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Topic: pKa of Amines  (Read 19605 times)

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Offline MorningStar

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pKa of Amines
« on: April 26, 2010, 02:50:59 AM »
Hi everyone,

I am wondering if my logic is correct on this problem.  I am asked to compare three molecules:
Conjugate acid of Isopropylamine
CH3CHCH3
      |
      NH3+  :larrow: pKa 10.78

Propanoic acid
CH3CH2COOH  :larrow: pKa 4.78

Conjugate acid of Alanine
CH3CHCOOH :larrow: pKa 2.35
      |
      NH3+ :larrow: pKa 9.87

These are the questions, followed by my responses.

1. How do you account for the fact that the --NH3+ group of the conjugate acid of alanine is a stronger acid than the --NH3+ group of the conjugate acid of isopropylamine?
My answer: The --NH3+ group of the conjugate acid of isopropylamine is adjacent to two methyl groups, which are electron donating, thus making it a weaker acid.  The --NH3+ group of the conjugate acid of alanine is adjacent to a carboxyl group, which is electron withdrawing, thus making it a stronger acid. 

2. How do you account for the fact that the --COOH group of the conjugate acid of alanine is a stronger acid than the --COOH group of propanoic acid?
My answer: The conjugate acid of alanine is a stronger acid because the --COO- group can easily deprotonate the --NH3+ group in an intramolecular reaction to stabilize the molecule. 

I am not very good at acid-base chemistry, so I appreciate the help. :)



Offline nj_bartel

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Re: pKa of Amines
« Reply #1 on: April 26, 2010, 12:56:39 PM »
I think 1 and 2 should have very similar explanations.

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