My book mentions that

Delocalization of $\pi$ electrons makes a compound stable as the process of delocalization involves loss in internal energy. This is the synopsis of resonance.

But how does shifting of $\pi$ electrons let the body lose internal energy? What is the cause?

While studying resonance, I observed that one structure possesses charge while in other no charge. But in the final structure, there is, according to the book, partial charge present.

Eg. In phenol, three resonating structures contain '+' charge on $\ce{O}$ . Thus, in the final structure, there is a slightly positive charge on oxygen and also, small negative charge on the main body.

Now my question is, why there is partial negative charge on the body? And why does oxygen carry partial positive charge? What is the cause?

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    $\begingroup$ @edkinsella The solution to the "particle in a box" problem shows that energy levels are a function of 1/L^2, where L is the length of the box. So even with just 1 electron in a molecular orbital, where electron repulsion doesn't come into play, the longer we make the box (e.g. the longer the delocalization path) the more stable (lower in energy) the electron. $\endgroup$ – ron Nov 26 '14 at 21:29

In simple terms electrons repel each other, so distributing them over a geater number of bonds will reduce the repulsive force between them so resulting in a lower energy state. In the case of phenol the oxygen p orbital lone pair at right angles to the ring can overlap with the delocalised p orbital above and below the ring enabling a degree of delocalisation into the ring system so leaving a slight + charge on the oxygen. This in turn will reduce the electron density in the OH bond so weakening it and accounting for the weak acidity of phenol.

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  • $\begingroup$ @user36790 You were probably thinking of a,b-unsaturated carbonyl that is commonly drawn as the oxygen carrying the partial charge. Now think of a,b-saturated carbonyl. It has the same amount of electrons but more space for the electrons to fly in. The oxygen does not draw more electrons than it can handle, the partial charge at the oxygen will not increase to infinity when you add double bonds to the other end. $\endgroup$ – Ich bin ein gäst Nov 26 '14 at 18:14

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