Aniline, C6H5NH2 is weakly basic in aqueous solution. in which solvent woult it be a strong base

1. CH3OH
2. CH3CH2OH
3. CF3CO2H
4.NH3
5. CH3(CH2)4CH3

Please explain your answer. I really need to understand the concept behind it. Does the same rules apply for all other compound?

The one which is the most acidic, CF3COOH because it will take lone pair of N from Aniline more better than all other and makes Aniline more basic.

Acidity of CF3COOH in preference of other acid.
1)Because OH group is less acidic, CH3 is a +R group donate electrons thus makes OH bond stronger so removal of H is depressed.
2)Same as 1st.
4)NH3 is a weak Lewis base.
5)Alkanes are not acidic.

So, CF3 group being -R decreses the e density make removal of H more favourable thus incresing its acidic strength.

Yes, same rule will apply for all other compound acidic or basic. Just like CH3COOH and H2SO4 appears almost same acidic strength in NH3 solvent, because NH3 having lone pair increses the acidic strength to almost same strength.

I hope this will help.

CF3COOH

CF3 group in it is an electron-donating. so it increases the negative electric charge around C of COOH group. as the OH is also negative electric charged, OH tries to break the bond with C. (2 negative electric charged groups try to repulse from each other) the more negative electric charge is increased the more OH tries to break the bond. so, when OH- are released to the solution it finds H+ to form H2O. aniline provides H+ to the solution. the more OH- are released to the the solution the more aniline releases H+. when the amount of H+ which is released by aniline gets high, it becomes a strong base.
so here, CF3 is the most powerful electrons donating group. so it increases the negative charge around the C of COOH more than the other groups. so CF3COOH releases more OH to the solution than other solvent. so it attract more H+ from aniline and make aniline a strong base

(so sorry for my bad english)