Molecules that are arranged in the shape of a ring can be described as one of three things: completely aromatic, anti-aromatic, and non-aromatic. Their descriptions are as follows.

Aromaticity has four major criteria. In order for a molecule to be considered aromatic, it must:

1.      Be cyclic

-          This means the molecule’s carbon atoms are arranged in the shape of a hexagon^1.

2.      Be planar

-          In short, this means that the molecule must be flat. In order for it to be flat, all atoms in the ring must have a p orbital, and all p orbitals must be aligned. A p orbital is kind of like a tiny atmosphere floating above the atom, which houses a π electron (a negatively charged particle). Because the electrons are negative, you might think of them as angry little birds. Just as you wouldn’t want too many angry little birds in the same cramped area of sky, you don’t want lots of negative charge from electrons too close together…neither make for a desirable environment! Therefore, it would be better to spread out the little birds (or electrons). When the p orbitals are aligned, the density of the negative charges can be spread out, or delocalized²; this makes the molecules more stable…this is a very good thing!

3.      Be completely conjugated

-          For a molecule to be conjugated, every atom must have a p orbital²…its own tiny atmosphere.

4.      Satisfy Huckel’s rule

-          Huckel’s rule says the compound must contain 4n + 2 electrons in order to be aromatic². That means that as long as the molecule contains a number of electrons (those found in the p orbital) that is a multiple of 4n + 2, it meets the fourth criteria for aromaticity.

If a compound meets the first three criteria, but contains 4n π electrons (or birds) rather than 4n + 2, it is said to be an antiaromatic compound.³ Should a compound fail to meet one or more of the requirements it is nonaromatic.³

¹Smith, J. Organic Chemistry. 2^nd ed.; McGraw-Hill, 2008, pp 616.

²Smith, J. Organic Chemistry. 2^nd ed.; McGraw-Hill, 2008, pp 617.

³Smith, J. Organic Chemistry. 2^nd ed.; McGraw-Hill, 2008, pp 618.

Exam I Questions

On the first exam, I anticipated that there would be a question requiring us to label important peaks for a given IR sectrum and the groups that produced them, rather than trying to identify the compound that best matches the spectrum. I expected the compound depicted in the IR spectrum to included either an alkene or alkyne, a hydroxyl group, and an amine or nitrile. With the table that was given, it would have been easy to identify these groups. Some of which can be easily identified without the table. The hydroxyl group would be easiest to identify, with or without the table; it is identified by one strong absorption between 3600cm-1 and 3200cm-1. It's the first thing I look for when trying to identify groups on the spectrum. The alkene could have been identified by the major Csp3-H peak at around 2900cm-1 and a small, sharp peak at 1650cm-1. If the compound had included an alkyne, it could have been identified by the major Csp3-H peak as well, but the second peak, being small and sharp, would have been at around 2250cm-1. If the compound shown had included an amine, however, I'm not sure whether or not the peak for the amine could be visible. This is because the amine shows two weak peaks at 3300cm-1 and 3400cm-1, requiring some of the same levels as the hydroxyl group (3600-3200cm-1). If a nitrile were present, it would have peaked at about 2250cm-1. This may have conflicted with the second peak of the alkyne, had it been present as well.