0
$\begingroup$

Nucleobases G, C and A carry an amine group ($\ce{-NH2}$). In amino acids, the amine groups are considered protonated at physiological pH. Why isn't it the case for the amine groups of nucleobases?

I found some references discussing the protonation of nucleobases, but I am not enough expert in either DFT (http://www.ncbi.nlm.nih.gov/pubmed/19049279) or Mass Spectrometry (http://www.ncbi.nlm.nih.gov/pubmed/17920289/) to get a rational from this. Plus, neither they nor the references they cite and that I have briefly examined actually consider protonation of the amine groups.

$\endgroup$
1
  • $\begingroup$ Amines are basic because of the lone pair on the nitrogen. In the nucleobases, the amine is adjacent to a double bond, which allows the lone pair to be delocalized. This reduces the basicity of the amine. $\endgroup$
    – f''
    Jul 23, 2016 at 16:42

1 Answer 1

1
$\begingroup$

There are 2 answers:

  1. As f' answered, the amino (NH2) group in a nucleobase is conjugated with the heterocycle. Therefore, its pKa is closer to the one of aniline. In other words, it is much less basic than an alkylamine.
  2. Any drawing of a molecule in chemistry or biology is only a mere representation (agreed on by scientists) of the reality. For example, amino-acids are most usually drawn as H2N-R-COOH. We all know that it does not reflect reality but we all use that representation because it is much more readable than +H3N-R-COO- (which itself is also wrong), and because the real form of the amino acid will greatly depend on the experiment conditions. So you must always consider in what medium the molecule is to determine, more or less, its major form.
$\endgroup$
1
  • $\begingroup$ Ah! I had indeed overlooked the delocalization of the lone pairs. Thanks a lot! $\endgroup$
    – The Quark
    Jul 24, 2016 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.