# Why is the cyclopropenium ion aromatic?

I don't understand how the 'cyclopropenium' ion is aromatic. According to my understanding, the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital - since each carbon is bonded to 3 other atoms and has 4 valence electrons. How does that lead to aromaticity, since this doesn't fullfil Huckel's rule?

• en.wikipedia.org/wiki/Cyclopropenium_ion#Bonding Jan 2, 2018 at 10:19
• Jan 2, 2018 at 10:21
• Jan 2, 2018 at 10:23
• It's because the of the small size of the molecule. The hydrogen on the cation does not interfere with pi electrons orbiting around the cyclic molecule.
– don
Apr 19, 2018 at 8:47

It does meet Huckel's rule. When you have $m$ carbon atoms in a conjugated ring with a positive charge of $q$, there are $m-q$ pi electrons. Here $m=3, q=1$ therefore the number of pi electrons is $m-q=2=4×0+2$.

A Huckel model calculation indicates that cyclopropenyl cation has roughly as much stabilization as benzene, but it's distributed over three instead of six carbon atoms. Thus aromaticity is especially powerful in a cyclopropenyl cation ring. This explains how a positive (formal) charge can be stabilized on as few as three carbon atoms and the ring holds together despite steric strain.

• Thanks very much Oscar, I supposed i overlooked the positive charge. You mention how stabilization is distributed over three carbons instead of six, are you referring to the positive charge? Am i right in thinking that in general a larger distribution of charge lead to greater stability? Jan 2, 2018 at 12:36
• I am comparing the pi-electron stability based on the "Frost circle" which is a picture of the Huckel model) versus ordinary valence bond structures. Benzene gives a difference equal to one "extra" pi bond in this model. Cycylopropenyl cation gives that amount of stabilization too, but each ring atom gets one third of it not one sixth. That bigger per-atom stabilization helps stabilize the charge and fight steric strain. Jan 2, 2018 at 12:41

the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital

This is incorrect.

The requirements for Huckel aromaticity are:

• There must be a continuous, planer ring of overlapping $p$ orbitals.

• There must be $4n+2$ $\pi$ electrons in the system.

The cyclopropenyl fulfills these requirements because it is planar, has overlapping $p$ orbitals, and has 2 $\pi$ electrons ($n = 0$).

The cyclopropenyl cation is aromatic because it is meeting all the definitions of Huckel's rule of aromaticity:

• All carbons are sp2 hybridised.
• There are 3 carbon atoms which form a conjugated system and moreover it has a positive charge therefore it has $3$-$1$=$2$ pi electrons.
• If we put $n=0$ in Huckel's expression ($4n+2$) pi electrons we are getting 2 as the result.
• The cation also has overlapping planar and parallel p orbitals. You may be knowing that for maximum overlapping, orbitals should be parallel to one another and planar.

Since the required criteria are fulfilled, it is aromatic.