Hi Grandma,
In class we have been talking about benzene rings, the functions of these compounds, as well as aromaticity. However, for you to even understand what this means I need to explain some of the terms. For compounds to be aromatic they must meet four conditions which include: it must be a ring, it must be flat(planar), it must have in each atom of the ring a p orbital that’s orthogonal to the plane of the ring. In other words, the atoms in ring are sp2 hybridized, and it must have a Huckel number of pi electrons, which must follow the 4n+2 rule.
With knowing these rules let’s look further in dept of ways to classify all other compounds which include: if the molecule meets the first three conditions, but only contains 4nπ electrons the molecule is considered to be anti-aromatic. However, if the molecule fails any or the first three conditions then the molecule is considered to be non-aromatic. Now with a little bit of background let’s now explore the conditions a little bit more in depth. Condition one it must be a ring means that only rings can be aromatic; acyclic (having an open chain structure) systems cannot be aromatic. The second condition it must be flat deals with the shape of the ring. Ring systems can be planar (flat) or three-dimensional. Most conjugated ring systems tend to be flat so that it maximizes the overlap between the p orbital’s. An example would be naphthalene which is planar, and cyclodecapentaene is nonplanar due to two of the hydrogen’s. Both examples are shown below.
(naphthalene) (cyclodecapentaene)
(planar showing aromactity) (nonplanar due to two H’s).
The third condition deals with the p orbital’s. An aromatic system must have an unbroken ring of p orbitals, so that any ring that contains a sp3 hybridized carbon will not be aromatic. For example cycloheptatriene is non-aromatic due to the fact that one of the ring carbons is sp3 hybridized. However, carbocations’ (which have a positively charged carbon) are sp2 hybridized (and contain an empty p orbital); with this cycloheptatriene cation has an unbroken ring of p orbitals and is an aromatic compound. An example of this is shown below.
The fourth condition deals with Huckels rule. A tricky aspect that comes into play with Huckels rule is that you must remember of counting the number of pi electrons in the pi system when the ring contains heteroatoms like O, S, N. So how do you know which lone pairs to count as part of the pi system and which to ignore. The general rule of counting substituent’s to determine the hybridization holds true, it does fail when the atom contains a lone pair which is adjacent to a double bond; which means when it is conjugated. A diagram of pi electrons is shown below:
Pi electron Counts
Integer(n) | Aromatic Numbers (4n+2) | Anti-aromatic numbers (4n) |
0 | 2 | -- |
1 | 6 | 4 |
2 | 10 | 8 |
3 | 14 | 12 |
4 | 18 | 16 |
After reading this letter grandma I hope that it gives you a little insight on what aromaticity means and how it relates to benzene rings.
for some reason the pictures wouldn't show up. sorry i'm not for sure on what happened
ReplyDeleteAfter reading your blog, I think you did a good job of describing what aromaticity is and the steps that are included with it. The only thing I would change in your description is basically giving an example of an aromatic compound like benzene. So I found this website: http://shodor.org/succeed-1.0/programs/compchem98/labs/benzene/ and I thought it might help you understand something to. This website helped me tremendously because it stated that aromaticity occurs when carbon forms bonds which are not exactly single or double, but rather are of intermediate length. Since the resonance structures are equal, benzene rings do not exist in exactly either state, but in a mixture of both. The bond lengths between carbons in a benzene ring are nearly halfway between the normal C-C and C=C bond lengths. This indicates that the electrons which would form the double bonds do not stay between any two carbons, but float or delocalize so that they spend equal amounts of time between all of the carbons. This makes all six bonds a little more than single bonds and a little less than single bonds. This delocalization makes these carbons aromatic, which make them exhibit aromaticity. Overall, you did a good job and I wish I could’ve seen your pictures. I tried going to the link to see them, but it still wouldn’t let me.
ReplyDelete-Alexia