Benzene has six delocalized electrons in its structure then where does it places the new electrons which it gets from other substituents like oh, etc. Why does it accepts those new ones? Please give an elaborated answer.
Why benzene takes in more delocalized electrons from its substituents even though it has six of them
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2$\begingroup$ That's not the best way to think about it; the benzene ring in phenol doesn't exist in isolation, so you have to consider the system as a whole. In benzene itself, there are no substituents and thus there's no problem having six π-electrons into six spaces. In phenol, there are eight π-electrons in total and eight spaces for them. The difference is that the spaces are distributed differently: there are >6 spaces on the benzene ring and <2 spaces on the oxygen atom for them. It still adds up to 8; there's no overcrowding. A proper answer would illustrate this using the molecular orbitals. $\endgroup$– orthocresolMar 30, 2020 at 4:29
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$\begingroup$ Please explain using molecular orbital $\endgroup$– ManasMar 30, 2020 at 5:30
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2$\begingroup$ I'm sorry, but that will have to be somebody else's job; I've got other stuff to do. Why not try looking it up in a book? $\endgroup$– orthocresolMar 30, 2020 at 5:50
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$\begingroup$ @orthocresol slight language issue. When you say eight places for the delocalized electrons in phenol I assume you mean four bonding ot nonbinding orbitals, but people might think it's eight conjugated atoms when there are really only seven. I think we should be more precise. $\endgroup$– Oscar LanziMar 30, 2020 at 10:10
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$\begingroup$ @OscarLanzi I appreciate the clarification. Indeed - four orbitals is what I meant, which collectively can hold eight electrons. $\endgroup$– orthocresolMar 30, 2020 at 10:28
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1 Answer
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Using atomic orbitals to construct molecular orbitals involves some shifting and redrawing because of overlaps. Benzene is symmetrical, so whether you draw six carbon atomic p-orbitals or squash and spread them out, you get the idea. Now, when you look at phenol, the C-O bond is unsymmetrical, so if you want to be more precise, you say that oxygen draws some electron density from the carbon thru the sigma system (because oxygen is more electronegative).
The oxygen becomes richly negative at carbon's expense, but this isn't fair(!), so it gives back a little electron density thru the pi system of molecular orbitals that it doesn't care about so much (because the pi lobes are farther from the oxygen nucleus). If you draw molecular orbitals, draw the sigma bond with more electron density toward oxygen, and draw the pi system with electron density a wee bit toward carbon.
The energy diagram of the molecular orbitals of benzene will be shifted by putting a hydroxyl on the ring. One one hand you can consider the six orbitals of the aromatic system merely being shifted in the space above the carbon ring, or, on the other hand, you could consider the three filled pi orbitals to remain essentially the same while oxygen donates some electron density to the unfilled (anti-bonding) orbitals (ignoring the sigma system, which just gets shoved around a little).
So the bottom line is that you are not putting whole electrons into someplace new, but swelling up some electron balloons while squashing some others.
answered Mar 30, 2020 at 14:22
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$\begingroup$ James gaidlis I just don't understand one thing that now the electrons flow in a joint p orbital then how does the charge develop on para and ortho $\endgroup$– ManasMar 31, 2020 at 0:53
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$\begingroup$ The molecular orbitals define electron density for symmetrical molecules as a symmetrical smooth function. When you perturb one end of a nice symmetrical orbital, it adjusts in such a way that more electron density builds up on certain atoms. Without going thru the mathematics, use resonance diagrams to view the charge repositioning in phenolate (PhO-) at ortho and para positions. These resonance diagrams are partial contributors to the whole electron density positioning, so the electron density on the oxygen is slightly less than 1.00, and slightly more than 0.00 on o and p carbons. $\endgroup$ Mar 31, 2020 at 14:12
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$\begingroup$ thanks for this answer, Got the intuition, Will learn rigorous theory in Quantum Mechanical Model Soon $\endgroup$ Sep 26, 2023 at 7:31