Question

Aromaticity help....?


Here,what is the hybridisation of the negative part...how can I explain it to be anti-aromatic...please briefly discuss how we get the number of loose electron?

Answer 1

Remember the Huckel criterion for aromaticity.....?

Explanation:

`4n+2` `pi` electrons....does this criterion hold here? Well, there are `3` `pi` bonds I can see, but there are an extra 2 electrons that constitute the negative charge.

`4n` `pi` electrons are a characteristic of `"antiaromaticity"`.

How many electrons you got in the anion? If you had a cation would the system be aromatic?

Answer 2

The tropylium anion really has two valence electrons on the negative carbon:

The criterion for aromaticity are:

• `4n+2` `pi` electrons IN the ring, where `n = 1, 2, 3, . . .`
• Planar structure
• Closed ring
• Electron conjugation all around the ring.

If, however, there are `4n` `pi` electrons instead, but the other three criteria are satisfied, the compound is antiaromatic.

In the case of tropylium anion, the lone pair of electrons is in the ring (i.e. they are electrons in the `pi` system) because there is a hydrogen atom parallel to the ring in an `sp^2` hybridization:

This structure may or may not be planar, but assuming that it is, the lone pair would have to be in the ring, being within a pure `p` orbital perpendicular to the ring, delocalizing `bbpi` electron density throughout the ring:

Under this assumption that it is planar, and the ring is obviously closed, we satisfy three out of four conditions.

However, with `8 xx pi` electrons in the ring,

`4n = 8` `pi` electrons in the `pi` system, with `n = 2`,

and you cannot write `4n + 2 = 8` unless `n` is not an integer.

Therefore, tropylium anion is antiaromatic. However, if it is not planar in reality (that is, if the lone pair is sufficient to push the hydrogen out of the plane), then it is non-aromatic.

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On the other hand, tropylium cation is aromatic: