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.

  • 3
    $\begingroup$ 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
    Commented Jan 23, 2019 at 10:54
  • 1
    $\begingroup$ @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
    Commented Jan 23, 2019 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
    Commented Jan 23, 2019 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
    Commented Jan 23, 2019 at 11:12

1 Answer 1


This is a problem of counting lone electron pairs as part of the delocalized pi system. In pyrrole you count the lone pair on the nitrogen, in pyridine you don't. What's the difference?

In pyrrole 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. A later question edit, however, does specify that the molecule is neutral $\ce{C3H2}$, cyclopropenylidene.

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 and at Saturn's moon Titan. The aromatic stabilization of this species may be one reason it is seen in greater quantities than a polar but acyclic isomer.

If it is cyclopropenyl anion

The question was edited, so this possibility is eliminated and will not be discussed here.

  • 2
    $\begingroup$ 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
    Commented Jan 23, 2019 at 11:07

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