I have a problem where I need to separate benzoic acid from aniline. My teacher gave me a slight hint by saying it involved HCl. I am guessing that aniline (Ph-NH2) will react with HCl in the reaction:
Ph-NH2 + HCl ---> Ph-NH3(+)Cl(-)
This will make the product soluble in water. Meanwhile, benzoic acid (Ph-COOH) will remain as an acid and therefore be soluble in a nonpolar organic solvent, such as ether or CH2Cl2.
The thing I am confused on is why Ph-NH2 would react with HCl to yield Ph-NH3(+). If we have a solution of HCl dissolved in water that has a higher pH than aniline's pKa, wouldn't it remain in a protonated form (Ph-NH2), whereas if we add a base with a lower pH than aniline's pKa, such as NaOH, it would yield a deprotonated form, Ph-NH(-). I was taught that when you add a substance in a solution with a lower pH than its pKa, you keep it the way it is instead of adding a proton.
Thanks so much.