What makes a compound aromatic?

Most people associate aromatic compounds with the word aroma. They think that all aromatic compounds smell nice instead of knowing the specific criteria that makes terms them as aromatic. To understand these characteristics properly, I will define them in terms that everyone can understand not just people with a background in chemistry.

First, the compound must be cyclic. But what does that mean?  To understand this, think of a ring.  A ring is a circle and for a compound to be aromatic it must be shaped like this.  This is the easiest way to comprehend the first rule.

Second, the molecule must be planar.  Think of a piece of cardboard with rubber balls attached to each side. All of the balls on the top side should be equivalent.  For example, the balls should all be the same color, shape, and size. The same goes for the underside. The piece of cardboard represents the molecule and the balls represent the orbital.  All of these aspects are on the same plane, which makes the molecule planar.

Third, the molecule should be completely conjugated.  What does it mean to be completely conjugate?  To understand this, think of a group of girls who are holding hands to form a circle.  This represents the molecule and each girl represents an atom.  Aromatic molecules must have a p orbital on every atom.  So think of every girl as having the same hat on, this could represent the p orbital. 

Fourth, the molecule must satisfy Huckel’s rule and contain a particular number of pi electrons.  An aromatic compound must contain 4n + 2 pi electrons.  This rule is the hardest to understand.  Pi refers to the p orbital as explained above. The electrons in pi bonds are referred to as pi electrons.  To understand this, relate the pi electrons that are needed to make a compound aromatic with the p orbital analogy in rule three.

If all four of these rules are satisfied then a compound can be termed as aromatic.  A compound can meet the first three rules but not have the appropriate number of pi electrons, therefore it is anti-aromatic.  These compounds have 4n pi electrons.  Those compounds without 4n or 4n +2 pi electrons or ones that disobey one of the first three rules are non-aromatic. After reading this, I hope you can understand better what makes a compound aromatic. Next time you smell something nice don’t assume it’s aromatic, determine whether it is based on this criteria.

Question Not Seen on Exam 1

I was expecting to see a question about determining an unknown compound from interpreting it's mass spectrum and IR spectrum on the exam. The example on page 484 of chapter 13 is a good representation of what I was expecting. The problem and solution is as follows:

Q: What information is obtained from the mass pectrum and IR spectrum of an unknown compound X? Assume X contains the elements C, H, and O.

Step 1: Use the molecular ion to determine possible molecular formulas. In this case, the molecular ion is 88. Use an exact mass to determine a molecular formula. Divide the mass by 12 (1 C atom) to determine the maximum amount of carbons. Replace 1 C by 12 H's for another possible molecular formula. Substitute 1 O for CH4 then repeat for a possible molecular formula with 2 O atoms. There will be three possible formulas: C5H12O, C4H8O2, and C3H4O3. The possible hydrocarbons are discounted because the compound X contains an O atom. If the molecular ion had an exact mass of 88.0580, the molecular formula of X is C4H8O2.

Step 2: Calculate the degrees of unsaturation. For a compound of molecular formula C4H8O2, the maximum number of H's = 2n + 2 = 2(4) + 2 = 10. Because the compound contains only 8 H's, it has 2 fewer H's than the maximum number. Since each degree of unsaturation removes 2 H's, X has one degree of unsaturation. This means that X has one ring or one pi bond.

Step 3: Determine what functional group is present from the IR spectrum. The two major absorptions in the IR spectrum above 1500 are due to sp3 hybridized C-H bonds (3000-2850) and a C-O double bond (1740). Thus, the one degree of unsaturation in X is due to the presence of the C-O double bond.

Mass spectrometry and IR spectrometry give valuable information when determining an unknown compound and that is why I expected a problem like this to be on the test. If the structure was wanted then NMR would be needed.