In the last post we introduced ourselves to a new concept in organic chemistry – Aromaticity. We will continue to discuss it in this post.
Before learning the Hückel’s rule for aromaticity we need to brush up our hybridization basics.We have learnt in post 61 that –
Type of hybridization
# unhybridized p orbitals
Four sp3 orbitals
Three sp2 orbitals
One p orbital
Two sp orbitals
Two p orbitals
While studying aromaticity we are going to deal with the sp2 hybrid orbitals and the unhybridized p- orbital exclusively.
Hückel’s rule .
Erich Hückel was a German physical chemist, who was born in Berlin, Germany. His father was a physician, who had built a laboratory in their house. In this laboratory, he used to conduct experiments in chemistry and physics with his sons. This is how Erich Hückel developed his interest in chemistry. He is one of the most celebrated scientist of the 19th century. He is the one who gave us the Hückel’s rule, Hückel’s molecular orbital theory and Debye- Hückel theory.
Erich Hückel formulated a simple expression of the relationship between the structure of a compound and aromaticity as follows –
Planar monocyclic completely conjugated hydrocarbons will be aromatic when the ring contains 4n+ 2π electrons, where n ⇒ zero or any integer. Thus, rings with 2,6,10,14 etc electrons are aromatic.
Benzene has 6 π electrons and is thus aromatic.
Erich Hückel developed this rule to predict which systems can be termed aromatic. The rule applies to structures which –
i) have a cyclic (ring) structure.
ii) are planar i.e all atoms lie in the same plane. Thus, the atoms are sp2 hybridised. This ensures that there are continuously overlapping p- orbitals.
iii) have 4n+2 π electrons
iv)π electrons delocalized over continuously overlapping p- orbitals.
e.g.- Benzene molecule – It has a planar ring structure. All carbon atoms are sp2 hybridised.The unhybridised p- orbitals lie perpendicular to the σ framework as studied in post no 106 and the electrons in them are delocalised (i.e they are not localised on one single carbon atom) .They form an electron cloud as shown in the figure below. The electrons can be in the upper lobe or lower lobe of of the p-orbital.The electrons in the upper lobe and lower lobes form upper and lower π electron clouds respectively as shown in the figure below.
Benzene has 6 π electrons , which means it satisfies the Huckel’s rule. Thus, all these features make it an aromatic structure.
Q: Which of the following structures are aromatic?
A : This is a planar ring, however there are no conjugated double bonds.
The no of π electrons =2×no of double bonds = 2× 2 =4. Thus, this doesnot follow the 4n+2 π electron rule. So, this structure is NOT aromatic.
A : This structure looks unfamiliar and a bit complicated. However, we just have to see whether or not it is aromatic. Let us see if it obeys the Huckel’s rule or not. As seen above all the atoms are in the same plane. Thus, this is a planar structure.
# of π electrons = 2× no of double bonds = 2× 5 = 10. If we substitute n=2, then,
4n+2 = 4(2)+2= 8+2=10.
So, this molecule has 4n+2 π electrons and it also has conjugated double bonds. Thus, this molecule is aromatic.
A: The above molecule is planar and also has conjugated double bonds. However, it does not have a cyclic structure and so is NOT aromatic.
In the next post we shall study the Huckel molecular orbital theory. Till then, be a perpetual student of life and keep learning….
Good day !
References and further reading –