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In this question, 2a1H-cyclopenta[cd]indene is said to be aromatic. My doubt is that the central C atom (connecting the three cyclic rings) is sp³ hybridized. So I think it should be non planar thus making compound non aromatic

Can someone please explain this compound’s aromaticity?

2a1H-cyclopenta[cd]indene

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    $\begingroup$ sp³ carbon can't and doesn't have to be planar. We just don't care about it at all. Otherwise that isopropyl tail would make cumene non-aromatic. $\endgroup$ – Ivan Neretin Dec 7 '17 at 13:57
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    $\begingroup$ @orthocresol Do you know whether this compound by any chance has a short traditional name which can be included in title?:) $\endgroup$ – andselisk Dec 7 '17 at 14:01
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    $\begingroup$ @andselisk, that crossed my mind, but I don't know of anything. A bit of research suggests that 7aH-cyclopent[cd]indene might be suitable, but I know nothing about this kind of nomenclature, so I opted to leave the title as is. $\endgroup$ – orthocresol Dec 7 '17 at 14:03
  • $\begingroup$ @orthocresol 7bH-cyclopenta[cd]indene corresponds to the former (1998) numbering, which is no longer recommended. The name 2a1H-cyclopenta[cd]indene is in accordance with current recommendations. $\endgroup$ – Loong Mar 24 '18 at 20:42
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It's a modified form of [10] annulene (https://en.wikipedia.org/wiki/Cyclodecapentaene) where internal bridges are used to hold the outer, conjugated ring in a planar geometry and thus with the $4n+2$ rule that outer ring is aromatic.

You may be more familiar with the methylene bridged version of [10] annulene that works the same way, but with a simpler bridging structure. See the above mentioned Wikipedia link.

Addendum:

As mentioned in comments, it is overly simplistic to use the $4n+2$ rule. IUPAC justifies it only for simple annulene systems. When you have internal bonds in the ring like the compound here, you no longer have the symmetry and degeneracy structure on which the rule is based.

When you have saturated interior atoms surrounded by an outer conjugated ring, the $\pi$ orbitals are only slightly altered and that is why we are tempted to still apply the $4n+2$ rule. But beware: if we unsaturate any internal atoms, in this case by removing the hydrogen from the central carbon, the $\pi$ orbitals are altered more considerably and the $4n+2$ rule is well past its use-by date.

Maybe in the end we can use the rule as a guide, but ultimately we need to back our conclusions up with hard evidence. In this case the conclusion of aromaticity is supported by https://doi.org/10.1039/C39800000691, in which the authors studied substituted versions of the compound involved here. They found proton n m r shifts consistent with an aromatic outer ring.

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    $\begingroup$ That Wikipedia page also gives this nice reference: doi:10.1039/C39800000691 where the authors made substituted versions of the compound in question. In their analogues, the central "tetrahedral" carbon bears a methyl group which comes in at –1.34 or –1.25 ppm in the proton NMR, indicating aromaticity. The peripheral hydrogens also appear between 7.7 and 8.5 ppm. This would probably be a helpful thing to include in the answer. $\endgroup$ – orthocresol Dec 7 '17 at 13:56
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    $\begingroup$ Hückel's rules cannot be employed for this compound! $\endgroup$ – Martin - マーチン Dec 7 '17 at 14:12
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    $\begingroup$ @Martin, I suspect they are so applied in the OP's textbook. I rolled with the punch. $\endgroup$ – Oscar Lanzi Dec 7 '17 at 14:21
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    $\begingroup$ I know - and I am very sad about that - that Hückel's rules are often misrepresented in (especially) undergraduate studies, and consequently taught wrong. It is one of these common problems that leads to wrong conclusions, and a completely distorted understanding of chemistry, especially aromaticity. And it is so hard to get rid of it... $\endgroup$ – Martin - マーチン Dec 7 '17 at 14:31

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