[rleung]

Hi,

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.

Ryan

[Yggdrasil]

The amino group of aniline has a lone pair capable of accepting a proton.  It is relatively easy to protonate the amino group to form an -NH₃⁺ group, but it is very difficult to deprotonate the amino group to form an -NH^-.

[movies]

You have to be careful when you look at pKas for amines because you will often see a pKa for the protonated form (R-NH₃⁺) and one for the deprotonation of the neutral for, (R-NH₂ --> R-NH^-).  For example, the pKa of aniline is 28 (Ph-NH₂ --> Ph-NH^-) while the pKa of anilinium is 4.6 (Ph-NH₃⁺ --> Ph-NH₂).