Amr Morsi Posted July 23, 2011 Posted July 23, 2011 Why are some aromatic compounds' molecules aren't loose enough, although their structure make them very liable to breaking?
hypervalent_iodine Posted July 23, 2011 Posted July 23, 2011 I'm not sure I quite understand your question. Are you asking why aromatic compounds tend to be planar? The rigidity of compounds such as benzene is in fact necessary for them to be aromatic. If benzene were bent into conformations such as those seen in cyclohexane, the pi orbitals would not lie in the same plane and thus the electrons in them would not be able to delocalise around the ring, which would cause for it to cease being aromatic. I'm not sure if that answers your question or not though, so perhaps your could rephrase it?
mississippichem Posted July 23, 2011 Posted July 23, 2011 (edited) The rotational barrier around a [imath] \pi [/imath]-bond is very high. In the case of an aromatic like benzene, the symmetry (D6h) is such that the conjugated [imath] \pi [/imath]- system is able be delocalized over the whole ring. So all the electron density in the HOMO is evenly distributed above and below the plane of the ring. The bonds in benzene are stronger than they would've been in the theoretical cyclohexatriene, however they are rigid, meaning that their vibrational and rotational modes are quite limited. There is a phenomenon called ring puckering though, so the ring isn't totally rigid. No molecules are totally rigid. It's interesting to note that benzene can actually have a good bit more conformational freedom in high excited states Edited July 23, 2011 by mississippichem
Amr Morsi Posted July 23, 2011 Author Posted July 23, 2011 Great!!!! Then some aromatic compounds are non planar so that they gain enough enforced structure. Is this correct? And, if yes, has a tri-atomic molecule been able to get analyzed totally before? i.e. angles, bond-lengths ..... etc.?
mississippichem Posted July 23, 2011 Posted July 23, 2011 Great!!!! Then some aromatic compounds are non planar so that they gain enough enforced structure. Is this correct? Huh? No, aromatics are more stable than their equivalent non-aromatic analogues. The planarity is what causes the delocalization and therefore the enhanced stability. Some aromatics are, in fact, not perfectly planar. However, these tend to not be as stable and display a lower "energy of aromaticity" which is the increase in bond dissociation enthalpy (or decrease in energy of formation, can't remember, sorry, the correlation is roughly the same either way) that is gained by being aromatic. And, if yes, has a tri-atomic molecule been able to get analyzed totally before? i.e. angles, bond-lengths ..... etc.? I don't see how this follows, but yes. X-ray diffraction can be used to determine structural feature to a high degree of certainty if you can get a good crystal. There is also powder x-ray diffraction which I don't know much about. Do I understand this question properly?
Amr Morsi Posted July 23, 2011 Author Posted July 23, 2011 "Aromatics are more stable than their equivalent non-aromatic analogues". A rule to me, man. I am just gonna last check to be a postulate to me. With respect to the analysis, I mean any way, especially by Quantum Field Theory Equations.
mississippichem Posted July 24, 2011 Posted July 24, 2011 With respect to the analysis, I mean any way, especially by Quantum Field Theory Equations. I'm comfortable with aspects of non-relativistic QM. I am confused by QED. When I hear QFT though, I prefer to run and hide .
Amr Morsi Posted July 24, 2011 Author Posted July 24, 2011 When the field gets changed, QED is a must. We don't know originally where are the nuclei with respect to each others. And, the bonds are very effective. I've searched this before.
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