₃⁺ has pKa of +4,6.

But it is still much more basic than water, or presumably phenylether.
How does an ether group across the aromatic ring affect pKa of an aromatic amine?