In one of my organic chemistry experiments, I synthesized tetracyclone (Tetraphenylcyclopentadienone), and we were supposed to dissolve it in DCM to do our TLC.

DCM is polar and looking at the structure of tetracyclone I think it is non polar, isn't there a saying that "like dissolves like"? Then how could tetracyclone dissolve in DCM?

Below is the structure of DCM and tetracyclone:

enter image description here enter image description here

Edit: After searching online, I found that due to the lack in ability to form hydrogen bonds, we treat DCM as a non polar solvent, even though it is weakly polar.

Website: https://www.quora.com/Is-CH2Cl2-both-polar-and-non-polar-Why

  • 1
    $\begingroup$ Tetraphenylcyclopentadienone is a polar moleule. Look at the resonance structure with a positive charge on oxygen and a negative charge delocalized within the 5-membered ring. This would make the ring aromatic with 6 pi electrons. Therefore such dipolar structures are significant descriptors of the molecule, making it polar. $\endgroup$ – ron Nov 25 '15 at 14:32
  • $\begingroup$ My comment is: surely it dissolved as the experiments are pre-checked? The molecule has a dipole moment, same as DCM.@Ron - I've seen the charge split you're talking about for cyclopropenone where the positive charge is on the carbon to give a ring with 2-pi electrons, with a negative charge on the oxygen. I've not seen anything with a positive charge on the oxygen and negative on the carbon. $\endgroup$ – Beerhunter Nov 25 '15 at 23:42
  • $\begingroup$ @Beerhunter Yes, tetracyclone dissolved in DCM no problem. I asked my professor and she said that DCM is in between polar and non polar, so it can dissolve both polar and non polar substances. $\endgroup$ – liya77 Nov 25 '15 at 23:47
  • $\begingroup$ @ron, but the tendency for oxygen to have a positive charge, plus octet deficiency isnt quite stabilized by aromaticity. Molecule might be polar, but not so polar. $\endgroup$ – Pritt says Reinstate Monica Apr 27 '17 at 3:30

Your Answer

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

Browse other questions tagged or ask your own question.