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The pK1 and 2 values for alanine are 2.3 and 9.7. In the dipeptide Ala-Ala, they are 3.1 and 8.3 and in tri-peptide Ala-Ala-Ala, they are 3.3 and 8.0. Why does this trend in pKa values exist? Why does the isoelectric range decrease? The reason I can think of is the extra Alanine dumps electrons towards the C terminal and so makes the O-H bond less polar and thus less acidic while it also pulls more electrons from the N terminus to make that more polar and thus more acidic. Thanks!

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It is because of the inter- and intramolecular associations that occur.

In Ala, the first loss of a proton forms a two point intermolecular association which is very favourable. Since the equilibrium tends to the right, more $\ce{H^+}$ are liberated and the p$K_a$ is lower.

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For the second step, this stable dimeric association must be destroyed, so now the equilibrium tends to the left and the p$K_a$ is higher then it would be without this association.

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In the case of Ala-Ala-Ala, this type of dimeric association can not occur because the distance is too long. There is intermolecular association but with different molecules, not in pairs. Here the first p$K_a$ is higher than in the case of Ala and the second lower because the equilibrium is not forced so much to the associated forms.

Finally, in the case of Ala-Ala, besides the intermolecular association between different molecules like in Ala-Ala-Ala, there is also an intramolecular association. This one is less stable than the association in pairs of the individual amino acids, but more than the random intermolecular ones. Therefore, the two p$K_a$ are in between of the ones for Ala and Ala-Ala-Ala.

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I would not say that the extra alanine unit dumps or pulls electrons, rather it "insulates" the inductive effect of charged functional groups that we can see in the monomer.

The postitively charged alanine monomer has the structure

$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{COOH}$

The dimer has this:

$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{C(O)NH}-\ce{C(H)(CH_3)}-\ce{C(H)(CH_3)}-\ce{COOH}$

When the carboxyl group ($\ce{COOH}$) is deprotonated the monomer gets some stabilization of the negative charge from the positively charged, electron withdrawing function $\ce{NH_3^+}$ two positions away on the chain. But the dimer has a neutral, less electron drawing peptide ($\ce{C(O)NH}$) function at the same position, making the deprotonation less favorable.

After the first deprotonation the molecules, now zwitterionic, are as follows:

$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{COO^-}$

$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{C(O)NH}-\ce{C(H)(CH_3)}-\ce{C(H)(CH_3)}-\ce{COO^-}$

So we move on to the second deprotonation which neutralizes the ammonium function. The double deprotonated monomer then has an amino group, $\ce{NH_2}$, two places from the carboxylate anion. The diner has a peptide function, $\ce{C(O)NH}$, at that location and this is more electron withdrawing, thus more stabilizing to the negative charge, than the monomer's amino group. So in this second stage the dimeric anion is more stable and forms under less strongly basic conditions.

Thus we see inductive effects working in opposite ways in the two amino acid functions. It's an example of how the two ends have opposing characteristics.

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