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Topic: separation of aniline from benzoic acid  (Read 11855 times)

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Offline rleung

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separation of aniline from benzoic acid
« on: September 24, 2005, 02:19:09 PM »
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

Offline Yggdrasil

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Re:separation of aniline from benzoic acid
« Reply #1 on: September 24, 2005, 04:31:48 PM »
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-.

Offline movies

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Re:separation of aniline from benzoic acid
« Reply #2 on: September 25, 2005, 01:34:05 PM »
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).
« Last Edit: September 25, 2005, 01:34:45 PM by movies »

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