Resonance/Aromaticity - Basic question

[MCATMadness]

its been a while since taking ochem.... can someone explain the difference between aromaticity and resonance and conjugation? i need a simply conceptual definition and im not finding what i need in my books or online.

also, does anyone remember the simple rules for identifying resonance and aromaticity and conjugation? i vaguely remember easy rules to follow to ID each ... but im not sure.

and also, can someone explain why the imine N on the his amino acid is protonated and not the amine N. I understand that resonance is lost (and therefore aromaticity is lost) after protonating the amine (NH2+). but when i first approached the question, i looked at the amine and imine and said both can contribute their electrons to the pi-bond system so why would one be protonated over the other.... can someone explain to me the logic for this question.

any help would be great. thanks!

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[loveoforganic]

The first thing that needs to be talked about is orbitals.

This is a representation of a ...C=C-C=C... system. The teardrop shaped images are your p-orbitals. Each p-orbital of each C=C bond is "locked" where it is relative to the other one (otherwise their wouldn't be a pi bond). For C=C1 to conjugate with C=C2, the 4 p orbitals must be in the same plane (as shown). When in the same plane, the electrons from each C=C pi bond can flow freely through the other C=C pi system. This increase in electron movement freedom leads to a decrease in energy of the system (favorable). Therefore, bonds that can potentially form resonance generally will, unless there's a major steric/electronic hinderance.

Aromaticity is a special situation of resonance that can sometimes occur in cyclic elements of molecules.

Here's benzene. You can see how the p orbitals are all in the same plane. However, aromaticity can only occur under certain instances.

To be aromatic, the system must

1. follow Huckel's rule, having 4n+2 electrons in the delocalized p-orbital cloud;
2. be able to be planar and are cyclic;
3. every atom in the circle is able to participate in delocalizing the electrons by having a p-orbital or an unshared pair of electrons.

For the example you gave with histidine,

this is the orbital layout (you can see which atom corresponds to which on the right. The amine nitrogen use its lone pair to contribute to histidine's aromaticity. It has only 3 sigma bonds, so it can hybridize as p,sp2 with the lone pair going into the p orbital. This is because the 120 degree sp2 bonds don't significantly hurt the bond strain in the ring structure.

However, for the imine nitrogen, it has two sigma bonds and one pi bond. It is already p,sp2 hybridized. Therefore, the lone pair electrons lie in an sp2 orbital. Theoretically, you could rehybridize to p,p,sp and try to put the lone pair electrons into the pi system, but this couldn't happen because they are stuck in an orbital out of the plane of the pi system (it couldn't happen anyway due to the enormous bond strain of sp hybridized atom in a small cyclic structure, but that isn't the important thing here).

 

[orgohacks]

Good job, loveoforganic, as always