I'm studying medicine, not chemistry, but I hope you can help me anyway.

I just had to solve a question where it is given that the drug Propranolol has a $\text{p}K_{\text{a}}$ of 9.5 and the un-ionized form is non-polar.

Is there any way to say if propranolol is an acid or base just from the $\text{p}K_{\text{a}}$?

Hope to hear from you, and sorry if it is a stupid question.

  • $\begingroup$ It's not a pKa of propranolol, but its pKbH+ - pKa of its conjugated acid. $\endgroup$ – Mithoron Jun 5 '17 at 14:43
  • $\begingroup$ Probably, but it only says "pka = 9,5". So if i don't have the information about propranolol being a base, how can i know? $\endgroup$ – Nanna Jun 5 '17 at 14:46
  • $\begingroup$ pKa of coniugated are often not properly distinguished, though luck kid... $\endgroup$ – Mithoron Jun 5 '17 at 14:54
  • $\begingroup$ Hi @Nanna, to answer your specific question, if all they tell you is 'Propranolol has a ${pKa}$ of 9.5 and the un-ionized form is non-polar', and you don't know anything else about it, then no, you can't tell if it's an acid or a base. Un-ionized molecules are generally non-polar, or at least less polar than ionized ones: for an acid the ionized form is an anion; for an amine, the ionized form is the protonated amine. Doesn't help, does it? I suspect what they actually meant to tell you was that in the acid-base reaction the loss of a proton produced a less polar species. Then it's a base. $\endgroup$ – user6376297 Jun 5 '17 at 16:36
  • $\begingroup$ You're welcome. And don't be afraid of asking questions, they are never stupid if you're someone who wants to learn. In fact, it's the exam/test question that is badly formulated, as it does not give you sufficient information to answer, or actually even misleads you. $\endgroup$ – user6376297 Jun 6 '17 at 17:24

Based on the structure, this drug is weakly basic. It has an amine moiety, which if protonated, would likely have a $\mathrm{p}K_{\mathrm{a}}$ value of around 9.5.

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    $\begingroup$ I know you can say a lot about a drug based on the structure - but my exam is without any forms of 'aid' - I can't just look it up on the internet, so I have to use the information I am given. :-) $\endgroup$ – Nanna Jun 6 '17 at 9:05
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    $\begingroup$ At a pKa of 9.5 for a drug, you can probably assume an anime as a first guess. $\endgroup$ – Zhe Jun 6 '17 at 13:11
  • $\begingroup$ @Zhe That must be one hell of a drug to "assume" anime from:D $\endgroup$ – andselisk Sep 5 '17 at 3:06

No, you cannot say because the terminology is, unfortunately, used interchangeably. For example, phenol has a pKa around 10. That means it is weakly acidic. However, amine pKa's are frequently referred to as being around 9-11, when in fact those numbers are the pKa of the conjugate acid. The pKa of secondary amines for example, is in the 20's as a very strong base is required to deprotonate them. So in your example, I would wager the pKa is of the salt form, not the neutral molecule.


We are taught (assuming the $\ce{pK}$ values are about aqueous solutions) \begin{align} \mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} &= 14& \text{so} && \mathrm{p}K_\mathrm{b} &\approx 14 - \mathrm{p}K_\mathrm{a}. \end{align} In this case, $\mathrm{p}K_\mathrm{b} \approx 3.5$, which is not a strong base ($\mathrm{p}K_\mathrm{b} < \approx 2-3$) but it's close. In medical literature a "drug" (i.e. a medicine) is a family of chemical compounds (or mixtures of compounds) where the exact chemical structure is often ignored. Propranolol may refer to the nonpolar compound, or one of several salts (such as Propranolol.HCl). There's no way to know without knowledge of how the drug is actually provided.

Anyone studying medicine should also study $\mathrm{p}K$ values, you may have done that in high school. As far as identifying the organic chemical groups which may ionize, the good news is that there are only a few, so you can learn them. At least, only a few at physiological pH.
Knowing that a group can ionize isn't as good as knowing it does ionize, but it's probably the best you can do from basic chemical knowledge. Experience will help some, but generally you'll have to rely on the literature since the exact electronic (and perhaps steric) effects that contribute are complex and difficult for the non-expert to analyze.

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    $\begingroup$ I have removed the unscientific banter at the beginning, which came off a bit rude. I invite you to review or be nice policy and act accordingly. $\endgroup$ – Martin - マーチン Jun 6 '17 at 6:26

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