I have been learning about zwitterions and the isoelectric point, and have been trying to apply this to hydrolysis of peptides. From the diagram in my textbook, I can see that when acid hydrolysis occurs, the amino acids formed from the hydrolysis act as bases because the pH is below their isoelectric point, so they accept the $\ce{H+}$ ions forming positively charged ions.
Now when it comes to alkaline hydrolysis, I don't understand why the peptide would not just be hydrolysed, and then act as an acid and donate the extra $\ce{H+}$ ion on its amine group to the $\ce{OH-}$ ions to form negative ions and a $\ce{H2O}$ molecule. Instead, in my text book it shows something like this:

$$\ce{-NH-CH(CH2SH)-C(=O)-NH-CH(CH3)-C(=O)- + NaOH \\-> -NH-CH(CH2SH)-C(=O)-O^{-}Na^{+} + H2N-CH(CH3)-C(=O)-}$$

I'm really sorry if this makes no sense whatsoever...

  • $\begingroup$ @Martin Thank you for editing my post. What program are you using for that writing so that I can use it in the future? $\endgroup$ – Meep Sep 9 '14 at 6:48
  • $\begingroup$ No problem at all. For first aid have a look at the help center. It is basically an implementation of LaTeX as a markup language. These two posts on meta might also be of help: here and here. However, even if you are tempted (everybody is) please do not use markup in the title field, see here for details. Welcome on board! $\endgroup$ – Martin - マーチン Sep 9 '14 at 6:52

In the drawing shown below, the first 3 lines show a small peptide and how it will exist under

  • acidic conditions: the basic amino nitrogen is protonated
  • neutral conditions
  • basic conditions: the carboxylic acid group is deprotonated

A peptide contains multiple amide bonds ($\ce{-RCO-NHR'-}$). When a peptide is hydrolysed under basic conditions, the amide bond is broken. Note that, as mentioned above, under basic conditions any acid groups in the peptide will exist as very stable (due to resonance delocalization) carboxylate groups. Line 4 in the drawing shows this in more detail.

The actual mechanism for the hydrolysis of the amide group under basic conditions is pictured in the bottom line of the drawing. The reaction occurs by a nucleophilic bimolecular substitution reaction ($\ce{S_{N}2}$). The nucleophilic hydroxide anion attacks the relatively electrophilic amide carbonyl carbon to form a tetrahedral intermediate. This intermediate can either return back to the amide or go forward to the carboxylic acid (hydrolysed amide). This same $\ce{S_{N}2}$ reaction can be repeated until all of the amide bonds in the peptide are hydrolysed.

enter image description here

  • $\begingroup$ Thank you for your reply! I have a quick question; in the mechanism shown, you have R-CO-NH2.Does it work the same way if you have R-CO-NH-R', as in an amino acid? Then at the end you would have R-COO(minus) + H2N-R' and Na+ attaches to the -COO(minus)... $\endgroup$ – Meep Sep 9 '14 at 8:15
  • $\begingroup$ Yes, that is correct. $\endgroup$ – ron Sep 9 '14 at 12:49

The unprotonated amino group ($\ce{R-NH2}$) does not protonate $\ce{OH-}$ because it is an extremely weak acid. Primary and secondary amines have a $pK_a$ of around 35, while the $pK_a$ of hydroxide is about 13-14 in aqueous solution. Deprotonating $\ce{NH2}$ groups requires much stronger bases in non-aqueous medium. Therefore, the carboxylic acid group of the amino acid is deprotonated in aqueous alkaline solution, as shown in the scheme from your textbook.


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