13.1: Allyl System

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We have previously described the molecular orbitals of ethylene, which contains only a single \(π\) bond. When we have consecutive double bonds (\(π\) systems), the molecular orbital description actually changes dramatically and we say that the system is conjugated.

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Conjugated systems are present in anything that has resonance delocalization, but it also has implications for simple neutral systems. Electrons are delocalized if \(π\) bonds are conjugated. The molecules below have electrons that can spread out over multiple atoms. Thus, our molecular orbital description must account for this.

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Allyl System

Let’s take a look at the simplest system that has conjugation – where electrons can be delocalized over three atoms. This is known as the allyl system.

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Notice that there is never any way to place the –, +, or * on the central carbon atom! Why? Let’s construct the molecular orbitals for the allyl group’s \(π\) system.

1. n atomic orbitals combine to form n molecular orbitals

2. lowest energy molecular orbital has zero nodal planes

3. any increase in energy level results in an additional nodal plane

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If we consider just the allyl anion, what you might notice is that in the HOMO, the coefficients are equal at the ends of the system (if symmetrical). The electron density is found at the ends, NEVER at the central carbon atom because there is a node there! This should tell you that if the allyl anion is reacting as a nucleophile, the electrons it will use will be the HOMO electrons, and the only atoms that have electron density in the HOMO are the ends!

What this means is that the molecular orbitals will tell us where all of the “chemistry” (reactivity) will occur. It is the same for the allyl cation and the allyl radical and is predicted by the molecular orbitals. All chemistry must occur in \(π\)₂.

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These are all examples of symmetrical systems – what about asymmetric systems? Well, the only thing that changes is the size of the coefficients in the molecular orbital diagram. Consider the asymmetric system below – which carbon would you expect to have more electron density in \(π\)₂? In other words, which end of the \(π\) system is more reactive? It turns out that the resonance form that contains the negative charge on the more highly substituted carbon atom is more reactive because of the extra \(σ\)-donation into the \(π\) system. This raises the coefficient on this carbon atom and makes it more reactive (better orbital overlap with an electrophile).

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What about a masked allyl system? In N,N-dimethylformamide (DMF), the electrons on nitrogen can delocalize into the \(π\) system of the carbonyl to create a dipolar resonance form with the negative charge on oxygen (making oxygen more basic). Because this delocalization occurs similarly to a simple allyl system, we say that the nitrogen atom is sp² hybridized and NOT sp³ hybridized, and the lone pairs on nitrogen are in an atomic p orbital that can align with the \(π\)_C-O bond. This means that there is partial double bond character in the C-N bond and the C-O bond. What effect does this have? Well, it means that in the dipolar resonance form, the methyl groups are in different magnetic environments since there is not rotation about the C-N bond. Thus, there are two peaks for the methyl groups in the ¹³C NMR, not one.

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