# Participation of Pi-bonds and lone pairs in resonance

Below is a quote from Klein's Organic Chemistry 3rd edition:

Here is the bottom line: Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

However, I am confused as a few pages prior it was shown how allylic lone pairs and pi-bonds participated in resonance:

The first curved arrow goes from the lone pair to form a π bond, while the second curved arrow goes from the π bond to form a lone pair:

What have I understood wrongly?

• In your example there is no atom that possesses both a π bond and a lone pair at once. Commented Apr 2, 2018 at 4:55
• In the 2-D resonance structure shown, no distinction is made between the pair of electrons in the pi-framework and the two pairs of electrons in the sp2 non-bonding orbitals. The choice of the electron pair on the left of the oxygen is arbitrary. One pair of electrons on oxygen and the pair of electrons in the double bond are in the pi-framework of the canonical resonance structure that is orthogonal the plane of the structure. A 3-D resonance representation of the enolate would have clarified this point. Commented Dec 18, 2018 at 18:35

You have probably misunderstood what the sentence actually means.

It means that only one, either $\pi$ bond or lone pair will participate in resonance if the atom has both.

For example, in pyridine, the nitrogen has lone pair as well as is attached with a $\pi$ bond.

While resonance, its $\pi$ bond participates in resonance because it (and not the lone pair) is in the plane perpendicular to the molecule.

Observe that the lone pair remains at the place it is meaning it doesn't participate.

Conclusion: The pair of electrons present in the orbital perpendicular to the plane of molecule participates in resonance if it is present.

What's meant is that at a time for an atom only one lone pair will participate in resonance.

Oxygen over here has three lone pairs but only one of them can participate in resonance since the other other two will be perpendicular to the $\pi$ orbitals of the carbons.