# Question #1183c

Mar 30, 2016

The phenyl cation is not involved in resonance because the vacant orbital is in the plane of the benzene ring.

#### Explanation:

The carbon atoms in benzene are $s {p}^{2}$ hybridized.

They are joined to each other by $s {p}^{2} \text{-} s {p}^{2}$ σ bonds and to the $\text{H}$ atoms by $s \text{-} s p 2$ σ bonds.

(from www.chemtube3d.com)

The unhybridized $p$ orbitals overlap to form the aromatic π system of benzene.

The phenyl cation is formed by removing an $\text{H}$ atom with its bonding pair of electrons.

$\text{C"_6"H"_6 → "C"_6"H"_5^+ + "H:"^"-}$

If we draw the phenyl cation as below, it looks as if we could draw curved arrows to get resonance structures.

But we can't do this!.

Remember that the $\text{H}$ was removed from an $s p 2$ σ bond, so the $s p 2$ orbital is the vacant orbital (in the plane of the benzene ring).

The major lobe of the orbital has everywhere a "+" amplitude.

Any overlap of the $s p 2$ orbital with the "+" amplitude of the π orbital on the top side of the ring (red) is cancelled by overlap with the "-" amplitude of the blue π orbital on the bottom side of the ring (blue).

There is no net overlap and therefore no resonance with the vacant orbital.

We say that the two orbitals are orthogonal to each other.

NOTE: The phenyl cation still has the normal aromatic resonance, because the π system is unchanged.