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I was wondering whether someone could explain me if nifedipine is a weak base or a weak acid. I barely can find any information about nifedipine's physicochemical properties in studies, however some suggested that it is a weak acid. If it is a weak acid, which hydrogen atom is dissociable and what would the pKa of that group be?

Thanks in advanced.

Nifedipine's structure, as shown on wikipedia:

Nifedipine

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  • $\begingroup$ Here is a reference which states that the "strongest basic" pKa = 5.33. As for which proton, I'm not sure. $\endgroup$ May 14, 2016 at 15:32
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    $\begingroup$ @ToddMinehardt perhaps the H atom at position 4 of the 1,4-dihydropyridine. $\endgroup$
    – user7951
    May 14, 2016 at 18:10
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    $\begingroup$ My guess is still the N-hydrogen. However a $pK_a$ of 4 is more acidic than I would expect, e.g. in comparison with phtalimide. But there may be higher order effects that affect the acidity. $\endgroup$
    – aventurin
    May 14, 2016 at 19:35

2 Answers 2

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Nifedipine is a weak acid.

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  • $\begingroup$ Removed the statement that the N-hydrogen might be the acidic one. $\endgroup$
    – aventurin
    May 14, 2016 at 17:07
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My speculation is a bit related to the answer provided by @aventurin. In absence of experimental evidence to support this, I think in a first step the proton in (then) para-postion to the nitrogen is abstracted:

enter image description here

leading to a protonated, pyridinium specis, then in equilibrium to the non-protonated form. This may be favourable by thecreation of an aromatic system. Again, provided as a speculation including the notice that the initial deprotonation might be irreversible.

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    $\begingroup$ The abstraction of a proton in the first step would result in a nifedipine anion. $\endgroup$
    – aventurin
    May 14, 2016 at 19:10
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    $\begingroup$ Your image shows a 3-nitrophenyl compound; however, nifedipine is a 2-nitrophenyl compound. $\endgroup$
    – user7951
    May 14, 2016 at 19:16
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    $\begingroup$ That nitrogen has a lone pair, your proposed "deprotonation" would lead to a nitrogen with 10 electrons around it (and as aventurin points out it should be negatively charged not positively). Your reaction is in fact a dehydrogenation and the intermediate results from a loss of hydride, not a deprotonation. $\endgroup$ May 14, 2016 at 20:03

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