Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: rleung on September 24, 2005, 02:19:09 PM
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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
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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-.
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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).