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This question already has an answer here:

If we look at carbons 1 and 2, they have 5 electrons. 3 are directly involved in bonding, and 2 are involved in delocalisation to create the 2 benzene rings. This would mean that they had 5 valence electrons, which isn't possible, and thus there should be only one benzene ring.

Naphthalene

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marked as duplicate by M.A.R., Todd Minehardt, Buttonwood, Wildcat, Jon Custer Apr 28 '17 at 16:57

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    $\begingroup$ These are not benzene rings, and even calling them aromatic because of Huckel's rule is a bit of a stretch. Yet still they are rings, and they contain delocalized bonds, so why not draw them with a circle. $\endgroup$ – Ivan Neretin Apr 28 '17 at 13:57
  • $\begingroup$ If you really need a Lewis structure, use the one without rings, with 5 double bonds, and have on mind that it os one of may so called resonance structures that are in reality "averaged". $\endgroup$ – mykhal Apr 29 '17 at 1:32
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Chemists are already accustomed to this type of formula, even though it is not absolutely correct. More exact would be drawing all bonds with double solid/dashed bonds to express that the bond order is between 1 and 2, approximately 1.5, but is almost never done

naphthalene picture 1

Rarely can be seen formula with the two circles merged (and maybe dashed also), like

naphthalene picture 2

But the electrons are in fact delocalized across all bonds of the polycycle structure in "electron cloud" that looks somewhat like the infinity sign ∞ in this case (naphthalene) above and below the planar molecule. What matters is that the count of π (those from the "second bond" of the double bond) electrons is 4n+2 for n is some natural number (here it is n=2). It is a special feature called aromaticity, which has something to do with advanced topic of quantum chemistry; and it greatly stabilizes the molecule. Any of these formulas is unable to describe reality.

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  • $\begingroup$ Do the two rings in this behave like benzene rings? $\endgroup$ – Saad Apr 28 '17 at 15:31
  • $\begingroup$ @Saad yes, they have similar properties, but not exactly.. thanks to symmetry, there are there three types of carbon centers (some even do not bear hydrogen atom), while in the benzene there is only one type. $\endgroup$ – mykhal Apr 28 '17 at 15:36
  • $\begingroup$ So the friedel-crafts reactions would work with this? $\endgroup$ – Saad Apr 28 '17 at 15:46
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    $\begingroup$ Instead of the historical "4n+2 pi electrons", I recommend thinking of Huckel's Rule as "an odd number of pi-electron pairs." Much more intuitive and accessible for new students. $\endgroup$ – electronpusher Apr 28 '17 at 16:30
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    $\begingroup$ Well, to nitpick, this one may carry the wrong impression that every carbon on the ring has equal electron density. I don't really think there's a way to be 100% accurate at describing the structure of naphthalene, without drawing out two resonance forms. So your concluding statement is certainly important. $\endgroup$ – orthocresol Apr 28 '17 at 19:33
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I find this representation to be kind of sloppy for the reason you mentioned. The $\pi$ system is delocalized throughout the molecule, but that doesn't mean it's two aromatic systems.

I find this representation much better:

Alternative representation of naphthalene

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    $\begingroup$ This one carries the wrong impression that the rings are different, which they are not. $\endgroup$ – Ivan Neretin Apr 28 '17 at 14:50
  • $\begingroup$ @Ivan Neretin Then how would one represent it? Two 'circles' leave no room to differentiate it from two benzene rings. Also, if these rings aren't benzene rings, then it implies that the one with the OH group attached cannot be called a phenol group? Do these rings behave like benzene rings though? $\endgroup$ – Saad Apr 28 '17 at 15:17
  • $\begingroup$ Two circles are fine. There is no such thing as benzene ring. These rings are delocalized; arguably, they are aromatic, albeit to a lesser extent than benzene. In some reactions naphthalene behaves like benzene, and in some it behaves differently. $\endgroup$ – Ivan Neretin Apr 28 '17 at 15:22
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    $\begingroup$ Any representation of anything is a simplification and misses some of the complexity. As long as we remember that, we'll be all right. $\endgroup$ – Ivan Neretin Apr 28 '17 at 15:35
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    $\begingroup$ Actually, if you treat it as one of two resonance structures, it's good depiction, showcased in Clar's rule. $\endgroup$ – Mithoron Apr 28 '17 at 17:57

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