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Aromatic? Here's a compound which we have to tell whether it's aromatic or not.

Drawing pi-orbital for the 3-membered ring, I think there are only 2 pi electrons delocalised all over the ring. The orbital having non-bonding electrons on the carbon having sp2 hybridization(carbon no. 1) is not having the correct orientation with the pi-orbital of the ring so these electrons are not delocalised over the ring. 2 being a Huckel's number, I think that this compound should be aromatic.

But the answer key says that it is ANTI-AROMATIC! Where am I going wrong? Perhaps the ring is not planar?

EDIT: The carbon no. 1 doesn't have a negative charge so it is not an anion. It is neutral.

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    $\begingroup$ Are you referring to the cyclopropenyl anion? If so, this has been answered here: chemistry.stackexchange.com/questions/51467/… If you mean cyclopropenylidene, it is aromatic, and you can read about it on the wiki en.wikipedia.org/wiki/Cyclopropenylidene $\endgroup$ – Blaise Jan 23 at 10:54
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    $\begingroup$ @Blaise Looks like title of my question is edited by someone else. The compound I'm talking about is not correctly mentioned in the title $\endgroup$ – Shivansh J Jan 23 at 10:59
  • $\begingroup$ @Blaise Thanks for that wiki link. I'm talking about cyclopropenylidene (which has a carbene), so is my explanation which says it should be aromatic and those non bonding electrons are not delocalized correct? $\endgroup$ – Shivansh J Jan 23 at 11:03
  • $\begingroup$ A quick google shows a couple papers and a book confirming the singlet carbene to be aromatic. Maybe your book intended to indicate the anion. $\endgroup$ – Blaise Jan 23 at 11:12
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This is a problem of counting lone electron pairs as part of the delocalized pi system. In pyrrhole you count the lone pair on the nitrogen, in pyridine you don't. What's the difference?

In pyrrhole you have the lone pair left over after you have given every ring a ligand (the outer hydrogen atoms). With the sigma bonding requirements thus satisfied, the lone pair goes into the pi system. In pyridine the nitrogen is missing its ligand, so the lone pair must replace that in the plane of the ring and can't go into the pi system.

Which situation corresponds to your compound depends on what compound is meant, and there seems to be confusion on this point. Cyclopropenylidene ($\ce{C3H2}$) and cyclopropenyl anion ($\ce{C3H3^-}$) differ only by a carbon-hydrogen bond, and the shorthand method commonly used to render organic structures thus fails to distinguish them. The problem statement should be checked to ascertain which of these compounds is meant because they give different outcomes, and the answer key is right for only one of them.

If it is cyclopropenylidene

In cyclopropenylidene, the carbon with the lone pair is missing a ligand like pyridine, so the lone pair has to replace the missing ligand in the plane of the ring. That leaves two pi electrons to be delocalized which makes the ring aromatic. Cyclopropenylidene is not only aromatic by the electron count, it is a sufficiently preferred three-carbon structure to be seen in interstellar space.

If it is cyclopropenyl anion

Cyclopropenyl anion has all the ligands in place, so the lone pair goes into the pi system like pyrrhole. Unfortunately in this case, adding that lone pair gets you four pi electrons in the ring and that's bad. In fact it turns out to be really had, even worse than larger (and thus more flexible) formally antiaromatic rings like cyclooctatetraene.

Cyclopropenyl anion goes through some serious contortions to get rid of this antiaromaticity. See this question.

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    $\begingroup$ Sorry I was not asking about Cyclopropenyl anion (someone edited the title of question wrongly). I already know that Cyclopropenyl anion is antiaromatic. $\endgroup$ – Shivansh J Jan 23 at 11:07
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The answer key is right. This is a simple compound,planar and conjugated.It has 1 lone pair (carbene) delocalised therefore contains 4 elctrons (2 pi and 2 from LP). Huckels rule for Anti Aromatic compounds say,if a compound is planar,conjugated and cyclic and has 4n pi electrons,properly delocalised then the compound is Anti-Aromatic. For non aromatic,the compound must be localised or non conjugated. For Aromatic, 4n+2 pi electrons with proper delocalisation. Even if there was a cyclopropenylidene anion,it would still be an anti aromatic. And it would be Aromatic,if it was a cation it would then follow 4n+2 pi where n =0, ie; 2pi system with proper delocalisation. SO THE COMPOUND IS ANTI-AROMATIC.

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    $\begingroup$ Did you see the picture I posted?, the non bonding electrons from carbene are not in the correct orientation to get delocalized. $\endgroup$ – Shivansh J Jan 23 at 11:45
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    $\begingroup$ Its orbital is perpendicular to the plane of ring. $\endgroup$ – Shivansh J Jan 23 at 11:46

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