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In case of small molecules like water things are more or less clear, but when I see a large complex molecule like Thyroid Hormone - how should I find out that this molecule is non-polar?

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    $\begingroup$ I've been taught about the "1:4"-rule; 1 non-polar atom "makes up" for 4 polar atoms. Water, consisting of only hydrogen and oxygen (where the difference in electronegativity is large) is rather polar. Octane, in the other hand, consists of 8 carbon and 18 hydrogen, making the ratio 4:9, hence quite non-polar. Obviously, this is just a rough estimate, although it's rather useful for a simple TLC-analysis for example in order to determine the right eluent. $\endgroup$ – Argo Mar 19 '14 at 18:38
  • $\begingroup$ For what reason you want this estimate? If it is water solubility, it is reasonably simple. If for other reason, you can try to roughly estimate the dipole moment of whole molecule (thinking little bit of 3D structure). $\endgroup$ – ssavec Mar 19 '14 at 20:03
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    $\begingroup$ You can be very certain that the larger a molecule is, the more likely it will be somewhat polar, esp. if it contains (so called) hetero atoms. $\endgroup$ – Martin - マーチン Mar 26 '14 at 7:16
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Once you move up to "larger" molecules like this, the definition of a "polar" molecule becomes a bit fuzzy. Rather, you will find that molecules contain polar and non-polar groups which all contribute to the overall characteristics of the molecule. It really depends on the context that you are interested in.

If you are curious about more macroscale properties, such as boiling point or solubility, a variety of group contribution methods have been developed to provide ballpark estimates based purely on the contributions of individual components of the molecule.

While group contribution methods will primarily give you an idea of bulk properties, you may also be interested in lower level behavior such as micellization or specific protein binding which will depend on the location and arrangements of all of the polar and non-polar groups on a molecular level.

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A practical tutorial with Avogadro

enter image description here

It is not so easy to understand from the structure because polarity in fact depends on the arrangement of the groups in tridimensional space relative to one another and not simply on the groups present (e.g. $\ce{trans-C2H2Cl2}$ is not polar because the individual bond dipoles cancel one another, $\ce{cis-C2H2Cl2}$ is polar) so you have to use a geometry optimization tool first. You can perform this optimization and make some assumptions from the model calculating the electrostatic potential with Avogadro. I've made this short tutorial.

  1. Draw you molecule.
  2. Check it (doing a lot of calculations with the wrong molecule is not fun...)
  3. Start optimize geometry tool, wait until dE=0 or near 0
  4. Stop optimize geometry tool
  5. Go to Visualize - Properties - Molecule properties read the polar momentum in Dalton
  6. If you want, you can create the potential surface from extension - surface
  7. Eventually you can try to search for other conformers and repeat the process

enter image description here

How accurate is this method?

From my attempts unfortunately not very. If you try with $\ce{C2H2Cl2}$ you get about 1.086 D more for the cis conformer and 0.22 Dalton, more sometimes depending on the optimization, for the trans conformer after the research of other conformers.

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  • $\begingroup$ This is a really excellent way to get a general idea of polarity in simple molecules. But as you pointed out, getting dipoles correct in molecular modeling is a tremendously challenging problem and an area of active research. $\endgroup$ – CTKlein Mar 25 '14 at 23:23
  • $\begingroup$ @CTKlein you are right, I would like to compare it with other software and methods! I am very grateful for the edit! Thanks a lot! $\endgroup$ – G M Mar 26 '14 at 6:50
  • $\begingroup$ Avogadro is a nice first estimate, but results are questionable. Anyway, the dipole moment is 'only' a number and might not say anything about polarity. For example $\ce{CO2}$ has no (permanent) dipole, but is very polar. $\endgroup$ – Martin - マーチン Mar 26 '14 at 7:13
  • $\begingroup$ @Martin I was thinking about it yesterday but I've read "A polar molecule has a net dipole as a result of the opposing charges" reading wikipedia article. $\endgroup$ – G M Mar 26 '14 at 7:26
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    $\begingroup$ @GM Unfortunately polarity, or in this case nonpolarity is not very well defined (goldbook). It usually is based on a relative scale as every molecule is somewhat polar (even $\ce{H2}$). When dealing with surfactants, one usually describes part of the molecule as nonpolar (tail, hydrophobic) and polar (head, hydrophilic). $\endgroup$ – Martin - マーチン Mar 26 '14 at 7:49
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According to me, Find out about the polarity of molecules is a difficult task. It is all the depends on the drawing of Lewis structure. Polar and non-polar molecules have many differences like polar molecules have positive and negative ends while non- polar doesn't. Molecules have different shapes like linear, tetrahedral, trigonal planar is symmetric and bent, or trigonal pyramid is an asymmetric shape.

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    $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$ – angussidney Mar 24 '18 at 7:07

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