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Cyanidine is not polar due to its make-up of only hydrogen atoms and carbon atoms, correct? But "like dissolve like", which means it would HAVE to be polar, or no? Is it then soluble solely because of its OH-groups being able to create hydrogen bonds with the water molecules, thereby making it miscible?

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    $\begingroup$ And what is it this "polarity" your asking about? I wonder if you know what you're talking about. $\endgroup$ – Mithoron Apr 28 '17 at 18:26
  • $\begingroup$ Well, I read it was soluble in water. It'd have to be polar, then? @Mithoron $\endgroup$ – javanewbie Apr 28 '17 at 18:41
  • $\begingroup$ Then ask why it is soluble in water... $\endgroup$ – Mithoron Apr 28 '17 at 18:50
  • $\begingroup$ @Mithoron Good point. I'll edit. $\endgroup$ – javanewbie Apr 28 '17 at 18:53
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    $\begingroup$ You say cyanidine is not polar due to being made only of hydrogen and carbon, but then talk about it having OH groups. It has oxygen. Not only that but it is a charged species (also clearly shown on your structure). So that is two reasons why your initial suspicion of non-polarity because it is s simple hydrocarbon are wrong. $\endgroup$ – matt_black Apr 28 '17 at 19:28
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Cyanidine actually looks to me to be a rather highly polar compound. It's essentially a $\ce{C15}$ hydrocarbon having 5 hydroxyl groups and a formal positive charge.

The ratio of hydroxyl groups to the number of carbons is the same as that of water-miscible isopropyl alcohol. (Of course smaller molecules like isopropyl alcohol also tend to be more soluble than larger molecules.)

The most important feature indicative of it's rather high polarity and water solubility is it's formal positive charge. This is analogous to placing a positive charge on a large amine by protination in order to turn the amine into a polar, water soluble compound.

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  • $\begingroup$ So the positive charge makes it polar? I suppose it is irrelevant that it is placed practically in the middle of the molecule? $\endgroup$ – javanewbie Apr 29 '17 at 8:05
  • $\begingroup$ Yes and yes. Usually when we talk about a molecule being polar, we're talking about regions of somewhat higher and lower electron density due to inductive or other effects. We usually call this "delta +" or "delta -" indicating a partial charge. These partial charges, or regions of higher or lower electron density, are far less pronounced than the case of having a full formal charge. The location of the charge can make some difference if it's very sterically "hidden", but that's not the case here. $\endgroup$ – airhuff Apr 29 '17 at 18:26

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