[orgopete]

So the scale of my compounds is:

N-methylbutylamine > cyclohexylamine > aniline > dimethyl ether > dimethyl sulfide
Am I right?

Thank you, things are gettin clearer to me now.
I always considered aniline stabilised by resonance, and talking about acidity I was right (because I considered the conjugated base). But now I'm considering the basicity, so I have to consider not the conjugated base, but the conjugated acid. And it is a little bit different. Ph-NH3+ doesn't have any resonance effect.

I agree with both accounts. I have a different opinion on the acidity than others and have been flagged for my contrarian answer in the past, but I'll give it another try. I think aniline or the anilinium cation is showing an inductive effect. I argue it is the same inductive effect as chloroform being more acidic than methane. The chlorine atoms pull the electrons away from the C-H bond making the proton more acidic. If you compare the acidity of pyrrole and pyrrolidine (the saturated heterocycle), pyrrole is much more acidic. It isn't resonance as the non-bonded electrons of pyrrole are already being delocalized. The sp2-carbons attached to the nitrogen are more electron withdrawing than the sp3-carbons of pyrrolidine.

The effect of delocalization can be an issue. That is why I point to the pyrrole example. It would appear as though the N-H electrons are not delocalized due to stereochemistry. Even though the electrons are not delocalized per se, there is still an effect. The non-bonded electrons of hydroxide are all more basic than those of water. Protonation of one pair affects the basicity of the others as well. After hydroxide is protonated, it takes a stronger acid to add another proton to form hydronium ion. Similarly, the H-O bond of hydroxide is less acidic than of water. The change in the electrons of water in becoming hydroxide also affect the H-O bond. It is now more difficult to remove that proton to form the dianion as all of the oxygen electrons were affected by the loss of a proton from water.

This isn't about charge per se. It is about basicity of the electrons. Oxygen can form a dianion just as sulfur does, but because oxygen is less electron withdrawing, it is not as acidic as H2S or HS(-). Remember, ammonia is a stronger base than fluoride ion. The charge of a pair of electrons is just the same in ammonia as in fluoride ion. The nuclear charge of fluoride is greater and its field has a greater affect on the electrons than ammonia. Although ammonia has one more proton, it is not in the nucleus, and by the inverse square law, its greater distance results in a smaller effect, hence the electrons of ammonia are more basic.

Once an anion is formed, it may or may not become delocalized, but that is a different question. If we compare the acidity of phthalimide, because the sp2-carbons have the very electron withdrawing oxygen atoms attached, the acidity of the NH is increased much more. We can draw resonance structures for a phthalimide anion. It will not matter if they are delocalized as the now non-bonded electrons are also more basic due to the deprotonation. Reactions of phthalimide anion (or other amide anions) occur on the nitrogen. Reactions with strong alkylating agents and protonation occur on oxygen (see acid hydrolysis of amides). You may have to think about that one on your own.

In short, the pKa table could tell you a benzene ring is more electron withdrawing than alkyl groups, hence PhNH3(+) is more acidic than cyclohexylNH3(+).

[Babcock_Hall]

Are there more than one resonance structures you can draw for aniline?

Yes, there are three of them.
So here's the big difference with the alkyl amines?

It is one of the differences.  If the electron pair on nitrogen is somewhat delocalized into the ring, then it is less available to accept a proton.  However, the situation is made more complex by the things that orgopete discussed.